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(2014-2016)
JEE Main Question Papers and Answer Keys
IntroductionCareers360 brings you a compilation of JEE Main Question Papers and Answer Keys from the year 2014 to 2016. This e-book helps you master questions which appeared in the national level engi-neering entrance examination in the last 3 years.
This booklet contains 40 printed pages.
ß‚ ¬ÈÁSÃ∑§Ê ◊¥ ◊ÈÁŒ˝Ã ¬Îc∆ 40 „Ò¥–
Do not open this Test Booklet until you are asked to do so.
ß‚ ¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê ∑§Ê Ã’ Ã∑§ Ÿ πÊ‹¥ ¡’ Ã∑§ ∑§„Ê Ÿ ¡Ê∞–Read carefully the Instructions on the Back Cover of this Test Booklet.
ß‚ ¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê ∑§ Á¬¿‹ •Êfl⁄áÊ ¬⁄ ÁŒ∞ ª∞ ÁŸŒ¸‡ÊÊ¥ ∑§Ê äÿÊŸ ‚ ¬…∏¥–
Name of the Candidate (in Capital letters ) :
¬⁄ˡÊÊÕ˸ ∑§Ê ŸÊ◊ (’«∏ •ˇÊ⁄Ê¥ ◊¥) —Roll Number : in figures
•ŸÈ∑§◊Ê¥∑§ — •¥∑§Ê¥ ◊¥: in words — ‡ÊéŒÊ¥ ◊¥
Examination Centre Number :¬⁄ˡÊÊ ∑§ãŒ˝ Ÿê’⁄U —Name of Examination Centre (in Capital letters) : ¬⁄UˡÊÊ ∑§ãŒ˝ ∑§Ê ŸÊ◊ (’«∏ •ˇÊ⁄UÊ¥ ◊¥ ) —Candidate’s Signature : 1. Invigilator’s Signature : ¬⁄ˡÊÊÕ˸ ∑§ „SÃÊˇÊ⁄ — ÁŸ⁄ˡÊ∑§ ∑§ „SÃÊˇÊ⁄ —
2. Invigilator’s Signature : ÁŸ⁄ˡÊ∑§ ∑§ „SÃÊˇÊ⁄ —
No. :
Important Instructions :
1. Immediately fill in the particulars on this page of the TestBooklet with only Blue / Black Ball Point Pen provided bythe Board.
2. The Answer Sheet is kept inside this Test Booklet. When youare directed to open the Test Booklet, take out the AnswerSheet and fill in the particulars carefully.
3. The test is of 3 hours duration.4. The Test Booklet consists of 90 questions. The maximum
marks are 360.5. There are three parts in the question paper A, B, C
consisting of Physics, Chemistry and Mathematics having30 questions in each part of equal weightage. Each questionis allotted 4 (four) marks for correct response.
6. Candidates will be awarded marks as stated above in instructionNo. 5 for correct response of each question. ¼ (one fourth) markswill be deducted for indicating incorrect response of each question.No deduction from the total score will be made if no response isindicated for an item in the answer sheet.
7. There is only one correct response for each question. Fillingup more than one response in any question will be treated aswrong response and marks for wrong response will bededucted accordingly as per instruction 6 above.
8. For writing particulars/marking responses on Side-1 andSide–2 of the Answer Sheet use only Blue/Black Ball PointPen provided by the Board.
9. No candidate is allowed to carry any textual material, printedor written, bits of papers, pager, mobile phone, any electronicdevice, etc. except the Admit Card inside the examinationroom/hall.
10. Rough work is to be done on the space provided for thispurpose in the Test Booklet only. This space is given at thebottom of each page and in one page (i.e. Page 39) at the endof the booklet.
11. On completion of the test, the candidate must hand over theAnswer Sheet to the Invigilator on duty in the Room/Hall.However, the candidates are allowed to take away this TestBooklet with them.
12. The CODE for this Booklet is E. Make sure that the CODEprinted on Side–2 of the Answer Sheet and also tally theserial number of the Test Booklet and Answer Sheet are thesame as that on this booklet. In case of discrepancy, thecandidate should immediately report the matter to theInvigilator for replacement of both the Test Booklet and theAnswer Sheet.
13. Do not fold or make any stray mark on the Answer Sheet.
◊„ûfl¬Íáʸ ÁŸŒ¸‡Ê —
1. ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê ∑§ ß‚ ¬Îc∆U ¬⁄U •Êfl‡ÿ∑§ Áflfl⁄UáÊ ∑§fl‹ ’Ê«¸U mÊ⁄UÊ©¬‹éœ ∑§⁄UÊÿ ªÿ ŸË‹ / ∑§Ê‹ ’ÊÚ‹ åflÊߥ≈U ¬Ÿ ‚ Ãà∑§Ê‹ ÷⁄¥–
2. ©ûÊ⁄U ¬òÊ ß‚ ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê ∑§ •ãŒ⁄U ⁄UπÊ „Ò– ¡’ •Ê¬∑§Ê ¬⁄UˡÊÊ ¬ÈÁSÃ∑§ÊπÊ‹Ÿ ∑§Ê ∑§„Ê ¡Ê∞, ÃÊ ©ûÊ⁄U ¬òÊ ÁŸ∑§Ê‹ ∑§⁄U ‚ÊflœÊŸË¬Ífl∑§ Áflfl⁄UáÊ ÷⁄¥U–
3. ¬⁄UˡÊÊ ∑§Ë •flÁœ 3 ÉÊ¥≈U „Ò–4. ß‚ ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê ◊¥ 90 ¬˝‡Ÿ „Ò¥– •Áœ∑§Ã◊ •¥∑§ 360 „Ò¥–5. ß‚ ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê ◊¥ ÃËŸ ÷ʪ A, B, C „Ò¥, Á¡‚∑§ ¬˝àÿ∑§ ÷ʪ ◊¥
÷ÊÒÁÃ∑§ ÁflôÊÊŸ, ⁄U‚ÊÿŸ ÁflôÊÊŸ ∞fl¥ ªÁáÊà ∑§ 30 ¬˝‡Ÿ „Ò¥ •ÊÒ⁄U ‚÷ˬ˝‡ŸÊ¥ ∑§ •¥∑§ ‚◊ÊŸ „Ò¥– ¬˝àÿ∑§ ¬˝‡Ÿ ∑§ ‚„Ë ©ûÊ⁄U ∑§ Á‹∞ 4 (øÊ⁄U)•¥∑§ ÁŸœÊ¸Á⁄Uà Á∑§ÿ ªÿ „Ò¥–
6. •èÿÁÕ¸ÿÊ¥ ∑§Ê ¬˝àÿ∑§ ‚„Ë ©ûÊ⁄U ∑§ Á‹∞ ©¬⁄UÊÄà ÁŸŒ¸‡ÊŸ ‚¥ÅÿÊ 5 ∑§ÁŸŒ¸‡ÊÊŸÈ‚Ê⁄U •¥∑§ ÁŒÿ ¡Êÿ¥ª– ¬˝àÿ∑§ ¬˝‡Ÿ ∑§ ª‹Ã ©ûÊ⁄U ∑§ Á‹ÿ¼ flÊ¥ ÷ʪ ∑§Ê≈U Á‹ÿÊ ¡ÊÿªÊ– ÿÁŒ ©ûÊ⁄U ¬òÊ ◊¥ Á∑§‚Ë ¬˝‡Ÿ ∑§Ê ©ûÊ⁄UŸ„Ë¥ ÁŒÿÊ ªÿÊ „Ê ÃÊ ∑ȧ‹ ¬˝Ê#Ê¥∑§ ‚ ∑§Ê߸ ∑§≈UÊÒÃË Ÿ„Ë¥ ∑§Ë ¡ÊÿªË–
7. ¬˝àÿ∑§ ¬˝‡Ÿ ∑§Ê ∑§fl‹ ∞∑§ „Ë ‚„Ë ©ûÊ⁄U „Ò– ∞∑§ ‚ •Áœ∑§ ©ûÊ⁄U ŒŸ ¬⁄U©‚ ª‹Ã ©ûÊ⁄U ◊ÊŸÊ ¡ÊÿªÊ •ÊÒ⁄U ©¬⁄UÊÄà ÁŸŒ¸‡Ê 6 ∑§ •ŸÈ‚Ê⁄U •¥∑§ ∑§Ê≈UÁ‹ÿ ¡Êÿ¥ª–
8. ©ûÊ⁄U ¬òÊ ∑§ ¬Îc∆U-1 ∞fl¥ ¬Îc∆U-2 ¬⁄U flÊ¥Á¿Uà Áflfl⁄UáÊ ∞fl¥ ©ûÊ⁄U •¥Á∑§Ã∑§⁄UŸ „ÃÈ ’Ê«¸U mÊ⁄UÊ ©¬‹éœ ∑§⁄UÊÿ ªÿ ∑§fl‹ ŸË‹/∑§Ê‹ ’ÊÚ‹ åflÊߥ≈UU¬Ÿ ∑§Ê „Ë ¬˝ÿʪ ∑§⁄¥U–
9. ¬⁄UˡÊÊÕË mÊ⁄UÊ ¬⁄UˡÊÊ ∑§ˇÊ/„ÊÚ‹ ◊¥ ¬fl‡Ê ∑§Ê«U ∑§ •‹ÊflÊ Á∑§‚Ë ÷Ë ¬∑§Ê⁄U∑§Ë ¬Ê∆Uÿ ‚Ê◊ªË, ◊ÈÁŒÃ ÿÊ „SÃÁ‹ÁπÃ, ∑§Êª¡ ∑§Ë ¬ÁøÿÊ°, ¬¡⁄U, ◊Ê’Êß‹»§ÊŸ ÿÊ Á∑§‚Ë ÷Ë ¬˝∑§Ê⁄U ∑§ ß‹Ä≈UÊÚÁŸ∑§ ©¬∑§⁄UáÊÊ¥ ÿÊ Á∑§‚Ë •ãÿ ¬˝∑§Ê⁄U ∑§Ë‚Ê◊ªË ∑§Ê ‹ ¡ÊŸ ÿÊ ©¬ÿʪ ∑§⁄UŸ ∑§Ë •ŸÈ◊Áà Ÿ„Ë¥ „Ò–
10. ⁄U»§ ∑§Êÿ¸ ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê ◊¥ ∑§fl‹ ÁŸœÊ¸Á⁄Uà ¡ª„ ¬⁄U „Ë ∑§ËÁ¡∞– ÿ„¡ª„ ¬˝àÿ∑§ ¬Îc∆U ¬⁄U ŸËø ∑§Ë •Ê⁄U •ÊÒ⁄U ¬ÈÁSÃ∑§Ê ∑§ •¥Ã ◊¥ ∞∑§ ¬Îc∆U ¬⁄U(¬Îc∆U 39) ŒË ªß¸ „Ò–
11. ¬⁄UˡÊÊ ‚◊Êåà „ÊŸ ¬⁄U, ¬⁄UˡÊÊÕ˸ ∑§ˇÊ/„ÊÚ‹ ¿UÊ«∏Ÿ ‚ ¬Ífl¸ ©ûÊ⁄U ¬òÊ ∑§ˇÊÁŸ⁄UˡÊ∑§ ∑§Ê •fl‡ÿ ‚ÊÒ¥¬ Œ¥– ¬⁄UˡÊÊÕ˸ •¬Ÿ ‚ÊÕ ß‚ ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê∑§Ê ‹ ¡Ê ‚∑§Ã „Ò¥–
12. ß‚ ¬ÈÁSÃ∑§Ê ∑§Ê ‚¥∑§Ã E „Ò– ÿ„ ‚ÈÁŸÁ‡øà ∑§⁄U ‹¥ Á∑§ ß‚ ¬ÈÁSÃ∑§Ê ∑§Ê‚¥∑§Ã, ©ûÊ⁄U ¬òÊ ∑§ ¬Îc∆U-2 ¬⁄U ¿U¬ ‚¥∑§Ã ‚ Á◊‹ÃÊ „Ò •ÊÒ⁄U ÿ„ ÷Ë‚ÈÁŸÁ‡øà ∑§⁄U ‹¥ Á∑§ ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê •ÊÒ⁄U ©ûÊ⁄U ¬òÊ ∑§Ë ∑˝§◊ ‚¥ÅÿÊÁ◊‹ÃË „Ò– •ª⁄U ÿ„ Á÷ÛÊ „Ê ÃÊ ¬⁄UˡÊÊÕ˸ ŒÍ‚⁄UË ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê •ÊÒ⁄U©ûÊ⁄U ¬òÊ ‹Ÿ ∑§ Á‹∞ ÁŸ⁄UˡÊ∑§ ∑§Ê ÃÈ⁄Uãà •flªÃ ∑§⁄UÊ∞°–
13. ©ûÊ⁄U ¬òÊ ∑§Ê Ÿ ◊Ê«∏¥ ∞fl¥ Ÿ „Ë ©‚ ¬⁄U •ãÿ ÁŸ‡ÊÊŸ ‹ªÊ∞°–
Test Booklet Code
¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê ‚¥∑§Ã
E
PAPER - 1 : PHYSICS, CHEMISTRY & MATHEMATICS
¬˝‡Ÿ¬ÈÁSÃ∑§Ê - 1 : ÷ÊÒÁÃ∑§ ÁflôÊÊŸ, ⁄U‚ÊÿŸ ÁflôÊÊŸ ÃÕÊ ªÁáÊÃ
S S O
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E/Page 2 SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„
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PART A — PHYSICS
ALL THE GRAPHS GIVEN ARE SCHEMATIC
AND NOT DRAWN TO SCALE.
1. A student measures the time period of 100
oscillations of a simple pendulum four
times. The data set is 90 s, 91 s, 95 s and
92 s. If the minimum division in the
measuring clock is 1 s, then the reported
mean time should be :
(1) 92±2 s
(2) 92±5.0 s
(3) 92±1.8 s
(4) 92±3 s
2. A particle of mass m is moving along the
side of a square of side ‘a’, with a uniform
speed v in the x-y plane as shown in the
figure :
Which of the following statements is false
for the angular momentum →
L about the
origin ?
(1)→ ∧
2
mvL R k=− when the particle is
moving from A to B.
(2)→ ∧
2
RL mv a k= − when the
particle is moving from C to D.
(3)→ ∧
2
RL mv a k= + when the
particle is moving from B to C.
(4)→ ∧
2
mvL R k= when the particle is
moving from D to A.
÷ʪ A — ÷ÊÒÁÃ∑§ ÁflôÊÊŸÁŒ∞ ªÿ ‚÷Ë ª˝Ê»§ •Ê⁄UπËÿ „Ò¥ •ÊÒ⁄US∑§‹ ∑§ •ŸÈ‚Ê⁄U ⁄UπÊ¥Á∑§Ã Ÿ„Ë¥ „Ò–
1. ∞∑§ ¿UÊòÊ ∞∑§ ‚⁄U‹-•Êflø-ŒÊ‹∑§ ∑§ 100 •ÊflÎÁûÊÿÊ¥∑§Ê ‚◊ÿ 4 ’Ê⁄U ◊ʬÃÊ „Ò •ÊÒ⁄U ©Ÿ∑§Ê 90 s, 91 s,
95 s •ÊÒ⁄U 92 s ¬ÊÃÊ „Ò– ßSÃ◊Ê‹ ∑§Ë ªß¸ ÉÊ«∏Ë ∑§ÊãÿÍŸÃ◊ •À¬Ê¥‡Ê 1 s „Ò– Ã’ ◊ʬ ªÿ ◊Êäÿ ‚◊ÿ ∑§Ê©‚ Á‹πŸÊ øÊÁ„ÿ —
(1) 92±2 s
(2) 92±5.0 s
(3) 92±1.8 s
(4) 92±3 s
2. ÁøòÊ ◊¥ ÷È¡Ê ‘a’ ∑§Ê flª¸ x-y Ë ◊¥ „Ò– m Œ˝√ÿ◊ÊŸ∑§Ê ∞∑§ ∑§áÊ ∞∑§‚◊ÊŸ ªÁÃ, v ‚ ß‚ flª¸ ∑§Ë ÷ȡʬ⁄U ø‹ ⁄U„Ê „Ò ¡Ò‚Ê Á∑§ ÁøòÊ ◊¥ Œ‡ÊʸÿÊ ªÿÊ „Ò–
Ã’ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ‚Ê ∑§ÕŸ, ß‚ ∑§áÊ ∑§ ◊Í‹Á’¥ŒÈ
∑§ ÁªŒ¸ ∑§ÊáÊËÿ •ÊÉÊÍáʸ →
L ∑§ Á‹ÿ, ª‹Ã „Ò?
(1)→ ∧
2
mvL R k=− , ¡’ ∑§áÊ A ‚ B ∑§Ë
•Ê⁄U ø‹ ⁄U„Ê „Ò–
(2)→ ∧
2
RL mv a k= − , ¡’ ∑§áÊ C ‚
D ∑§Ë •Ê⁄U ø‹ ⁄U„Ê „Ò–
(3)→ ∧
2
RL mv a k= + , ¡’ ∑§áÊ B ‚
C ∑§Ë •Ê⁄U ø‹ ⁄U„Ê „Ò–
(4)→ ∧
2
mvL R k= , ¡’ ∑§áÊ D ‚ A ∑§Ë •Ê⁄U
ø‹ ⁄U„Ê „Ò–
SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„E/Page 3
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3. ‘m’ Œ˝√ÿ◊ÊŸ ∑§Ê ∞∑§ Á’¥ŒÈ ∑§áÊ ∞∑§ πÈ⁄UŒ⁄U ¬Õ PQR
(ÁøòÊ ŒÁπÿ) ¬⁄U ø‹ ⁄U„Ê „Ò– ∑§áÊ •ÊÒ⁄U ¬Õ ∑§ ’ËøÉÊ·¸áÊ ªÈáÊÊ¥∑§ µ „Ò– ∑§áÊ P ‚ ¿UÊ«∏ ¡ÊŸ ∑§ ’ÊŒ R ¬⁄U¬„È°ø ∑§⁄U L§∑§ ¡ÊÃÊ „Ò– ¬Õ ∑§ ÷ʪ PQ •ÊÒ⁄U QR ¬⁄Uø‹Ÿ ◊¥ ∑§áÊ mÊ⁄UÊ πø¸ ∑§Ë ªß¸ ™§¡Ê¸∞° ’⁄UÊ’⁄U „Ò¥–PQ ‚ QR ¬⁄U „ÊŸ flÊ‹ ÁŒ‡ÊÊ ’Œ‹Êfl ◊¥ ∑§Ê߸ ™§¡Ê¸πø¸ Ÿ„Ë¥ „ÊÃË–Ã’ µ •ÊÒ⁄U ŒÍ⁄UË x(=QR) ∑§ ◊ÊŸ ‹ª÷ª „Ò¥ ∑˝§◊‡Ê— —
(1) 0.2 •ÊÒ⁄U 6.5 m
(2) 0.2 •ÊÒ⁄U 3.5 m
(3) 0.29 •ÊÒ⁄U 3.5 m
(4) 0.29 •ÊÒ⁄U 6.5 m
3. A point particle of mass m, moves along
the uniformly rough track PQR as shown
in the figure. The coefficient of friction,
between the particle and the rough track
equals µ. The particle is released, from rest,
from the point P and it comes to rest at a
point R. The energies, lost by the ball, over
the parts, PQ and QR, of the track, are
equal to each other, and no energy is lost
when particle changes direction from PQ
to QR.
The values of the coefficient of friction µ
and the distance x(=QR), are, respectively
close to :
(1) 0.2 and 6.5 m
(2) 0.2 and 3.5 m
(3) 0.29 and 3.5 m
(4) 0.29 and 6.5 m
E/Page 4 SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„
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4. ∞∑§ ÷Ê⁄UÊûÊÊ‹∑§ ÷Ê⁄U ∑§Ê ¬„‹ ™§¬⁄U •ÊÒ⁄U Á»§⁄U ŸËøÃ∑§ ‹ÊÃÊ „Ò– ÿ„ ◊ÊŸÊ ¡ÊÃÊ „Ò Á∑§ Á‚»¸§ ÷Ê⁄U ∑§Ê™§¬⁄U ‹ ¡ÊŸ ◊¥ ∑§Êÿ¸ „ÊÃÊ „Ò •ÊÒ⁄U ŸËø ‹ÊŸ ◊¥ ÁSÕÁᙧ¡Ê¸ ∑§Ê OÊ‚ „ÊÃÊ „Ò– ‡Ê⁄UË⁄U ∑§Ë fl‚Ê ™§¡Ê¸ ŒÃË „Ò ¡ÊÿÊ¥ÁòÊ∑§Ëÿ ™§¡Ê¸ ◊¥ ’Œ‹ÃË „Ò– ◊ÊŸ ‹¥ Á∑§ fl‚Ê mÊ⁄UÊŒË ªß¸ ™§¡Ê¸ 3.8×107 J ¬˝Áà kg ÷Ê⁄U „Ò, ÃÕÊ ß‚∑§Ê◊ÊòÊ 20% ÿÊ¥ÁòÊ∑§Ëÿ ™§¡Ê¸ ◊¥ ’Œ‹ÃÊ „Ò– •’ ÿÁŒ∞∑§ ÷Ê⁄UÊûÊÊ‹∑§ 10 kg ∑§ ÷Ê⁄U ∑§Ê 1000 ’Ê⁄U 1 m
∑§Ë ™°§øÊ߸ Ã∑§ ™§¬⁄U •ÊÒ⁄U ŸËø ∑§⁄UÃÊ „Ò Ã’ ©‚∑§‡Ê⁄UË⁄U ‚ fl‚Ê ∑§Ê ˇÊÿ „Ò — (g=9.8 ms−2 ‹¥)
(1) 2.45×10−3 kg
(2) 6.45×10−3 kg
(3) 9.89×10−3 kg
(4) 12.89×10−3 kg
5. ŒÊ ‡Ê¥∑ȧ ∑§Ê ©Ÿ∑§ ‡ÊË·¸ O ¬⁄U ¡Ê«∏∑§⁄U ∞∑§ ⁄UÊ‹⁄U’ŸÊÿÊ ªÿÊ „Ò •ÊÒ⁄U ©‚ AB fl CD ⁄U‹ ¬⁄U •‚◊Á◊Ã⁄UπÊ ªÿÊ „Ò (ÁøòÊ ŒÁπÿ)– ⁄UÊ‹⁄U ∑§Ê •ˇÊ CD ‚‹ê’flà „Ò •ÊÒ⁄U O ŒÊŸÊ¥ ⁄U‹ ∑§ ’ËøÊ’Ëø „Ò– „À∑§ ‚œ∑§‹Ÿ ¬⁄U ⁄UÊ‹⁄U ⁄U‹ ¬⁄U ß‚ ¬˝∑§Ê⁄U ‹È…∏∑§ŸÊ •Ê⁄Uê÷∑§⁄UÃÊ „Ò Á∑§ O ∑§Ê øÊ‹Ÿ CD ∑§ ‚◊Ê¥Ã⁄U „Ò (ÁøòÊŒÁπÿ)– øÊÁ‹Ã „Ê ¡ÊŸ ∑§ ’ÊŒ ÿ„ ⁄UÊ‹⁄U —
(1) ’Ê°ÿË¥ •Ê⁄U ◊È«∏ªÊ–
(2) ŒÊÿË¥ •Ê⁄U ◊È«∏ªÊ–
(3) ‚ËœÊ ø‹ÃÊ ⁄U„ªÊ–
(4) ’Êÿ¥ ÃÕÊ ŒÊÿ¥ ∑˝§◊‡Ê— ◊È«∏ÃÊ ⁄U„ªÊ–
4. A person trying to lose weight by burning
fat lifts a mass of 10 kg upto a height of
1 m 1000 times. Assume that the potential
energy lost each time he lowers the mass
is dissipated. How much fat will he use
up considering the work done only when
the weight is lifted up ? Fat supplies
3.8×107 J of energy per kg which is
converted to mechanical energy with a
20% efficiency rate. Take g=9.8 ms−2 :
(1) 2.45×10−3 kg
(2) 6.45×10−3 kg
(3) 9.89×10−3 kg
(4) 12.89×10−3 kg
5. A roller is made by joining together two
cones at their vertices O. It is kept on two
rails AB and CD which are placed
asymmetrically (see figure), with its axis
perpendicular to CD and its centre O at
the centre of line joining AB and CD (see
figure). It is given a light push so that it
starts rolling with its centre O moving
parallel to CD in the direction shown. As
it moves, the roller will tend to :
(1) turn left.
(2) turn right.
(3) go straight.
(4) turn left and right alternately.
SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„E/Page 5
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6. ¬ÎâflË ∑§Ë ‚Ä ‚ ‘h’ ™°§øÊ߸ ¬⁄U ∞∑§ ©¬ª˝„ flÎûÊÊ∑§Ê⁄U¬Õ ¬⁄U øÄ∑§⁄U ∑§Ê≈U ⁄U„Ê „Ò (¬ÎâflË ∑§Ë ÁòÊíÿÊ R ÃÕÊh<<R)– ¬ÎâflË ∑§ ªÈL§àfl ˇÊòÊ ‚ ¬‹ÊÿŸ ∑§⁄UŸ ∑§Á‹ÿ ß‚∑§Ë ∑§ˇÊËÿ ªÁà ◊¥ •Êfl‡ÿ∑§ ãÿÍŸÃ◊ ’Œ‹Êfl„Ò — (flÊÿÈ◊¥«U‹Ëÿ ¬˝÷Êfl ∑§Ê Ÿªáÿ ‹ËÁ¡∞–)
(1) 2 gR
(2) gR
(3) 2gR /
(4) ( ) 2 1gR −
7. ∞∑§ ¬ãU«ÈU‹◊ ÉÊ«∏Ë 408C Ãʬ◊ÊŸ ¬⁄U 12 s ¬˝ÁÃÁŒŸœË◊Ë „Ê ¡ÊÃË „Ò ÃÕÊ 208C Ãʬ◊ÊŸ ¬⁄U 4 s ¬˝ÁÃÁŒŸÃ$¡ „Ê ¡ÊÃË „Ò– Ãʬ◊ÊŸ Á¡‚ ¬⁄U ÿ„ ‚„Ë ‚◊ÿŒ‡ÊʸÿªË ÃÕÊ ¬ãU«ÈU‹◊ ∑§Ë œÊÃÈ ∑§Ê ⁄UπËÿ-¬˝‚Ê⁄U ªÈáÊÊ¥∑§(α) ∑˝§◊‡Ê— „Ò¥ —
(1) 258C; α=1.85×10−5/8C
(2) 608C; α=1.85×10−4/8C
(3) 308C; α=1.85×10−3/8C
(4) 558C; α=1.85×10−2/8C
6. A satellite is revolving in a circular orbit at
a height ‘h’ from the earth’s surface (radius
of earth R ; h<<R). The minimum increase
in its orbital velocity required, so that the
satellite could escape from the earth’s
gravitational field, is close to : (Neglect
the effect of atmosphere.)
(1) 2 gR
(2) gR
(3) 2gR /
(4) ( ) 2 1gR −
7. A pendulum clock loses 12 s a day if the
temperature is 408C and gains 4 s a day if
the temperature is 208C. The temperature
at which the clock will show correct time,
and the co-efficient of linear expansion
(α) of the metal of the pendulum shaft are
respectively :
(1) 258C; α=1.85×10−5/8C
(2) 608C; α=1.85×10−4/8C
(3) 308C; α=1.85×10−3/8C
(4) 558C; α=1.85×10−2/8C
E/Page 6 SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„
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8. ∞∑§ •ÊŒ‡Ê¸ ªÒ‚ ©à∑˝§◊áÊËÿ SÕÒÁÃ∑§-∑§À¬ ¬˝∑˝§◊ ‚ªÈ$¡⁄UÃË „Ò ÃÕÊ ©‚∑§Ë ◊Ê‹⁄U-™§c◊Ê-œÊÁ⁄UÃÊ C ÁSÕ⁄U⁄U„ÃË „Ò– ÿÁŒ ß‚ ¬˝∑˝§◊ ◊¥ ©‚∑§ ŒÊ’ P fl •ÊÿßV ∑§ ’Ëø ‚¥’¥œ PVn=constant „Ò– (C
P ÃÕÊ
CV ∑˝§◊‡Ê— ÁSÕ⁄U ŒÊ’ fl ÁSÕ⁄U •Êÿß ¬⁄U ™§c◊Ê-
œÊÁ⁄UÃÊ „Ò) Ã’ ‘n’ ∑§ Á‹ÿ ‚◊Ë∑§⁄UáÊ „Ò —
(1) P
V
Cn
C=
(2)
P
V
C Cn
C C
−
=
−
(3)
P
V
C Cn
C C
−
=
−
(4)
V
P
C Cn
C C
−
=
−
9. ‘n’ ◊Ê‹ •ÊŒ‡Ê¸ ªÒ‚ ∞∑§ ¬˝∑˝§◊ A→B ‚ ªÈ$¡⁄UÃË „Ò(ÁøòÊ ŒÁπÿ)– ß‚ ¬˝∑˝§◊ ∑§ ŒÊÒ⁄UÊŸ ©‚∑§Ê •Áœ∑§Ã◊Ãʬ◊ÊŸ „ÊªÊ —
(1) 0 09
4
P V
nR
(2) 0 03
2
P V
nR
(3) 0 09
2
P V
nR
(4) 0 09 P V
nR
8. An ideal gas undergoes a quasi static,
reversible process in which its molar heat
capacity C remains constant. If during this
process the relation of pressure P and
volume V is given by PVn=constant, then
n is given by (Here CP and C
V are molar
specific heat at constant pressure and
constant volume, respectively) :
(1) P
V
Cn
C=
(2)
P
V
C Cn
C C
−
=
−
(3)
P
V
C Cn
C C
−
=
−
(4)
V
P
C Cn
C C
−
=
−
9. ‘n’ moles of an ideal gas undergoes a
process A→B as shown in the figure. The
maximum temperature of the gas during
the process will be :
(1) 0 09
4
P V
nR
(2) 0 03
2
P V
nR
(3) 0 09
2
P V
nR
(4) 0 09 P V
nR
SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„E/Page 7
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10. ∞∑§ ∑§áÊ ‘A’ •ÊÿÊ◊ ‚ ‚⁄U‹-•Êflø ŒÊ‹Ÿ ∑§⁄U ⁄U„Ê
„Ò– ¡’ ÿ„ •¬Ÿ ◊Í‹-SÕÊŸ ‚ 23
A ¬⁄U ¬„È°øÃÊ „Ò
Ã’ •øÊŸ∑§ ß‚∑§Ë ªÁà ÁÃªÈŸË ∑§⁄U ŒË ¡ÊÃË „Ò– Ã’
ß‚∑§Ê ŸÿÊ •ÊÿÊ◊ „Ò —
(1) 413
A
(2) 3A
(3) 3A
(4)7
3
A
11. 20 m ‹ê’Ê߸ ∑§Ë ∞∑§‚◊ÊŸ «UÊ⁄UË ∑§Ê ∞∑§ ŒÎ…∏ •ÊœÊ⁄U‚ ‹≈U∑§ÊÿÊ ªÿÊ „Ò– ß‚∑§ ÁŸø‹ Á‚⁄U ‚ ∞∑§ ‚͡◊Ã⁄¥Uª-S¬¥Œ øÊÁ‹Ã „ÊÃÊ „Ò– ™§¬⁄U •ÊœÊ⁄U Ã∑§ ¬„È°øŸ◊¥ ‹ªŸ flÊ‹Ê ‚◊ÿ „Ò —(g = 10 ms−2 ‹¥)
(1) 2 2 sπ
(2) 2 s
(3) 2 2 s
(4) 2 s
10. A particle performs simple harmonic
motion with amplitude A. Its speed is
trebled at the instant that it is at a distance
2
3
A from equilibrium position. The new
amplitude of the motion is :
(1) 413
A
(2) 3A
(3) 3A
(4)7
3
A
11. A uniform string of length 20 m is
suspended from a rigid support. A short
wave pulse is introduced at its lowest end.
It starts moving up the string. The time
taken to reach the support is :
(take g = 10 ms−2)
(1) 2 2 sπ
(2) 2 s
(3) 2 2 s
(4) 2 s
E/Page 8 SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„
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12. ÁòÊíÿÊ ‘a’ ÃÕÊ ‘b’ ∑§ ŒÊ ∞∑§-∑§ãŒ˝Ë ªÊ‹Ê¥ ∑§ (ÁøòÊŒÁπÿ) ’Ëø ∑§ SÕÊŸ ◊¥ •Êÿß •Êfl‡Ê-ÉÊŸàfl
A
r =ρ „Ò, ¡„Ê° A ÁSÕ⁄UÊ¥∑§ „Ò ÃÕÊ r ∑§ãŒ˝ ‚ ŒÍ⁄UË „Ò–
ªÊ‹Ê¥ ∑§ ∑§ãŒ˝ ¬⁄U ∞∑§ Á’ãŒÈ-•Êfl‡Ê Q „Ò– ‘A’ ∑§Êfl„ ◊ÊŸ ’ÃÊÿ¥ Á¡‚‚ ªÊ‹Ê¥ ∑§ ’Ëø ∑§ SÕÊŸ ◊¥∞∑§‚◊ÊŸ flÒlÈÃ-ˇÊòÊ „Ê —
(1)2
2
Q
aπ
(2)( )2 2
2
Q
b a−π
(3)( )2 2
2
Q
a b−π
(4)2
2Q
aπ
12. The region between two concentric spheres
of radii ‘a’ and ‘b’, respectively (see figure),
has volume charge density A
r =ρ , where
A is a constant and r is the distance from
the centre. At the centre of the spheres is
a point charge Q. The value of A such
that the electric field in the region between
the spheres will be constant, is :
(1)2
2
Q
aπ
(2)( )2 2
2
Q
b a−π
(3)( )2 2
2
Q
a b−π
(4)2
2Q
aπ
SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„E/Page 9
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13. ‚¥œÊÁ⁄UòÊÊ¥ ‚ ’Ÿ ∞∑§ ¬Á⁄U¬Õ ∑§Ê ÁøòÊ ◊¥ ÁŒπÊÿÊ ªÿÊ„Ò– ∞∑§ Á’ãŒÈ-•Êfl‡Ê Q (Á¡‚∑§Ê ◊ÊŸ 4 µF ÃÕÊ9 µF flÊ‹ ‚¥œÊÁ⁄UòÊÊ¥ ∑§ ∑ȧ‹ •Êfl‡ÊÊ¥ ∑§ ’⁄UÊ’⁄U „Ò)∑§ mÊ⁄UÊ 30 m ŒÍ⁄UË ¬⁄U flÒlÈÃ-ˇÊòÊ ∑§Ê ¬Á⁄U◊ÊáÊ „ÊªÊ —
(1) 240 N/C
(2) 360 N/C
(3) 420 N/C
(4) 480 N/C
14. ÃÊ°’Ê ÃÕÊ •◊ÊÁŒÃ (undoped) Á‚Á‹∑§ÊŸ ∑§¬˝ÁÃ⁄Uʜʥ ∑§Ë ©Ÿ∑§ Ãʬ◊ÊŸ ¬⁄U ÁŸ÷¸⁄UÃÊ, 300-400 K
Ãʬ◊ÊŸ •¥Ã⁄UÊ‹ ◊¥, ∑§ Á‹ÿ ‚„Ë ∑§ÕŸ „Ò —
(1) ÃÊ°’Ê ∑§ Á‹ÿ ⁄UπËÿ ’…∏Êfl ÃÕÊ Á‚Á‹∑§ÊŸ ∑§Á‹ÿ ⁄UπËÿ ’…∏Êfl–
(2) ÃÊ°’Ê ∑§ Á‹ÿ ⁄UπËÿ ’…∏Êfl ÃÕÊ Á‚Á‹∑§ÊŸ ∑§Á‹ÿ ø⁄UUÉÊÊÃÊ¥∑§Ë ’…∏Êfl–
(3) ÃÊ°’Ê ∑§ Á‹ÿ ⁄UπËÿ ’…∏Êfl ÃÕÊ Á‚Á‹∑§ÊŸ ∑§Á‹ÿ ø⁄UUÉÊÊÃÊ¥∑§Ë ÉÊ≈UÊfl–
(4) ÃÊ°’Ê ∑§ Á‹ÿ ⁄UπËÿ ÉÊ≈UÊfl ÃÕÊ Á‚Á‹∑§ÊŸ ∑§Á‹ÿ ⁄UπËÿ ÉÊ≈UÊfl–
13. A combination of capacitors is set up as
shown in the figure. The magnitude of
the electric field, due to a point charge Q
(having a charge equal to the sum of the
charges on the 4 µF and 9 µF capacitors),
at a point distant 30 m from it, would
equal :
(1) 240 N/C
(2) 360 N/C
(3) 420 N/C
(4) 480 N/C
14. The temperature dependence of resistances
of Cu and undoped Si in the temperature
range 300-400 K, is best described by :
(1) Linear increase for Cu, linear
increase for Si.
(2) Linear increase for Cu, exponential
increase for Si.
(3) Linear increase for Cu, exponential
decrease for Si.
(4) Linear decrease for Cu, linear
decrease for Si.
E/Page 10 SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„
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15. ŒÊ ∞∑§‚◊ÊŸ ÃÊ⁄U A fl B ¬˝àÿ∑§ ∑§Ë ‹ê’Ê߸ ‘l’, ◊¥‚◊ÊŸ œÊ⁄UÊ I ¬˝flÊÁ„à „Ò– A ∑§Ê ◊Ê«∏∑§⁄U R ÁòÊíÿÊ ∑§Ê∞∑§ flÎûÊ •ÊÒ⁄U B ∑§Ê ◊Ê«∏∑§⁄U ÷È¡Ê ‘a’ ∑§Ê ∞∑§ flª¸’ŸÊÿÊ ¡ÊÃÊ „Ò– ÿÁŒ B
A ÃÕÊ B
B ∑˝§◊‡Ê— flÎûÊ ∑§
∑§ãŒ˝ ÃÕÊ flª¸ ∑§ ∑§ãŒ˝ ¬⁄U øÈê’∑§Ëÿ ˇÊòÊ „Ò ¥,
Ã’ •ŸÈ¬Êà A
B
B
B
„ÊªÊ —
(1)2
8
π
(2)2
16 2
π
(3)2
16
π
(4)2
8 2
π
16. ŒÊ øÈê’∑§Ëÿ ¬ŒÊÕ¸ A ÃÕÊ B ∑§ Á‹ÿ Á„S≈⁄UÁ‚‚-‹Í¬ ŸËø ÁŒπÊÿ ªÿ „Ò¥ —
ߟ ¬ŒÊÕÊZ ∑§Ê øÈê’∑§Ëÿ ©¬ÿʪ ÁfllÈÃ-¡Ÿ⁄U≈U⁄U ∑§øÈê’∑§, ≈˛UÊã‚»§ÊÚ◊¸⁄U ∑§Ë ∑˝§Ê«U ∞fl¥ ÁfllÈÃ-øÈê’∑§ ∑§Ë∑˝§Ê«U •ÊÁŒ ∑§ ’ŸÊŸ ◊¥ Á∑§ÿÊ ¡ÊÃÊ „Ò– Ã’ ÿ„ ©ÁøÃ„Ò Á∑§ —
(1) A ∑§Ê ßSÃ◊Ê‹ ÁfllÈÃ-¡Ÿ⁄U≈U⁄U ÃÕÊ ≈˛UÊã‚»§ÊÚ◊¸⁄UŒÊŸÊ¥ ◊¥ Á∑§ÿÊ ¡Ê∞–
(2) A ∑§Ê ßSÃ◊Ê‹ ÁfllÈÃ-øÈê’∑§ ◊¥ ÃÕÊ B ∑§ÊÁfllÈÃ-¡Ÿ⁄U≈U⁄U ◊¥ Á∑§ÿÊ ¡Ê∞–
(3) A ∑§Ê ßSÃ◊Ê‹ ≈˛UÊã‚»§ÊÚ◊¸⁄U ◊¥ ÃÕÊ B ∑§ÊÁfllÈÃ-¡Ÿ⁄U≈U⁄U ◊¥ Á∑§ÿÊ ¡Ê∞–
(4) B ∑§Ê ßSÃ◊Ê‹ ÁfllÈÃ-øÈê’∑§ ÃÕÊ ≈˛UÊã‚»§ÊÚ◊¸⁄UŒÊŸÊ¥ ◊¥ Á∑§ÿÊ ¡Ê∞–
15. Two identical wires A and B, each of length
‘l’, carry the same current I. Wire A is bent
into a circle of radius R and wire B is bent
to form a square of side ‘a’. If BA
and BB
are the values of magnetic field at the
centres of the circle and square
respectively, then the ratio A
B
B
B
is :
(1)2
8
π
(2)2
16 2
π
(3)2
16
π
(4)2
8 2
π
16. Hysteresis loops for two magnetic materials
A and B are given below :
These materials are used to make magnets
for electric generators, transformer core
and electromagnet core. Then it is proper
to use :
(1) A for electric generators and
transformers.
(2) A for electromagnets and B for
electric generators.
(3) A for transformers and B for electric
generators.
(4) B for electromagnets and
transformers.
SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„E/Page 11
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17. ∞∑§ •Ê∑¸§ ‹Òê¬ ∑§Ê ¬˝∑§ÊÁ‡Êà ∑§⁄UŸ ∑§ Á‹ÿ 80 V ¬⁄U10 A ∑§Ë ÁŒc≈U œÊ⁄UÊ (DC) ∑§Ë •Êfl‡ÿ∑§ÃÊ „ÊÃË „Ò–©‚Ë •Ê∑¸§ ∑§Ê 220 V (rms) 50 Hz ¬˝àÿÊflÃ˸ œÊ⁄UÊ(AC) ‚ ø‹ÊŸ ∑§ Á‹ÿ üÊáÊË ◊¥ ‹ªŸ flÊ‹ ¬˝⁄U∑§àfl∑§Ê ◊ÊŸ „Ò —
(1) 80 H
(2) 0.08 H
(3) 0.044 H
(4) 0.065 H
18. ÁŸêŸ ¬˝Áà ÄflÊ¥≈U◊ flÒlÈÃ-øÈê’∑§Ëÿ ÁflÁ∑§⁄UáÊÊ¥ ∑§Ê ©Ÿ∑§Ë™§¡Ê¸ ∑§ ’…∏à „È∞ ∑˝§◊ ◊¥ ‹ªÊÿ¥ —
A : ŸË‹Ê ¬˝∑§Ê‡Ê B : ¬Ë‹Ê ¬˝∑§Ê‡Ê
C : X - Á∑§⁄UáÊ¥ D : ⁄UÁ«UÿÊ Ã⁄¥Uª
(1) D, B, A, C
(2) A, B, D, C
(3) C, A, B, D
(4) B, A, D, C
19. ŒÍ⁄U ÁSÕà 10 m ™°§ø ¬«∏ ∑§Ê ∞∑§ 20 •Êflœ¸Ÿ ˇÊ◊ÃÊflÊ‹ ≈UÁ‹S∑§Ê¬ ‚ ŒπŸ ¬⁄U ÄÿÊ ◊„‚Í‚ „ʪÊ?
(1) ¬«∏ 10 ªÈŸÊ ™°§øÊ „Ò–
(2) ¬«∏ 10 ªÈŸÊ ¬Ê‚ „Ò–
(3) ¬«∏ 20 ªÈŸÊ ™°§øÊ „Ò–
(4) ¬«∏ 20 ªÈŸÊ ¬Ê‚ „Ò–
17. An arc lamp requires a direct current of
10 A at 80 V to function. If it is connected
to a 220 V (rms), 50 Hz AC supply, the
series inductor needed for it to work is
close to :
(1) 80 H
(2) 0.08 H
(3) 0.044 H
(4) 0.065 H
18. Arrange the following electromagnetic
radiations per quantum in the order of
increasing energy :
A : Blue light B : Yellow light
C : X-ray D : Radiowave.
(1) D, B, A, C
(2) A, B, D, C
(3) C, A, B, D
(4) B, A, D, C
19. An observer looks at a distant tree of
height 10 m with a telescope of magnifying
power of 20. To the observer the tree
appears :
(1) 10 times taller.
(2) 10 times nearer.
(3) 20 times taller.
(4) 20 times nearer.
E/Page 12 SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„
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20. ∞∑§ Á¬Ÿ-„Ê‹ ∑Ò§◊⁄UÊ ∑§Ë ÀÊê’Ê߸ ‘L’ „Ò ÃÕÊ Á¿UŒ˝ ∑§ËÁòÊíÿÊ a „Ò– ©‚ ¬⁄U λ Ã⁄¥UªŒÒÉÿ¸ ∑§Ê ‚◊Ê¥Ã⁄U ¬˝∑§Ê‡Ê•Ê¬ÁÃà „Ò– Á¿UŒ˝ ∑§ ‚Ê◊Ÿ flÊ‹Ë ‚Ä ¬⁄U ’Ÿ S¬ÊÚ≈U∑§Ê ÁflSÃÊ⁄U Á¿UŒ˝ ∑§ íÿÊÁ◊ÃËÿ •Ê∑§Ê⁄U ÃÕÊ ÁflfløŸ∑§ ∑§Ê⁄UáÊ „È∞ ÁflSÃÊ⁄U ∑§Ê ∑ȧ‹ ÿʪ „Ò– ß‚ S¬ÊÚ≈U ∑§ÊãÿÍŸÃ◊ •Ê∑§Ê⁄U b
min Ã’ „ÊªÊ ¡’ —
(1)2
aL=
λÃÕÊ b
min=
22
L
λ
(2) a L= λ ÃÕÊ bmin
=
22
L
λ
(3) a L= λ ÃÕÊ bmin
= 4 Lλ
(4)2
aL=
λÃÕÊ b
min= 4 Lλ
21. ∞∑§ »§Ê≈UÊ-‚‹ ¬⁄U λ Ã⁄¥UªŒÒÉÿ¸ ∑§Ê ¬˝∑§Ê‡Ê •Ê¬ÁÃÄҖ ©à‚Á¡¸Ã ß‹Ä≈˛UÊÚŸ ∑§Ë •Áœ∑§Ã◊ ªÁà ‘v’ „Ò–
ÿÁŒ Ã⁄¥UªŒÒÉÿ¸ 3
4
λ „Ê Ã’ ©à‚Á¡¸Ã ß‹Ä≈˛UÊÚŸ ∑§Ë
•Áœ∑§Ã◊ ªÁà „ÊªË —
(1)
1
24
3v
>
(2)
1
24
3v
<
(3)
1
24
3v =
(4)
1
23
4v =
20. The box of a pin hole camera, of length L,
has a hole of radius a. It is assumed that
when the hole is illuminated by a parallel
beam of light of wavelength λ the spread
of the spot (obtained on the opposite wall
of the camera) is the sum of its geometrical
spread and the spread due to diffraction.
The spot would then have its minimum
size (say bmin
) when :
(1)2
aL=
λand b
min=
22
L
λ
(2) a L= λ and bmin
=
22
L
λ
(3) a L= λ and bmin
= 4 Lλ
(4)2
aL=
λand b
min= 4 Lλ
21. Radiation of wavelength λ, is incident on
a photocell. The fastest emitted electron
has speed v. If the wavelength is changed
to 3
4
λ, the speed of the fastest emitted
electron will be :
(1)
1
24
3v
>
(2)
1
24
3v
<
(3)
1
24
3v =
(4)
1
23
4v =
SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„E/Page 13
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22. ŒÊ ⁄UÁ«UÿÊœ◊˸ Ãàfl A ÃÕÊ B ∑§Ë •h¸•ÊÿÈ ∑˝§◊‡Ê—20 min ÃÕÊ 40 min „Ò¥– ¬˝Ê⁄¥U÷ ◊¥ ŒÊŸÊ¥ ∑§ Ÿ◊ÍŸÊ¥ ◊¥ŸÊÁ÷∑§Ê¥ ∑§Ë ‚¥ÅÿÊ ’⁄UÊ’⁄U „Ò– 80 min ∑§ ©¬⁄UÊ¥ÃA ÃÕÊ B ∑§ ˇÊÿ „È∞ ŸÊÁ÷∑§Ê¥ ∑§Ë ‚¥ÅÿÊ ∑§Ê •ŸÈ¬ÊÃ„ÊªÊ —
(1) 1 : 16
(2) 4 : 1
(3) 1 : 4
(4) 5 : 4
23. ∞∑§ ª≈U ◊¥ a, b, c, d ߟ¬È≈U „Ò¥ •ÊÒ⁄U x •Ê™§≈U¬È≈U „Ò–Ã’ ÁŒÿ ªÿ ≈UÊß◊-ª˝Ê»§ ∑§ •ŸÈ‚Ê⁄U ª≈U „Ò —
(1) NOT
(2) AND
(3) OR
(4) NAND
22. Half-lives of two radioactive elements
A and B are 20 minutes and 40 minutes,
respectively. Initially, the samples have
equal number of nuclei. After 80 minutes,
the ratio of decayed numbers of A and B
nuclei will be :
(1) 1 : 16
(2) 4 : 1
(3) 1 : 4
(4) 5 : 4
23. If a, b, c, d are inputs to a gate and x is its
output, then, as per the following time
graph, the gate is :
(1) NOT
(2) AND
(3) OR
(4) NAND
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24. ‚„Ë ∑§ÕŸ øÈÁŸÿ —(1) •ÊÿÊ◊ ◊Ê«È‹Ÿ ◊¥ ©ìÊ •ÊflÎÁûÊ ∑§Ë flÊ„∑§U
Ã⁄¥Uª ∑§ •ÊÿÊ◊ ◊¥ ’Œ‹Êfl äflÁŸ Á‚ÇŸ‹ ∑§•ÊÿÊ◊ ∑§ •ŸÈ¬ÊÃË „Ò–
(2) •ÊÿÊ◊ ◊Ê«ÈU‹Ÿ ◊¥ ©ìÊ •ÊflÎÁûÊ ∑§Ë flÊ„∑§Ã⁄¥Uª ∑§Ë •ÊflÎÁûÊ ◊¥ ’Œ‹Êfl äflÁŸ Á‚ÇãÊ‹ ∑§•ÊÿÊ◊ ∑§ •ŸÈ¬ÊÃË „Ò–
(3) •ÊflÎÁûÊ ◊Ê«ÈU‹Ÿ ◊¥ ©ìÊ •ÊflÎÁûÊ ∑§Ë flÊ„∑§Ã⁄¥Uª ∑§ •ÊÿÊ◊ ◊¥ ’Œ‹Êfl äflÁŸ Á‚ÇãÊ‹ ∑§•ÊÿÊ◊ ∑§ •ŸÈ¬ÊÃË „Ò–
(4) •ÊflÎÁûÊ ◊Ê«ÈU‹Ÿ ◊¥ ©ìÊ-•ÊflÎÁûÊ ∑§Ë flÊ„∑§Ã⁄¥Uª ∑§Ë •ÊÿÊ◊ ◊¥ ’Œ‹Êfl äflÁŸ Á‚ÇŸ‹ ∑§Ë•ÊflÎÁûÊ ∑§ •ŸÈ¬ÊÃË „Ò–
25. ∞∑§ S∑ͧ-ª¡ ∑§Ê Á¬ø 0.5 mm „Ò •ÊÒ⁄U ©‚∑§ flÎûÊËÿ-S∑§‹ ¬⁄U 50 ÷ʪ „Ò¥– ß‚∑§ mÊ⁄UÊ ∞∑§ ¬Ã‹Ë•ÀÿÈ◊ËÁŸÿ◊ ‡ÊË≈U ∑§Ë ◊Ê≈UÊ߸ ◊Ê¬Ë ªß¸– ◊ʬ ‹Ÿ ∑§¬Ífl¸ ÿ„ ¬ÊÿÊ ªÿÊ Á∑§ ¡’ S∑˝Í§-ª¡ ∑§ ŒÊ ¡ÊÚflÊ¥ ∑§ÊSÊê¬∑¸§U ◊¥ ‹ÊÿÊ ¡ÊÃÊ „Ò Ã’ 45 flÊ¥ ÷ʪ ◊ÈÅÿ S∑§‹‹Ê߸Ÿ ∑§ ‚¥¬ÊÃË „ÊÃÊ „Ò •ÊÒ⁄U ◊ÈÅÿ S∑§‹ ∑§Ê ‡ÊÍãÿ (0)
◊ÈÁ‡∑§‹ ‚ ÁŒπÃÊ „Ò– ◊ÈÅÿ S∑§‹ ∑§Ê ¬Ê∆KÊ¥∑§ ÿÁŒ0.5 mm ÃÕÊ 25 flÊ¥ ÷ʪ ◊ÈÅÿ S∑§‹ ‹Ê߸Ÿ ∑§‚¥¬ÊÃË „Ê, ÃÊ ‡ÊË≈U ∑§Ë ◊Ê≈UÊ߸ ÄÿÊ „ʪË?
(1) 0.75 mm
(2) 0.80 mm
(3) 0.70 mm
(4) 0.50 mm
24. Choose the correct statement :
(1) In amplitude modulation the
amplitude of the high frequency
carrier wave is made to vary in
proportion to the amplitude of the
audio signal.
(2) In amplitude modulation the
frequency of the high frequency
carrier wave is made to vary in
proportion to the amplitude of the
audio signal.
(3) In frequency modulation the
amplitude of the high frequency
carrier wave is made to vary in
proportion to the amplitude of the
audio signal.
(4) In frequency modulation the
amplitude of the high frequency
carrier wave is made to vary in
proportion to the frequency of the
audio signal.
25. A screw gauge with a pitch of 0.5 mm and
a circular scale with 50 divisions is used to
measure the thickness of a thin sheet of
Aluminium. Before starting the
measurement, it is found that when the
two jaws of the screw gauge are brought
in contact, the 45th division coincides with
the main scale line and that the zero of
the main scale is barely visible. What is
the thickness of the sheet if the main scale
reading is 0.5 mm and the 25th division
coincides with the main scale line ?
(1) 0.75 mm
(2) 0.80 mm
(3) 0.70 mm
(4) 0.50 mm
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26. ŒÊŸÊ¥ Á‚⁄UÊ¥ ¬⁄U πÈ‹ ∞∑§ ¬Ê߬ ∑§Ë flÊÿÈ ◊¥◊Í‹-•ÊflÎÁûÊ ‘f ’ „Ò– ¬Ê߬ ∑§Ê ™§äflʸœ⁄U ©‚∑§Ë•ÊœË-‹ê’Ê߸ Ã∑§ ¬ÊŸË ◊¥ «ÈU’ÊÿÊ ¡ÊÃÊ „Ò– Ã’ ß‚◊¥’ø flÊÿÈ-∑§Ê‹◊ ∑§Ë ◊Í‹ •ÊflÎÁûÊ „ÊªË —
(1)2
f
(2)3
4
f
(3) 2 f
(4) f
27. ∞∑§ ªÒÀflŸÊ◊Ë≈U⁄U ∑§ ∑§Êß‹ ∑§Ê ¬˝ÁÃ⁄UÊœ 100 Ω „Ò–1 mA œÊ⁄UÊ ¬˝flÊÁ„à ∑§⁄UŸ ¬⁄U ß‚◊¥ »È§‹-S∑§‹ ÁflˇÊ¬Á◊‹ÃÊ „Ò– ß‚ ªÒÀflŸÊ◊Ë≈U⁄U ∑§Ê 10 A ∑§ ∞◊Ë≈U⁄U ◊¥’Œ‹Ÿ ∑§ Á‹ÿ ¡Ê ¬˝ÁÃ⁄UÊœ ‹ªÊŸÊ „ÊªÊ fl„ „Ò —
(1) 0.01 Ω
(2) 2 Ω
(3) 0.1 Ω
(4) 3 Ω
28. ∞∑§ ¬˝ÿʪ ∑§⁄U∑§ ÃÕÊ i− δ ª˝Ê»§ ’ŸÊ∑§⁄U ∞∑§ ∑§Ê°ø‚ ’Ÿ Á¬˝ï◊ ∑§Ê •¬fløŸÊ¥∑§ ÁŸ∑§Ê‹Ê ¡ÊÃÊ „Ò– ¡’∞∑§ Á∑§⁄UáÊ ∑§Ê 358 ¬⁄U •Ê¬ÁÃà ∑§⁄UŸ ¬⁄U fl„ 408 ‚ÁfløÁ‹Ã „ÊÃË „Ò ÃÕÊ ÿ„ 798 ¬⁄U ÁŸª¸◊ „ÊÃË „Ò– ß‚ÁSÕÁà ◊¥ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ‚Ê ◊ÊŸ •¬fløŸÊ¥∑§ ∑§•Áœ∑§Ã◊ ◊ÊŸ ∑§ ‚’‚ ¬Ê‚ „Ò?
(1) 1.5
(2) 1.6
(3) 1.7
(4) 1.8
26. A pipe open at both ends has a
fundamental frequency f in air. The pipe
is dipped vertically in water so that half of
it is in water. The fundamental frequency
of the air column is now :
(1)2
f
(2)3
4
f
(3) 2 f
(4) f
27. A galvanometer having a coil resistance
of 100 Ω gives a full scale deflection, when
a current of 1 mA is passed through it.
The value of the resistance, which can
convert this galvanometer into ammeter
giving a full scale deflection for a current
of 10 A, is :
(1) 0.01 Ω
(2) 2 Ω
(3) 0.1 Ω
(4) 3 Ω
28. In an experiment for determination of
refractive index of glass of a prism by
i− δ, plot, it was found that a ray incident
at angle 358, suffers a deviation of 408 and
that it emerges at angle 798⋅ Ιn that case
which of the following is closest to the
maximum possible value of the refractive
index ?
(1) 1.5
(2) 1.6
(3) 1.7
(4) 1.8
E/Page 16 SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„
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29. ÁøòÊ (a), (b), (c), (d) Œπ∑§⁄U ÁŸœÊ¸Á⁄Uà ∑§⁄¥U Á∑§ ÿÁøòÊ ∑˝§◊‡Ê— Á∑§Ÿ ‚◊Ë∑§ã«UÄ≈U⁄U Á«UflÊ߸‚ ∑§•Á÷‹ˇÊÁáÊ∑§ ª˝Ê»§ „Ò¥?
(1) ‚ÊœÊ⁄UáÊ «UÊÿÊ«U, ¡ËŸ⁄U «UÊÿÊ«, ‚Ê‹⁄U ‚‹,LDR (‹Ê߸≈U Á«U¬ã«Uã≈U ⁄UÁ¡S≈Uã‚)
(2) ¡ËŸ⁄U «UÊÿÊ«U, ‚ÊœÊ⁄UáÊ «UÊÿÊ«U, LDR (‹Ê߸≈UÁ«U¬ã«Uã≈U ⁄UÁ¡S≈Uã‚), ‚Ê‹⁄U ‚‹
(3) ‚Ê‹⁄U ‚‹, LDR (‹Êß≈U Á«U¬ã«Uã≈U ⁄UÁ¡S≈Uã‚),¡ËŸ⁄U «UÊÿÊ«U, ‚ÊœÊ⁄UáÊ «UÊÿÊ«U
(4) ¡ËŸ⁄U «UÊÿÊ«U, ‚Ê‹⁄U ‚‹, ‚ÊœÊ⁄UáÊ «UÊÿÊ«U,LDR (‹Ê߸≈U Á«U¬ã«Uã≈U ⁄UÁ¡S≈Uã‚)
30. ©÷ÿÁŸc∆U-©à‚¡¸∑§ ÁflãÿÊ‚ ∑§ Á‹ÿ α ÃÕÊ β ∑§’Ëø ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ‚Ê ‚¥’¥œ ª‹Ã „Ò? α ÃÕÊ βÁøq ‚Ê◊Êãÿ ◊Ë’ flÊ‹ „Ò¥ —
(1)1 1
1= +
α β
(2) 1
=
−
βα
β
(3) 1
=
+
βα
β
(4)2
2
1
=
+
βα
β
29. Identify the semiconductor devices whose
characteristics are given below, in the
order (a), (b), (c), (d) :
(1) Simple diode, Zener diode, Solar cell,
Light dependent resistance
(2) Zener diode, Simple diode, Light
dependent resistance, Solar cell
(3) Solar cell, Light dependent
resistance, Zener diode, Simple
diode
(4) Zener diode, Solar cell, Simple diode,
Light dependent resistance
30. For a common emitter configuration, if
α and β have their usual meanings, the
incorrect relationship between α and β
is :
(1)1 1
1= +
α β
(2) 1
=
−
βα
β
(3) 1
=
+
βα
β
(4)2
2
1
=
+
βα
β
SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„E/Page 17
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÷ʪ B — ⁄U‚ÊÿŸ ÁflôÊÊŸ
31. 300 K ÃÕÊ 1 atm ŒÊ’ ¬⁄ U, 15 mL ªÒ‚Ëÿ„Êß«˛UÊ∑§Ê’¸Ÿ ∑§ ¬Íáʸ Œ„Ÿ ∑§ Á‹ÿ 375 mL flÊÿÈÁ¡‚◊¥ •Êÿß ∑§ •ÊœÊ⁄U ¬⁄U 20% •ÊÚÄ‚Ë¡Ÿ „Ò, ∑§Ë•Êfl‡ÿ∑§ÃÊ „ÊÃË „Ò– Œ„Ÿ ∑§ ’ÊŒ ªÒ‚¥ 330 mL
ÉÊ⁄UÃË „Ò– ÿ„ ◊ÊŸÃ „È∞ Á∑§ ’ŸÊ „È•Ê ¡‹ Œ˝fl M§¬ ◊¥„Ò ÃÕÊ ©‚Ë Ãʬ◊ÊŸ ∞fl¥ ŒÊ’ ¬⁄U •Êÿßʥ ∑§Ë ◊ʬ ∑§Ëªß¸ „Ò ÃÊ „Êß«˛UÊ∑§Ê’¸Ÿ ∑§Ê »§Ê◊¸Í‹Ê „Ò —
(1) C3H
6
(2) C3H
8
(3) C4H
8
(4) C4H
10
32. ‚◊ÊŸ •Êÿß (V) ∑§ ŒÊ ’¥Œ ’À’, Á¡Ÿ◊¥ ∞∑§ •ÊŒ‡Ê¸ªÒ‚ ¬˝Ê⁄UÁê÷∑§ ŒÊ’ pi ÃÕÊ Ãʬ T
1 ¬⁄U ÷⁄UË ªß¸ „Ò, ∞∑§Ÿªáÿ •Êÿß ∑§Ë ¬Ã‹Ë ≈˜UÿÍ’ ‚ ¡È«∏ „Ò¥ ¡Ò‚Ê Á∑§ŸËø ∑§ ÁøòÊ ◊¥ ÁŒπÊÿÊ ªÿÊ „Ò– Á»§⁄U ߟ◊¥ ‚ ∞∑§’À’ ∑§Ê Ãʬ ’…∏Ê∑§⁄U T
2 ∑§⁄U ÁŒÿÊ ¡ÊÃÊ „Ò– •¥ÁÃ◊ŒÊ’ pf „Ò —
(1)1 2
1 2
i
T Tp
T T
+
(2)1
1 2
2 i
Tp
T T
+
(3)2
1 2
2 i
Tp
T T
+
(4)1 2
1 2
2 i
T Tp
T T
+
PART B — CHEMISTRY
31. At 300 K and 1 atm, 15 mL of a gaseous
hydrocarbon requires 375 mL air
containing 20% O2 by volume for complete
combustion. After combustion the gases
occupy 330 mL. Assuming that the water
formed is in liquid form and the volumes
were measured at the same temperature
and pressure, the formula of the
hydrocarbon is :
(1) C3H
6
(2) C3H
8
(3) C4H
8
(4) C4H
10
32. Two closed bulbs of equal volume (V)
containing an ideal gas initially at pressure
pi and temperature T1 are connected
through a narrow tube of negligible
volume as shown in the figure below. The
temperature of one of the bulbs is then
raised to T2. The final pressure pf is :
(1) 1 2
1 2
i
T Tp
T T
+
(2)1
1 2
2 i
Tp
T T
+
(3)2
1 2
2 i
Tp
T T
+
(4)1 2
1 2
2 i
T Tp
T T
+
E/Page 18 SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„
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33. ∞∑§ ª◊¸ Á»§‹Ê◊¥≈U ‚ ÁŸ∑§‹Ë ß‹Ä≈˛UÊÚŸ œÊ⁄UÊ ∑§ÊV esu ∑§ Áfl÷flÊãÃ⁄U ¬⁄ ⁄Uπ ŒÊ •ÊflÁ‡Êà åÀÊ≈UÊ¥ ∑§’Ëø ‚ ÷¡Ê ¡ÊÃÊ „Ò– ÿÁŒ ß‹Ä≈˛UÊÚŸ ∑§ •Êfl‡Ê ÃÕÊ‚¥„Áà ∑˝§◊‡Ê— e ÃÕÊ m „Ê¥ ÃÊ h/λ ∑§Ê ◊ÊŸ ÁŸêŸ ◊¥ ‚Á∑§‚∑§ mÊ⁄UÊ ÁŒÿÊ ¡ÊÿªÊ? (¡’ ß‹Ä≈˛UÊÚŸ Ã⁄¥Uª ‚‚ê’ÁãœÃ Ã⁄¥UªŒÒäÿ¸ λ „Ò)
(1) meV
(2) 2meV
(3) meV
(4) 2 meV
34. fl„ S¬Ë‡ÊË$¡, Á¡‚◊¥ N ¬⁄U◊ÊáÊÈ sp ‚¥∑§⁄UáÊ ∑§Ë •flSÕÊ◊¥ „Ò, „ÊªË —
(1)2
NO+
(2)2
NO−
(3)3NO−
(4) NO2
35. ∑§Ê’Ÿ ÃÕÊ ∑§Ê’Ÿ ◊ÊŸÊÄ‚ÊÚß«U ∑§Ë Œ„Ÿ ™§c◊Êÿ¥ ∑§◊‡Ê—−393.5 ÃÕÊ −283.5 kJ mol−1 „Ò¥– ∑§Ê’¸Ÿ◊ÊŸÊÄ‚Êß«U ∑§Ë ‚¥÷flŸ ™§c◊Ê (kJ ◊¥) ¬Áà ◊Ê‹ „ÊªË —
(1) 110.5
(2) 676.5
(3) −676.5
(4) −110.5
33. A stream of electrons from a heated
filament was passed between two charged
plates kept at a potential difference V esu.
If e and m are charge and mass of an
electron, respectively, then the value of
h/λ (where λ is wavelength associated
with electron wave) is given by :
(1) meV
(2) 2meV
(3) meV
(4) 2 meV
34. The species in which the N atom is in a
state of sp hybridization is :
(1)2
NO+
(2)2
NO−
(3)3NO−
(4) NO2
35. The heats of combustion of carbon and
carbon monoxide are −393.5 and
−283.5 kJ mol−1, respectively. The heat
of formation (in kJ) of carbon monoxide
per mole is :
(1) 110.5
(2) 676.5
(3) −676.5
(4) −110.5
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36. 18 g Ç‹È∑§Ê‚ (C6H
12O
6) ∑§Ê 178.2 g ¬ÊŸË ◊¥
Á◊‹ÊÿÊ ¡ÊÃÊ „Ò– ß‚ ¡‹Ëÿ Áfl‹ÿŸ ∑§ Á‹∞ ¡‹∑§Ê flÊc¬ ŒÊ’ (torr ◊¥) „ÊªÊ —
(1) 7.6
(2) 76.0
(3) 752.4
(4) 759.0
37. Ãʬ◊ÊŸ 298 K ¬⁄U, ∞∑§ •Á÷Á∑˝§ÿÊ A+BC+D
∑§ Á‹∞ ‚Êêÿ ÁSÕ⁄UÊ¥∑§ 100 „Ò– ÿÁŒ ¬˝Ê⁄UÁê÷∑§ ‚ÊãŒ˝ÃÊ‚÷Ë øÊ⁄UÊ¥ S¬Ë‡ÊË¡ ◊¥ ‚ ¬˝àÿ∑§ ∑§Ë 1 M „ÊÃË, ÃÊ D∑§Ë ‚Êêÿ ‚ÊãŒ˝ÃÊ (mol L−1 ◊¥) „ÊªË —
(1) 0.182
(2) 0.818
(3) 1.818
(4) 1.182
38. ªÒÀflŸÊß¡‡ÊŸ ÁŸêŸ ◊¥ ‚ Á∑§‚∑§ ∑§Ê≈U ‚ „ÊÃÊ „Ò?
(1) Pb
(2) Cr
(3) Cu
(4) Zn
36. 18 g glucose (C6H
12O
6) is added to
178.2 g water. The vapor pressure of
water (in torr) for this aqueous solution
is :
(1) 7.6
(2) 76.0
(3) 752.4
(4) 759.0
37. The equilibrium constant at 298 K for a
reaction A+BC+D is 100. If the initial
concentration of all the four species were
1 M each, then equilibrium concentration
of D (in mol L−1) will be :
(1) 0.182
(2) 0.818
(3) 1.818
(4) 1.182
38. Galvanization is applying a coating of :
(1) Pb
(2) Cr
(3) Cu
(4) Zn
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39. H2O
2 ∑§Ê ÁflÉÊ≈UŸ ∞∑§ ¬˝Õ◊ ∑§ÊÁ≈U ∑§Ë •Á÷Á∑˝§ÿÊ
„Ò– ¬øÊ‚ Á◊Ÿ≈U ◊¥ ß‚ ¬˝∑§Ê⁄U ∑§ ÁflÉÊ≈UŸ ◊¥ H2O
2
∑§Ë ‚ÊãŒ˝ÃÊ ÉÊ≈U∑§⁄U 0.5 ‚ 0.125 M „Ê ¡ÊÃË „Ò– ¡’H
2O
2 ∑§Ë ‚ÊãŒ˝ÃÊ 0.05 M ¬„È°øÃË „Ò, ÃÊ O
2 ∑§
’ŸŸ ∑§Ë Œ⁄U „ÊªË —
(1) 6.93×10−2 mol min−1
(2) 6.93×10−4 mol min−1
(3) 2.66 L min−1 (STP ¬⁄U)
(4) 1.34×10−2 mol min−1
40. » ˝ § Ê Úÿã« U Á‹∑§ •Áœ‡ÊÊ ·áÊ ‚◊ÃÊ¬Ë fl∑ ˝ § ◊ ¥log (x/m) ÃÕÊ log p ∑§ ’Ëø πË¥ø ªÿ ⁄UπËÿå‹Ê≈U ∑§ Á‹∞ ÁãÊêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ‚„Ë „Ò?(k ÃÕÊ n ÁSÕ⁄UÊ¥∑§ „Ò¥)
(1) k ÃÕÊ 1/n ŒÊŸÊ¥ „Ë S‹Ê¬ ¬Œ ◊¥ •Êà „Ò¥–
(2) 1/n ßã≈U⁄U‚å≈U ∑§ M§¬ •ÊÃÊ „Ò–
(3) ◊ÊòÊ 1/n S‹Ê¬ ∑§ M§¬ ◊¥ •ÊÃÊ „Ò–
(4) log (1/n) ßã≈U⁄U‚å≈U ∑§ M§¬ ◊¥ •ÊÃÊ „Ò–
41. ÁŸêŸ ¬⁄U◊ÊáÊÈ•Ê¥ ◊¥ Á∑§‚∑§Ë ¬˝Õ◊ •ÊÿŸŸ ™§¡Ê¸ ©ëøÃ◊„Ò?
(1) Rb
(2) Na
(3) K
(4) Sc
39. Decomposition of H2O
2 follows a first
order reaction. In fifty minutes the
concentration of H2O
2 decreases from
0.5 to 0.125 M in one such decomposition.
When the concentration of H2O
2 reaches
0.05 M, the rate of formation of O2 will
be :
(1) 6.93×10−2 mol min−1
(2) 6.93×10−4 mol min−1
(3) 2.66 L min−1 at STP
(4) 1.34×10−2 mol min−1
40. For a linear plot of log (x/m) versus log p
in a Freundlich adsorption isotherm,
which of the following statements is
correct ? (k and n are constants)
(1) Both k and 1/n appear in the slope
term.
(2) 1/n appears as the intercept.
(3) Only 1/n appears as the slope.
(4) log (1/n) appears as the intercept.
41. Which of the following atoms has the
highest first ionization energy ?
(1) Rb
(2) Na
(3) K
(4) Sc
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42. »˝§ÊÚÕ ç‹Ê≈U‡ÊŸ ÁflÁœ mÊ⁄UÊ ÁŸêŸ ◊¥ ‚ fl„ ∑§ÊÒŸ ‚Ê•ÿS∑§ ‚flʸÁœ∑§ M§¬ ‚ ‚ÊÁãŒ˝Ã Á∑§ÿÊ ¡Ê ‚∑§ÃÊ„Ò?
(1) ◊ÒÇŸ≈UÊß≈U
(2) Á‚«U⁄UÊß≈U
(3) ªÒ‹ŸÊ
(4) ◊Ò‹Ê∑§Êß≈U
43. ¡‹ ∑§ ‚ê’㜠◊¥ ÁŸêŸ ∑§ÕŸÊ¥ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∞∑§ª‹Ã „Ò?
(1) ¬˝∑§Ê‡Ê‚¥‡‹·áÊ ◊¥ ¡‹ •ÊÄ‚Ë∑Χà „Ê∑§⁄U•ÊÄ‚Ë$¡Ÿ ŒÃÊ „Ò–
(2) ¡‹, •ê‹ ÃÕÊ ˇÊÊ⁄U∑§ ŒÊŸÊ¥ „Ë M§¬ ◊¥ ∑§Êÿ¸∑§⁄U ‚∑§ÃÊ „Ò–
(3) ß‚∑§ ‚¥ÉÊÁŸÃ ¬˝ÊflSÕÊ ◊¥ ÁflSÃËáʸ •¥Ã—•áÊÈ∑§„Êß«˛UÊ¡Ÿ •Ê’㜠„Êà „Ò¥–
(4) ÷Ê⁄UË ¡‹ mÊ⁄UÊ ’ŸÊ ’»¸§ ‚Ê◊Êãÿ ¡‹ ◊¥ «ÍU’ÃÊ„Ò–
44. „flÊ ∑§ •ÊÁœÄÿ ◊¥ Li, Na •ÊÒ⁄U K ∑§ Œ„Ÿ ¬⁄U’ŸŸflÊ‹Ë ◊ÈÅÿ •ÊÄ‚Êß«¥U ∑˝§◊‡Ê— „Ò¥ —
(1) Li2O, Na
2O ÃÕÊ KO
2
(2) LiO2, Na
2O
2 ÃÕÊ K
2O
(3) Li2O
2, Na
2O
2 ÃÕÊ KO
2
(4) Li2O, Na
2O
2 ÃÕÊ KO
2
42. Which one of the following ores is best
concentrated by froth floatation method ?
(1) Magnetite
(2) Siderite
(3) Galena
(4) Malachite
43. Which one of the following statements
about water is FALSE ?
(1) Water is oxidized to oxygen during
photosynthesis.
(2) Water can act both as an acid and
as a base.
(3) There is extensive intramolecular
hydrogen bonding in the condensed
phase.
(4) Ice formed by heavy water sinks in
normal water.
44. The main oxides formed on combustion of
Li, Na and K in excess of air are,
respectively :
(1) Li2O, Na
2O and KO
2
(2) LiO2, Na
2O
2 and K
2O
(3) Li2O
2, Na
2O
2 and KO
2
(4) Li2O, Na
2O
2 and KO
2
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45. ÃŸÈ ÃÕÊ ‚ÊãŒ˝ ŸÊßÁ≈˛U∑§ ∞Á‚«U ∑§ ‚ÊÕ Á¡¥∑§ ∑§Ë•Á÷Á∑˝§ÿÊ mÊ⁄UÊ ∑˝§◊‡Ê— ©à¬ãŸ „Êà „Ò¥ —
(1) N2O ÃÕÊ NO
2
(2) NO2 ÃÕÊ NO
(3) NO ÃÕÊ N2O
(4) NO2 ÃÕÊ N
2O
46. fl„ ÿÈÇ◊ Á¡Ÿ◊¥ »§ÊS»§Ê⁄U‚ ¬⁄U◊ÊáÊÈ•Ê¥ ∑§Ë »§Ê◊¸‹•ÊÚÄ‚Ë∑§⁄UáÊ •flSÕÊ +3 „Ò, „Ò —
(1) •ÊÕʸ»§ÊS»§Ê⁄U‚ ÃÕÊ ¬Êÿ⁄UÊ»§ÊS»§Ê⁄U‚ ∞Á‚«U
(2) ¬Êÿ⁄UÊ»§ÊS»§Ê⁄U‚ ÃÕÊ „Ê߬ʻ§ÊS»§ÊÁ⁄U∑§ ∞Á‚«U
(3) •ÊÕʸ»§ÊS»§Ê⁄U‚ ÃÕÊ „Ê߬ʻ§ÊS»§ÊÁ⁄U∑§ ∞Á‚«U
(4) ¬Êÿ⁄UÊ»§ÊS»§Ê⁄U‚ ÃÕÊ ¬Êÿ⁄UÊ»§ÊS»§ÊÁ⁄U∑§ ∞Á‚«U
47. ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ÿÊÒÁª∑§ œÊÁàfl∑§ ÃÕÊ »§⁄UÊ◊ÒªŸÁ≈U∑§(‹ÊÒ„ øÈê’∑§Ëÿ) „Ò?
(1) TiO2
(2) CrO2
(3) VO2
(4) MnO2
45. The reaction of zinc with dilute and
concentrated nitric acid, respectively,
produces :
(1) N2O and NO
2
(2) NO2 and NO
(3) NO and N2O
(4) NO2 and N
2O
46. The pair in which phosphorous atoms
have a formal oxidation state of +3 is :
(1) Orthophosphorous and
pyrophosphorous acids
(2) Pyrophosphorous and
hypophosphoric acids
(3) Orthophosphorous and
hypophosphoric acids
(4) Pyrophosphorous and
pyrophosphoric acids
47. Which of the following compounds is
metallic and ferromagnetic ?
(1) TiO2
(2) CrO2
(3) VO2
(4) MnO2
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48. ∞∑§„Ë øÈê’∑§Ëÿ •ÊÉÊÍáʸ ∑§Ê ÿÈÇ◊ „Ò —
[At. No. : Cr=24, Mn=25, Fe=26, Co=27]
(1) [Cr(H2O)
6]2+ ÃÕÊ [CoCl
4]2−
(2) [Cr(H2O)
6]2+ ÃÕÊ [Fe(H
2O)
6]2+
(3) [Mn(H2O)
6]2+ ÃÕÊ [Cr(H
2O)
6]2+
(4) [CoCl4]2− ÃÕÊ [Fe(H
2O)
6]2+
49. ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÊÚêå‹Ä‚ ¬˝∑§ÊÁ‡Ê∑§ ‚◊ÊflÿflÃʬ˝ŒÁ‡Ê¸Ã ∑§⁄UªÊ?
(1) [Co(NH3)3Cl
3]
(2) cis[Co(en)2Cl
2]Cl
(3) trans[Co(en)2Cl
2]Cl
(4) [Co(NH3)4Cl
2]Cl
(en=ethylenediamine)
50. ÷ÍÁ◊ªÃ ¤ÊË‹ ‚ ¬˝Êåà ¡‹ ¬˝ÁÃŒ‡Ê¸ ◊¥ ç‹Ê⁄UÊß«U, ‹«U,ŸÊß≈˛U≈U ÃÕÊ •Êÿ⁄UŸ ∑§Ë ‚ÊãŒ˝ÃÊ ∑˝§◊‡Ê— 1000 ppb,
40 ppb, 100 ppm ÃÕÊ 0.2 ppm ¬Ê߸ ªß¸– ÿ„¡‹ ÁŸêŸ ◊¥ ‚ Á∑§‚∑§Ë ©ìÊ ‚ÊãŒ˝ÃÊ ‚ ¬ËŸ ÿÊÇÿŸ„Ë¥ „Ò?
(1) ç‹Ê⁄UÊß«U
(2) ‹«U
(3) ŸÊß≈˛U≈U
(4) •Êÿ⁄UŸ
48. The pair having the same magnetic
moment is :
[At. No. : Cr=24, Mn=25, Fe=26, Co=27]
(1) [Cr(H2O)
6]2+ and [CoCl
4]2−
(2) [Cr(H2O)
6]2+ and [Fe(H
2O)
6]2+
(3) [Mn(H2O)
6]2+ and [Cr(H
2O)
6]2+
(4) [CoCl4]2− and [Fe(H
2O)
6]2+
49. Which one of the following complexes
shows optical isomerism ?
(1) [Co(NH3)3Cl
3]
(2) cis[Co(en)2Cl
2]Cl
(3) trans[Co(en)2Cl
2]Cl
(4) [Co(NH3)4Cl
2]Cl
(en=ethylenediamine)
50. The concentration of fluoride, lead, nitrate
and iron in a water sample from an
underground lake was found to be
1000 ppb, 40 ppb, 100 ppm and 0.2 ppm,
respectively. This water is unsuitable for
drinking due to high concentration of :
(1) Fluoride
(2) Lead
(3) Nitrate
(4) Iron
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51. ‚Ê’ÈŸ ©lʪ ◊¥ ÷ÈÄÇʷ ‹Êß (S¬ã≈U ‹Êß) ‚ ÁÇ‹‚⁄UÊÚ‹¬ÎÕ∑§ ∑§⁄UŸ ∑§ Á‹∞ ‚’‚ ©¬ÿÈÄà •Ê‚flŸ ÁflÁœ„Ò —
(1) ‚Ê◊Êãÿ •Ê‚flŸ
(2) ¬˝÷Ê¡Ë •Ê‚flŸ
(3) ’Êc¬ •Ê‚flŸ
(4) ‚◊ÊŸËà ŒÊ’ ¬⁄U •Ê‚flŸ
52. ŸËø ŒË ªß¸ •Á÷Á∑˝§ÿÊ ∑§ Á‹∞ ©à¬ÊŒ „ÊªÊ —
(1)
(2)
(3)
(4)
51. The distillation technique most suited for
separating glycerol from spent-lye in the
soap industry is :
(1) Simple distillation
(2) Fractional distillation
(3) Steam distillation
(4) Distillation under reduced pressure
52. The product of the reaction given below
is :
(1)
(2)
(3)
(4)
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53. ÁŒ∞ ªÿ ÿÊÒÁª∑§ ∑§Ê ÁŸ⁄U¬ˇÊ ÁflãÿÊ‚ „Ò —
CO2H
OHH
ClH
CH3
(1) (2R, 3S)
(2) (2S, 3R)
(3) (2S, 3S)
(4) (2R, 3R)
54. ◊ÕŸÊÚ‹ ◊ ¥ 2-Ä‹Ê⁄UÊ -2-◊ÁÕ‹¬ã≈ UŸ, ‚ÊÁ«Uÿ◊◊ÕÊÄ‚Êß«U ∑§ ‚ÊÕ •Á÷Á∑˝§ÿÊ ∑§⁄U∑§ ŒÃË „Ò —
(a)
(b)
(c)
(1) ߟ◊¥ ‚ ‚÷Ë
(2) (a) ÃÕÊ (c)
(3) ◊ÊòÊ (c)
(4) (a) ÃÕÊ (b)
53. The absolute configuration of
CO2H
OHH
ClH
CH3
is :
(1) (2R, 3S)
(2) (2S, 3R)
(3) (2S, 3S)
(4) (2R, 3R)
54. 2-chloro-2-methylpentane on reaction
with sodium methoxide in methanol
yields :
(a)
(b)
(c)
(1) All of these
(2) (a) and (c)
(3) (c) only
(4) (a) and (b)
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55. ¬˝Ê¬ËŸ ∑§Ë HOCl (Cl2+H
2O) ∑§ ‚ÊÕ •Á÷Á∑˝§ÿÊ
Á¡‚ ◊äÿflÃ˸ ‚ „Ê∑§⁄U ‚ê¬ãŸ „ÊÃË „Ò, fl„ „Ò —
(1) CH3−CH+−CH
2−OH
(2) CH3−CH+−CH
2−Cl
(3)3 2CH CH(OH) CH+− −
(4)3 2
CH CHCl CH+
− −
56. „Ê»§◊ÊŸ ’˝Ê◊Ê◊Êß«U ÁŸêŸË∑§⁄UáÊ •Á÷Á∑˝§ÿÊ ◊¥, NaOH
ÃÕÊ Br2 ∑§ ¬˝ÿÈÄà ◊Ê‹Ê¥ ∑§Ë ‚¥ÅÿÊ ¬˝ÁÃ◊Ê‹ •◊ËŸ
∑§ ’ŸŸ ◊¥ „ÊªË —
(1) ∞∑§ ◊Ê‹ NaOH ÃÕÊ ∞∑§ ◊Ê‹ Br2–
(2) øÊ⁄U ◊Ê‹ NaOH ÃÕÊ ŒÊ ◊Ê‹ Br2–
(3) ŒÊ ◊Ê‹ NaOH ÃÕÊ ŒÊ ◊Ê‹ Br2–
(4) øÊ⁄U ◊Ê‹ NaOH ÃÕÊ ∞∑§ ◊Ê‹ Br2–
57. ÁŸêŸ ÉÊãÊàfl ∑§ ¬Ê‹ËÕËŸ ∑§ ‚ê’㜠◊¥ ÁŸêŸ ◊¥ ‚∑§ÊÒŸ ‚Ê ∑§ÕŸ ª‹Ã „Ò?
(1) ß‚∑§ ‚¥‡‹·áÊ ◊¥ ©ìÊ ŒÊ’ ∑§Ë •Êfl‡ÿ∑§ÃÊ„ÊÃË „Ò–
(2) ÿ„ ÁfllÈà ∑§Ê „ËŸ øÊ‹∑§ „Ò–
(3) ß‚◊¥ «UÊ߸•ÊÄ‚Ë¡Ÿ •ÕflÊ ¬⁄U•ÊÄ‚Êß«UߟËÁ‚ÿ≈U⁄ (¬˝Ê⁄Uê÷∑§) ©à¬˝⁄U∑§ ∑§ M§¬ ◊¥øÊÁ„∞–
(4) ÿ„ ’∑§≈U (’ÊÀ≈UË), «US≈U-Á’Ÿ, •ÊÁŒ ∑§©à¬ÊŒŸ ◊¥ ¬˝ÿÈÄà „ÊÃË „Ò–
55. The reaction of propene with HOCl
(Cl2+H
2O) proceeds through the
intermediate :
(1) CH3−CH+−CH
2−OH
(2) CH3−CH+−CH
2−Cl
(3)3 2CH CH(OH) CH+− −
(4)3 2
CH CHCl CH+
− −
56. In the Hofmann bromamide degradation
reaction, the number of moles of NaOH
and Br2
used per mole of amine produced
are :
(1) One mole of NaOH and one mole of
Br2
.
(2) Four moles of NaOH and two moles
of Br2
.
(3) Two moles of NaOH and two moles
of Br2
.
(4) Four moles of NaOH and one mole
of Br2
.
57. Which of the following statements about
low density polythene is FALSE ?
(1) Its synthesis requires high pressure.
(2) It is a poor conductor of electricity.
(3) Its synthesis requires dioxygen or a
peroxide initiator as a catalyst.
(4) It is used in the manufacture of
buckets, dust-bins etc.
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58. ÕÊÿÊ‹ ª˝È¬ Á¡‚◊¥ ©¬ÁSÕà „Ò, fl„ „Ò —
(1) ‚Êß≈UÊ‚ËŸ
(2) Á‚ÁS≈UŸ (Cystine)
(3) Á‚S≈UËŸ (Cysteine)
(4) ◊ÕÊß•ÊŸËŸ
59. ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∞ŸÊßÁŸ∑§ Á«U≈U⁄U¡¥≈U „Ò?
(1) ‚ÊÁ«Uÿ◊ S≈UË•⁄U≈U
(2) ‚ÊÁ«Uÿ◊ ‹ÊÁ⁄U‹ ‚À»§≈U
(3) ‚Á≈U‹≈˛UÊß◊ÁÕ‹ •◊ÊÁŸÿ◊ ’˝Ê◊Êß«U
(4) ÁÇ‹‚Á⁄U‹ •ÊÁ‹∞≈U
60. ŸËø ŒË ªß¸ Á»§ª⁄U ◊¥ ’Èã‚Ÿ ç‹◊ ∑§Ê ‚flʸÁœ∑§ ª◊¸÷ʪ „Ò —
(1) ⁄UË¡Ÿ 1
(2) ⁄UË¡Ÿ 2
(3) ⁄UË¡Ÿ 3
(4) ⁄UË¡Ÿ 4
58. Thiol group is present in :
(1) Cytosine
(2) Cystine
(3) Cysteine
(4) Methionine
59. Which of the following is an anionic
detergent ?
(1) Sodium stearate
(2) Sodium lauryl sulphate
(3) Cetyltrimethyl ammonium bromide
(4) Glyceryl oleate
60. The hottest region of Bunsen flame shown
in the figure below is :
(1) region 1
(2) region 2
(3) region 3
(4) region 4
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÷ʪ C — ªÁáÊÃ
61. ÿÁŒ f (x)+2f
1
x
=3x, 0x ≠ „Ò, ÃÕÊ
S : ( ) ( )x f x f xR= = − „Ò ; ÃÊ S :
(1) ∞∑§ Á⁄UÄà ‚◊ÈìÊÿ „Ò–
(2) ◊¥ ∑§fl‹ ∞∑§ •flÿfl „Ò–
(3) ◊¥ Ãâÿ× ŒÊ •flÿfl „Ò¥–
(4) ◊¥ ŒÊ ‚ •Áœ∑§ •flÿfl „Ò¥–
62. θ ∑§Ê fl„ ∞∑§ ◊ÊŸ Á¡‚∑§ Á‹∞ 2 3 sin
1 2 sin
i
i
+
−
θ
θ
¬Íáʸ×
∑§ÊÀ¬ÁŸ∑§ „Ò, „Ò —
(1)3
π
(2)6
π
(3)1 3
sin4
−
(4)1 1
sin3
−
63. x ∑§ ©Ÿ ‚÷Ë flÊSÃÁfl∑§ ◊ÊŸÊ¥ ∑§Ê ÿʪ ¡Ê ‚◊Ë∑§⁄UáÊ
( )2
4 602
5 5 1x x
x x
+ −
− + = ∑§Ê ‚¥ÃÈc≈U ∑§⁄UÃ
„Ò¥, „Ò —
(1) 3
(2) −4
(3) 6
(4) 5
PART C — MATHEMATICS
61. If f (x)+2f
1
x
=3x, 0x ≠ , and
S : ( ) ( )x f x f xR= = − ; then S :
(1) is an empty set.
(2) contains exactly one element.
(3) contains exactly two elements.
(4) contains more than two elements.
62. A value of θ for which 2 3 sin
1 2 sin
i
i
+
−
θ
θ
is
purely imaginary, is :
(1)3
π
(2)6
π
(3)1 3
sin4
−
(4)1 1
sin3
−
63. The sum of all real values of x satisfying
the equation
( )2
4 602
5 5 1x x
x x
+ −
− + = is :
(1) 3
(2) −4
(3) 6
(4) 5
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64. ÿÁŒ 5
A 3 2
a b
−
= ÃÕÊ A adj A=A AT „Ò¥,
ÃÊ 5a+b ’⁄UÊ’⁄U „Ò —
(1) −1
(2) 5
(3) 4
(4) 13
65. ⁄ÒUÁπ∑§ ‚◊Ë∑§⁄UáÊ ÁŸ∑§Êÿ
x+λy−z=0
λx−y−z=0
x+y−λz=0
∑§Ê ∞∑§ •ÃÈë¿U „‹ „ÊŸ ∑§ Á‹∞ —
(1) λ ∑§ •Ÿ¥Ã ◊ÊŸ „Ò¥–
(2) λ ∑§Ê Ãâÿ× ∞∑§ ◊ÊŸ „Ò–
(3) λ ∑§ Ãâÿ× ŒÊ ◊ÊŸ „Ò¥–
(4) λ ∑§ Ãâÿ× ÃËŸ ◊ÊŸ „Ò¥–
66. ‡ÊéŒ SMALL ∑§ •ˇÊ⁄UÊ¥ ∑§Ê ¬˝ÿʪ ∑§⁄U∑§, ¬Ê°ø •ˇÊ⁄UÊ¥flÊ‹ ‚÷Ë ‡ÊéŒÊ¥ (•Õ¸¬Íáʸ •ÕflÊ •Õ¸„ËŸ) ∑§Ê‡ÊéŒ∑§Ê‡Ê ∑§ ∑˝§◊ÊŸÈ‚Ê⁄U ⁄UπŸ ¬⁄U, ‡ÊéŒ SMALL ∑§ÊSÕÊŸ „Ò —
(1) 46 flʥ
(2) 59 flʥ
(3) 52 flʥ
(4) 58 flʥ
64. If 5
A 3 2
a b
−
= and A adj A=A AT, then
5a+b is equal to :
(1) −1
(2) 5
(3) 4
(4) 13
65. The system of linear equations
x+λy−z=0
λx−y−z=0
x+y−λz=0
has a non-trivial solution for :
(1) infinitely many values of λ.
(2) exactly one value of λ.
(3) exactly two values of λ.
(4) exactly three values of λ.
66. If all the words (with or without meaning)
having five letters, formed using the letters
of the word SMALL and arranged as in a
dictionary; then the position of the word
SMALL is :
(1) 46th
(2) 59th
(3) 52nd
(4) 58th
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67. ÿÁŒ 2
2 41 , 0
n
x
x x
≠
− + ∑§ ¬˝‚Ê⁄U ◊¥ ¬ŒÊ¥
∑§Ë ‚¥ÅÿÊ 28 „Ò, ÃÊ ß‚ ¬˝‚Ê⁄U ◊¥ •ÊŸ flÊ‹ ‚÷Ë ¬ŒÊ¥
∑§ ªÈáÊÊ¥∑§Ê¥ ∑§Ê ÿʪ „Ò —
(1) 64
(2) 2187
(3) 243
(4) 729
68. ÿÁŒ ∞∑§ •ø⁄UÃ⁄U ‚◊Ê¥Ã⁄U üÊ…∏Ë ∑§Ê ŒÍ‚⁄UÊ, 5 flÊ¥ ÃÕÊ9 flÊ¥ ¬Œ ∞∑§ ªÈáÊÊûÊ⁄U üÊ…∏Ë ◊¥ „Ò¥, ÃÊ ©‚ ªÈáÊÊûÊ⁄U üÊ…∏Ë∑§Ê ‚Êfl¸ •ŸÈ¬Êà „Ò —
(1)8
5
(2)4
3
(3) 1
(4)7
4
69. ÿÁŒ üÊáÊË
2 2 2 223 2 1 4
1 2 3 4 4 ...... ,5 5 5 5
+ + + + +
∑§ ¬˝Õ◊ Œ‚ ¬ŒÊ¥ ∑§Ê ÿʪ 16
5m „Ò, ÃÊ m ’⁄UÊ’⁄U
„Ò —
(1) 102
(2) 101
(3) 100
(4) 99
67. If the number of terms in the expansion of
2
2 41 , 0,
n
x
x x
≠
− + is 28, then the sum
of the coefficients of all the terms in this
expansion, is :
(1) 64
(2) 2187
(3) 243
(4) 729
68. If the 2nd, 5th and 9th terms of a
non-constant A.P. are in G.P., then the
common ratio of this G.P. is :
(1)8
5
(2)4
3
(3) 1
(4)7
4
69. If the sum of the first ten terms of the series
2 2 2 223 2 1 4
1 2 3 4 4 ...... ,5 5 5 5
+ + + + +
is 16
5m , then m is equal to :
(1) 102
(2) 101
(3) 100
(4) 99
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70. ◊ÊŸÊ ( )1
22
0
lim 1 tanx
x
p x→ += + „Ò, ÃÊ log p
’⁄UÊ’⁄U „Ò —
(1) 2
(2) 1
(3)1
2
(4)1
4
71. x e R ∑ § Á‹∞ f (x)=?log2−sinx? ÃÕÊg(x)=f (f (x)) „Ò¥, ÃÊ —
(1) x=0 ¬⁄U g •fl∑§‹ŸËÿ Ÿ„Ë¥ „Ò–
(2) g9(0)=cos(log2) „Ò–
(3) g9(0)=−cos(log2) „Ò–
(4) x=0 ¬⁄ U g •fl∑§‹ŸËÿ „ Ò ÃÕÊg9(0)=−sin(log2) „Ò–
72.1 1 sin
( ) tan , 0, 1 sin 2
xf x x
x
−+
=
−
π
¬⁄U ÁfløÊ⁄U ∑§ËÁ¡∞– y=f (x) ∑§ Á’¥ŒÈ 6
x=
π ¬⁄U
πË¥øÊ ªÿÊ •Á÷‹¥’ ÁŸêŸ Á’¥ŒÈ ‚ ÷Ë „Ê∑§⁄U ¡ÊÃÊ „Ò —
(1) (0, 0)
(2)2
0, 3
π
(3) , 06
π
(4) , 04
π
70. Let ( )1
22
0
lim 1 tanx
x
p x→ += + then log p
is equal to :
(1) 2
(2) 1
(3)1
2
(4)1
4
71. For x e R, f (x)=?log2−sinx? and
g(x)=f (f (x)), then :
(1) g is not differentiable at x=0
(2) g9(0)=cos(log2)
(3) g9(0)=−cos(log2)
(4) g is differentiable at x=0 and
g9(0)=−sin(log2)
72. Consider
1 1 sin( ) tan , 0,
1 sin 2
xf x x .
x
−+
=
−
π
A normal to y=f (x) at 6
x=
π
also passes
through the point :
(1) (0, 0)
(2)2
0, 3
π
(3) , 06
π
(4) , 04
π
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73. 2 ß∑§Ê߸ ‹¥’Ë ∞∑§ ÃÊ⁄U ∑§Ê ŒÊ ÷ʪʥ ◊¥ ∑§Ê≈U ∑§⁄U ©ã„¥∑˝§◊‡Ê— x ß∑§Ê߸ ÷È¡Ê flÊ‹ flª¸ ÃÕÊ r ß∑§Ê߸ ÁòÊíÿÊflÊ‹ flÎûÊ ∑§ M§¬ ◊¥ ◊Ê«∏Ê ¡ÊÃÊ „Ò– ÿÁŒ ’ŸÊÿ ªÿ flª¸ÃÕÊ flÎûÊ ∑§ ˇÊòÊ»§‹Ê¥ ∑§Ê ÿʪ ãÿÍŸÃ◊ „Ò, ÃÊ —
(1) 2x=(π+4)r
(2) (4−π)x=πr
(3) x=2r
(4) 2x=r
74. ‚◊Ê∑§‹ ( )
12 9
35 3
2 5
1
x xdx
x x
+
+ +
’⁄UÊ’⁄U „Ò —
(1)
( )
5
25 3
1
xC
x x
−+
+ +
(2)
( )
10
25 3
2 1
xC
x x
+
+ +
(3)
( )
5
25 3
2 1
xC
x x
+
+ +
(4)
( )
10
25 3
2 1
xC
x x
−+
+ +
¡„Ê° C ∞∑§ Sflë¿U •ø⁄U „Ò–
73. A wire of length 2 units is cut into two
parts which are bent respectively to form
a square of side=x units and a circle of
radius=r units. If the sum of the areas of
the square and the circle so formed is
minimum, then :
(1) 2x=(π+4)r
(2) (4−π)x=πr
(3) x=2r
(4) 2x=r
74. The integral
( )
12 9
35 3
2 5
1
x xdx
x x
+
+ +
is equal
to :
(1)
( )
5
25 3
1
xC
x x
−+
+ +
(2)
( )
10
25 3
2 1
xC
x x
+
+ +
(3)
( )
5
25 3
2 1
xC
x x
+
+ +
(4)
( )
10
25 3
2 1
xC
x x
−+
+ +
where C is an arbitrary constant.
SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„E/Page 33
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75. ( ) ( )1
2
1 2 . . . 3lim
n
nn
n n n
n→
∞
+ + ’⁄UÊ’⁄U „Ò —
(1) 4
18
e
(2) 2
27
e
(3) 2
9
e
(4) 3 log3−2
76. ˇÊòÊ
2 2 2( , ): 2 4 0, 0x y y x x y x, x y + ≤ ÃÕÊ
∑§Ê ˇÊòÊ»§‹ (flª¸ ß∑§ÊßÿÊ¥ ◊¥) „Ò —
(1)4
3−π
(2)8
3−π
(3)4 2
3
−π
(4)2 2
2 3−
π
75. ( ) ( )1
2
1 2 . . . 3lim
n
nn
n n n
n→
∞
+ + is equal
to :
(1) 4
18
e
(2) 2
27
e
(3) 2
9
e
(4) 3 log3−2
76. The area (in sq. units) of the region
2 2 2( , ): 2 and 4 0, 0x y y x x y x, x y + ≤
is :
(1)4
3−π
(2)8
3−π
(3)4 2
3
−π
(4)2 2
2 3−
π
E/Page 34 SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„
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77. ÿÁŒ ∞∑§ fl∑˝§ y=f (x) Á’¥ŒÈ (1, −1) ‚ „Ê∑§⁄U ¡ÊÃÊ„Ò ÃÕÊ •fl∑§‹ ‚◊Ë∑§⁄UáÊ y(1+xy) dx=x dy
∑§Ê ‚¥ÃÈc≈U ∑§⁄UÃÊ „Ò, ÃÊ 1 2
f − ’⁄UÊ’⁄U „Ò —
(1)2
5−
(2)4
5−
(3)2
5
(4)4
5
78. ÿÁŒ ∞∑§ ‚◊øÃÈ÷¸È¡ ∑§Ë ŒÊ ÷È¡Ê∞°, ⁄UπÊ•Ê¥x−y+1=0 ÃÕÊ 7x−y−5=0 ∑§Ë ÁŒ‡ÊÊ ◊¥ „Ò¥ÃÕÊ ß‚∑§ Áfl∑§áʸ Á’¥ŒÈ (−1, −2) ¬⁄U ¬˝ÁÃë¿UŒ∑§⁄Uà „Ò¥, ÃÊ ß‚ ‚◊øÃÈ÷¸È¡ ∑§Ê ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚ʇÊË·¸ „Ò?
(1) (−3, −9)
(2) (−3, −8)
(3)
1 8,
3 3
−
(4)
10 7,
3 3
− −
79. ©Ÿ flÎûÊÊ¥ ∑§ ∑§ãŒ˝, ¡Ê flÎûÊ x2+y2−8x−8y−4=0
∑§Ê ’Ês M§¬ ‚ S¬‡Ê¸ ∑§⁄Uà „Ò¥ ÃÕÊ x-•ˇÊ ∑§Ê ÷ËS¬‡Ê¸ ∑§⁄Uà „Ò¥, ÁSÕà „Ò¥ —
(1) ∞∑§ flÎûÊ ¬⁄U–
(2) ∞∑§ ŒËÉʸflÎûÊ ¬⁄U ¡Ê flÎûÊ Ÿ„Ë¥ „Ò–
(3) ∞∑§ •Áì⁄Ufl‹ÿ ¬⁄U–
(4) ∞∑§ ¬⁄Ufl‹ÿ ¬⁄U–
77. If a curve y=f (x) passes through the point
(1, −1) and satisfies the differential
equation, y(1+xy) dx=x dy, then 1
2
f −
is equal to :
(1)2
5−
(2)4
5−
(3)2
5
(4)4
5
78. Two sides of a rhombus are along the lines,
x−y+1=0 and 7x−y−5=0. If its
diagonals intersect at (−1, −2), then
which one of the following is a vertex of
this rhombus ?
(1) (−3, −9)
(2) (−3, −8)
(3)
1 8,
3 3
−
(4)
10 7,
3 3
− −
79. The centres of those circles which touch
the circle, x2+y2−8x−8y−4=0,
externally and also touch the x-axis, lie
on :
(1) a circle.
(2) an ellipse which is not a circle.
(3) a hyperbola.
(4) a parabola.
SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„E/Page 35
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80. ÿÁŒ ‚◊Ë∑§⁄UáÊ x2+y2−4x+6y−12=0 mÊ⁄Uʬ˝ŒûÊ ∞∑§ flÎûÊ ∑§Ê ∞∑§ √ÿÊ‚ ∞∑§ •ãÿ flÎûÊ S, Á¡‚∑§Ê∑§ãŒ˝ (−3, 2) „Ò, ∑§Ë ¡ËflÊ „Ò, ÃÊ flÎûÊ S ∑§Ë ÁòÊíÿÊ„Ò —
(1) 5 2
(2) 5 3
(3) 5
(4) 10
81. ◊ÊŸÊ ¬⁄Ufl‹ÿ y2=8x ∑§Ê P ∞∑§ ∞‚Ê Á’¥ŒÈ „Ò ¡ÊflÎûÊ x2+(y+6)2=1, ∑§ ∑§ãŒ˝ C ‚ ãÿÍŸÃ◊ ŒÍ⁄Uˬ⁄U „Ò, ÃÊ ©‚ flÎûÊ ∑§Ê ‚◊Ë∑§⁄UáÊ ¡Ê C ‚ „Ê∑§⁄U ¡ÊÃÊ„Ò ÃÕÊ Á¡‚∑§Ê ∑§ãŒ˝ P ¬⁄U „Ò, „Ò —
(1) x2+y2−4x+8y+12=0
(2) x2+y2−x+4y−12=0
(3) x2+y2−4
x
+2y−24=0
(4) x2+y2−4x+9y+18=0
82. ©‚ •Áì⁄Ufl‹ÿ, Á¡‚∑§ ŸÊÁ÷‹¥’ ∑§Ë ‹¥’Ê߸ 8 „ÒÃÕÊ Á¡‚∑§ ‚¥ÿÈÇ◊Ë •ˇÊ ∑§Ë ‹¥’Ê߸ ©‚∑§Ë ŸÊÁ÷ÿÊ¥∑§ ’Ëø ∑§Ë ŒÍ⁄UË ∑§Ë •ÊœË „Ò, ∑§Ë ©à∑§ãŒ˝ÃÊ „Ò —
(1)4
3
(2)4
3
(3)2
3
(4) 3
80. If one of the diameters of the circle, given
by the equation, x2+y2−4x+6y−12=0,
is a chord of a circle S, whose centre is at
(−3, 2), then the radius of S is :
(1) 5 2
(2) 5 3
(3) 5
(4) 10
81. Let P be the point on the parabola, y2=8x
which is at a minimum distance from the
centre C of the circle, x2+(y+6)2=1.
Then the equation of the circle, passing
through C and having its centre at P is :
(1) x2+y2−4x+8y+12=0
(2) x2+y2−x+4y−12=0
(3) x2+y2−4
x
+2y−24=0
(4) x2+y2−4x+9y+18=0
82. The eccentricity of the hyperbola whose
length of the latus rectum is equal to 8 and
the length of its conjugate axis is equal to
half of the distance between its foci, is :
(1)4
3
(2)4
3
(3)2
3
(4) 3
E/Page 36 SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„
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83. Á’¥ŒÈ (1, −5, 9) ∑§Ë ‚◊Ë x−y+z=5 ‚ fl„ŒÍ⁄UË ¡Ê ⁄UπÊ x=y=z ∑§Ë ÁŒ‡ÊÊ ◊¥ ◊Ê¬Ë ªß¸ „Ò, „Ò —
(1) 3 10
(2) 10 3
(3)10
3
(4)20
3
84. ÿÁŒ ⁄UπÊ 2 3 4
2 1 3
yx z+− += =
−
, ‚◊Ë
lx+my−z=9 ◊¥ ÁSÕà „Ò, ÃÊ l2+m2 ’⁄UÊ’⁄U „Ò —
(1) 26
(2) 18
(3) 5
(4) 2
85. ◊ÊŸÊ , a b→ →
ÃÕÊ c→
ÃËŸ ∞‚ ◊ÊòÊ∑§ ‚ÁŒ‡Ê „Ò¥ Á∑§
3 .
2a b c b c
→ → → → → × × = + „ Ò– ÿÁŒ
b→
, c→
∑§ ‚◊Ê¥Ã⁄U Ÿ„Ë¥ „Ò, ÃÊ a→
ÃÕÊ b→
∑§ ’Ëø
∑§Ê ∑§ÊáÊ „Ò —
(1)3
4
π
(2)2
π
(3)2
3
π
(4)5
6
π
83. The distance of the point (1, −5, 9) from
the plane x−y+z=5 measured along the
line x=y=z is :
(1) 3 10
(2) 10 3
(3)10
3
(4)20
3
84. If the line, 2 3 4
2 1 3
yx z+− += =
−
lies in
the plane, lx+my−z=9, then l2+m2 is
equal to :
(1) 26
(2) 18
(3) 5
(4) 2
85. Let , and a b c→ → →
be three unit vectors such
that 3
.2
a b c b c
→ → → → → × × = + If
b→
is not parallel to c→
, then the angle
between a→
and b→
is :
(1)3
4
π
(2)2
π
(3)2
3
π
(4)5
6
π
SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„E/Page 37
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86. ÿÁŒ ‚¥ÅÿÊ•Ê¥ 2, 3, a ÃÕÊ 11 ∑§Ê ◊ÊŸ∑§ Áflø‹Ÿ3.5 „Ò, ÃÊ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚Ê ‚àÿ „Ò?
(1) 3a2−26a+55=0
(2) 3a2−32a+84=0
(3) 3a2−34a+91=0
(4) 3a2−23a+44=0
87. ◊ÊŸÊ ŒÊ •ŸÁ÷ŸÃ ¿U— »§‹∑§Ëÿ ¬Ê‚ A ÃÕÊ B ∞∑§‚ÊÕ ©¿UÊ‹ ªÿ– ◊ÊŸÊ ÉÊ≈UŸÊ E
1 ¬Ê‚ A ¬⁄U øÊ⁄U
•ÊŸÊ Œ‡ÊʸÃË „Ò, ÉÊ≈UŸÊ E2 ¬Ê‚ B ¬⁄U 2 •ÊŸÊ Œ‡ÊʸÃË
„Ò ÃÕÊ ÉÊ≈UŸÊ E3 ŒÊŸÊ¥ ¬Ê‚Ê¥ ¬⁄U •ÊŸ flÊ‹Ë ‚¥ÅÿÊ•Ê¥
∑§Ê ÿʪ Áfl·◊ Œ‡ÊʸÃË „Ò, ÃÊ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚Ê∑§ÕŸ ‚àÿ Ÿ„Ë¥ „Ò?
(1) E1 ÃÕÊ E
2 SflÃ¥òÊ „Ò¥–
(2) E2 ÃÕÊ E
3 SflÃ¥òÊ „Ò¥–
(3) E1 ÃÕÊ E
3 SflÃ¥òÊ „Ò¥–
(4) E1, E2 ÃÕÊ E
3 SflÃ¥òÊ „Ò¥–
88. ÿÁŒ 0≤x<2π „Ò, ÃÊ x ∑§ ©Ÿ flÊSÃÁfl∑§ ◊ÊŸÊ¥ ∑§Ë‚¥ÅÿÊ ¡Ê ‚◊Ë∑§⁄UáÊcosx+cos2x+cos3x+cos4x=0 ∑§Ê ‚¥ÃÈc≈U∑§⁄Uà „Ò¥, „Ò —
(1) 3
(2) 5
(3) 7
(4) 9
86. If the standard deviation of the numbers
2, 3, a and 11 is 3.5, then which of the
following is true ?
(1) 3a2−26a+55=0
(2) 3a2−32a+84=0
(3) 3a2−34a+91=0
(4) 3a2−23a+44=0
87. Let two fair six-faced dice A and B be
thrown simultaneously. If E1 is the event
that die A shows up four, E2 is the event
that die B shows up two and E3 is the event
that the sum of numbers on both dice is
odd, then which of the following
statements is NOT true ?
(1) E1 and E
2 are independent.
(2) E2 and E
3 are independent.
(3) E1 and E
3 are independent.
(4) E1, E2 and E
3 are independent.
88. If 0≤x<2π, then the number of real values
of x, which satisfy the equation
cosx+cos2x+cos3x+cos4x=0, is :
(1) 3
(2) 5
(3) 7
(4) 9
E/Page 38 SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„
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SPACE FOR ROUGH WORK / ⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„
89. ∞∑§ √ÿÁÄà ∞∑§ ™§äflʸœ⁄U π¥÷ ∑§Ë •Ê⁄U ∞∑§ ‚Ëœ ¬Õ¬⁄U ∞∑§ ‚◊ÊŸ øÊ‹ ‚ ¡Ê ⁄U„Ê „Ò– ⁄UÊSà ¬⁄U ∞∑§ Á’¥ŒÈA ‚ fl„ π¥÷ ∑§ Á‡Êπ⁄U ∑§Ê ©ÛÊÿŸ ∑§ÊáÊ 308 ◊ʬÃÊ„Ò– A ‚ ©‚Ë ÁŒ‡ÊÊ ◊¥ 10 Á◊Ÿ≈U •ÊÒ⁄U ø‹Ÿ ∑§ ’ÊŒÁ’¥ŒÈ B ‚ fl„ π¥÷ ∑§ Á‡Êπ⁄U ∑§Ê ©ÛÊÿŸ ∑§ÊáÊ 608
¬ÊÃÊ „Ò, ÃÊ B ‚ π¥÷ Ã∑§ ¬„È°øŸ ◊¥ ©‚ ‹ªŸ flÊ‹Ê‚◊ÿ (Á◊Ÿ≈UÊ¥ ◊¥) „Ò —
(1) 6
(2) 10
(3) 20
(4) 5
90. ’Í‹ ∑§ √ÿ¥¡∑§ (Boolean Expression)
(p∧~q)∨q∨(~p∧q) ∑§Ê ‚◊ÃÈÀÿ „Ò —
(1) ~ p ∧ q
(2) p ∧ q
(3) p ∨ q
(4) p ∨ ~ q
- o O o -
89. A man is walking towards a vertical pillar
in a straight path, at a uniform speed. At
a certain point A on the path, he observes
that the angle of elevation of the top of
the pillar is 308. After walking for 10
minutes from A in the same direction, at a
point B, he observes that the angle of
elevation of the top of the pillar is 608.
Then the time taken (in minutes) by him,
from B to reach the pillar, is :
(1) 6
(2) 10
(3) 20
(4) 5
90. The Boolean Expression (p∧~q)∨q∨(~p∧q)
is equivalent to :
(1) ~ p ∧ q
(2) p ∧ q
(3) p ∨ q
(4) p ∨ ~ q
- o O o -
Read the following instructions carefully :
1. The candidates should fill in the required particularson the Test Booklet and Answer Sheet (Side–1) withBlue/Black Ball Point Pen.
2. For writing/marking particulars on Side–2 of theAnswer Sheet, use Blue/Black Ball Point Pen only.
3. The candidates should not write their Roll Numbersanywhere else (except in the specified space) on theTest Booklet/Answer Sheet.
4. Out of the four options given for each question, onlyone option is the correct answer.
5. For each incorrect response, one–fourth (¼) of the totalmarks allotted to the question would be deducted fromthe total score. No deduction from the total score,however, will be made if no response is indicated foran item in the Answer Sheet.
6. Handle the Test Booklet and Answer Sheet with care,as under no circumstances (except for discrepancy inTest Booklet Code and Answer Sheet Code), another setwill be provided.
7. The candidates are not allowed to do any rough workor writing work on the Answer Sheet. All calculations/writing work are to be done in the space provided forthis purpose in the Test Booklet itself, marked ‘Spacefor Rough Work’. This space is given at the bottom ofeach page and in one page (i.e. Page 39) at the end ofthe booklet.
8. On completion of the test, the candidates must handover the Answer Sheet to the Invigilator on duty in theRoom/Hall. However, the candidates are allowed totake away this Test Booklet with them.
9. Each candidate must show on demand his/her AdmitCard to the Invigilator.
10. No candidate, without special permission of theSuperintendent or Invigilator, should leave his/herseat.
11. The candidates should not leave the Examination Hallwithout handing over their Answer Sheet to theInvigilator on duty and sign the Attendance Sheetagain. Cases where a candidate has not signed theAttendance Sheet second time will be deemed not tohave handed over the Answer Sheet and dealt with asan unfair means case. The candidates are also requiredto put their left hand THUMB impression in the spaceprovided in the Attendance Sheet.
12. Use of Electronic/Manual Calculator and anyElectronic device like mobile phone, pager etc. isprohibited.
13. The candidates are governed by all Rules andRegulations of the JAB/Board with regard to theirconduct in the Examination Hall. All cases of unfairmeans will be dealt with as per Rules and Regulationsof the JAB/Board.
14. No part of the Test Booklet and Answer Sheet shall bedetached under any circumstances.
15. Candidates are not allowed to carry any textualmaterial, printed or written, bits of papers, pager,mobile phone, electronic device or any other materialexcept the Admit Card inside the examinationroom/hall.
ÁŸêŸÁ‹Áπà ÁŸŒ¸‡Ê äÿÊŸ ‚ ¬…∏¥ —
1. ¬⁄UˡÊÊÁÕ¸ÿÊ¥ ∑§Ê ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê •ÊÒ⁄U ©ûÊ⁄U ¬òÊ (¬ÎD -1) ¬⁄U flÊ¥Á¿UÃÁflfl⁄UáÊ ŸË‹/∑§Ê‹ ’ÊÚ‹ åflÊߥ≈U ¬Ÿ ‚ „Ë ÷⁄UŸÊ „Ò–
2. ©ûÊ⁄U ¬òÊ ∑§ ¬ÎD-2 ¬⁄U Áflfl⁄UáÊ Á‹πŸ/•¥Á∑§Ã ∑§⁄UŸ ∑§ Á‹∞ ∑§fl‹ŸË‹/∑§Ê‹ ’ÊÚ‹ åflÊߥ≈U ¬Ÿ ∑§Ê ¬˝ÿʪ ∑§⁄¥U–
3. ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê/©ûÊ⁄U ¬òÊ ¬⁄U ÁŸœÊ¸Á⁄Uà SÕÊŸ ∑§ •‹ÊflÊ ¬⁄UˡÊÊÕ˸•¬ŸÊ •ŸÈ∑˝§◊Ê¥∑§ •ãÿ ∑§„Ë¥ Ÿ„Ë¥ Á‹π¥–
4. ¬˝àÿ∑§ ¬˝‡Ÿ ∑§ Á‹ÿ ÁŒÿ ªÿ øÊ⁄U Áfl∑§À¬Ê¥ ◊¥ ‚ ∑§fl‹ ∞∑§ Áfl∑§À¬‚„Ë „Ò–
5. ¬˝àÿ∑§ ª‹Ã ©ûÊ⁄U ∑§ Á‹∞ ©‚ ¬˝‡Ÿ ∑§ Á‹∞ ÁŸœÊ¸Á⁄Uà ∑ȧ‹ •¥∑§Ê¥◊¥ ‚ ∞∑§-øÊÒÕÊ߸ (¼) •¥∑§ ∑ȧ‹ ÿʪ ◊¥ ‚ ∑§Ê≈U Á‹∞ ¡Ê∞°ª–ÿÁŒ ©ûÊ⁄U ¬òÊ ◊¥ Á∑§‚Ë ¬˝‡Ÿ ∑§Ê ∑§Ê߸ ©ûÊ⁄U Ÿ„Ë¥ ÁŒÿÊ ªÿÊ „Ò, ÃÊ∑ȧ‹ ÿʪ ◊¥ ‚ ∑§Ê߸ •¥∑§ Ÿ„Ë¥ ∑§Ê≈U ¡Ê∞°ª–
6. ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê ∞fl¥ ©ûÊ⁄U ¬òÊ ∑§Ê äÿÊŸ¬Ífl¸∑§ ¬˝ÿʪ ∑§⁄¥U ÄÿÊ¥Á∑§Á∑§‚Ë ÷Ë ¬Á⁄UÁSÕÁà ◊¥ (∑§fl‹ ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê ∞fl¥ ©ûÊ⁄U ¬òÊ ∑§‚¥∑§Ã ◊¥ Á÷ÛÊÃÊ ∑§Ë ÁSÕÁà ∑§Ê ¿UÊ«∏∑§⁄U), ŒÍ‚⁄UË ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê©¬‹éœ Ÿ„Ë¥ ∑§⁄UÊÿË ¡Ê∞ªË–
7. ©ûÊ⁄U ¬òÊ ¬⁄U ∑§Ê߸ ÷Ë ⁄U»§ ∑§Êÿ¸ ÿÊ Á‹πÊ߸ ∑§Ê ∑§Ê◊ ∑§⁄UŸ ∑§Ë•ŸÈ◊Áà Ÿ„Ë¥ „Ò– ‚÷Ë ªáÊŸÊ ∞fl¥ Á‹πÊ߸ ∑§Ê ∑§Ê◊, ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê◊¥ ÁŸœÊ¸Á⁄Uà ¡ª„ ¡Ê Á∑§ “⁄U»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„” mÊ⁄UÊ ŸÊ◊Ê¥Á∑§Ã„Ò, ¬⁄U „Ë Á∑§ÿÊ ¡Ê∞ªÊ– ÿ„ ¡ª„ ¬˝àÿ∑§ ¬ÎD ¬⁄U ŸËø ∑§Ë •Ê⁄U •ÊÒ⁄U¬ÈÁSÃ∑§Ê ∑§ •¥Ã ◊¥ ∞∑§ ¬ÎD ¬⁄U (¬ÎD 39) ŒË ªß¸ „Ò–
8. ¬⁄ˡÊÊ ‚ê¬ÛÊ „ÊŸ ¬⁄U, ¬⁄UˡÊÊÕ˸ ∑§ˇÊ/„ÊÚ‹ ¿UÊ«∏Ÿ ‚ ¬Ífl¸ ©ûÊ⁄U ¬òÊ∑§ˇÊ ÁŸ⁄UˡÊ∑§ ∑§Ê •fl‡ÿ ‚ÊÒ¥¬ Œ¥– ¬⁄UˡÊÊÕ˸ •¬Ÿ ‚ÊÕ ß‚¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê ∑§Ê ‹ ¡Ê ‚∑§Ã „Ò¥–
9. ◊Ê¥ª ¡ÊŸ ¬⁄U ¬˝àÿ∑§ ¬⁄UˡÊÊÕ˸ ÁŸ⁄UˡÊ∑§ ∑§Ê •¬ŸÊ ¬˝fl‡Ê ∑§Ê«¸ •fl‡ÿÁŒπÊ∞°–
10. •œËˇÊ∑§ ÿÊ ÁŸ⁄UˡÊ∑§ ∑§Ë Áfl‡Ê· •ŸÈ◊Áà ∑§ Á’ŸÊ ∑§Ê߸ ¬⁄UˡÊÊÕ˸•¬ŸÊ SÕÊŸ Ÿ ¿UÊ«∏¥–
11. ∑§Êÿ¸⁄Uà ÁŸ⁄UˡÊ∑§ ∑§Ê •¬ŸÊ ©ûÊ⁄U ¬òÊ ÁŒ∞ Á’ŸÊ ∞fl¥ ©¬ÁSÕÁà ¬òʬ⁄U ŒÈ’Ê⁄UÊ „SÃÊˇÊ⁄U Á∑§∞ Á’ŸÊ ∑§Ê߸ ¬⁄UˡÊÊÕ˸ ¬⁄UˡÊÊ „ÊÚ‹ Ÿ„Ë¥ ¿UÊ«∏¥ª–ÿÁŒ Á∑§‚Ë ¬⁄UˡÊÊÕ˸ Ÿ ŒÍ‚⁄UË ’Ê⁄U ©¬ÁSÕÁà ¬òÊ ¬⁄U „SÃÊˇÊ⁄U Ÿ„Ë¥Á∑§∞ ÃÊ ÿ„ ◊ÊŸÊ ¡Ê∞ªÊ Á∑§ ©‚Ÿ ©ûÊ⁄U ¬òÊ Ÿ„Ë¥ ‹ÊÒ≈UÊÿÊ „Ò Á¡‚•ŸÈÁøà ‚ÊœŸ ¬˝ÿʪ üÊáÊË ◊¥ ◊ÊŸÊ ¡Ê∞ªÊ– ¬⁄UˡÊÊÕ˸ •¬Ÿ ’Êÿ¥„ÊÕ ∑§ •¥ªÍ∆U ∑§Ê ÁŸ‡ÊÊŸ ©¬ÁSÕÁà ¬òÊ ◊¥ ÁŒ∞ ª∞ SÕÊŸ ¬⁄U•fl‡ÿ ‹ªÊ∞°–
12. ß‹Ä≈˛UÊÚÁŸ∑§/„SÃøÊÁ‹Ã ¬Á⁄U∑§‹∑§ ∞fl¥ ◊Ê’Êß‹ »§ÊŸ, ¬¡⁄U ßàÿÊÁŒ¡Ò‚ Á∑§‚Ë ß‹Ä≈˛UÊÚÁŸ∑§ ©¬∑§⁄UáÊ ∑§Ê ¬˝ÿʪ flÁ¡¸Ã „Ò–
13. ¬⁄UˡÊÊ „ÊÚ‹ ◊¥ •Êø⁄UáÊ ∑§ Á‹∞ ¬⁄UˡÊÊÕ˸ ¡.∞.’./’Ê«¸U ∑§ ‚÷ËÁŸÿ◊Ê¥ ∞fl¥U ÁflÁŸÿ◊Ê¥ mÊ⁄UÊ ÁŸÿÁ◊à „Ê¥ª– •ŸÈÁøà ‚ÊœŸ ¬˝ÿʪ ∑§‚÷Ë ◊Ê◊‹Ê¥ ∑§Ê »Ò§‚‹Ê ¡.∞.’./’Ê«¸U ∑§ ÁŸÿ◊Ê¥ ∞fl¥ ÁflÁŸÿ◊Ê¥ ∑§•ŸÈ‚Ê⁄U „ʪʖ
14. Á∑§‚Ë ÷Ë ÁSÕÁà ◊¥ ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê ÃÕÊ ©ûÊ⁄U ¬òÊ ∑§Ê ∑§Ê߸ ÷Ë ÷ʪ•‹ª Ÿ„Ë¥ Á∑§ÿÊ ¡Ê∞ªÊ–
15. ¬⁄UˡÊÊÕ˸ mÊ⁄UÊ ¬⁄UˡÊÊ ∑§ˇÊ/„ÊÚ‹ ◊¥ ¬˝fl‡Ê ∑§Ê«¸U ∑§ •‹ÊflÊÁ∑§‚Ë ÷Ë ¬˝∑§Ê⁄U ∑§Ë ¬Ê∆˜Uÿ ‚Ê◊ª˝Ë, ◊ÈÁŒ˝Ã ÿÊ „SÃÁ‹ÁπÃ,∑§Êª¡ ∑§Ë ¬Áø¸ÿÊ°, ¬¡⁄U, ◊Ê’Êß‹ »§ÊŸ ÿÊ Á∑§‚Ë ÷Ë ¬˝∑§Ê⁄U∑§ ß‹Ä≈˛UÊÚÁŸ∑§ ©¬∑§⁄UáÊÊ¥ ÿÊ Á∑§‚Ë •ãÿ ¬˝∑§Ê⁄U ∑§Ë ‚Ê◊ª˝Ë∑§Ê ‹ ¡ÊŸ ÿÊ ©¬ÿʪ ∑§⁄UŸ ∑§Ë •ŸÈ◊Áà Ÿ„Ë¥ „Ò–
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E E
E E
Question and Answer KeyQuestion No. Answer Key Question No. Answer Key Question No. Answer Key
Q1 1 Q31 2 Q61 3Q2 8 Q32 3 Q62 4Q3 3 Q33 4 Q63 1Q4 4 Q34 1 Q64 2Q5 1 Q35 4 Q65 4Q6 4 Q36 3 Q66 4Q7 1 Q37 3 Q67 4Q8 2 Q38 4 Q68 2Q9 1 Q39 2 Q69 2
Q10 4 Q40 3 Q70 3Q11 3 Q41 4 Q71 2Q12 1 Q42 3 Q72 2Q13 3 Q43 3 Q73 3Q14 3 Q44 4 Q74 2Q15 4 Q45 1 Q75 2Q16 4 Q46 1 Q76 2Q17 4 Q47 2 Q77 4Q18 1 Q48 2 Q78 3Q19 4 Q49 2 Q79 4Q20 3 Q50 3 Q80 2Q21 1 Q51 4 Q81 1Q22 4 Q52 2 Q82 3Q23 3 Q53 2 Q83 2Q24 1 Q54 1 Q84 4Q25 2 Q55 2 Q85 4Q26 4 Q56 4 Q86 2Q27 1 Q57 4 Q87 4Q28 1 Q58 3 Q88 3Q29 1 Q59 2 Q89 4Q30 8 Q60 2 Q90 3
Note:- 8 indicates that answer option 2 and 4 both are correct.
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 1
1. In the following ‘I’ refers to current and
other symbols have their usual meaning.
Choose the option that corresponds to the
dimensions of electrical conductivity :
(1) ML−3
T−3
I2
(2) M−1
L3
T3
I
(3) M−1
L−3
T3
I2
(4) M−1
L−3
T3
I
1. ÁŸêŸÁ‹Áπà ◊¥ ‘I’ ÁfllÈà œÊ⁄UÊ ∑§Ê ∞fl¥ •ãÿ Áøq•¬Ÿ ‚Ê◊Êãÿ •Õ¸ ∑§Ê ߥÁªÃ ∑§⁄Uà „Ò¥– ÁŸêŸÁ‹Áπà ◊¥‚ ∑§ÊÒŸ-‚Ê Áfl∑§À¬ flÒlÈà øÊ‹∑§ÃÊ ∑§Ë ‚„Ë Áfl◊Ê ∑§Ê’ÃÊÃÊ „Ò?(1) ML
−3 T
−3 I2
(2) M−1
L3
T3
I
(3) M−1
L−3
T3
I2
(4) M−1
L−3
T3
I
1. r_ç_rgrMsdp„ ‘I’ A¡ rhÛysâhpl v$ip®h¡ R>¡ s\p AÞek„¿epAp¡ s¡_p¡ âQrgs A\®. _uQ¡ Ap ¡g rhL$ë`p¡dp„\urhÛyshplL$sp_y„ kpQy„ `qfdpZ v$ip®hsp¡ rhL$ë` `k„v$L$fp¡.(1) ML
−3 T
−3 I2
(2) M−1
L3
T3
I
(3) M−1
L−3
T3
I2
(4) M−1
L−3
T3
I
JEE Main 2016 Question Paper 1 Online (April 9, 2016)JEE Main 2016 Question Paper 1 Online (April 9, 2016)JEE Main 2016 Question Paper 1 Online (April 9, 2016)
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 2
2. Which of the following option correctly
describes the variation of the speed v and
acceleration ‘a’ of a point mass falling
vertically in a viscous medium that applies
a force F =−kv, where ‘k’ is a constant, on
the body ? (Graphs are schematic and not
drawn to scale)
(1)
(2)
(3)
(4)
2. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ-‚Ê Áfl∑§À¬ ©‚ Á’ãŒÈ-Œ˝√ÿ◊ÊŸ∑§Ë ªÁà ‘v’ •ÊÒ⁄U àfl⁄UáÊ ‘a’ ∑§ ’Œ‹Êfl ∑§Ê ‚„Ë Ã⁄U„ ‚Œ‡ÊʸÃÊ „Ò ¡Ê Á∑§ Á∑§‚Ë ‡ÿÊŸ ◊Êäÿ◊ ◊¥ ™§äflʸœ⁄U ÁŒ‡ÊÊ◊¥ ŸËø ∑§Ë •Ê⁄U Áª⁄Uà „È∞ ◊Êäÿ◊ ∑§ ∑§Ê⁄UáÊ ∞∑§ ’‹F =−kv, ¡„Ê° ¬⁄U ‘k’ ∞∑§ ÁŸÿÃÊ¥∑§ „Ò, ∑§Ê •ŸÈ÷fl∑§⁄UÃÊ „Ò– (ª˝Ê»§Ê¥ ∑§Ê √ÿflSÕÊà◊∑§ ÁŸM§¬áÊ ◊ʬ ∑§•ŸÈ‚Ê⁄U Ÿ„Ë¥ „Ò–)
(1)
(2)
(3)
(4)
2. r_ç_rgrMs rhL$ë`p¡dp„\u L$ep¡ rhL$ë` tbvy$ Öìedp__uNrs v A_¡ âh¡N a _p¡ kpQp¡ k„b„ hZ®h¡ R>¡. Alutbvy$ Öìedp_ A¡ õ\p_ dpÝed L¡$ S>¡ F =−kv S>¡V$gp¡`v$p\® f bm gNpX¡$ R>¡ s¡hp dpÝeddp„ DÝh® _uQ¡ sfa`X¡$ R>¡. ‘k’ A¡ AQmp„L$ R>¡. (N°pa_y„ ìehõ\pÐdL$ r_ê$ ZR>¡ A_¡ dp` A_ykpf _\u.)
(1)
(2)
(3)
(4)
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 3
3. A rocket is fired vertically from the earth
with an acceleration of 2g, where g is the
gravitational acceleration. On an inclined
plane inside the rocket, making an angle θ
with the horizontal, a point object of mass
m is kept. The minimum coefficient of
friction µmin
between the mass and the
inclined surface such that the mass does not
move is :
(1) tanθ
(2) 2 tanθ
(3) 3 tanθ
(4) tan2θ
4. A car of weight W is on an inclined road
that rises by 100 m over a distance of 1 km
and applies a constant frictional force • ••
on the car. While moving uphill on the road
at a speed of 10 ms−1
, the car needs power
P. If it needs power ••
while moving
downhill at speed v then value of v is :
(1) 20 ms−1
(2) 15 ms−1
(3) 10 ms−1
(4) 5 ms−1
3. ∞∑§ ⁄UÊÚ∑§≈U ∑§Ê ¬ÎâflË ‚ ™§äflʸœ⁄U ÁŒ‡ÊÊ ◊¥ 2g ∑§ àfl⁄UáÊ‚ ¬˝ˇÊÁ¬Ã Á∑§ÿÊ ªÿÊ „Ò– ß‚ ⁄UÊÚ∑§≈U ∑§ •¥Œ⁄ U ˇÊÒÁá‚ θ ∑§ÊáÊ ’ŸÊà „È∞ ∞∑§ •ÊŸÃ-Ë ¬⁄U ∞∑§m Œ˝√ÿ◊ÊŸ ∑§Ê Á’¥ŒÈ ∑§áÊ ÁSÕà „Ò– ÿÁŒ ⁄UÊÚ∑§≈U ∑§¬˝ˇÊÁ¬Ã „ÊŸ ¬⁄U Á’¥ŒÈ-∑§áÊ ÁSÕ⁄U •flSÕÊ ◊¥ „Ë ⁄U„ÃÊ „ÒÃ’ Œ˝√ÿ◊ÊŸ ∞fl¥ •ÊŸÃ Ë ∑§ ’Ëø ÉÊ·¸áÊ-ªÈáÊÊ¥∑§µ
min ∑§Ê ◊ÊŸ ÄÿÊ „ʪÊ? (‘g’ ªÈL§àflËÿ àfl⁄UáÊ „Ò) —
(1) tanθ
(2) 2 tanθ
(3) 3 tanθ
(4) tan2θ
4. ∞∑§ W ÷Ê⁄U ∑§Ë ∑§Ê⁄U ∞∑§ ∞‚Ë •ÊŸÃ-‚«∏∑§ ¬⁄U ø‹⁄U„Ë „Ò ¡Ê Á∑§ 1 km ŒÍ⁄UË ¬⁄U 100 m ™°§øË „Ê ¡ÊÃË „Ò,
•ÊÒ⁄U ∑§Ê⁄U ¬⁄U • ••
◊ÊŸ ∑§Ê ÁŸÿà ÉÊ·¸áÊ ’‹ ‹ªÊÃË „Ò–
ÿÁŒ ∑§Ê⁄U ∑§Ê ‚«∏∑§ ¬⁄U ™§¬⁄U ∑§Ë •Ê⁄U 10 ms−1 ∑§Ë
ªÁà „ÃÈ P ‡ÊÁÄàÊ ∑§Ë •Êfl‡ÿ∑§ÃÊ „Ò ∞fl¥ ŸËø ∑§Ë •Ê⁄U
v ªÁà ‚ ø‹ÊŸ „ÃÈ ••
‡ÊÁÄàÊ ∑§Ë •Êfl‡ÿ∑§ÃÊ ¬«∏ÃË
„Ò, ÃÊ v ∑§Ê ◊ÊŸ „ÊªÊ —(1) 20 ms
−1
(2) 15 ms−1
(3) 10 ms−1
(4) 5 ms−1
3. A¡L$ fp¡L¡$V$_¡ ©Õhu\u DÝh® qv$ipdp„ 2g âh¡N\u ân¡r`sL$fhpdp„ Aph¡ R>¡. Al] g NyfyÐhâh¡N R>¡. Ap fp¡L¡$V$dp„kdrnrsS> kp\¡ θ S>¡V$gp¡ MyZp¡ b_phsp Y$msp pqV$ep`f m - v$m_p¡ A¡L$ tbvy$hs `v$p\® dyL¡$g R>¡. Ap v$m[õ\f fl¡ (Mk¡ _lu) s¡ dpV¡$_p¡ v$m A_¡ Y$msp pqV$ep_uk`pV$u hÃQ¡_p¡ gOyÑd ^j®Zp¯L$ µ
min R>¡.
(1) tanθ
(2) 2 tanθ
(3) 3 tanθ
(4) tan2θ
4. 1 km _p A„sf ky udp„ 100 m S>¡V$gp¡ KQp¡ \C S>spY$p¡mphhpmp fp¡X$ D`f W hS>__u A¡L$ L$pf R>¡.
Y$p¡mphhpmp¡ Ap fp¡X$ L$pf `f • ••
S>¡V$gy„ AQm Oj®Z
bm gNpX¡$ R>¡. 10 ms−1 _u Nrs\u fp¡X$ `f D`f
sfa S>sp„ L$pf_¡ phf P _u S>ê$f X¡$ R>¡. Å¡ L$pf Y$p¡mph
`f\u v S>¡V$gu TX$`\u _uQ¡ Dsfsp phf ••
_u S>ê$f
`X¡$ sp¡ v R>¡ :(1) 20 ms
−1
(2) 15 ms−1
(3) 10 ms−1
(4) 5 ms−1
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 4
5. A cubical block of side 30 cm is moving
with velocity 2 ms−1
on a smooth
horizontal surface. The surface has a bump
at a point O as shown in figure. The angular
velocity (in rad/s) of the block immediately
after it hits the bump, is :
(1) 5.0
(2) 6.7
(3) 9.4
(4) 13.3
6. Figure shows elliptical path abcd of a planet
around the sun S such that the area of
triangle csa is
•• the area of the ellipse. (See
figure) With db as the semimajor axis, and
ca as the semiminor axis. If t1
is the time
taken for planet to go over path abc and t2
for path taken over cda then :
(1) t1=t
2
(2) t1=2t
2
(3) t1=3t
2
(4) t1=4t
2
5. ∞∑§ 30 cm ÷È¡Ê flÊ‹Ê ÉÊŸËÿ é‹ÊÚ∑§ ∞∑§ Áø∑§Ÿ ÊÒÁáË ¬⁄U 2 ms
−1 ∑§ flª ‚ ªÁÃ◊ÊŸ „Ò– ¡Ò‚Ê Á∑§ ÁøòÊ◊¥ ÁŒπÊÿÊ ªÿÊ „Ò, O ¬⁄U ∞∑§ •fl⁄UÊœ ÁSÕà „Ò– •fl⁄UÊœ‚ ≈U∑§⁄UÊŸ ∑§ ÃÈ⁄¥Uà ’ÊŒ é‹ÊÚ∑§ ∑§Ê ∑§ÊáÊËÿ flª (⁄UÁ«UÿŸ/
‚∑¥§«U ◊¥) „ÊªÊ —
(1) 5.0
(2) 6.7
(3) 9.4
(4) 13.3
6. ∞∑§ ª˝„ ‚Íÿ¸ S ∑§ øÊ⁄UÊ¥ •Ê⁄U ∞∑§ ŒËÉʸflÎûÊËÿ ∑§ˇÊabcd ◊¥ ß‚ Ã⁄U„ ‚ øÄ∑§⁄U ‹ªÊÃÊ „Ò Á∑§ csa ÁòÊ÷È¡∑§Ê ˇÊòÊ»§‹ ŒËÉʸflÎûÊ ∑§ ˇÊòÊ»§‹ ∑§Ê ∞∑§-øÊÒÕÊ߸ „Ò(ÿ„Ê° ¬⁄U ac ‹ÉÊÈ-•ˇÊ ∞fl¥ bd ŒËÉʸ-•ˇÊ „Ò)– ÿÁŒª˝„ abc ÃÕÊ cda ∑§ˇÊËÿ ¬ÕÊ¥ ∑§ Á‹∞ ∑˝§◊‡Ê—t1 ÃÕÊ t
2 ∑§Ê ‚◊ÿ ‹ÃÊ „Ò, Ã’ —
(1) t1=t
2
(2) t1=2t
2
(3) t1=3t
2
(4) t1=4t
2
5. 30 cm bpSy>hpmp¡ A¡L$ kdO_ 2 ms−1 _p h¡N\u
kdrnrsS> k`pV$u `f Nrs L$f¡ R>¡. ApL©$rÑdp„ v$ip®ìepâdpZ¡ O tbvy$ pk¡ A¡L$ bç` R>¡. Ap bç`_¡ A\X$peL¡$ sfs S> kdO__p¡ L$p¡Zueh¡N (rad/sdp„) R>¡ :
(1) 5.0
(2) 6.7
(3) 9.4
(4) 13.3
6. k|e® (S) t1 _u afs¡ Nrs L$fsp A¡L$ D`N°l_p¡ D`hgeu
dpN® abcd A¡ ApL©$rsdp„ v$ip®h¡g R>¡. Äep„ rÓcyS> csa
_y„ n¡Óam A¡ D`hge_p n¡Óam_y„ •• R>¡, db A¡ s¡_u
v$uO®-An s\p ca s¡_u gOy-An R>¡. Å¡ D`N°l s¡_pL$nue \ abc A_¡ cda dpV¡$ g¡hpsp¡ kde A_y¾$d¡ t
1
A_¡ t2 lp¡e sp¡ :
(1) t1=t
2
(2) t1=2t
2
(3) t1=3t
2
(4) t1=4t
2
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 5
7.
Consider a water jar of radius R that has
water filled up to height H and is kept on a
stand of height h (see figure). Through a
hole of radius r (r << R) at its bottom, the
water leaks out and the stream of water
coming down towards the ground has a
shape like a funnel as shown in the figure.
If the radius of the cross-section of water
stream when it hits the ground is x. Then :
(1)
•• •••• • •
•
•• •
(2)
•••• •••
• • ••
•• •
(3)
•••• •••
• • ••
•• •
(4)
••• •••• • •
•
•• •
7.
∞∑§ R ÁòÊíÿÊ ∑§ ¬ÊŸË ∑§ ¡Ê⁄, Á¡‚ ¬ÊŸË ‚ H ™°§øÊ߸Ã∑§ ÷⁄UÊ ªÿÊ „Ò, ∑§Ê h ™°§øÊ߸ ∑§ S≈Ò¥U«U ¬⁄U ⁄UπÊ ªÿÊ „Ò(ÁøòÊ Œπ¥)– Ë ◊¥ ∞∑§ ¿UÊ≈U Á¿UŒ˝, Á¡‚∑§Ë ÁòÊíÿÊ r„Ò (r << R), ‚ ŸËø Áª⁄Uà „È∞ ¬ÊŸË ∑§Ë œÊ⁄U ∞∑§“∑§Ë¬” ∑§Ê •Ê∑§Ê⁄U œÊ⁄UáÊ ∑§⁄UÃË „Ò– ÿÁŒ ÷ÍÁ◊ ∑§ ˬ⁄U ¬ÊŸË ∑§Ë œÊ⁄U ∑§ •ŸÈ¬˝SÕ ∑§Ê≈U ∑§Ë ÁòÊíÿÊ x „Ò,Ã’ —
(1)
•• •••• • •
•
•• •
(2)
•••• •••
• • ••
•• •
(3)
•••• •••
• • ••
•• •
(4)
••• •••• • •
•
•• •
7.
R rÓÄep fphsp A¡L$ pZu_p Åfdp„ H KQpC ky u_y„`pZu cf¡gy A_¡ s¡_¡ h KQpC_p õV¡$ÞX$ `f fpM¡g R>¡s¡d ^pfp¡. (ApL©$rs Sy>Ap¡) Åf_p smue¡ r -rÓÄep_pA¡L$ L$pZpdp„\u (r << R) `pZu r_L$m¡ R>¡ S>¡ N°pDÞX$sfa _pmQp_p ApL$pfdp„ Å¡hp dm¡ R>¡. Ap `pZu_u^pfp Äepf¡ S>du__¡ AX¡$ R>¡ Ðepf¡ s¡_p ApX$R>¡v$_u rÓÄepx R>¡ sp¡ :
(1)
•• •••• • •
•
•• •
(2)
•••• •••
• • ••
•• •
(3)
•••• •••
• • ••
•• •
(4)
••• •••• • •
•
•• •
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 6
8. 200 g water is heated from 40C to 60C.
Ignoring the slight expansion of water, the
change in its internal energy is close to
(Given specific heat of
water=4184 J/kg/K) :
(1) 8.4 kJ
(2) 4.2 kJ
(3) 16.7 kJ
(4) 167.4 kJ
9. The ratio of work done by an ideal
monoatomic gas to the heat supplied to it
in an isobaric process is :
(1)
••
(2)
••
(3)
••
(4)
••
8. ¬ÊŸË ∑§ ÁflSÃÊ⁄U ∑§Ê Ÿªáÿ ◊ÊŸÃ „È∞, 200 g ¬ÊŸË ∑§Ê40C ‚ 60C Ã∑§ ª⁄U◊ ∑§⁄UŸ ¬⁄U ©‚∑§Ë •Ê¥ÃÁ⁄U∑§™§¡Ê¸ ◊¥ •ŸÈ◊ÊÁŸÃ ¬Á⁄UfløŸ „ÊªÊ (¬ÊŸË ∑§Ê ÁflÁ‡Êc≈UÃʬ = 4184 J/kg/K ‹¥) —(1) 8.4 kJ
(2) 4.2 kJ
(3) 16.7 kJ
(4) 167.4 kJ
9. Á∑§‚Ë ‚◊÷ÊÁ⁄U∑§ ¬˝Á∑˝§ÿÊ ◊¥ ∞∑§ •ÊŒ‡Ê¸ ∞∑§¬⁄U◊ÊáÊÈ∑§ªÒ‚ ∑§ mÊ⁄UÊ Á∑§ÿ ª∞ ∑§Êÿ¸ ÃÕÊ ©‚ ŒË ªß¸ ™§c◊Ê ∑§Ê•ŸÈ¬Êà „ÊªÊ —
(1)
••
(2)
••
(3)
••
(4)
••
8. 200 g `pZu_¡ 40C \u 60C ky u Nfd L$fhpdp„Aph¡ R>¡. `pZu_y„ rhõsfZ AhNZsp, s¡_u Ap„sqfL$EÅ®dp„ \sp¡ a¡fapf A¡ gNcN li¡ :
(`pZu_u rhriô$ Dódp=4184 J/kg/K)(1) 8.4 kJ
(2) 4.2 kJ
(3) 16.7 kJ
(4) 167.4 kJ
9. L$p¡C kdcpqfL$ âq¾$epdp„ A¡L$ Apv$i® A¡L$ `fdpÎhuehpey Üpfp \sp L$pe® s\p s¡_¡ Ap`hpdp„ Aphsu Dódp_p¡NyZp¡Ñf R>¡ :
(1)
••
(2)
••
(3)
••
(4)
••
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 7
10. Two particles are performing simple
harmonic motion in a straight line about
the same equilibrium point. The amplitude
and time period for both particles are same
and equal to A and T, respectively. At time
t=0 one particle has displacement A while
the other one has displacement •••
and
they are moving towards each other. If they
cross each other at time t, then t is :
(1) ••
(2)
•••
(3) ••
(4) ••
11. Two engines pass each other moving in
opposite directions with uniform speed of
30 m/s. One of them is blowing a whistle
of frequency 540 Hz. Calculate the
frequency heard by driver of second engine
before they pass each other. Speed of sound
is 330 m/sec :
(1) 450 Hz
(2) 540 Hz
(3) 648 Hz
(4) 270 Hz
10. ŒÊ ∑§áÊ ∞∑§ ‚⁄U‹ ⁄UπËÿ ¬Õ ¬⁄U ÁSÕà ∞∑§ „Ë ◊ÊäÿÁ’¥ŒÈ ∑§ ‚ʬ Ê ß‚ Ã⁄U„ ‚ ‚⁄U‹ •Êflø ªÁÃ◊ÊŸ •flSÕÊ◊¥ „Ò Á∑§ ©Ÿ∑§ •ÊÿÊ◊ (A) ÃÕÊ •Êflø-∑§Ê‹ (T) ∞∑§‚◊ÊŸ „Ò¥– ÿÁŒ t=0 ‚◊ÿ ¬⁄U ∞∑§-ŒÍ‚⁄U ∑§Ë Ã⁄U»§•Êà „È∞, ∞∑§ ∑§áÊ ∑§Ê ÁflSÕʬŸ A „Ò ÃÕÊ ŒÍ‚⁄U ∑§áÊ
∑§Ê ÁflSÕʬŸ •••
„Ê, ÃÊ t ‚◊ÿ ¬⁄U fl ∞∑§ ŒÍ‚⁄U ∑§Ê
¬Ê⁄U ∑§⁄Uà „Ò¥– t ∑§Ê ◊ÊŸ „ÊªÊ —
(1) ••
(2)
•••
(3) ••
(4) ••
11. ŒÊ ⁄U‹-ߥ¡Ÿ ∞∑§-ŒÍ‚⁄U ∑§Ê ¬Ê⁄U ∑§⁄Uà „È∞ Áfl¬⁄UËÃÁŒ‡ÊÊ ◊¥ 30 m/s ∑§Ë ∞∑§ ‚◊ÊŸ ªÁà ‚ ø‹ ⁄U„ „Ò¥–©Ÿ◊¥ ‚ ∞∑§ ߥ¡Ÿ ÿÁŒ 540 Hz •ÊflÎÁûÊ ‚ ‚Ë≈UË ’¡Ê⁄U„Ê „Ò, ÃÊ ŒÍ‚⁄U ߥ¡Ÿ ∑§ «˛UÊ߸fl⁄U mÊ⁄UÊ ‚ÈŸË ªß¸ äflÁŸ ∑§Ë•Êfl Î Áû Ê „Ê ªË (äflÁŸ ∑§Ë ªÁà ∑§Ê ◊ÊŸ330 m/sec ‹¥) —(1) 450 Hz
(2) 540 Hz
(3) 648 Hz
(4) 270 Hz
10. A¡L$S> kdsygus tbvy$_u kp ¡n¡ b¡ L$Zp¡ ku^u f¡Mpdp„kfm Aphs®Nrs L$f¡ R>¡. Ap b„_¡ L$Zp¡ dpV¡$ L„$`rhõspfs\p Aphs® L$pm kdp_ R>¡ A_¡ s¡ ¾$di: A s\p T R>¡.t=0 kde¡, A¡L$buÅ sfa Aphsp A¡L$ L$Z_y„ õ\p_p„sf
A s\p buÅ_y„ õ\p_p„sf •••
R>¡. Å¡ t kde¡ s¡
A¡L$buÅ_¡ `pf L$f¡ sp¡ t R>¡ :
(1) ••
(2)
•••
(3) ••
(4) ••
11. b¡ A¡ [ÞS>_ A¡L $buÅ_¡ `pf L $fsp rh`fusqv$ipdp„ 30 m/s _u kdp_ Nrs\u Qpg¡ R> ¡.Apdp_y „ A¡L$ A¡[ÞS>_ Å¡ 540 Hz Aph©rs\ukuV$u hNpX$u füy „ lp ¡e, sp ¡ buÆ A¡[ÞS>__pX² $pehf hX¡$ k„cmpsp AhpS>_u Aph©rs li¡ :(AhpS>_u TX$` 330 m/sec gp¡)(1) 450 Hz
(2) 540 Hz
(3) 648 Hz
(4) 270 Hz
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 8
12. The potential (in volts) of a charge
distribution is given by
V(z)=30−5z2 for z 1 m
V(z)=35−10 z for z 1 m.
V(z) does not depend on x and y. If this
potential is generated by a constant charge
per unit volume ρ0 (in units of
0 ) which is
spread over a certain region, then choose
the correct statement.
(1) ρ0=10
0 for z 1 m and ρ
0=0
elsewhere
(2) ρ0=20
0 in the entire region
(3) ρ0=40
0 in the entire region
(4) ρ0=20
0 for z 1 m and ρ
0=0
elsewhere
13. Three capacitors each of 4 µF are to be
connected in such a way that the effective
capacitance is 6 µF. This can be done by
connecting them :
(1) all in series
(2) two in series and one in parallel
(3) all in parallel
(4) two in parallel and one in series
12. ∞∑§ •Êfl‡Ê-ÁflÃ⁄UáÊ ∑§ mÊ⁄UÊ ÁŸêŸÁ‹Áπà Áfl÷fl (flÊÀ≈U◊¥) ©à¬ÛÊ „ÊÃÊ „Ò —
V(z)=30−5z2, z 1 m ◊¥
V(z)=35−10 z , z 1 m ◊¥
V(z), x ∞fl¥ y ¬⁄U ÁŸ÷¸⁄U Ÿ„Ë¥ ∑§⁄UÃÊ– ÿÁŒ ÿ„ Áfl÷fl∞∑§ ÁŸÿà •Êfl ‡Ê ¡Ê ¬ ˝ Áà ß∑§Êß ¸ •Êÿßρ
0 (
0 ß∑§ÊßÿÊ¥ ◊¥) „Ò ÃÕÊ ∞∑§ ÁŒÿ „È∞ ˇÊòÊ ◊¥ »Ò§‹Ê
„È•Ê „Ò, ‚ ©à¬ÊÁŒÃ „Ò, Ã’ ÁŸêŸÁ‹Áπà ◊¥ ‚ ‚„ËÁfl∑§À¬ ∑§Ê øÿŸ ∑§⁄¥U —
(1) ρ0=10
0, z1 m ◊¥ ÃÕÊ ρ
0=0 •ãÿòÊ
(2) ρ0=20
0, ‚fl¸òÊ
(3) ρ0=40
0, ‚fl¸òÊ
(4) ρ0=20
0, z1 m ◊¥ ÃÕÊ ρ
0=0 •ãÿòÊ
13. 4 µF œÊÁ⁄UÃÊ ∑§ ÃËŸ ‚¥œÊÁ⁄UòÊÊ¥ ‚ ß‚ Ã⁄U„ ‚ ‚¥ÿÊ¡Ÿ’ŸÊŸÊ „Ò Á∑§ ¬˝÷ÊflË œÊÁ⁄UÃÊ 6 µF „Ê ¡Ê∞– ÿ„ ÁŸêŸ‚¥ÿÊ¡Ÿ ‚ ¬˝ÊåàÊ Á∑§ÿÊ ¡Ê ‚∑§ÃÊ „Ò —
(1) ÃËŸÊ¥ üÊáÊË ∑˝§◊ ◊¥
(2) ŒÊ üÊáÊË ∑˝§◊ ◊¥ ÃÕÊ ÃË‚⁄UÊ ¬Ê‡fl¸∑˝§◊ ◊¥
(3) ÃËŸÊ¥ ¬Ê‡fl¸∑˝§◊ ◊¥
(4) ŒÊ ¬Ê‡fl¸∑˝§◊ ◊¥ ÃÕÊ ÃË‚⁄UÊ üÊáÊË ∑˝§◊ ◊¥
12. A¡L$ rhÛyscpf rhsfZdp„ rhch (volts dp„)V(z)=30−5z
2, z 1 m dpV¡$
V(z)=35−10 z, z 1 m dpV¡$ Ap`hpdp„ Aph¡R>¡.
V(z) A¡ x A_¡ y `f Ap^pf fpMsp¡ _\u. Å¡ Ap[õ\rsdp_ A¡ r_es rhS>cpf ârs L$v$ ρ
0 (
0 _p
A¡L$ddp„) hX¡$ DÐ`Þ_ L$fhpdp„ Aph¡ L¡$ S>¡ L$p¡C A¡L$Qp¡½$k n¡Ódp„ rhõsf¡g lp¡e, sp¡ kpQy„ rh^p_ `k„v$L$fp¡.
(1) ρ0=10
0, z1 m dpV¡$ s\p ρ
0=0 AÞeÓ
(2) ρ0=20
0 kh®n¡Ódp„
(3) ρ0=40
0 kh®n¡Ódp„
(4) ρ0=20
0, z1 m dpV¡$ A_¡ ρ
0=0 AÞeÓ
13. 4 µF _p¡ A¡L$ A¡hp ÓZ L¸$`p rkV$fp¡ A¡ fus¡ Å¡X$hpdp„Aph¡g R>¡ L¡$ s¡d_p¡ kdsyëe L¸$`prkV$Þk 6 µF R>¡.Aphy„ s¡d_¡ :
(1) î¡Zudp„ Å¡X$hp\u \C iL¡$
(2) b¡_¡ î¡Zudp„ A_¡ A¡L$_¡ kdp„sf Å¡X$hp\u \CiL¡$
(3) kdp„sf Å¡X$hp\u \C iL¡$
(4) b¡_¡ kdp„sf A_¡ A¡L$_¡ î¡Zudp„ Å¡X$hp\u \CiL¡$
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 9
14.
In the circuit shown, the resistance r is a
variable resistance. If for r=f R, the heat
generation in r is maximum then the value
of f is :
(1)
••
(2)
••
(3)
••
(4) 1
15. A magnetic dipole is acted upon by two
magnetic fields which are inclined to each
other at an angle of 75. One of the fields
has a magnitude of 15 mT. The dipole
attains stable equilibrium at an angle of 30
with this field. The magnitude of the other
field (in mT ) is close to :
(1) 11
(2) 36
(3) 1
(4) 1060
14.
ÁŒÿ „È∞ ¬Á⁄U¬Õ ◊¥ r ∞∑§ ø⁄U-¬˝ÁÃ⁄UÊœ „Ò– ÿÁŒr=f R, Ã’ r ◊¥ ™§c◊Ê ©à¬ÊŒŸ •Áœ∑§Ã◊ „ÊŸ ∑§Á‹ÿ f ∑§Ê ◊ÊŸ „ÊªÊ —
(1)
••
(2)
••
(3)
••
(4) 1
15. ∞∑§ øÈê’∑§Ëÿ Ámœ˝Èfl ¬⁄U ŒÊ øÈê’∑§Ëÿ ˇÊòÊ, ¡Ê •Ê¬‚◊¥ 75 ∑§ÊáÊ ’ŸÊà „Ò¥, ∞∑§ ‚ÊÕ Á∑˝§ÿÊ ∑§⁄Uà „Ò¥– ÿÁŒÿ„ Ámœ˝Èfl ‚¥ÃÈ‹Ÿ ∑§Ë •flSÕÊ ◊¥ øÈê’∑§Ëÿ ¬˝⁄UáÊ15 mT ∑§ ∞∑§ øÈê’∑§Ëÿ ˇÊòÊ ‚ 30 ∑§Ê ∑§ÊáÊ ’ŸÊÃÊ„Ò, ÃÊ ŒÍ‚⁄U øÈê’∑§Ëÿ ÊòÊ ∑§ øÈê’∑§Ëÿ ¬⁄UáÊ ∑§Ê ‹ª÷ª◊ÊŸ ( mT ◊¥ ) „ÊªÊ —(1) 11
(2) 36
(3) 1
(4) 1060
14.
Ap ¡g `qf`\dp„ r A¡L$ Qg Ahfp¡ R>¡. Å¡ r=f R
dpV¡$ r dp„ dlÑd Dódp DÐ`Þ_ \pe sp¡ f _u qL„$ds\i¡ :
(1)
••
(2)
••
(3)
••
(4) 1
15. A¡L$buÅ_u kpd¡ 75 _p MyZp¡ Y$msp b¡ Qy„bL$ue n¡Óp¡`f A¡L$ Qy„bL$ue qÜ- y°h (Dipole) L$pe®fs R>¡. L$p¡CA¡L$ n¡Ó_y„ dyëe 15 mT R>¡ A_¡ Ap n¡Ó kp\¡ 30 _pMyZp\u X$pe`p¡g õ\peu k„sygus Ahõ\p âpá L$f¡ R>¡.buÅ n¡Ó_y„ dyëe (mT dp„) Apif¡ li¡ :(1) 11
(2) 36
(3) 1
(4) 1060
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 10
16. A 50 Ω resistance is connected to a battery
of 5 V. A galvanometer of resistance
100 Ω is to be used as an ammeter to
measure current through the resistance, for
this a resistance rs
is connected to the
galvanometer. Which of the following
connections should be employed if the
measured current is within 1% of the
current without the ammeter in the
circuit ?
(1) rs=0.5 Ω in parallel with the
galvanometer
(2) rs=0.5 Ω in series with the
galvanometer
(3) rs
= 1 Ω in series with galvanometer
(4) rs=1 Ω in parallel with galvanometer
16. ∞∑§ 50 Ω ∑§Ê ¬˝ÁÃ⁄UÊœ ∞∑§ 5 V ∑§Ë ’Ò≈U⁄UË ‚ ¡È«∏Ê „È•Ê„Ò– ∞∑§ ªÒÀflŸÊ◊Ë≈U⁄U Á¡‚∑§Ê ¬˝ÁÃ⁄UÊœ 100 Ω „Ò, ∑§Ê∞ê¬Ëÿ⁄U◊Ë≈U⁄U ∑§ M§¬ ◊¥ ¬˝ÿʪ Á∑§ÿÊ ¡ÊŸÊ „Ò–ªÒÀflŸÊ◊Ë≈U⁄U ∑§ ‚ÊÕ ∞∑§ ¬˝ÁÃ⁄UÊœ r
s ‚¥ÿÊÁ¡Ã „Ò– ÿÁŒ
ß‚ ‚¥ÿÊ¡Ÿ ◊¥ ◊ÊÁ¬Ã œÊ⁄UÊ ∞ê¬Ëÿ⁄U◊Ë≈U⁄U ∑§Ê „≈UÊŸ ¬⁄U◊ÊÁ¬Ã œÊ⁄UÊ ∑§ ◊ÊŸ ‚ 1% ∑§ ÷ËÃ⁄U „Ê ÃÊ ÁŸêŸÁ‹ÁπÃ◊¥ ‚ ∑§ÊÒŸ-‚Ê ‚¥ÿÊ¡Ÿ ©Áøà „ʪÊ?
(1) rs=0.5 Ω ªÒÀflŸÊ◊Ë≈U⁄U ∑§ ‚ÊÕ ¬Ê‡fl¸∑˝§◊ ◊¥
(2) rs=0.5 Ω ªÒÀflŸÊ◊Ë≈U⁄U ∑§ ‚ÊÕ üÊáÊË ∑˝§◊ ◊¥
(3) rs
= 1 Ω ªÒÀflŸÊ◊Ë≈U⁄U ∑§ ‚ÊÕ üÊáÊË ∑˝§◊ ◊¥
(4) rs=1 Ω ªÒÀflŸÊ◊Ë≈U⁄U ∑§ ‚ÊÕ ¬Ê‡fl¸∑˝§◊ ◊¥
16. 5 V _u b¡V$fu kp\¡ 50 Ω _p¡ A¡L$ Ahfp¡ gNpX¡$g R>¡.Ap Ahfp ¡^dp „\u `kpf \sp âhpl dp`hp100 Ω _p Ahfp¡ fphsy„ N¡ëh¡_p¡duV$f_¡ A¡rdV$f sfuL¡$hp`fhpdp„ Aph¡ R>¡. Ap dpV¡$ A¡L$ Ahfp¡ r
s N¡ëh¡_p¡duV$f
kp\¡ Å¡X$hpdp„ Aph¡ R>¡.
_uQ¡ Ap ¡g L$ey Å¡X$pZ hp`fhy„ Å¡CA¡ L¡$ S>¡\u dp ¡gâhpl A¡ A¡rdV$f hNf `qf`\_p dp„_p âhpl_u1% _u Ahr^dp„ lp¡e.
(1) rs=0.5 Ω N¡ëh¡_p¡duV$f_¡ kdp„sf
(2) rs=0.5 Ω N¡ëh¡_p¡duV$f_u î¡Zudp„
(3) rs
= 1 Ω N¡ëh¡_p¡duV$f_u î¡Zudp„
(4) rs=1 Ω N¡ëh¡_p¡duV$f_¡ kdp„sf
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 11
17. A series LR circuit is connected to a voltage
source with V(t)=V0
sinΩt. After very
large time, current I(t) behaves as
••••
>>
:
(1)
(2)
(3)
(4)
17. ∞∑§ üÊ áÊË LR ¬Á⁄U¬Õ ∑§Ê ∞∑§ flÊÀ≈UËÿ dÊÃV(t)=V
0 sinΩt ‚ ¡Ê«∏Ê ¡ÊÃÊ „Ò ∑§Ê»§Ë ‹¥’ ‚◊ÿ
’ÊŒ ÁfllÈà œÊ⁄UÊ I(t) ∑§Ê ‚„Ë ÁøòÊáÊ Á∑§‚ Ã⁄U„ ∑§Ê
„ʪÊ? ••••
>>
¡„Ê°
(1)
(2)
(3)
(4)
17. A¡L$ LR î¡Zu qf`\ V(t)=V0
sinΩt hp¡ëV¡$S> öp¡skp\¡ Å¡X¡$g R>¡. Ap gp„bp kde A„sfpg bpv$, rhÛys
âhpl I(t) hs®i¡ ••••
>>
:
(1)
(2)
(3)
(4)
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 12
18. Microwave oven acts on the principle of :
(1) transferring electrons from lower to
higher energy levels in water
molecule
(2) giving rotational energy to water
molecules
(3) giving vibrational energy to water
molecules
(4) giving translational energy to water
molecules
18. ◊Êß∑˝§Êflfl •ÊflŸ Á∑§‚ ¬˝Á∑˝§ÿÊ ¬⁄U •ÊœÊÁ⁄Uà „Ò?
(1) ¡‹ •áÊÈ•Ê¥ ◊¥ ÁSÕà ߋÄ≈˛UÊÚŸÊ¥ ∑§ ∑§◊ ™§¡Ê¸‚ •Áœ∑§ ™§¡Ê¸ flÊ‹ ‹fl‹ ◊¥ SÕÊŸÊ¥ÃÁ⁄Uà ∑§⁄UŸ∑§Ë ¬˝Á∑˝§ÿÊ ¬⁄U–
(2) ¡‹ •áÊÈ•Ê¥ ∑§Ê ÉÊÍáʸŸ ™§¡Ê¸ ¬˝ŒÊŸ ∑§⁄UŸ ∑§Ë¬˝Á∑˝§ÿÊ ¬⁄U–
(3) ¡‹ •áÊÈ•Ê¥ ∑§Ê ∑¥§¬Ÿ ™§¡Ê¸ ¬˝ŒÊŸ ∑§⁄UŸ ∑§Ë¬˝Á∑˝§ÿÊ ¬⁄U–
(4) ¡‹ •áÊÈ•Ê¥ ∑§Ê SÕÊŸÊ¥Ã⁄UË ™§¡Ê¸ ¬˝ŒÊŸ ∑§⁄UŸ∑§Ë ¬˝Á∑˝§ÿÊ ¬⁄U–
18. dpC¾$p¡h¡h Ap¡h_ L$ep rkÝ^p„s `f L$pe® L$f¡ R>¡ ?
(1) `pZu_p AÏ„Ap¡dp„ Ap¡R>u \u h y EÅ® õsfsfa Cg¡¼V²$p¡__y„ õ\p_p„sfZ
(2) `pZu_p AÏ„Ap¡_¡ O|Z®_ EÅ® Ap`hp_y„
(3) `pZu_p AÏ„Ap¡_¡ L„$`_ EÅ® Ap`hp_y„
(4) `pZu_p AÏ„Ap¡_¡ õ\p_p„sqfs EÅ® Ap`hp_y„
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 13
19. A convex lens, of focal length 30 cm, a
concave lens of focal length 120 cm, and a
plane mirror are arranged as shown. For
an object kept at a distance of 60 cm from
the convex lens, the final image, formed by
the combination, is a real image, at a
distance of :
(1) 60 cm from the convex lens
(2) 60 cm from the concave lens
(3) 70 cm from the convex lens
(4) 70 cm from the concave lens
19. ∞∑§ ©ûÊ‹ ‹Ò¥‚ fl •flË ‹Ò¥‚, Á¡Ÿ∑§Ë $»§Ê∑§‚ ŒÍ⁄UË∑˝§◊‡Ê— 30 cm ∞fl¥ 120 cm „Ò, ÃÕÊ ‚◊Ë Œ¬¸áÊÁŸêŸ ÁøòÊ ∑§ •ŸÈ‚Ê⁄U ⁄Uπ ªÿ „Ò– ∞∑§ Á’ê’ ©ûÊ‹‹Ò¥‚ ‚ 60 cm ∑§Ë ŒÍ⁄UË ¬⁄U ÁSÕà „Ò– ß‚ ‚¥ÿÊ¡Ÿ mÊ⁄UÊÁŸÁ◊¸Ã •¥ÁÃ◊ ¬˝ÁÃÁ’ê’ ∞∑§ flÊSÃÁfl∑§ ¬˝ÁÃÁ’ê’ „ÒÁ¡‚∑§Ë ÁSÕÁà ÁŸêŸÁ‹Áπà „ÊªË —
(1) ©ûÊ‹ ‹Ò¥‚ ‚ 60 cm ∑§Ë ŒÍ⁄UË ¬⁄U–
(2) •flË ‹Ò¥‚ ‚ 60 cm ∑§Ë ŒÍ⁄UË ¬⁄U–
(3) ©ûÊ‹ ‹Ò¥‚ ‚ 70 cm ∑§Ë ŒÍ⁄UË ¬⁄U–
(4) •flË ‹Ò¥‚ ‚ 70 cm ∑§Ë ŒÍ⁄UË ¬⁄U–
19. 30 cm L¡$ÞÖg„bpC ^fphsp¡ A¡L$ brlNp£m L$pQ,120 cm L¡$ÞÖg„bpC ^fphsp¡ A¡L$ A„sNp£m L$pQ A_¡A¡L$ kpv$p¡ Afukp¡ ApL©$rÑdp„ bspìep âdpZ¡ Np¡W$$h¡gR>¡.
brlNp£m L$pQ\u 60 cm vy$f fpM¡g hõsy_y„ _uQ¡ v$ip®h¡gL$C [õ\rsdp„ Ap Np¡W$hhp\u L¡$V$gp A„sf¡ hõsy_y„ hpõsrhL$ârstbb dmi¡.
(1) brlNp£m L$pQ\u 60 cm
(2) A„sNp£m L$pQ\u 60 cm
(3) brlNp£m L$pQ\u 70 cm
(4) A„sNp£m L$pQ\u 70 cm
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 14
20. In Young’s double slit experiment, the
distance between slits and the screen is
1.0 m and monochromatic light of 600 nm
is being used. A person standing near the
slits is looking at the fringe pattern. When
the separation between the slits is varied,
the interference pattern disappears for a
particular distance d0 between the slits. If
the angular resolution of the eye is
•• •
•,
the value of d0 is close to :
(1) 1 mm
(2) 2 mm
(3) 4 mm
(4) 3 mm
20. ÿ¥ª ∑§ Ám-Á! Ê⁄UË ¬˝ÿʪ ◊¥, Á¡‚◊¥ ¬Œ¸ ∞fl¥ Á! Ê⁄UË ∑§’Ëø ∑§Ë ŒÍ⁄UË 1.0 m ÃÕÊ 600 nm Ã⁄¥UªŒÒäÿ¸ ∑§∞∑§fláÊ˸ÿ ¬˝∑§Ê‡Ê ∑§Ê ©¬ÿʪ Á∑§ÿÊ ªÿÊ „Ò– Á! ÊÁ⁄UÿÊ¥∑§ ‚◊ˬ π«∏Ê „È•Ê ∞∑§ √ƒÊÁÄàÊ √ÿÁÃ∑§⁄UáÊ ¬Ò≈UŸ¸ ∑§ÊŒπ ⁄U„Ê „Ò– ŒÊŸÊ¥ Á! ÊÁ⁄UÿÊ¥ ∑§ ’Ëø ∑§Ë ŒÍ⁄UË ∑§Ê ¬Á⁄UflÁøÃ∑§⁄Ÿ ¬⁄U ∞∑§ Áfl‡Ê· ŒÍ⁄UË d
0 ¬⁄U √ÿÁàÊ∑§⁄UáÊ ¬Ò≈UŸ¸ ‹ÈåàÊ
„Ê ¡ÊÃÊ „Ò– ÿÁŒ √ÿÁÄàÊ ∑§Ë •Ê°π ∑§Ê ∑§ÊáÊËÿ ÁflÿÊ¡Ÿ
•• •
•„Ê, ÃÊ d
0 ∑§Ê ◊ÊŸ ‹ª÷ª „ÊªÊ —
(1) 1 mm
(2) 2 mm
(3) 4 mm
(4) 3 mm
20. e„Ó_p X$bg [õgV$_p âep¡N, [õgV$$ A_¡ `X$v$p hÃQ¡_y„A„sf 1.0 m R>¡ s\p 600 nm A¡L$ f„Nue âL$piD`ep¡Ndp„ g¡hpdp„ Aph¡g R>¡. [õgV$_u _ÆL$ Dc¡g A¡L$ìe[¼s igpL$pcps sfa Å¡h¡ R>¡. Äepf¡ [õgV$ hÃQ¡_y„A„sf bv$ghpdp„ Aph¡ R>¡ Ðepf¡ b¡ [õgV$ hÃQ¡_p rhi¡jA„sf d
0 dpV¡$ ìersL$fZ cps gyá \pe R>¡. Å¡ Ap„M_y
L$p¡Zue rhc¡v$_ •• •
• R>¡, sp¡ d
0 _y„ dyëe _________
_u _ÆL$_y„ \i¡.(1) 1 mm
(2) 2 mm
(3) 4 mm
(4) 3 mm
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 15
21. When photons of wavelength λ1
are
incident on an isolated sphere, the
corresponding stopping potential is found
to be V. When photons of wavelength λ2
are used, the corresponding stopping
potential was thrice that of the above value.
If light of wavelength λ3 is used then find
the stopping potential for this case :
(1)
• • •
• • •• ••• •• •• •
•• • •• •• • •
(2)
• • •
• • •• ••• •• •• •
•• • •• •• • •
(3)
• • •
• • •• •
• ••• •• •• •
•• • •• •• • •
(4)
• • •
• • ••
• ••• •• •• •
•• • •• •• • •
21. ¡’ λ1 Ã⁄¥UªŒÒäÿ¸ ∑§ »§Ê≈ÊŸ ∞∑§ Áfl‹ÁªÃ ªÊ‹ ∑§Ê
¬˝ŒËåàÊ ∑§⁄Uà „Ò¥, ÃÊ ‚¥ªÃ “ÁŸ⁄UÊœË-Áfl÷fl” ∑§Ê ◊ÊŸ V¬ÊÿÊ ¡ÊÃÊ „Ò– ¡’ λ
2 Ã⁄¥UªŒÒäÿ¸ ∑§ »§Ê≈ÊŸ ©¬ÿʪ ◊¥
‹Êÿ ¡Êà „Ò¥ ÃÊ ÁŸ⁄UÊœË-Áfl÷fl ∑§Ê ◊ÊŸ ÁÃªÈŸÊ (3V)
„Ê ¡ÊÃÊ „Ò– •ª⁄U λ3 Ã⁄¥UªŒÒäÿ¸ ∑§ »§Ê≈UÊŸ ‚ ªÊ‹ ∑§Ê
¬˝ŒËåàÊ Á∑§ÿÊ ¡Ê∞ ÃÊ ÁŸ⁄ÊœË-Áfl÷fl ∑§Ê ◊ÊŸ „ÊªÊ —
(1)
• • •
• • •• ••• •• •• •
•• • •• •• • •
(2)
• • •
• • •• ••• •• •• •
•• • •• •• • •
(3)
• • •
• • •• •
• ••• •• •• •
•• • •• •• • •
(4)
• • •
• • ••
• ••• •• •• •
•• • •• •• • •
21. Äepf¡ λ1 sf„Ng„bpC_p ap¡V$p¡Þk_¡ A¡L$ AgN L$f¡g Np¡mp
`f Ap`ps L$fhpdp„ Aph¡ R>¡ Ðepf¡ A_yê$` r_fp¡ u-rhch (õV$p¡t`N `p¡V¡$[Þieg) V S>¡V$gp¡ \pe R>¡. Äepf¡λ
2 sf„Ng„bpC_p ap¡V$p¡Þk_p¡ D`ep¡N L$fhpdp„ Aph¡ R>¡
Ðepf¡ s¡_¡ A_yê$` r_fp¡ u-rhch l¡gp L$fsp„ ÓZ NZp¡\pe R>¡. Äepf¡ λ
3 sf„Ng„bpC_p¡ âL$pi hp`fhpdp„ Aph¡
R>¡ Ðepf¡ r_fp¡ u-rhch_y„ dyëe \i¡ :
(1)
• • •
• • •• ••• •• •• •
•• • •• •• • •
(2)
• • •
• • •• ••• •• •• •
•• • •• •• • •
(3)
• • •
• • •• •
• ••• •• •• •
•• • •• •• • •
(4)
• • •
• • ••
• ••• •• •• •
•• • •• •• • •
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 16
22. A hydrogen atom makes a transition from
n=2 to n=1 and emits a photon. This
photon strikes a doubly ionized lithium
atom (z=3) in excited state and completely
removes the orbiting electron. The least
quantum number for the excited state of the
ion for the process is :
(1) 2
(2) 3
(3) 4
(4) 5
23. The truth table given in fig. represents :
• • •• • •• • •• • •• • •
(1) AND - Gate
(2) OR - Gate
(3) NAND - Gate
(4) NOR - Gate
22. ∞∑§ „Êß«˛UÊ¡Ÿ ¬⁄U◊ÊáÊÈ n=2 Ä√ÊÊ¥≈U◊ ‚¥ÅÿÊ flÊ‹ ™§¡Ê¸‹fl‹ ‚ n=1 Ä√ÊÊ¥≈U◊ ‚¥ÅÿÊ flÊ‹ ™§¡Ê¸ ‹fl‹ ◊¥‚¥∑˝§◊áÊ ∑§⁄UŸ ¬⁄U ∞∑§ »§Ê≈UÊŸ ©à‚Á¡¸Ã ∑§⁄UÃÊ „Ò– ÿ„»§Ê≈UÊŸ ∞∑§ Ám-•ÊÿÁŸÃ Á‹ÁÕÿ◊ ¬⁄U◊ÊáÊÈ (z=3) (¡ÊÁ∑§ ©ûÊÁ¡Ã •flSÕÊ ◊¥ „Ò) ‚ ≈U∑§⁄UÊÃÊ „Ò •ÊÒ⁄U ∑§ˇÊËÿß‹Ä≈˛UÊÚŸ (orbiting electron) ∑§Ê ¬Í⁄UË Ã⁄U„ ‚ ’Ê„⁄UÁŸ∑§Ê‹ ŒÃÊ „Ò– ß‚ ¬˝Á∑˝§ÿÊ ∑§ Á‹∞ •ÊÿŸ ∑§Ë ©ûÊÁ¡Ã•flSÕÊ ∑§Ë ãÿÍŸÃ◊ Ä√ÊÊ¥≈U◊ ‚¥ÅÿÊ „ÊªË —(1) 2
(2) 3
(3) 4
(4) 5
23. ÁøòÊ ◊¥ ÁŒπÊ߸ ªß¸ ‚àÿ◊ÊŸ-‚Ê⁄UáÊË ÁŸêŸÁ‹Áπà ◊¥ ‚Á∑§‚ ª≈U ∑§Ê Œ‡ÊʸÃË „Ò?
• • •• • •• • •• • •• • •
(1) AND ª≈U
(2) OR ª≈U
(3) NAND ª≈U
(4) NOR ª≈U
22. A¡L$ lpCX²$p¡S>_ `fdpÏ„ n=2 \u n=1 EÅ® õsfdp„k„¾$dZ L$f¡ R>¡ A_¡ A¡L$ ap¡V$p¡_ DÐkÆ®s L$f¡ R>¡. Apap¡V$p¡_ qÜ-Apep¡_pCÈX$ rg\ued `fdpÏ„ (z=3) _¡s¡_u DÑ¡Æs Ahõ\pdp„ A\X$pe R>¡ A_¡ L$nueCg¡¼V²$p¡__¡ k„ |Z® blpf r_L$pm¡ R>¡. Ap âq¾$ep dpV¡$Ape__u DÑ¡Æs Ahõ\p_u Þe|_sd ¼hpÞV$d _„bf\i¡ :(1) 2
(2) 3
(3) 4
(4) 5
23. ApL©$rsdp„ Ap`¡g V³$\ V¡$bg _uQ¡ Ap`¡g L$ey N¡V$ bsph¡R>¡ ?
• • •• • •• • •• • •• • •
(1) AND - Gate
(2) OR - Gate
(3) NAND - Gate
(4) NOR - Gate
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 17
24. A¡L$ Ýhr_ rkÁ_g b¡ Sy>v$p Sy>v$p AhÅ¡ ^fph¡ R>¡. A¡L$dp_h-cprjs rkÁ_g R>¡ S>¡ 200 Hz \u 2700 Hz
Aph©rÑ A„sfpgdp„ R>¡. Äepf¡ buSy> A¡ DÃQ Aph©rÑk„Nus_y„ rkÁ_g R>¡ S>¡ 10200 Hz \u 15200 Hz
Aph©rÑ A„sfpgdp„ R>¡. bÞ_¡ rkÁ_gp¡_¡ kp\¡ k„QfhpdpV¡$ S>ê$fu AM rkÁ_g_u b¢X$ rhX¹$\ s\p a¼s dp_h-cprjs rkÁ_g k„QfZ dpV¡$ S>ê$fu AM rkÁ_g b¢X$rhX¹$\_p¡ NyZp¡Ñf R>¡.(1) 3
(2) 5
(3) 6
(4) 2
25. m Öìedp_ hpmp Np¡mp ^fphsp s\p Öìedp_ frls^psy_p spf\u b_¡g A¡L$ kpv$p gp¡gL$_p¡ T=0C A¡Aphs®L$pm 2 s R>¡. Å¡ spf_p sp`dp_dp„ h^pfp¡ L$fsp,Aphs®L$pmdp„ \sp a¡fapf_¡ Apg¡M Üpfp v$ip®hpe, sp¡`qfZpdu Apg¡M A¡ A¡L$ f¡Mp R>¡ S>¡_p¡ Y$pm S R>¡. Å¡^psy_p¡ f¡Mue-âkfZp„L$ α lp¡e, sp¡ S _y„ dyëe R>¡ :(1) α
(2) ••
(3) 2α
(4)
••
24. ∞∑§ äflÁŸ-Á‚ÇŸ‹ ŒÊ S¬c≈U äflÁŸÿÊ¥ ‚ ÁŸÁ◊¸Ã „Җߟ◊¥ ‚ ∞∑§ ◊ŸÈcÿ mÊ⁄UÊ ÷ÊÁ·Ã Á‚ÇãÊ‹ „Ò ¡Ê200 Hz ‚ 2700 Hz ∑§Ë •ÊflÎÁûÊ •¥Ã⁄UÊ‹ ∑§Ê „Ò, ÃÕʌ͂⁄UÊ Á‚ÇŸ‹ 10200 Hz ‚ 15200 Hz ©ìÊ •ÊflÎÁûÊflÊ‹ ‚¥ªËà ∑§Ê „Ò– ŒÊŸÊ¥ Á‚ÇŸ‹Ê¥ ∑§ ‚¥ø⁄UáÊ ∑§ Á‹∞•Êfl‡ÿ∑§ AM Á‚ÇãÊ‹ ∑§Ë ’Ò¥«U-øÊÒ«∏Ê߸ •ÊÒ⁄U ∑§fl‹◊ŸÈcÿ mÊ⁄UÊ ÷ÊÁ·Ã Á‚ÇŸ‹ ∑§ ‚¥ø⁄UáÊ ∑§ Á‹∞ •Êfl‡ÿ∑§AM Á‚ÇãÊ‹ ∑§Ë ’Ò¥«U-øÊÒ«∏Ê߸ ∑§Ê •ŸÈ¬Êà ÄÿÊ „ʪÊ?(1) 3
(2) 5
(3) 6
(4) 2
25. T=0C ¬⁄U ∞∑§ ‚⁄U‹-‹Ê‹∑§, ¡Ê Á∑§ m Œ˝√ÿ◊ÊŸ∑§ ªÊ‹∑§ •ÊÒ⁄U Œ˝√ÿ◊ÊŸ ⁄UÁ„à œÊÃÈ ∑§ ÃÊ⁄U ‚ ÁŸÁ◊¸Ã „Ò,∑§Ê •Êflûʸ-∑§Ê‹ 2 s „Ò– •ª⁄U ÃÊ⁄U ∑§ Ãʬ◊ÊŸ ∑§Ê’…∏ÊŸ ‚, •Êflûʸ-∑§Ê‹ ◊¥ „È߸ flÎÁh ∑§Ê ª˝Ê»§ mÊ⁄UÊŒ‡ÊʸÿÊ ¡Êÿ, ÃÊ ¬Á⁄UáÊÊ◊Ë ª˝Ê»§ ∑§Ë …Ê‹-◊ʬ ( slope)
S „Ò– ÿÁŒ ÃÊ⁄U ∑§Ê ⁄ÒUÁπ∑§-¬˝‚Ê⁄U ªÈáÊÊ¥∑§ α „Ò ÃÊ S ∑§Ê◊ÊŸ „ÊªÊ —(1) α
(2) ••
(3) 2α
(4)
••
24. An audio signal consists of two distinct
sounds : one a human speech signal in the
frequency band of 200 Hz to 2700 Hz, while
the other is a high frequency music signal
in the frequency band of 10200 Hz to
15200 Hz. The ratio of the AM signal
bandwidth required to send both the
signals together to the AM signal
bandwidth required to send just the human
speech is :
(1) 3
(2) 5
(3) 6
(4) 2
25. A simple pendulum made of a bob of
mass m and a metallic wire of negligible
mass has time period 2 s at T=0C. If the
temperature of the wire is increased and the
corresponding change in its time period is
plotted against its temperature, the
resulting graph is a line of slope S. If the
coefficient of linear expansion of metal is α
then the value of S is :
(1) α
(2) ••
(3) 2α
(4)
••
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 18
26. A¡L$ eyr_ap¡d® i„Ly$ ApL$pf_p¡ spf Y e„N dp¡X$éygk hpmpÖìedp„\u b_ph¡g R>¡ S>¡_u sZphdy¼s g„bpC L R>¡.Ap i„Ly$ ApL$pf_p spf_p D`f_p R>¡X$p_u rÓÄep R A_¡_uQ¡_p R>¡X$p_u rÓÄep 3R R>¡ D`f_p¡ R>¡X$p¡ ×Y$ Ap^pfkp\¡ Å¡X¡$g R>¡ A_¡ _uQ¡_p R>¡X¡$ M v$m Å¡X¡$g R>¡.k„s ygus Ahõ\pdp „ Ap spf_u sZph g„bpC\i¡ :
(1) ••• • • ••
• ••
• •
•π
(2) ••• • • • ••
• ••
• •
•π
(3) ••• • • • ••
• ••
• •
•π
(4) ••• • • • ••
• ••
• •
•π
27. A ®Aphs®__u fus¡ N¡ëh¡_p¡duV$f_p¡ Ahfp¡ G ÅZhp,V
E fphsu b¡V$fu A_¡ Ahfp¡ R. A¡ N¡ëh¡_p¡duV$fdp„
θ M|Zp_y„ Aphs®_ d¡mhhp h`fpe R>¡. Å¡ S Ahfp¡ _p¡i„V$ A¡ A ® Aphs®_ dpV¡$ S>ê$fu lp¡e, sp¡ G, R A_¡ SA¡ hÃQ¡_p¡ k„b„ R>¡ :(1) 2S (R+G)=RG
(2) S (R+G)=RG
(3) 2S=G
(4) 2G=S
26. •ÁflSÃÊÁ⁄Uà L ‹ê’Ê߸ ∑§Ë ∞∑§‚◊ÊŸ ‡Ê¥∑ȧŸÈ◊Ê ÃÊ⁄U ∑§Á‚⁄UÊ¥ ∑§Ë ÁòÊíÿÊ ∑˝§◊‡Ê— R ÃÕÊ 3R „Ò¥– ©‚∑§Ë œÊÃÈ ∑§Êÿ¥ª-◊Ê«ÈUU‹‚ Y „Ò– R ÁòÊíÿÊ flÊ‹ Á‚⁄U ∑§Ê ∞∑§ ŒÎ…∏•ÊœÊ⁄U ¬⁄U ¡Á«∏à Á∑§ÿÊ ªÿÊ „Ò ÃÕÊ ŒÍ‚⁄U Á‚⁄U ¬⁄U MŒ˝√ÿ◊ÊŸ ‹≈U∑§ÊÿÊ ªÿÊ „Ò– ‚¥ÃÈ‹Ÿ-•flSÕÊ ◊¥ ÃÊ⁄U∑§Ë ÀÊê’Ê߸ „ÊªË —
(1) ••• • • ••
• ••
• •
•π
(2) ••• • • • ••
• ••
• •
•π
(3) ••• • • • ••
• ••
• •
•π
(4) ••• • • • ••
• ••
• •
•π
27. ∞∑§ ªÒÀflŸÊ◊Ë≈U⁄U ∑§Ê ¬˝ÁÃ⁄UÊœ G ◊ʬŸ ∑§ Á‹ÿ •h¸-ÁflˇÊ¬ Ã⁄UË∑§ ∑§Ê ßSÃ◊Ê‹ Á∑§ÿÊ ªÿÊ Á¡‚◊¥ ’Ò≈U⁄UË ∑§Ëemf V
E „Ò– ¬˝ÁÃ⁄UÊœ R ∑§ Á‹ÿ θ ÁflˇÊ¬ Á◊‹Ê–
‡Ê¥≈U-¬˝ÁÃ⁄UÊœ S ∑§ Á‹ÿ •ÊœÊ ÁflˇÊ¬ Á◊‹Ê– Ã’G, R ÃÕÊ S Á∑§‚ ‚◊Ë∑§⁄UáÊ ‚ ‚¥’¥ÁœÃ „Ò¥?(1) 2S (R+G)=RG
(2) S (R+G)=RG
(3) 2S=G
(4) 2G=S
26. A uniformly tapering conical wire is made
from a material of Young’s modulus Y and
has a normal, unextended length L. The
radii, at the upper and lower ends of this
conical wire, have values R and 3R,
respectively. The upper end of the wire is
fixed to a rigid support and a mass M is
suspended from its lower end. The
equilibrium extended length, of this wire,
would equal :
(1) ••• • • ••
• ••
• •
•π
(2) ••• • • • ••
• ••
• •
•π
(3) ••• • • • ••
• ••
• •
•π
(4) ••• • • • ••
• ••
• •
•π
27. To know the resistance G of a galvanometer
by half deflection method, a battery of emf
VE
and resistance R is used to deflect the
galvanometer by angle θ. If a shunt of
resistance S is needed to get half deflection
then G, R and S are related by the
equation :
(1) 2S (R+G)=RG
(2) S (R+G)=RG
(3) 2S=G
(4) 2G=S
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 19
28. brlNp£m Afukp_u L¡$ÞÖg„bpC dp`hp, A¡L$ rhÛp\_uQ¡_p Ahgp¡L$_p¡ _p¢ ¡ R>¡.
hõsy u_ brlNp£m L$pQ brlNp£m Afukp¡ ârstbb u_• • •• •• • • • •• •• • • • •• •• • • • •• •• •
brlNp£m L$pQ_u L¡$ÞÖg„bpC f1 A_¡ Afukp_u L¡$ÞÖg„bpC
f2 R>¡. index correction _NÎe g¡sp f
1 A_¡ f
2 _y„
dyëe_u _ÆL$ li¡ :(1) f
1=12.7 cm f
2=7.8 cm
(2) f1=7.8 cm f
2=12.7 cm
(3) f1=7.8
cm f
2=25.4 cm
(4) f1=15.6 cm f
2=25.4 cm
29. T¡_fX$pep¡X$_u I - V gpnrZL$sp_p Aæepk dpV¡$ A¡L$âep¡N L$fhpdp„ Aph¡ R>¡. Äepf: `p¡V¡$[Þiep¡duV$f L¡$ S>¡_p¡Äep„ Ahfp¡ R=100 Ω R>¡ A_¡ 1 W dlÑd `phfX¡$ku`¡i_ R>¡. Ap `qf`\dp„ gNphhpdp„ Aphsp DC
õÓp¡s_u Þe|_sd hp¡ëV¡$S>_u Ahr^ R>¡ :(1) 0 – 5 V
(2) 0 – 8 V
(3) 0 – 12 V
(4) 0 – 24 V
28. ©ûÊ‹-Œ¬¸áÊ ∑§Ë »§Ê∑§‚ ŒÍ⁄UË ÁŸ∑§Ê‹Ÿ ∑§ ∞∑§ ¬˝ÿʪ◊¥ ÁŸêŸ «UÊ≈UÊ ¬˝ÊåàÊ „È•Ê
Á’¥’ ©ûÊ‹ ‹Ò¥‚ ©ûÊ‹ Œ¬áÊ ¬ÁÃÁ’¥’
• • •• •• • • • •• •• • • • •• •• • • • •• •• •
©ûÊ‹ ‹Ò¥‚ ∑§Ë »§Ê∑§‚ ŒÍ⁄UË f1 ÃÕÊ ©ûÊ‹-Œ¬¸áÊ ∑§Ë
»§Ê∑§‚ ŒÍ⁄UË f2 „Ò– index correction Ÿªáÿ „Ò–
Ã’ —(1) f
1=12.7 cm f
2=7.8 cm
(2) f1=7.8 cm f
2=12.7 cm
(3) f1=7.8
cm f
2=25.4 cm
(4) f1=15.6 cm f
2=25.4 cm
29. ∞∑§ ¡ËŸ⁄U «UÊÿÊ«U ∑§Ê •Á÷‹ˇÊÁáÊ∑§ I - V ª˝Ê»§ ’ŸÊŸ∑§ Á‹ÿ ∞∑§ ¬˝ÿʪ Á∑§ÿÊ ªÿÊ Á¡‚◊¥ R=100 Ω ∑§Ê¬˝Ê≈UÁÄ≈Ufl ¬˝ÁÃ⁄UÊœ •ÊÒ⁄U •Áœ∑§Ã◊ ¬Êfl⁄U 1 W ŒË ªß¸–Ã’ ¬Á⁄U¬Õ ◊¥ ‹ªÊÿ ªÿ DC dÊà ∑§Ë ãÿÍŸÃ◊ flÊÀ≈UÃÊ„Ò —(1) 0 – 5 V
(2) 0 – 8 V
(3) 0 – 12 V
(4) 0 – 24 V
28. To find the focal length of a convex mirror,
a student records the following data :
Object Pin Convex Lens Convex Mirror Image Pin
22.2 cm 32.2 cm 45.8 cm 71.2 cm
The focal length of the convex lens is f1 and
that of mirror is f2. Then taking index
correction to be negligibly small, f1 and f
2
are close to :
(1) f1=12.7 cm f
2=7.8 cm
(2) f1=7.8 cm f
2=12.7 cm
(3) f1=7.8
cm f
2=25.4 cm
(4) f1=15.6 cm f
2=25.4 cm
29. An experiment is performed to determine
the I - V characteristics of a Zener diode,
which has a protective resistance of
R=100 Ω, and a maximum power of
dissipation rating of 1 W. The minimum
voltage range of the DC source in the circuit
is :
(1) 0 – 5 V
(2) 0 – 8 V
(3) 0 – 12 V
(4) 0 – 24 V
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 20
30. ∞∑§ •ôÊÊà ≈˛UÊ¥Á¡S≈U⁄U ∑§Ê npn •ÕflÊ pnp ∑§ ¬˝∑§Ê⁄U◊¥ ¬„øÊŸ ∑§⁄UŸÊ „Ò– ∞∑§ pnp ≈˛UÊ¥Á¡S≈U⁄U ∑§Ê ≈UÁ◊¸Ÿ‹ 2©‚∑§Ê ’‚ „Ò– ∞∑§ êÊÀ≈UË◊Ë≈U⁄U ∑§ +ve fl −ve
≈UÁ◊¸Ÿ‹ ß‚ ≈˛UÊ¥Á¡S≈U⁄U ∑§ ÁflÁ÷ÛÊ ≈UÁ◊¸Ÿ‹Ê¥ 1, 2 ÿÊ 3 ∑§’Ëø ‹ªÊ∑§⁄U ¬˝ÁÃ⁄UÊœ ◊ʬ ªÿ Ã’ ß‚ ≈˛UÊ¥Á¡S≈U⁄U ∑§Á‹∞ ∑§ÊÒŸ-‚Ê ÁŸêŸ ∑§ÕŸ ‚àÿ „Ò?
(1) +ve ‚ ≈UÁ◊Ÿ‹ 1, −ve ‚ ≈UÁ◊Ÿ‹ 2, ¬ÁÃ⁄UÊœíÿʌʖ
(2) +ve ‚ ≈UÁ◊Ÿ‹ 2, −ve ‚ ≈UÁ◊Ÿ‹ 1, ¬ÁÃ⁄UÊœíÿʌʖ
(3) +ve ‚ ≈UÁ◊Ÿ‹ 3, −ve ‚ ≈UÁ◊Ÿ‹ 2, ¬ÁÃ⁄UÊœíÿʌʖ
(4) +ve ‚ ≈UÁ◊Ÿ‹ 2, −ve ‚ ≈UÁ◊Ÿ‹ 3, ¬ÁÃ⁄UÊœ∑§◊
31. 35.5 g •Ê⁄U‚ÁŸ∑§ •ê‹ ∑§Ê, ‚Ê¥Œ˝ HCl ∑§Ë ©¬ÁSÕÁÃ◊¥ H
2S ∑§Ë •Áœ∑§ ◊ÊòÊÊ ‚ ÁflfløŸ ∑§⁄UŸ ¬⁄U •Ê⁄U‚ÁŸ∑§
¬ã≈UÊ‚À$»§Êß«U ∑§Ë ¬˝Ê# „ÊŸ flÊ‹Ë ◊ÊòÊÊ „Ò (ÿÁŒ 100%
¬Á⁄UfløŸ ◊ÊŸ¥ ÃÊ)—
(1) 0.50 ◊Ê‹
(2) 0.25 ◊Ê‹
(3) 0.125 ◊Ê‹
(4) 0.333 ◊Ê‹
30. An unknown transistor needs to be
identified as a npn or pnp type. A
multimeter, with +ve and −ve terminals,
is used to measure resistance between
different terminals of transistor. If terminal
2 is the base of the transistor then which of
the following is correct for a pnp
transistor ?
(1) +ve terminal 1, −ve terminal
2, resistance high
(2) +ve terminal 2, −ve terminal
1, resistance high
(3) +ve terminal 3, −ve terminal
2, resistance high
(4) +ve terminal 2, −ve terminal
3, resistance low
31. The amount of arsenic pentasulphide that
can be obtained when 35.5 g arsenic acid is
treated with excess H2S in the presence of
conc. HCl ( assuming 100% conversion)
is :
(1) 0.50 mol
(2) 0.25 mol
(3) 0.125 mol
(4) 0.333 mol
30. A¡L$ npn A\hp pnp V²$p[ÞTõV$f_¡ Ap¡mMhp_p¡ R>¡.Ap dpV¡$ V²$p[ÞTõV$f_p AgN AgN V$rd®_ëk hÃQ¡_p¡Ahfp¡ +ve A_¡ −ve dëV$uduV$f\u dp`hpdp„ Aph¡R>¡. Å¡ V$rd®_g 2 A¡ V²$p[ÞTõV$f_p¡ b¡T lp¡e sp¡ pnp
V²$p[ÞTõV$f dpV¡ L$ey„ rh^p_ kpQy„ R>¡ ?(1) +ve terminal 1, −ve terminal
2, Ahfp¡ h^pf¡(2) +ve terminal 2, −ve terminal
1, Ahfp¡ h^pf¡(3) +ve terminal 3, −ve terminal
2, Ahfp¡ h^pf¡(4) +ve terminal 2, −ve terminal
3, Ahfp¡ Ap¡R>p¡
31. 35.5 g Apk£r_L$ A¡rkX$_¡, kp„Ö HCl _u lpS>fudp„h^pf¡ `X$sp H
2S kp\¡ âq¾$ep L$fsp„ Apk£r_L$
`¡ÞV$pkëapCX$ L¡ $V$gu dpÓpdp„ âpá \C iL$i¡ ?(100% `qfhs®_ dp_u gp¡)
(1) 0.50 dp¡g
(2) 0.25 dp¡g
(3) 0.125 dp¡g
(4) 0.333 dp¡g
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 21
32. M|b S> KQp v$bpZ¡, A¡L$ dp¡g hpey_p¡ v$b_ue Aheh(compressibility factor) _uQ¡ Ap ¡gdp„\u S>Zphp¡.
(1)
• •• •
(2) • •• •• •
• •
(3) • •• •• •
• •
(4) • • • ••• • • •
−
33. dy¿e ¼hp¡ÞV$d Ap„L$ 5 kp\¡ k„L$mpe¡g L$nL$p¡_u Ly$gk„¿ep ip¡ p¡.(1) 5
(2) 10
(3) 20
(4) 25
34. _uQ¡_pdp„\u L$ep Ap„sfApÎhue bmp¡ L¡$ S>¡ T¡_p¡_ hpey_pâhpluL$fZ dpV¡$ kp¥\u h^pf¡ S>hpbv$pf R>¡ ?
(1) qÜ y°h - qÜ °yh
(2) Ape_ - qÜ °yh
(3) Ðhqfs qÜ y°h - â¡qfs qÜ y°h (Instantaneous
dipole - induced dipole)
(4) Aper_L$
32. •àÿÁœ∑§ ŒÊ’ ¬⁄U ∞∑§ ◊Ê‹ ªÒ‚ ∑§Ê ‚¥¬Ë«˜ÿÃÊ ªÈáÊÊ¥∑§„ÊªÊ —
(1)
• •• •
(2) • •• •• •
• •
(3) • •• •• •
• •
(4) • • • ••• • • •
−
33. ◊ÈÅÿ ÄflÊ¥≈U◊ •¥∑§ 5 ‚ ¡È«∏ „È∞ ∑§ˇÊ∑§Ê¥ (•ÊÚÁ’¸≈U‹Ê¥)∑§Ë ∑ȧ‹ ‚¥ÅÿÊ „Ò —(1) 5
(2) 10
(3) 20
(4) 25
34. ߟ◊¥ ‚ ∑§ÊÒŸ‚Ë •¥Ã⁄UÊ-•ÊÁáfl∑§ ’‹ ¡ËŸÊÚŸ ∑§Œ˝flË∑§⁄UáÊ ∑§ Á‹∞ ‚’‚ •Áœ∑§ ©ûÊ⁄UŒÊÿË „Ò ?
(1) ÁmœÈ˝fl - Ámœ˝Èfl
(2) •ÊÿŸ - Ámœ˝Èfl
(3) ÃÊà∑§ÊÁ‹∑§ Ámœ˝Èfl - ¬˝Á⁄Uà Ámœ˝Èfl
(4) •ÊÿÁŸ∑§
32. At very high pressures, the compressibility
factor of one mole of a gas is given by :
(1)
• •• •
(2) • •• •• •
• •
(3) • •• •• •
• •
(4) • • • ••• • • •
−
33. The total number of orbitals associated with
the principal quantum number 5 is :
(1) 5
(2) 10
(3) 20
(4) 25
34. Which intermolecular force is most
responsible in allowing xenon gas to
liquefy ?
(1) Dipole - dipole
(2) Ion - dipole
(3) Instantaneous dipole - induced dipole
(4) Ionic
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 22
35. 1 bar f A¡L$ fpkperZL$ âq¾$ep _uQp sp`dp_¡ Ap`d¡m¡(õhe„c|) \su _\u `f„sy KQp sp`dp_¡ Ap`d¡m¡(õhe„c|) \pe R>¡. âq¾$ep A„N¡ _uQ¡ Ap ¡gp rh^p_p¡dp„\ukpQy„ rh^p_ ip¡ p¡.
(1) ∆H s\p ∆S b„_¡ F>Z R>¡.
(2) ∆H s\p ∆S b„_¡ ^_ R>¡.
(3) ∆H ^_ R>¡ Äepf¡ ∆S F>Z R>¡.
(4) ∆H F>Z R>¡ Äepf¡ ∆S ^_ R>¡.
36. 300 K A_¡ 500 torr (V$p¡f) Ap„riL$ v$bpZ¡ N2 _u
Öpìesp 0.01 g L−1 R>¡. 750 V$p¡f (torr) Ap„riL$
v$bpZ `f Öpìesp (g L−1 dp„) iy„ li¡ ?
(1) 0.0075
(2) 0.015
(3) 0.02
(4) 0.005
37. âq¾$ep dpV¡$, A(g)+B(g) → C(g)+D(g) 298 K
`f, ∆H A_¡ ∆S A_y¾$d¡,−29.8 kJ mol−1 A_¡
−0.100 kJ K−1
mol−1 R>¡. 298 K `f âq¾$ep
dpV¡$_p¡ k„syg_ AQmp„L$ _uQ¡_pdp„\u ip¡ p¡.(1) 1.0×10
−10
(2) 1.0×1010
(3) 10
(4) 1
35. ∞∑§ ⁄UÊ‚ÊÿÁŸ∑§ •Á÷Á∑§ÿÊ ÁŸêŸ Ãʬ ¬⁄U •Sfl× ¬flÁÃÃ„Ò Á∑§ãÃÈ ©ìÊ Ãʬ ¬⁄U Sfl× ¬˝flÁøà „Ê ¡ÊÃË „Ò– ß‚•Á÷Á∑˝§ÿÊ ∑§ ’Ê⁄U ◊¥ ÁŸêŸÁ‹Áπà ∑§ÕŸÊ¥ ◊¥ ‚ ‚„Ë∑§ÕŸ ∑§Ê ¬„øÊÁŸÿ —
(1) ∆H ÃÕÊ ∆S, ŒÊŸÊ¥ ´§áÊÊà◊∑§ „Ò¥–
(2) ∆H ÃÕÊ ∆S, ŒÊŸÊ¥ œŸÊà◊∑§ „Ò¥–
(3) ∆H œŸÊà◊∑§ ÃÕÊ ∆S ´§áÊÊà◊∑§ „Ò–
(4) ∆H ´§áÊÊà◊∑§ ÃÕÊ ∆S œŸÊà◊∑§ „Ò–
36. N2 ∑§Ë ¡‹ ◊¥ Áfl‹ÿÃÊ 300 K ÃÕÊ 500 torr •Ê¥Á‡Ê∑§
ŒÊ’ ¬⁄U 0.01 g L−1 „Ò– ß‚∑§Ë Áfl‹ÿÃÊ (g L
−1
◊¥) 750 torr •Ê¥Á‡Ê∑§ ŒÊ’ ¬⁄U „ÊªË —(1) 0.0075
(2) 0.015
(3) 0.02
(4) 0.005
37. ⁄UÊ‚ÊÿÁŸ∑§ •Á÷Á∑˝§ÿÊA(g)+B(g) → C(g)+D(g), ∑ Á‹∞ 298 K ¬⁄U∆H ÃÕÊ ∆S ∑§ ◊ÊŸ ∑˝§◊‡Ê— −29.8 kJ mol
−1
ÃÕÊ−0.100 kJ K−1
mol−1 „Ò¥– ß‚ •Á÷Á∑˝§ÿÊ ∑§Ê
298 K ¬⁄U ‚Êêÿ ÁSÕ⁄UÊ¥∑§ „Ò —(1) 1.0×10
−10
(2) 1.0×1010
(3) 10
(4) 1
35. A reaction at 1 bar is non-spontaneous at
low temperature but becomes spontaneous
at high temperature. Identify the correct
statement about the reaction among the
following :
(1) Both ∆H and ∆S are negative.
(2) Both ∆H and ∆S are positive.
(3) ∆H is positive while ∆S is negative.
(4) ∆H is negative while ∆S is positive.
36. The solubility of N2 in water at 300 K and
500 torr partial pressure is 0.01 g L−1
. The
solubility (in g L−1
) at 750 torr partial
pressure is :
(1) 0.0075
(2) 0.015
(3) 0.02
(4) 0.005
37. For the reaction,
A(g)+B(g) → C(g)+D(g), ∆H and ∆S are,
respectively, −29.8 kJ mol−1
and
−0.100 kJ K−1
mol−1
at 298 K. The
equilibrium constant for the reaction at
298 K is :
(1) 1.0×10−10
(2) 1.0×1010
(3) 10
(4) 1
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 23
38. Å¡ L$p¡ f ^psy_p¡ A¡L$ ågp¡L$ (block) A¡L$ buL$fdp„ `X$uÅe L¡$ S>¡ 1M ZnSO
4 _y„ ÖphZ ^fph¡ R>¡ sp¡ iy„
b_i¡ ?
(1) L$p¡ f ^psy Ap¡Nmu S>i¡ A_¡ T]L$ ^psy S>dp\i¡.
(2) lpCX²$p¡S>_ hpey _uL$mhp_u kp\¡ L$p¡`f ^psyAp¡Nmu S>i¡.
(3) Ap¡[¼kS>_ hpey _uL$mhp_u kp\¡ L$p¡`f ^psyAp¡Nmu S>i¡.
(4) L$p¡C âq¾$ep \i¡ _l].
39. ¼gp¡qf_ `fdpÏAp¡_u lpS>fudp„, Ap¡Tp¡_ kp\¡_uAp¡[¼kS>_ `fdpÏAp¡_u âq¾$ep, _uQ¡ âdpZ¡ b¡sb½$pdp„ â¾$d v$ip®h¡g R>¡ :
O3(g)+Cl
•(g) → O
2(g)+ClO
•(g) ____ (i)
ki= 5.2×10
9 L mol
−1 s
−1
ClO•(g)+O
•(g) → O
2(g)+Cl
•(g) ____ (ii)
kii= 2.6×10
10 L mol
−1 s
−1
Ly$g âq¾$ep O3(g)+O
•(g) → 2 O
2(g) _p¡ kp¥\u
_ÆL$_p¡ h¡N AQmp„L$ ip¡ p¡.(1) 5.2×10
9 L mol
−1 s
−1
(2) 2.6×1010
L mol−1
s−1
(3) 3.1×1010
L mol−1
s−1
(4) 1.4×1020
L mol−1
s−1
38. ÿÁŒ ∑§ÊÚ¬⁄U ∑§ ∞∑§ é‹ÊÚ∑§ (block) ∑§Ê ∞∑§ ’Ë∑§⁄U ◊¥«UÊ‹Ê ¡Êÿ Á¡‚◊¥ 1M ZnSO
4 ∑§Ê Áfl‹ÿŸ „Ê ÃÊ
ÄÿÊ „ÊªÊ ?
(1) ∑§ÊÚ¬⁄U œÊÃÈ ÉÊÈ‹ ¡ÊÿªË ÃÕÊ Á$¡¥∑§ œÊÃÈ ÁŸˇÊÁ¬Ã„Ê ¡ÊÿªË–
(2) „Êß«˛UÊ¡Ÿ ªÒ‚ ∑§ ÁŸ∑§‹Ÿ ∑§ ‚ÊÕ-‚ÊÕ ∑§ÊÚ¬⁄UœÊÃÈ ÉÊÈ‹ ¡ÊÿªË–
(3) •ÊÚÄ‚Ë¡Ÿ ªÒ‚ ∑§ ÁŸ∑§‹Ÿ ∑§ ‚ÊÕ-‚ÊÕ ∑§ÊÚ¬⁄UœÊÃÈ ÉÊÈ‹ ¡ÊÿªË–
(4) ∑§Ê߸ •Á÷Á∑˝§ÿÊ Ÿ„Ë¥ „ʪ˖
39. Ä‹Ê⁄ËUŸ ¬⁄U◊ÊáÊÈ•Ê¥ ∑§Ë ©¬ÁSÕÁà ◊¥, •Ê$¡ÊŸ ∑§Ë•ÊÚÄ‚Ë¡Ÿ ¬⁄U◊ÊáÊÈ•Ê¥ ‚ •Á÷Á∑˝§ÿÊ ÁŸêŸÁ‹ÁπÃÁm¬ŒËÿ ¬˝∑˝§◊ mÊ⁄UÊ „ÊÃË „Ò —
O3(g)+Cl
•(g) → O
2(g)+ClO
•(g) ____ (i)
ki= 5.2×10
9 L mol
−1 s
−1
ClO•(g)+O
•(g) → O
2(g)+Cl
•(g) ____ (ii)
kii= 2.6×10
10 L mol
−1 s
−1
∑ȧ‹ •Á÷Á∑˝§ÿÊ O3(g)+O
•(g) → 2 O
2(g) ∑§Ê
ÁŸ∑§≈UÃ◊ flª ÁŸÿÃÊ¥∑§ „Ò —(1) 5.2×10
9 L mol
−1 s
−1
(2) 2.6×1010
L mol−1
s−1
(3) 3.1×1010
L mol−1
s−1
(4) 1.4×1020
L mol−1
s−1
38. What will occur if a block of copper metal
is dropped into a beaker containing a
solution of 1M ZnSO4 ?
(1) The copper metal will dissolve and
zinc metal will be deposited.
(2) The copper metal will dissolve with
evolution of hydrogen gas.
(3) The copper metal will dissolve with
evolution of oxygen gas.
(4) No reaction will occur.
39. The reaction of ozone with oxygen atoms
in the presence of chlorine atoms can occur
by a two step process shown below :
O3(g)+Cl
•(g) → O
2(g)+ClO
•(g) ____ (i)
ki= 5.2×10
9 L mol
−1 s
−1
ClO•(g)+O
•(g) → O
2(g)+Cl
•(g) ____ (ii)
kii= 2.6×10
10 L mol
−1 s
−1
The closest rate constant for the overall
reaction O3(g)+O
•(g) → 2 O
2(g) is :
(1) 5.2×109 L mol
−1 s
−1
(2) 2.6×1010
L mol−1
s−1
(3) 3.1×1010
L mol−1
s−1
(4) 1.4×1020
L mol−1
s−1
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 24
40. A¡L$ rhi¡j Ar^ip¡jZ â¾$d_¡ _uQ¡_u gpnrZL$spAp¡R>¡ : (i) s¡ hpÞX$fhpg bmp¡_¡ L$pfZ¡ Dv¹$ch¡ R>¡. A_¡(ii) s¡ ârshse R>¡.D`f hZ®h¡g Ar^ip¡jZ â¾$d dpV¡$ kpQy„ rh^p_ip¡^p¡.
(1) Ar^ip¡jZ_u A¡Þ\pë`u 100 kJ mol−1 \u
h^pf¡ lp¡e R>¡.
(2) kq¾$eL$fZ EÅ® (i[¼s) _uQu R>¡.
(3) Ar^ip¡jZ A¡ A¡L$ õsfue R>¡.
(4) sp`dp_ h^hp_u kp\¡ Ar^ip¡jZ h ¡ R>¡.
41. A¡L$ A^psy S>¡ ^_ (positive) Ap¡[¼kX¡$i_ Ahõ\pv$ip®hsu _\u s¡ ip¡ p¡.
(1) Ap¡[¼kS>_
(2) Apep¡qX$_
(3) ¼gp¡qf_
(4) ãgp¡qf_
40. ∞∑§ Áfl‡Ê· •Áœ‡ÊÊ·áÊ ¬Á∑§ÿÊ ∑§ Áfl‡Ê· ªÈáÊœ◊ „Ò¥ —(i) ÿ„ flÊ¥«U⁄U flÊÀ‚ ’‹ ∑§ ∑§Ê⁄áÊ „ÊÃË „Ò ÃÕÊ (ii) ÿ„©à∑˝§◊áÊËÿ „Ò– ÁŸêŸÁ‹Áπà ◊¥ ‚ fl„ ‚„Ë ∑§ÕŸ¬„øÊÁŸÿ ¡Ê ß‚ •Áœ‡ÊÊ·áÊ ¬˝Á∑˝§ÿÊ ∑§Ê ‚„Ë fláʸŸ∑§⁄UÃÊ „Ò —
(1) •Áœ‡ÊÊ·áÊ ∑§Ë ∞ãÕÒÀ¬Ë 100 kJ mol−1 ‚
•Áœ∑§ „Ò–
(2) ‚Á∑˝§ÿáÊ ™§¡Ê¸ ÁŸêŸ „Ò–
(3) •Áœ‡ÊÊ·áÊ ∞∑§‹ •áÊÈ∑§ ¬⁄UÃËÿ „Ò–
(4) Ãʬ ’…∏Ÿ ¬⁄U •Áœ‡ÊÊ·áÊ ’…∏ÃÊ „Ò–
41. fl„ •œÊÃÈ ¡Ê œŸÊà◊∑§ •ÊÚÄ‚Ë∑§⁄UáÊ •flSÕÊ Ÿ„Ë¥Œ‡ÊʸÃË, „ÊªË —
(1) •ÊÚÄ‚Ë¡Ÿ
(2) •ÊÿÊ«UËŸ
(3) Ä‹Ê⁄UËŸ
(4) ç‹È•Ê⁄UËŸ
40. A particular adsorption process has the
following characteristics : (i) It arises due
to van der Waals forces and (ii) it is
reversible. Identify the correct statement
that describes the above adsorption
process :
(1) Enthalpy of adsorption is greater than
100 kJ mol−1
.
(2) Energy of activation is low.
(3) Adsorption is monolayer.
(4) Adsorption increases with increase in
temperature.
41. The non-metal that does not exhibit
positive oxidation state is :
(1) Oxygen
(2) Iodine
(3) Chlorine
(4) Fluorine
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 25
42. Apg¡Mdp„ _uQ¡ Ap ¡g b¡ âq¾$epAp¡ dpV¡$ −ln Kp
rhf yÝ^ sp`dp__u rcÞ_sp v $ip ®h ¡g R> ¡ .
••• •••• •• •• •• •• • ••••
→• • A_¡
••• •••• •• •• •• •• • ••••
→• •
kpQy„ rh^p_ ip¡ p¡.
(1) T>1200 K `f L$pb®_, MO(s) dp„\u M(s)
dp„ qfX$¼n_ L$fi¡.
(2) T<1200 K `f, âq¾$ep MO(s)+C(s) →
M(s)+CO(g) õhe„c| R>¡.
(3) T<1200 K `f, L$pb®__y„ Ap¡[¼kX¡$i_ ârsL|$m(unfavourable) R>¡.
(4) b^p S> sp`dp_ `f L$pb®__y„ Ap¡[¼kX¡$i_A_yLy$m (favourable) R>¡.
42. ‚ÊÕ ÁŒÿ „Èÿ •Ê‹π ◊¥ ÁŸêŸÁ‹Áπà ŒÊ •Á÷Á∑˝§ÿÊ•Ê¥∑§ Á‹ÿ −ln K
p ∑§Ê Ãʬ ∑§ ‚ÊÕ ¬Á⁄UfløŸ Œ‡ÊʸÿÊ
ªÿÊ „Ò–
••• •••• •• •• •• •• • ••••
→• ÃÕÊ
•• •••••
• O2(g) → CO(s)
ÁŸêŸÁ‹Áπà ∑§ÕŸÊ¥ ◊¥ ‚ ‚„Ë ∑§ÕŸ ¬„øÊÁŸÿ —
(1) T>1200 K, ¬⁄U ∑§Ê’¸Ÿ MO(s) ∑§Ê •¬øÁÿÃ∑§⁄U∑§ M(s) ŒªÊ–
(2) T<1200 K, ¬⁄U •Á÷Á∑˝§ÿÊ, MO(s)+C(s)
→ M(s)+CO(g) Sfl× ¬˝flÁøà „Ò –
(3) T<1200 K ¬⁄U ∑§Ê’¸Ÿ ∑§Ê ©¬øÿŸ ¬˝ÁÃ∑ͧ‹„Ò–
(4) ∑§Ê’¸Ÿ ∑§Ê ©¬øÿŸ ‚÷Ë Ãʬ ¬⁄U •ŸÈ∑ͧ‹ „Ò–
42. The plot shows the variation of −ln Kp
versus temperature for the two reactions.
→
→
•
•
•• •••• •• •• •• •• • ••••• • • •••• •••• •• •• •• •• • ••••
• •
• •
Identify the correct statement :
(1) At T>1200 K, carbon will reduce
MO(s) to M(s).
(2) At T<1200 K, the reaction
MO(s)+C(s) → M(s)+CO(g) is
spontaneous.
(3) At T<1200 K, oxidation of carbon is
unfavourable.
(4) Oxidation of carbon is favourable at
all temperatures.
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 26
43. cpf¡ `pZu k„b„r^s AkÐe rh^p_ ip¡ p¡.(1) s¡ Al
4C
3 kp\¡ âq¾$ep L$fu, CD
4 A_¡
Al(OD)3 b_ph¡ R>¡.
(2) L¡$ÞÖue fuA¡¼V$fp¡dp„ s¡_p¡ iusgL$ (coolant)
sfuL¡$ D`ep¡N \pe R>¡.(3) s¡ CaC
2 kp\¡ âq¾$ep L$fu, C
2D
2 A_¡
Ca(OD)2 b_ph¡ R>¡.
(4) s¡ SO3 kp\ ¡ âq¾ $ep L $fu_ ¡, X $é yV ¡ $ qfs ¹
(deuterated) këãeyqfL$ A¡rkX$ (D2SO
4)
b_ph¡ R>¡.$
44. ApëL$pgpC_-A\® psy këa¡V$p¡_u pZudp„_u Öpìesp_p¡kpQp¡ ¾$d ip¡ p¡.(1) Mg < Ca < Sr < Ba
(2) Mg < Sr < Ca < Ba
(3) Mg > Sr > Ca > Ba
(4) Mg > Ca > Sr > Ba
45. L$p¡gd - I dp„ Ap ¡gu ApCV$dp¡ (items) _¡ s¡_pL$p¡gd - II dp„ _p¢ ¡g dy¿e D`ep¡Np¡ kp\¡ Å¡X$p¡.
(A) rkrgL$p S>¡g (i) V²$p[ÞTõV$f(B) rkrgL$_•(Silicon) (ii) Ape_-rhr_deL$(C) rkrgL$p¡_•(Silicone) (iii) iyóL$ L$sp®(D) rkrgL¡$V$ (iv) kugÞV
L$ p ¡gd•••• L$ p ¡gd•••••
(1) (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)
(2) (A)-(iv), (B)-(i), (C)-(ii), (D)-(iii)
(3) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)
(4) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)
43. ÷Ê⁄UË ¬ÊŸË ∑§ ’Ê⁄U ◊¥ ÁŒÿ ªÿ ∑§ÕŸÊ¥ ◊¥ ‚ •‚àÿ∑§ÕŸ ¬„øÊÁŸÿ —(1) ÿ„ Al
4C
3 ‚ •Á÷Á∑˝§ÿÊ ∑§⁄U∑§ CD
4 ÃÕÊ
Al(OD)3 ’ŸÊÃÊ „Ò –
(2) ß‚∑§Ê ©¬ÿʪ ŸÊÁ÷∑§Ëÿ Á⁄U∞Ä≈U⁄U ◊¥ ‡ÊËË∑§∑§ M§¬ ◊¥ Á∑§ÿÊ ¡ÊÃÊ „Ò–
(3) ÿ„ CaC2 ‚ •Á÷Á∑˝§ÿÊ ∑§⁄U∑§ C
2D
2 ÃÕÊ
Ca(OD)2’ŸÊÃÊ „Ò–
(4) ÿ„ SO3 ‚ •Á÷Á∑˝§ÿÊ ∑§⁄U∑§ «KÍ≈UÁ⁄UÃ
‚ÀçÿÍÁ⁄U∑§ •ê‹ (D2SO
4) ’ŸÊÃÊ „Ò–
44. ˇÊÊ⁄UËÿ ◊ÎŒÊ œÊÃÈ ‚À$»§≈UÊ¥ ∑§Ë ¡‹ ◊¥ Áfl‹ÿÃÊ ∑§Ê ‚„Ë∑˝§◊ „Ò —(1) Mg < Ca < Sr < Ba
(2) Mg < Sr < Ca < Ba
(3) Mg > Sr > Ca > Ba
(4) Mg > Ca > Sr > Ba
45. ∑§ÊÚ‹◊ I ◊¥ ÁŒÿ ªÿ ¬ŒÊÕÊZ (items) ∑§Ê ∑§ÊÚ‹◊ II
◊¥ ÁŒÿ ªÿ ©¬ÿʪʥ ‚ ‚È◊Á‹Ã ∑§ËÁ¡ÿ —
•• • Á‚Á‹∑§Ê ¡ÒÒ‹ ••• ≈UUÊ¥Á‚S≈U⁄U•• • Á‚Á‹∑§Ÿ •••• •ÊÿŸ ÁflÁŸ◊ÿ∑§•• • Á‚Á‹∑§ÊŸ ••••• ‡ÊÈc∑§Ÿ ∑§◊∑§•• • Á‚Á‹∑§≈U ••• • ‚Ë‹∑§ ••• • •• • ••
∑§ÊÚ‹◊ • ∑§ÊÚ‹◊ •••
(1) (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)
(2) (A)-(iv), (B)-(i), (C)-(ii), (D)-(iii)
(3) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)
(4) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)
43. Identify the incorrect statement regarding
heavy water :
(1) It reacts with Al4C
3 to produce CD
4
and Al(OD)3.
(2) It is used as a coolant in nuclear
reactors.
(3) It reacts with CaC2 to produce C
2D
2
and Ca(OD)2.
(4) It reacts with SO3 to form deuterated
sulphuric acid (D2SO
4).
44. The correct order of the solubility of
alkaline-earth metal sulphates in water is :
(1) Mg < Ca < Sr < Ba
(2) Mg < Sr < Ca < Ba
(3) Mg > Sr > Ca > Ba
(4) Mg > Ca > Sr > Ba
45. Match the items in Column I with its main
use listed in Column II :
(A) Silica gel (i) Transistor
(B) Silicon (ii) Ion-exchanger
(C) Silicone (iii) Drying agent
(D) Silicate (iv) Sealant
• • •• • • •• • • •• • • •••
(1) (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)
(2) (A)-(iv), (B)-(i), (C)-(ii), (D)-(iii)
(3) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)
(4) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 27
46. L$ep AÏAp¡_p¡ kd|l kh®kd A\hp ArcÞ_ ApL$pf(Identical shape) ^fph¡ R>¡ ?(1) SF
4
, XeF4
, CCl4
(2) ClF3
, XeOF2
, •• • ••
(3) BF3
, PCl3
, XeO3
(4) PCl5
, IF5
, XeO2F
2
47. _uQ¡ Ap`¡gp [õ`rkTp¡ (ÅrsAp¡) dp„\u L$C A¡L$ S>gueÖphZdp„ õ\peu R>¡ ?(1) Cr
2+
(2) Cu+
(3)••• • • •
(4)••• • • •
48. _uQ¡ Ap ¡gp k„L$uZp£dp„\u L$ey„ A¡L$ Ag(NO3) _p S>gue
ÖphZ_u Ar^L$ syëesp_p¡ D`ep¡N L$fi¡ ?(1) Na
3[CrCl
6]
(2) [Cr(H2O)
5Cl]Cl
2
(3) [Cr(H2O)
6]Cl
3
(4) Na2[CrCl
5(H
2O)]
46. ∑§ÊÒŸ ‚Ê ‚◊Í„ ‚◊M§¬ •áÊÈ•Ê¥ ∑§Ê ‚◊Í„ „Ò —(1) SF
4
, XeF4
, CCl4
(2) ClF3
, XeOF2
, •• • ••
(3) BF3
, PCl3
, XeO3
(4) PCl5
, IF5
, XeO2F
2
47. ÁŸêŸÁ‹Áπà ◊¥ ‚ ¡‹Ëÿ Áfl‹ÿŸ ◊¥ SÕÊÿË S¬Ë‡ÊË$¡∑§ÊÒŸ ‚Ë „Ò?(1) Cr
2+
(2) Cu+
(3)••• • • •
(4)••• • • •
48. ÁŸêŸÁ‹Áπà ‚¥∑ȧ‹Ê¥ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ‚¥∑ȧ‹ Ag(NO3)
∑§ ¡‹Ëÿ Áfl‹ÿŸ ∑§ •Áœ∑§ ‚◊ÃÈÀÿ π¬ÊÿªÊ ?(1) Na
3[CrCl
6]
(2) [Cr(H2O)
5Cl]Cl
2
(3) [Cr(H2O)
6]Cl
3
(4) Na2[CrCl
5(H
2O)]
46. The group of molecules having identical
shape is :
(1) SF4
, XeF4
, CCl4
(2) ClF3
, XeOF2
, •• • ••
(3) BF3
, PCl3
, XeO3
(4) PCl5
, IF5
, XeO2F
2
47. Which one of the following species is stable
in aqueous solution ?
(1) Cr2+
(2) Cu+
(3)••• • • •
(4)••• • • •
48. Which one of the following complexes will
consume more equivalents of aqueous
solution of Ag(NO3) ?
(1) Na3[CrCl
6]
(2) [Cr(H2O)
5Cl]Cl
2
(3) [Cr(H2O)
6]Cl
3
(4) Na2[CrCl
5(H
2O)]
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 28
49. _uQ¡ Ap`¡gpdp„\u kpQu qv$ip (trend) ip¡^p¡.(`fdpÎhue ¾$dp„L$=Ti : 22, Cr=24 A_¡ Mo=42)
(1) ∆o _p¡ [Cr(H2O)
6]2+
>
[Mo(H2O)
6]2+
A_¡∆o _p¡ [Ti(H
2O)
6]3+
> [Ti(H2O)
6]2+
(2) ∆o _p¡ [Cr(H2O)
6]2+
> [Mo(H2O)
6]2+ A_¡
∆o _p¡ [Ti(H2O)
6]3+
< [Ti(H2O)
6]2+
(3) ∆o _p¡ [Cr(H2O)
6]2+
< [Mo(H2O)
6]2+
A_¡∆o _p¡ [Ti(H
2O)
6]3+
> [Ti(H2O)
6]2+
(4) ∆o _p¡ [Cr(H2O)
6]2+
< [Mo(H2O)
6]2+
A_¡∆o _p¡ [Ti(H
2O)
6]3+
< [Ti(H2O)
6]2+
50. BOD v$ip®h¡ R>¡.
(1) bpep¡gp¡ÆL$g Ap¡[¼kS>_ qX$dpÞX$
(2) b¡¼V$¡qfAg Ap¡[¼kX¡$i_ qX$dpÞX$
(3) bpep¡L¡$rdL$g Ap¡[¼kS>_ qX$dpÞX$
(4) bpep¡L¡$rdL$g Ap¡[¼kX¡$i_ qX$dpÞX$
49. ÁŸêŸÁ‹Áπà ◊¥ ‚ ‚„Ë ¬˝flÎÁûÊ ¬„øÊÁŸÿ —(¬⁄U◊ÊáÊÈ ∑˝§◊Ê¥∑§=Ti : 22, Cr : 24 ÃÕÊ Mo : 42)
(1) [Cr(H2O)
6]2+
∑§Ê
∆o > [Mo(H2O)
6]2+
ÃÕÊ
[Ti(H2O)
6]3+
∑§Ê ∆o > [Ti(H2O)
6]2+
(2) [Cr(H2O)
6]2+
∑§Ê ∆o >
[Mo(H2O)
6]2+
ÃÕÊ [Ti(H2O)
6]3+∑§Ê
∆o < [Ti(H2O)
6]2+
(3) [Cr(H2O)
6]2+
∑§Ê
∆o < [Mo(H2O)
6]2+
ÃÕÊ
[Ti(H2O)
6]3+
∑§Ê ∆o > [Ti(H2O)
6]2+
(4) [Cr(H2O)
6]2+
∑§Ê
∆o < [Mo(H2O)
6]2+
ÃÕÊ
[Ti(H2O)
6]3+ ∑§Ê ∆o < [Ti(H
2O)
6]2+
50. ’Ë.•Ê.«UË. (BOD) Œ‡ÊʸÃÊ „Ò —
(1) ’ÊÿÊ‹Ê¡Ë∑§‹ •ÊÚÄ‚Ë¡Ÿ Á«U◊Ê¥«U
(2) ’ÒÄ≈UËÁ⁄Uÿ‹ •ÊÚÄ‚Ë«U‡ÊŸ Á«U◊Ê¥«U
(3) ’ÊÿÊ∑Ò§Á◊∑§‹ •ÊÚÄ‚Ë¡Ÿ Á«U◊Ê¥«U
(4) ’ÊÿÊ∑Ò§Á◊∑§‹ •ÊÚÄ‚Ë«U‡ÊŸ Á«U◊Ê¥«U
49. Identify the correct trend given below :
(Atomic No.=Ti : 22, Cr : 24 and Mo : 42)
(1) ∆o of [Cr(H2O)
6]2+
>
[Mo(H2O)
6]2+
and
∆o of [Ti(H2O)
6]3+
> [Ti(H2O)
6]2+
(2) ∆o of [Cr(H2O)
6]2+
> [Mo(H2O)
6]2+
and
∆o of [Ti(H2O)
6]3+
< [Ti(H2O)
6]2+
(3) ∆o of [Cr(H2O)
6]2+
< [Mo(H2O)
6]2+
and
∆o of [Ti(H2O)
6]3+
> [Ti(H2O)
6]2+
(4) ∆o of [Cr(H2O)
6]2+
< [Mo(H2O)
6]2+
and
∆o of [Ti(H2O)
6]3+
< [Ti(H2O)
6]2+
50. BOD stands for :
(1) Biological Oxygen Demand
(2) Bacterial Oxidation Demand
(3) Biochemical Oxygen Demand
(4) Biochemical Oxidation Demand
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 29
51. A¡L$ L$pb®r_L$ k„ep¡S>_ C, H A_¡ S ^fph¡ R>¡. Å¡ s¡dp„8% këaf lp¡e sp¡ k„ep¡S>__p¡ Þe|_sd (minimum)
AÏcpfip¡ p¡. (`fdpÎhue hS>_ S _y„ =32 amu)(1) 200 g mol
−1
(2) 400 g mol−1
(3) 600 g mol−1
(4) 300 g mol−1
52. kps L$pb®_ `fdpÏAp¡hpmp A¡L$ lpCX²$p¡L$pb®_ L$ep¡ li¡L¡$ S>¡dp„ A¡L$ r_ep¡`¡ÞV$pCg A_¡ A¡L$ rh_pCg kd|l^fph¡ R>¡ ?
(1) 2, 2-X$perd\pCg-4-`¡ÞV$u_
(2) ApCkp¡âp¡`pCg-2-åe|V$u_
(3) 4, 4-X$perd\pCg`¡ÞV$u_
(4) 2, 2-X$perd\pCg-3-`¡ÞV$u_
53. 5 L A¡L$ ApëL¡$__p k„ |Z® v$l_ dpV¡$ 25 L Ap¡[¼kS>__uAphíeL$sp R>¡. b^p S> L$v$p¡ (volumes) AQmsp`dp_ A_¡ v$bpZ dp`hpdp„ Aph¡g lp¡e sp¡ ApëL¡$_L$ep¡ R>¡ ?
(1) C\¡_
(2) âp¡`¡_
(3) åe|V¡$_
(4) ApCkp¡åe|V¡_
51. ∞∑§ ∑§Ê’¸ÁŸ∑§ ÿÊÒÁª∑§ ◊¥ C, H ÃÕÊ S Áfll◊ÊŸ „Ò¥–ÿÁŒ ß‚ ÿÊÒÁª∑§ ◊¥ 8% ‚À$»§⁄U „Ê ÃÊ ß‚∑§Ê ãÿÍŸÃ◊•áÊÈ ÷Ê⁄U „ÊªÊ —
( S ∑§Ê ¬⁄U◊ÊáÊÈ ÷Ê⁄U =32 amu)
(1) 200 g mol−1
(2) 400 g mol−1
(3) 600 g mol−1
(4) 300 g mol−1
52. ‚Êà ∑§Ê’¸Ÿ ¬⁄U◊ÊáÊÈ•Ê¥ flÊ‹Ê ∞∑§ „Êß«˛UÊ∑§Ê’¸Ÿ ∑§ÊÒŸ„ÊªÊ Á¡‚◊¥ ∞∑§ ÁŸ•Ê¬Áã≈U‹ ‚◊Í„ ÃÕÊ ∞∑§ flÊßÁŸ‹‚◊Í„ „Ê —
(1) 2, 2-«UÊ߸◊ÁÕ‹-4-¬ã≈UËŸ
(2) •Ê߂ʬ˝ÊÁ¬‹-2-éÿÍÁ≈UŸ
(3) 4, 4-«UÊ߸◊ÁÕ‹¬ã≈UËŸ
(4) 2, 2-«UÊ߸◊ÁÕ‹-3-¬ã≈UËŸ
53. ∞∑§ ∞À∑§Ÿ ∑§Ë 5 L ◊ÊòÊÊ ∑§ ¬Íáʸ Œ„Ÿ ∑§ Á‹ÿ 25 L
•ÊÚÄ‚Ë¡Ÿ ∑§Ë •Êfl‡ÿ∑§ÃÊ „ÊÃË „Ò– ÿÁŒ ‚÷Ë •Êÿß◊ÊŸ∑§ Ãʬ ÃÕÊ ŒÊ’ ¬⁄U ◊ʬ ªÿ „Ê¥, ÃÊ ∞À∑§Ÿ„ÊªË —
(1) ∞ÕŸ
(2) ¬˝Ê¬Ÿ
(3) éÿÍ≈UŸ
(4) •Êß‚ÊéÿÍ≈UŸ
51. An organic compound contains C, H and
S. The minimum molecular weight of the
compound containing 8% sulphur is :
(atomic weight of S=32 amu)
(1) 200 g mol−1
(2) 400 g mol−1
(3) 600 g mol−1
(4) 300 g mol−1
52. The hydrocarbon with seven carbon atoms
containing a neopentyl and a vinyl group
is :
(1) 2, 2-dimethyl-4-pentene
(2) Isopropyl-2-butene
(3) 4, 4-dimethylpentene
(4) 2, 2-dimethyl-3-pentene
53. 5 L of an alkane requires 25 L of oxygen for
its complete combustion. If all volumes are
measured at constant temperature and
pressure, the alkane is :
(1) Ethane
(2) Propane
(3) Butane
(4) Isobutane
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 30
54. CH3MgBr _¡ rd\¡_p¡gdp„ Nfd L$fsp„ DÐ`Þ_ \sp¡
hpey ip¡ p¡.(1) HBr
(2) rd\¡_
(3) C\¡_
(4) âp¡`¡_
55. b|hp¡-åg¡ÞL$ (Bouveault-Blanc) fuX$n_ âq¾$epdp„\pe R>¡ :
(1) A¡kpCg l¡gpCX$_y„ H2/Pd kp\¡ qfX$n_.
(2) A¡õV$f_y„ Na/C2H
5OH kp\¡ qfX$n_.
(3) L$pbp¡®r_g k„ep¡S>__y„ Na/Hg A_¡ HCl kp\¡qfX$n_.
(4) A¡_lpCX²$pCX$_y„ LiAlH4 kp\¡ qfX$n_.
56. âp\rdL$, qÜrseL$ A_¡ s©rseL$ A¡dpC__¡ âc¡qv$s L$fsuL$kp¡V$u _uQ¡_pdp„\u ip¡^p¡.
(1) L$pbp®gA¡dpC_ âq¾$ep(2) C
6H
5SO
2Cl
(3) k¡ÞXd¡ef âq¾$ep
(4) dõV$X®$ Ap¡Cg L$kp¡V$u
54. CH3MgBr ∑§Ê ◊ÕÒŸÊÚ‹ ◊¥ ª◊¸ ∑§⁄UŸ ¬⁄U ©à¬ÛÊ „ÊŸ
flÊ‹Ë ªÒ‚ „Ò —(1) HBr
(2) ◊ÕÒŸ
(3) ∞ÕŸ
(4) ¬˝Ê¬Ÿ
55. ’ÍflÊ-é‹Ê¥∑§ Á⁄U«Ućʟ ¬˝Á∑˝§ÿÊ ◊¥ „ÊÃÊ „Ò —
(1) ∞Á‚‹ „Ò‹Êß«U ∑§Ê H2/Pd ‚ •¬øÿŸ–
(2) ∞S≈U⁄U ∑§Ê Na/C2H
5OH ‚ •¬øÿŸ–
(3) ∑§Ê’ʸÁŸ‹ ÿÊÒÁª∑§ ∑§Ê Na/Hg ÃÕÊ HCl ‚•¬øÿŸ–
(4) ∞∑§ ∞Ÿ„Êß«˛UÊß«U ∑§Ê LiAlH4 ‚ •¬øÿŸ–
56. ¬˝ÊÕÁ◊∑§, ÁmÃËÿ∑§ ÃÕÊ ÃÎÃËÿ∑§ ∞◊ËŸÊ¥ ◊¥ •ãÃ⁄U ∑§⁄UŸ∑§ Á‹ÿ ¬˝ÿÈÄà „ÊŸ flÊ‹Ê ¬⁄UˡÊáÊ „Ò —
(1) ∑§ÊÁ’¸‹∞◊ËŸ •Á÷Á∑˝§ÿÊ(2) C
6H
5SO
2Cl
(3) ‚Òã«U◊Êÿ⁄U •Á÷Á∑˝§ÿÊ
(4) ◊S≈U«¸U •ÊÚÿ‹ ¬⁄UˡÊáÊ
54. The gas evolved on heating CH3MgBr in
methanol is :
(1) HBr
(2) Methane
(3) Ethane
(4) Propane
55. Bouveault-Blanc reduction reaction
involves :
(1) Reduction of an acyl halide with
H2/Pd.
(2) Reduction of an ester with
Na/C2H
5OH.
(3) Reduction of a carbonyl compound
with Na/Hg and HCl.
(4) Reduction of an anhydride with
LiAlH4.
56. The test to distinguish primary, secondary
and tertiary amines is :
(1) Carbylamine reaction
(2) C6H
5SO
2Cl
(3) Sandmeyer’s reaction
(4) Mustard oil test
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 31
57. L$\_ : f¡ep¡_ A¡L$ A^®kp„ïg¡rjs blºgL$ R>¡. S>¡_pNyZ^dp£ Ly$v$fsu L$`pk (cotton) L$fsp h y kpfpR>¡.
L$pfZ : A¡rkqV$g¡i_ hX¡$ k¡ëeygp¡T_p ep„rÓL$ A_¡ kp¦v$e®^fphsp NyZ^dp£_¡ h y kpfp iL$pe R>¡.
(1) L$\_ A_¡ L$pfZ b„_¡ kpQp R>¡. L$pfZ A¡ L$\_dpV¡$_u kpQu kdS|>su R>¡.
(2) L$\_ A_¡ L$pfZ b„_¡ kpQp R>¡, `Z L$pfZ A¡L$\_ dpV¡$_u kpQu kdS|>su _\u.
(3) L$\_ A¡ AkÐe rh^p_ R>¡ `Z L$pfZ A¡ kÐeR>¡.
(4) L$\_ A_¡ L$pfZ b„_¡ AkÐe R>¡.
58. A¡õ`pqV®$L$ A¡rkX$ dpV¡$ _uQ¡_p¡ ¾$d Ýep_dp„ gp¡.
A¡õ`pqV®$L$ A¡rkX$_y„ pI (kdrhch tbvy$) ip¡ p¡.(1) 1.88
(2) 3.65
(3) 5.74
(4) 2.77
57. ∑§ÕŸ : ⁄UÿÊÚŸ ∞∑§ •h‚¥Á‡‹C ’„È‹∑§ „Ò Á¡‚∑§ªÈáÊœ◊ ¬Ê∑ΧÁÃ∑§ ∑§¬Ê‚ ‚ •Áœ∑§ •ë¿U„Ò¥–
∑§Ê⁄UáÊ : ∞‚ËÁ≈U‹Ë∑§⁄UáÊ ‚ ‚‹È‹Ê‚ ∑§ ÿÊ¥ÁòÊ∑§fl ‚ÊÒ¥Œÿ¸¬⁄U∑§ ªÈáÊœ◊ÊZ ∑§Ê ‚ÈœÊ⁄UÊ ¡Ê‚∑§ÃÊ „Ò–
(1) “∑§ÕŸ” fl “∑§Ê⁄UáÊ” ŒÊŸÊ¥ ‚„Ë „Ò¥ ÃÕÊ “∑§Ê⁄áÊ”,“∑§ÕŸ” ∑§Ë ‚„Ë √ÿÊÅÿÊ „Ò–
(2) “∑§ÕŸ” fl “∑§Ê⁄UáÊ” ŒÊŸÊ¥ ‚„Ë „Ò¥ Á∑§ãÃÈ “∑§Ê⁄UáÊ”,“∑§ÕŸ” ∑§Ë ‚„Ë √ÿÊÅÿÊ Ÿ„Ë¥ „Ò–
(3) “∑§ÕŸ” ª‹Ã „Ò Á∑§ãÃÈ “∑§Ê⁄UáÊ” ‚„Ë „Ò–
(4) “∑§ÕŸ” fl “∑§Ê⁄UáÊ” ŒÊŸÊ¥ ª‹Ã „Ò¥–
58. ∞S¬ÊÁ≈¸U∑§ •ê‹ ∑§ ÁŸêŸÁ‹Áπà •ŸÈ∑˝§◊ ¬⁄U ÁfløÊ⁄U∑§ËÁ¡∞ —
∞S¬ÊÁ≈U¸∑§ •ê‹ ∑§Ê pI (‚◊Áfl÷fl Á’¥ŒÈ ) „Ò —(1) 1.88
(2) 3.65
(3) 5.74
(4) 2.77
57. Assertion : Rayon is a semisynthetic
polymer whose properties
are better than natural cotton.
Reason : Mechanical and aesthetic
properties of cellulose can be
improved by acetylation.
(1) Both assertion and reason are correct,
and the reason is the correct
explanation for the assertion.
(2) Both assertion and reason are correct,
but the reason is not the correct
explanation for the assertion.
(3) Assertion is incorrect statement, but
the reason is correct.
(4) Both assertion and reason are
incorrect.
58. Consider the following sequence for
aspartic acid :
The pI (isoelectric point) of aspartic acid
is :
(1) 1.88
(2) 3.65
(3) 5.74
(4) 2.77
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 32
59. _uQ¡_pdp„\u L$ep L©$rÓd Nþep `v$p\p£_p¡ Nm`Z Ap„L$(sweetness value) L¡$_ kyNf (cane sugar) _usyg_pdp„ kp¥\u h^pf¡ R>¡.
(1) A¡õ`pV£_
(2) k¡L¡$fu_
(3) ky¾$pgp¡T
(4) A¡rgV¡$d
60. A¡N-Apëbydu_ kp¡g (egg-albumin sol) b_phhpdpV¡$_u kp¥\u kpfu b„ b¡ksu Ý^rs _uQ¡_pdp„\u ip¡ p¡.
(1) IX$p_¡ kpQhu_¡ sp¡X$p¡ A_¡ s¡_p `pfv$i cpN_¡100 mL 5% w/V _p ghZue (saline)
ÖphZdp„ d¡mhu_¡ kpfu fus¡ lgphp¡.
(2) IX$p_¡ kpQhu_¡ sp¡X$p¡ A_¡ s¡_p a¼s ump cpN_¡100 mL 5% w/V _p ghZue (saline)
ÖphZdp„ d¡mhu_¡ kpfu fus¡ lgphp¡.
(3) IX$p_¡ DL$msp `pZudp„ 10 rdr_V$ dpV¡$ fpMp¡s¡_u R>pg v|$f L$ep® bpv$ s¡_p¡ ka¡v$ cpN100 mL 5% w/V _p ghZue (saline)
ÖphZdp„ ep„rÓL$ i¡L$f (shaker) hX¡$ kdp„NuL©$s(homogenize) L$fp¡.
(4) IX$p_¡ DL$msp `pZudp„ 10 rdr_V$ dpV¡$ fpMp¡s¡_u R>pg v|$f L$ep® bpv$ s¡_p `ump¡ cpN100 mL 5% w/V _p ghZue (saline)
ÖphZdp„ ep„rÓL$ i¡L$f (shaker) hX¡$ kdp„NuL©$s(homogenize) L$fp¡.
59. fl„ ∑ΧÁòÊ◊ ◊œÈ⁄U∑§ Á¡‚∑§Ê ߡÊÈ ‡Ê∑¸§⁄UÊ ∑§Ë ÃÈ‹ŸÊ ◊¥◊ÊœÈÿ¸◊ÊŸ ‚’‚ •Áœ∑§ „Ò —
(1) ∞‚¬Ê≈¸UŸ
(2) ‚Ò∑§⁄UËŸ
(3) ‚È∑˝§Ê‹Ê‚
(4) ∞‹Ë≈U◊
60. •¥«U ∞ÀéÿÍÁ◊Ÿ ∑§Ê ‚ÊÚ‹ ’ŸÊŸ ∑§Ë ‚’‚ ©Áøà ÁflÁœ„Ò —
(1) •¥«U ∑§Ê äÿÊŸ¬Ífl¸∑§ ÃÊ«∏¥ •ÊÒ⁄U ©‚∑§ ¬Ê⁄UŒ‡Ê˸÷ʪ ∑§Ê 100 mL 5% w/V ‹fláÊ ¡‹ ◊¥Á◊‹Ê ∑§⁄U •ë¿UË Ã⁄U„ Á„‹Êÿ¥–
(2) •¥«U ∑§Ê äÿÊŸ¬Ífl¸∑§ ÃÊ«∏¥ •ÊÒ⁄U ©‚∑§ ¬Ë‹ ÷ʪ∑§Ê 100 mL 5% w/V ‹fláÊ ¡‹ ◊¥ Á◊‹Ê∑§⁄U •ë¿UË Ã⁄U„ Á„‹Êÿ¥–
(3) •¥«U ∑§Ê ©’‹Ã ¡‹ ◊¥ 10 Á◊Ÿ≈U Ã∑§ ⁄Uπ¥ –©‚∑§Ê Á¿U‹∑§Ê ©ÃÊ⁄UŸ ∑§ ¬‡øÊØ ©‚∑§ ‚»§Œ÷ʪ ∑§Ê 100 mL 5% w/V ‹fláÊ ¡‹ ◊¥Á◊‹Êÿ¥ •ÊÒ⁄U ÿÊ¥ÁòÊ∑§ „ÁÀ‹òÊ ◊¥ ‚◊Ê¥ªË∑Χà ∑§⁄¥U–
(4) •¥«U ∑§Ê ©’‹Ã ¡‹ ◊¥ 10 Á◊Ÿ≈U Ã∑§ ⁄Uπ¥–©‚∑§Ê Á¿U‹∑§Ê ©ÃÊ⁄UŸ ∑§ ¬‡øÊØ ©‚∑§ ¬Ë‹÷ʪ ∑§Ê 100 mL 5% w/V ‹fláÊ ¡‹ ◊¥Á◊‹Êÿ¥ •ÊÒ⁄U ÿÊ¥ÁòÊ∑§ „ÁÀ‹òÊ ◊¥ ‚◊Ê¥ªË∑Χà ∑§⁄¥U–
59. The artificial sweetener that has the highest
sweetness value in comparison to cane
sugar is :
(1) Aspartane
(2) Saccharin
(3) Sucralose
(4) Alitame
60. The most appropriate method of making
egg-albumin sol is :
(1) Break an egg carefully and transfer
the transparent part of the content to
100 mL of 5% w/V saline solution
and stir well.
(2) Break an egg carefully and transfer
only the yellow part of the content to
100 mL of 5% w/V saline solution
and stir well.
(3) Keep the egg in boiling water for 10
minutes. After removing the shell,
transfer the white part of the content
to 100 mL of 5% w/V saline solution
and homogenize with a mechanical
shaker.
(4) Keep the egg in boiling water for 10
minutes. After removing the shell,
transfer the yellow part of the content
to 100 mL of 5% w/V saline solution
and homogenize with a mechanical
shaker.
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 33
61. ^pfp¡ L¡$ x R, x ≠ 0, x ≠ 1 dpV¡$ ••• • ••
• •• •
•• •
• •A_¡ f
n+1 (x)=f
0 (f
n(x)), n=0, 1, 2, . . . sp¡ lh¡
• • • • •• ••• ••• •
• • •
• • •• • =
(1)
••
(2)
••
(3)
••
(4)
••
62. ApN®ÞX$ kdsgdp„ 2+i \u v$ip®h¡g tbvy$ A¡L$ A¡L$d|h® sfa Mk¡ R>¡, Ðepfbpv$ 2 A¡L$d DÑf sfa Mk¡ R>¡
A_¡ Ðep„\u R>¡hV¡$ • • A¡L$d _¥F>Ðe sfa Mk¡ R>¡.ApN®ÞX$ kdsgdp„ Ap tbvy$_y„ _hy„ õ\p_ L$C k„¿ep\uv$ip®hhpdp„ Aph¡ ?(1) 2+2i
(2) 1+i
(3) −1−i
(4) −2−2i
61. x R, x ≠ 0, x ≠ 1 ∑§ Á‹∞ ◊ÊŸÊ ••• • ••
• •• •
•• •
• •ÃÕÊ f
n+1 (x)=f
0 (f
n(x)), n=0, 1, 2, . . . „Ò, ÃÊ
• • • • •• ••• ••• •
• • •
• • •• • ’⁄UÊ’⁄U „Ò —
(1)
••
(2)
••
(3)
••
(4)
••
62. •Ê⁄Uªá«U ‚◊Ë ◊¥ 2+i mÊ⁄UÊ ÁŸÁŒ¸c≈U Á’¥ŒÈ, 1 ß∑§Ê߸¬Ífl¸ ÁŒ‡ÊÊ ◊¥ ø‹ÃÊ „Ò •ÊÒ⁄U Á»§⁄U 2 ß∑§Ê߸ ©ûÊ⁄U ÁŒ‡ÊÊ ◊¥
ø‹ÃÊ „Ò ÃÕÊ •ãà ◊¥ • • ß∑§Ê߸ ŒÁˇÊáÊ-¬Á‡ø◊ÁŒ‡ÊÊ ◊¥ ¡ÊÃÊ „Ò– ÃÊ •Ê⁄Uªá«U ‚◊Ë ◊¥ ß‚∑§Ê ŸÿÊSÕÊŸ Á¡‚ Á’¥ŒÈ mÊ⁄UÊ ÁŸÁŒ¸c≈U „ÊÃÊ „Ò, fl„ „Ò —(1) 2+2i
(2) 1+i
(3) −1−i
(4) −2−2i
61. For x R, x ≠ 0, x ≠ 1, let ••• • ••
• •• •
•• •
• • and
fn+1
(x)=f0
(fn(x)), n=0, 1, 2, . . . Then the
value of • • • • •• ••• ••• •
• • •
• • •• • is equal
to :
(1)
••
(2)
••
(3)
••
(4)
••
62. The point represented by 2+i in the
Argand plane moves 1 unit eastwards, then
2 units northwards and finally from there
• • units in the south-westwards
direction. Then its new position in the
Argand plane is at the point represented
by :
(1) 2+2i
(2) 1+i
(3) −1−i
(4) −2−2i
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 34
63. Å¡ kduL$fZp¡ x2+bx−1=0 A_¡ x
2+x+b=0
_¡ −1 rkhpe_y„ A¡L$ kpdpÞe buS> lp¡e sp¡ |b|
bfpbf :
(1) •(2) 2
(3) 3
(4) •
64. Å¡
• •• •• •• ••• •
• •• •• •• •• •• •
• ••
,
• •• •
• •• •• •• •
• • A_¡
Q=PAPT lp¡e sp¡ PT
Q2015
P bfpbf :
(1)
• • • • •• •• •• •• •
(2)
• • • • •• • • • •
• •• •• •
(3)
• • • • •• • • • •
• •• •• •
(4)
• • • • •• •• •• •• •
63. ÿÁŒ ‚◊Ë∑§⁄UáÊÊ¥ x2+bx−1=0 ÃÕÊ x2
+x+b=0
∑§Ê −1 ‚ Á÷㟠∞∑§ ‚Ê¥! ÊÊ ◊Í‹ „Ò, ÃÊ |b| ’⁄UÊ’⁄U„Ò —
(1) •(2) 2
(3) 3
(4) •
64. ÿÁŒ
• •• •• •• ••• •
• •• •• •• •• •• •
• ••
,
• •• •
• •• •• •• •
• • ÃÕÊ
Q=PAPT „Ò, ÃÊ PT
Q2015
P „Ò —
(1)
• • • • •• •• •• •• •
(2)
• • • • •• • • • •
• •• •• •
(3)
• • • • •• • • • •
• •• •• •
(4)
• • • • •• •• •• •• •
63. If the equations x2+bx−1=0 and
x2+x+b=0 have a common root different
from −1, then |b| is equal to :
(1) •(2) 2
(3) 3
(4) •
64. If
• •• •• •• ••• •
• •• •• •• •• •• •
• ••
,
• •• •
• •• •• •• •
• • and
Q=PAPT, then P
T Q
2015 P is :
(1)
• • • • •• •• •• •• •
(2)
• • • • •• • • • •
• •• •• •
(3)
• • • • •• • • • •
• •• •• •
(4)
• • • • •• •• •• •• •
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 35
65. kduL$fZ •• •• ••• • ••• •••• • •• •• ••• • •••• • ••• • •• ••
• • •• • •• • •
•• • _p A„sfpg
•• ••• •
• •• •π π
•• dp„ Aph¡g rcÞ_ buÅ¡_u k„¿ep L¡$V$gu
li¡ ?(1) 4
(2) 3
(3) 2
(4) 1
66. iåv$ “MEDITERRANEAN” _p„ Anfp¡_p¡ D`ep¡NL$fu Qpf Anfp¡ hpmp (A\® lp¡hp¡ S>ê$fu _\u) L¡$ S>¡_p¡â\d Anf R A_¡ Qp¡\p¡ Anf E lp¡e s¡hp sdpdiåv$p¡_u k„¿ep L¡$V$gu \pe ?
(1)( )
•• ••• •
(2) 110
(3) 56
(4) 59
65. ‚◊Ë∑§⁄UáÊ •• •• ••• • ••• •••• • •• •• ••• • •••• • ••• • •• ••
• • •• • •• • •
•• • , ∑§ •¥Ã⁄UÊ‹
•• ••• •
• •• •π π
•• ◊¥ Á÷㟠flÊSÃÁfl∑§ ◊Í‹Ê¥ ∑§Ë ‚¥ÅÿÊ „Ò —
(1) 4
(2) 3
(3) 2
(4) 1
66. ‡ÊéŒ “MEDITERRANEAN” ∑§ •ˇÊ⁄UÊ¥ ‚ øÊ⁄U•ˇÊ⁄UÊ¥ ∑§ ∞‚ ‡ÊéŒ (øÊ„ •Õ¸„ËŸ „Ê¥) ’ŸÊŸ „Ò¥ Á¡Ÿ∑§Ê¬„‹Ê •ˇÊ⁄U R ÃÕÊ øÊÒÕÊ •ˇÊ⁄U E „Ê, ÃÊ ∞‚ ‚÷ˇÊéŒÊ¥ ∑§Ë ∑ȧ‹ ‚¥ÅÿÊ „Ò —
(1)( )
•• ••• •
(2) 110
(3) 56
(4) 59
65. The number of distinct real roots of the
equation,
•• •• ••• • ••• •••• • •• •• ••• • •••• • ••• • •• ••
• • •• • •• • •
•• • in the
interval •• ••• •
• •• •π π
•• is :
(1) 4
(2) 3
(3) 2
(4) 1
66. If the four letter words (need not be
meaningful ) are to be formed using the
letters from the word
“MEDITERRANEAN” such that the first
letter is R and the fourth letter is E, then
the total number of all such words is :
(1)( )
•• ••• •
(2) 110
(3) 56
(4) 59
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Set - 04 36
67. Å¡ (1+x)2016
+x(1+x)2015
+x2(1+x)
2014
+......+x2016
=
• • • •
• •• •
••
• ••• •
, x R, x ≠−1 ; sp¡
a17
=
(1)
• • • • •• • •••• • • • •
(2)
• • • • •• • •••• • • • •
(3)
• • • • •• • • • •
(4)
• • • • •• • •
68. ^pfp¡ L¡$ _ hpõsrhL$ k„¿epAp¡ x, y, z A¡hu R>¡ L¡$ S>¡\ux+y+z=12 A_¡ x
3y
4z
5=(0.1) (600)
3. sp¡
x3+y
3+z
3=
(1) 270
(2) 258
(3) 342
(4) 216
67. x R, x ≠−1 ∑§ Á‹∞, ÿÁŒ
(1+x)2016
+x(1+x)2015
+x2(1+x)
2014
+......+x2016
=
• • • •
• •• •
••
• ••• •
„Ò, ÃÊ a17
’⁄UÊ’⁄U „Ò —
(1)
• • • • •• • •••• • • • •
(2)
• • • • •• • •••• • • • •
(3)
• • • • •• • • • •
(4)
• • • • •• • •
68. ◊ÊŸÊ x, y, z ∞‚Ë œŸÊà◊∑§ flÊSÃÁfl∑§ ‚¥ÅÿÊ∞° „Ò¥ Á∑§,x+y+z=12 ÃÕÊ x
3y
4z
5=(0.1) (600)
3 „Ò, ÃÊ
x3+y
3+z
3 ’⁄UÊ’⁄U „Ò —
(1) 270
(2) 258
(3) 342
(4) 216
67. For x R, x ≠−1, if
(1+x)2016
+x(1+x)2015
+x2(1+x)
2014
+......+x2016
=
• • • •
• •• •
••
• ••• •
, then a17
is equal
to :
(1)
• • • • •• • •••• • • • •
(2)
• • • • •• • •••• • • • •
(3)
• • • • •• • • • •
(4)
• • • • •• • •
68. Let x, y, z be positive real numbers
such that x+y+z=12 and
x3y
4z
5=(0.1) (600)
3. Then x
3+y
3+z
3 is
equal to :
(1) 270
(2) 258
(3) 342
(4) 216
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 37
69.
• •• ••
• ••• •
••
•
•••
•−=
=
(1) 560
(2) 680
(3) 1240
(4) 1085
70. Å¡ •
•••••• • •
•
•
• • •• •→∞
+ − = sp¡ ‘a’ =
(1) 2
(2)
••
(3)
••
(4)
••
69.
• •• ••
• ••• •
••
•
•••
•−=
∑§Ê ◊ÊŸ „Ò —
(1) 560
(2) 680
(3) 1240
(4) 1085
70. ÿÁŒ
••
••••• • •
•
•
• •• •→∞
+ − = „Ò, ÃÊ ‘a’ ’⁄UÊ’⁄U
„Ò —(1) 2
(2)
••
(3)
••
(4)
••
69. The value of
• •• ••
• ••• •
••
•
•••
•−=
is equal
to :
(1) 560
(2) 680
(3) 1240
(4) 1085
70. If
••
••••• • •
•
•
• • •• •→∞
+ − = then ‘a’ is
equal to :
(1) 2
(2)
••
(3)
••
(4)
••
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71. Å¡ rh ¡e
f(x)=
( )•
••••••••••••••••••••••••••••• •
• •• • • •• •
• • •
• • • • •
<•••
≤ ≤•••
•
• • • •
x=1 ApNm rhL$g_ue lp¡e sp¡ •• =
(1)
••
π •• •
(2)
••
π • • • •
(3)
••
π • •
(4) −1−cos−1
(2)
72. Å¡ h¾$ x=4t2+3, y=8t
3−1, t R, _p¡ P tbvy$
ApNm_p¡ õ`i®L$, âpQg t kp\¡, h¾$_¡ afuhpf Q tbvy$A¡dm¡ R>¡ sp¡ Q tbvy$_p„ epd R>¡:(1) (t
2+3, −t
3−1)
(2) (4t2+3, −8t
3−1)
(3) (t2+3, t
3−1)
(4) (16t2+3, −64t
3−1)
71. ÿÁŒ »§‹Ÿ
f(x)=
( )•
••••••••••••••••••••••••••••• •
• •• • • •• •
• • •
• • • • •
<•••
≤ ≤•••
•
• • • •
x=1 ¬⁄U •fl∑§‹ŸËÿ „Ò, ÃÊ •• ∑§Ê ◊ÊŸ „Ò —
(1)
••
π •• •
(2)
••
π • • • •
(3)
••
π • •
(4) −1−cos−1
(2)
72. ÿÁŒ fl∑˝§ x=4t2+3, y=8t
3−1, t R ∑§ Á’¥ŒÈ P,
t ¬˝Êø‹ ∑§ ‚ÊÕ, ¬⁄U S¬‡Ê¸ ⁄UπÊ, fl∑˝§ ∑§Ê ŒÈ’Ê⁄UÊ Á’¥ŒÈQ ¬⁄U Á◊‹ÃË „Ò, ÃÊ Q ∑§ ÁŸŒ¸‡ÊÊ¥∑§ „Ò¥ —(1) (t
2+3, −t
3−1)
(2) (4t2+3, −8t
3−1)
(3) (t2+3, t
3−1)
(4) (16t2+3, −64t
3−1)
71. If the function
f(x)=
( )•
••••••••••••••••••••••••••••• •
• •• • • •• •
• • •
• • • • •
<•••
≤ ≤•••
•
• • • •
is differentiable at x=1, then
•• is equal
to :
(1)
••
π •• •
(2)
••
π • • • •
(3)
••
π • •
(4) −1−cos−1
(2)
72. If the tangent at a point P, with parameter
t, on the curve x=4t2+3, y=8t
3−1, t R,
meets the curve again at a point Q, then the
coordinates of Q are :
(1) (t2+3, −t
3−1)
(2) (4t2+3, −8t
3−1)
(3) (t2+3, t
3−1)
(4) (16t2+3, −64t
3−1)
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73. h¾$ y=x2−4 D`f_p„ tbvy$Ap¡\u ENdtbvy$_y„ Þe|_Ñd
A„sf L¡$V$gy„ \pe ?
(1)
• ••
(2)
• ••
(3)
• ••
(4)
• ••
74. Å¡
( ) ( )• •
• •• • • •• • ••• • • •••• •
• • • • •• •
• • •
Äep„ k A¡ k„L$g__p¡ AQmp„L$ R>¡, sp¡ A+B+C=
(1)
• ••
(2)
• ••
(3)
•• •
(4)
• •• •
73. fl∑˝§ y=x2−4 ∑§ ∞∑§ Á’¥ŒÈ ‚ ◊Í‹ Á’¥ŒÈ ∑§Ë ãÿÍŸÃ◊
ŒÍ⁄UË „Ò —
(1)
• ••
(2)
• ••
(3)
• ••
(4)
• ••
74. ÿÁŒ
( ) ( )• •
• •• • • •• ••• • • •••• •
• • • • •• •
• • •
„Ò, ¡„Ê° k ‚◊Ê∑§‹Ÿ •ø⁄U „Ò, ÃÊ A+B+C ’⁄UÊ’⁄U„Ò —
(1)
• ••
(2)
• ••
(3)
•• •
(4)
• •• •
73. The minimum distance of a point on the
curve y=x2−4 from the origin is :
(1)
• ••
(2)
• ••
(3)
• ••
(4)
• ••
74. If
( ) ( )• •
• •• • • •• • ••• • • •••• •
• • • • •• •
• • •
where k is a constant of integration, then
A+B+C equals :
(1)
• ••
(2)
• ••
(3)
•• •
(4)
• •• •
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75. Å¡
• • • • • •• •• • •• •
• • •• • •• • •• •• • • • • • • •• •• •• • •
sp¡ • • •••• • •• ••• • • •• • • • =
(1) log4
(2) •• • ••
+π
(3) log2
(4) •• • ••π •• •
76. âv$¡iA=(x, y)|y ≥ x
2−5x+4, x+y ≥ 1, y ≤ 0 _y„
n¡Óam (Qp¡.A¡L$ddp„) L¡$V$gy„ \i¡ ?
(1)
••
(2)
• ••
(3)
• ••
(4)
• ••
75. ÿÁŒ
• • • • • •• •• • •• •
• • •• • •• • •• •• • • • • • •• •• •= − + „Ò,
ÃÊ
• • •••• • •• ••• • • •• • • • ’⁄UÊ’⁄U „Ò —
(1) log4
(2) •• • ••
+π
(3) log2
(4) •• • ••π •• •
76. A = (x, y)|y ≥ x2−5x+4, x+y ≥ 1, y ≤ 0
mÊ⁄UÊ ÁŸœÊ¸Á⁄Uà ˇÊòÊ ∑§Ê ˇÊòÊ»§‹ (flª¸ ß∑§ÊßÿÊ¥ ◊¥) „Ò —
(1)
••
(2)
• ••
(3)
• ••
(4)
• ••
75. If • • • • • •• •• • •• •
• • •• • •• • •• •• • • • • • • •• •• •• • •
then
• • •••• • •• ••• • • •• • • • is equal to :
(1) log4
(2) •• • ••
+π
(3) log2
(4) •• • ••π •• •
76. The area (in sq. units) of the region
described by
A=(x, y)|y ≥ x2−5x+4, x+y ≥ 1, y ≤ 0
is :
(1)
••
(2)
• ••
(3)
• ••
(4)
• ••
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77. Å¡ rh ¡e f (x) A„sfpg (0, ∞) D`f rhL$g_ue lp¡e
s\p f (1) = 1 A_¡ •••• •→
• •• •• • • • • ••• •
• • • • • •• •• ••
âÐe¡L$ x>0 dpV¡$ lp¡e sp¡ ( )••• =
(1)
• ••
(2)
• •• •
(3)
• ••
(4)
• •• •
78. f ¡MpAp ¡ • • • ••• •
•• • • A_¡ • • • ••• •
•• • • _p „
R>¡v$dp„\u kpf \su A¡L$ Qrgs f¡Mp epdpnp¡_¡ tbvy$Ap¡A A_¡ B, (A ≠ B) dp„ dm¡ R>¡. f¡MpM„X$ AB _pdÝetbvy$_p¡ tbvy$`\ L$ep kduL$fZ_y„ kdp^p_ L$fi¡ ?(1) 6xy=7(x+y)
(2) 4(x+y)2−28(x+y)+49=0
(3) 7xy=6(x+y)
(4) 14(x+y)2−97(x+y)+168=0
77. ÿÁŒ f (x), •¥Ã⁄UÊ‹ (0, ∞) ◊¥ ∞∑§ ∞‚Ê •fl∑§‹ŸËÿ»§‹Ÿ „Ò Á∑§ f (1) = 1 ÃÕÊ ¬˝àÿ∑§ x>0 ∑§ Á‹∞,
•••• •→
• •• •• • • • • ••• •
• • • • • •• •• ••
„Ò, ÃÊ ( )••• ’⁄UÊ’⁄U
„Ò —
(1)
• ••
(2)
• •• •
(3)
• ••
(4)
• •• •
78. ÿÁŒ ⁄ UπÊ•Ê ¥ • • • ••• •
•• • • ÃÕÊ • • • ••• •
•• • •
∑§ ¬˝ÁÃë¿UŒŸ ‚ „Ê∑§⁄U ¡ÊŸ flÊ‹Ë ∞∑§ ø⁄U ⁄UπÊ ß‚¬˝∑§Ê⁄U πË¥øË ªß¸ „Ò Á∑§ ÿ„ ÁŸŒ¸‡ÊÊ¥∑§ •ˇÊÊ¥ ∑§ÊA ÃÕÊ B, (A ≠ B) ¬⁄U Á◊‹ÃË „Ò, ÃÊ AB ∑§ ◊äÿÁ’¥ŒÈ∑§Ê Á’¥ŒÈ¬Õ „Ò —(1) 6xy=7(x+y)
(2) 4(x+y)2−28(x+y)+49=0
(3) 7xy=6(x+y)
(4) 14(x+y)2−97(x+y)+168=0
77. If f (x) is a differentiable function in the
interval (0, ∞) such that f (1) = 1 and
•••• •→
• •• •• • • • • ••• •
• • • • • •• •• ••
, for each x>0,
then ( )••• is equal to :
(1)
• ••
(2)
• •• •
(3)
• ••
(4)
• •• •
78. If a variable line drawn through the
intersection of the lines • • • ••• •
•• • • and
• • • ••• •
•• • • , meets the coordinate axes at
A and B, (A ≠ B), then the locus of the
midpoint of AB is :
(1) 6xy=7(x+y)
(2) 4(x+y)2−28(x+y)+49=0
(3) 7xy=6(x+y)
(4) 14(x+y)2−97(x+y)+168=0
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79. tbvy$ (2, 1) _y„ • • A¡L$d\u f¡Mp L : x−y=4 _¡
kdp„sf õ\p_p„sfZ L$fhpdp„ Aph¡ R>¡. Å¡ Ap _hy„ tbvy$Q ÓuÅ QfZdp„ Aph¡g lp¡e sp¡ Q dp„\u `kpf \suA_¡ L _¡ g„b lp¡e s¡hu f¡Mp_y„ kduL$fZ R>¡:
(1) x+y= • ••
(2) x+y= • • ••
(3) x+y= • • ••
(4) 2x+2y= • ••
80. A¡L$ hsy®m (−2, 4) dp„\u `kpf \pe R>¡ A_¡y-An_¡ tbvy$ (0, 2) ApNm õ`i£ R>¡. _uQ¡_pdp„\uL$ey„ kduL$fZ Ap hsy®m_p„ ìepk_¡ kdphsu f¡Mpv$ip®hi¡ ?(1) 4x+5y−6=0
(2) 2x−3y+10=0
(3) 3x+4y−3=0
(4) 5x+2y+4=0
81. ^pfp¡ L¡$ a A_¡ b A_y¾$d¡ A¡ Arshge_p„ A ® dy¿eAnA_¡ A ® A_ybÝ^ An_p„ dp` R>¡ S>¡_u DÐL¡$ÞÖspkduL$fZ 9e
2−18e+5=0 _y„ kdp^p_ L$f¡ R>¡. Å¡
S(5, 0) A¡ _prc A_¡ A_yê$` r_eprdL$p 5x=9 lp¡esp¡ Ap Arshge dpV¡$ a2
−b2 bfpbf :
(1) 7
(2) −7
(3) 5
(4) −5
79. Á’¥ŒÈ (2, 1) ∑§Ê ⁄UπÊ L : x−y=4 ∑§ ‚◊Ê¥Ã⁄U,
• • ß∑§Ê߸ SÕÊŸÊãÃÁ⁄Uà Á∑§ÿÊ ªÿÊ– ÿÁŒ ŸÿÊ Á’¥ŒÈ
Q ÃË‚⁄U øÃÈÕÊZ‡Ê ◊¥ ÁSÕà „Ò, ÃÊ Á’¥ŒÈ Q ‚ „Ê∑§⁄U ¡ÊŸflÊ‹Ë ÃÕÊ L ∑§ ‹¥’flà ⁄UπÊ ∑§Ê ‚◊Ë∑§⁄UáÊ „Ò —
(1) x+y= • ••
(2) x+y= • • ••
(3) x+y= • • ••
(4) 2x+2y= • ••
80. ∞∑§ flÎûÊ Á’¥ŒÈ (−2, 4) ‚ „Ê ∑§⁄U ¡ÊÃÊ „Ò ÃÕÊ y-•ˇÊ∑§Ê (0, 2) ¬⁄U S¬‡Ê¸ ∑§⁄UÃÊ „Ò– ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∞∑§‚◊Ë∑§⁄UáÊ ß‚ flÎûÊ ∑§ √ÿÊ‚ ∑§Ê ÁŸM§Á¬Ã ∑§⁄UÃÊ „Ò?(1) 4x+5y−6=0
(2) 2x−3y+10=0
(3) 3x+4y−3=0
(4) 5x+2y+4=0
81. ◊ÊŸÊ a ÃÕÊ b ∑˝§◊‡Ê—, ∞∑§ •Áì⁄Ufl‹ÿ Á¡‚∑§Ë©à∑¥§Œ˝ÃÊ ‚◊Ë∑§⁄UáÊ 9e
2−18e+5=0 ∑§Ê ‚¥ÃÈc≈U ∑§⁄UÃË
„Ò, ∑§ •œ¸•ŸÈ¬˝SÕ •ˇÊ ÃÕÊ •œ¸‚¥ÿÈÇ◊Ë •ˇÊ „Ò¥–ÿÁŒ S(5, 0) ß‚ •Áì⁄Ufl‹ÿ ∑§Ë ∞∑§ ŸÊÁ÷ ÃÕÊ5x=9 ‚¥ªÃ ÁŸÿãÃÊ (directrix) „Ò, ÃÊ a2
−b2 ’⁄UÊ’⁄U
„Ò —(1) 7
(2) −7
(3) 5
(4) −5
79. The point (2, 1) is translated parallel to the
line L : x−y=4 by • • units. If the new
point Q lies in the third quadrant, then the
equation of the line passing through Q and
perpendicular to L is :
(1) x+y= • ••
(2) x+y= • • ••
(3) x+y= • • ••
(4) 2x+2y= • ••
80. A circle passes through (−2, 4) and touches
the y-axis at (0, 2). Which one of the
following equations can represent a
diameter of this circle ?
(1) 4x+5y−6=0
(2) 2x−3y+10=0
(3) 3x+4y−3=0
(4) 5x+2y+4=0
81. Let a and b respectively be the semi-
transverse and semi-conjugate axes of a
hyperbola whose eccentricity satisfies the
equation 9e2−18e+5=0. If S(5, 0) is a focus
and 5x=9 is the corresponding directrix of
this hyperbola, then a2−b
2 is equal to :
(1) 7
(2) −7
(3) 5
(4) −5
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82. Å¡ D`hge ••
• • • ••• • •
•• • • _p¡ õ`i®L$, epdpnp¡_¡
A A_¡ B tbvy$dp„ dmsp¡ lp¡s s\p Å¡ O ENdtbvy$lp¡e sp¡ rÓL$p¡Z OAB _y„ gOyÑd n¡Óam (Qp¡.A¡L$ddp„)L¡$V$gy„ \pe ?
(1)
••
(2) • •
(3) • •(4) 9
83. f¡MpAp¡ • • • •• • •
•• •• • A_¡
• ••• •• • ••• • • •• •• • •
•• •−+ −
= =
−
hÃQ¡_y„ gOysd A„sf
L$ep A„sfpgdp„ Aph¡g R>¡ ?(1) [0, 1)
(2) [1, 2)
(3) (2, 3]
(4) (3, 4]
82. ÿÁŒ ŒËÉʸflÎûÊ ••
• • • ••• • •
•• • • ∑§ ∞∑§ Á’¥ŒÈ ¬⁄U πË¥øË
ªß¸ S¬‡Ê¸ ⁄UπÊ, ÁŸŒ¸‡ÊÊ¥∑§ •ˇÊÊ¥ ∑§Ê A ÃÕÊ B ¬⁄U Á◊‹ÃË„Ò ÃÕÊ O ◊Í‹ Á’¥ŒÈ „Ò, ÃÊ ÁòÊ÷È¡ OAB ∑§Ê ãÿÍŸÃ◊ˇÊòÊ»§‹ (flª¸ ß∑§ÊßÿÊ¥ ◊¥) „Ò —
(1)
••
(2) • •
(3) • •(4) 9
83. ⁄UπÊ•Ê¥ • • • •• • •
•• •• • ÃÕÊ
• ••• •• • ••• • • •• •• • •
•• •−+ −
= =
−
∑§ ’Ëø ∑§Ë ãÿÍŸÃ◊
ŒÍ⁄UË, Á¡‚ •¥Ã⁄UÊ‹ ◊¥ „Ò, fl„ „Ò —(1) [0, 1)
(2) [1, 2)
(3) (2, 3]
(4) (3, 4]
82. If the tangent at a point on the ellipse
••• • • ••
• • ••• • • meets the coordinate axes at
A and B, and O is the origin, then the
minimum area (in sq. units) of the triangle
OAB is :
(1)
••
(2) • •
(3) • •(4) 9
83. The shortest distance between the lines
• • • •• • •
•• •• • and
• ••• •• • ••• • • •• •• • •
•• •−+ −
= =
−
lies in the interval :
(1) [0, 1)
(2) [1, 2)
(3) (2, 3]
(4) (3, 4]
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84. tbvy$ (1, 2, 2) dp„\u `kpf \sp„ A_¡ kdsgp¡x−y+2z=3 A_¡ 2x−2y+z+12=0 _¡ g„blp¡e s¡hp kdsg_y„ tbv$y$ (1, −2, 4) \u A„sf L¡$V$gy„\pe ?
(1) • •(2) 2
(3) •
(4)
••
85. ^pfp¡ L¡$ rÓL$p¡Z ABC dp„ rifp¡tbvy$ A ApNm L$pV$M|Zp¡fQpe R>¡. Å¡ tbvy$ A, B A_¡ C _p„ õ\p_ kqv$ip¡
A_ y¾ $d ¡ • • •i j k i j p k
∧ ∧ ∧ ∧ ∧ ∧
• • • • • • • •• • •• • • A_¡
• •i q j k
∧ ∧ ∧
•• • •• • lp¡e sp¡, tbvy$ (p, q) S>¡ f¡Mp D`f
Aph¡g R>¡ s¡
(1) x-An_¡ kdp„sf R>¡.
(2) y-An_¡ kdp„sf R>¡.
(3) ^_ x-An kp\¡ gOyL$p¡Z b_ph¡ R>¡.
(4) ^_ x-An kp\¡ NyfyL$p¡Z b_ph¡ R>¡.
84. Á’¥ŒÈ (1, −2, 4) ∑§Ë ©‚ ‚◊Ë ‚ ŒÍ⁄UË, ¡Ê Á’¥ŒÈ(1, 2, 2) ‚ „Ê ∑§⁄ U ¡ÊÃÊ „ Ò ÃÕÊ ‚◊Ã‹Ê ¥x−y+2z=3 ÃÕÊ 2x−2y+z+12=0 ∑§ ‹¥’flÄÒ, „Ò —
(1) • •(2) 2
(3) •
(4)
••
85. ∞∑§ ÁòÊ÷È¡ ABC, ¡Ê Á∑§ ‡ÊË·¸ A ¬⁄U ‚◊∑§ÊáÊ „Ò, ◊¥A, B ÃÕÊ C ∑ § ÁSÕÁà ‚ÁŒ‡Ê ∑ ˝ §◊‡Ê—
• • •i j k i j p k
∧ ∧ ∧ ∧ ∧ ∧
• • • • • • • •• • •• • • ÃÕÊ
• •i q j k
∧ ∧ ∧
•• • •• • „Ò¥, ÃÊ Á’¥ŒÈ (p, q) Á¡‚ ⁄UπÊ ¬⁄U
ÁSÕÃ „Ò, fl„ —
(1) x-•ˇÊ ∑§ ‚◊Ê¥Ã⁄U „Ò–
(2) y-•ˇÊ ∑§ ‚◊Ê¥Ã⁄U „Ò–
(3) x-•ˇÊ ∑§Ë œŸÊà◊∑§ ÁŒ‡ÊÊ ‚ ãÿÍŸ ∑§ÊáÊ ’ŸÊÃË„Ò–
(4) x-•ˇÊ ∑§Ë œŸÊà◊∑§ ÁŒ‡ÊÊ ‚ •Áœ∑§ ∑§ÊáÊ’ŸÊÃË „Ò–
84. The distance of the point (1, −2, 4) from
the plane passing through the point
(1, 2, 2) and perpendicular to the planes
x−y+2z=3 and 2x−2y+z+12=0, is :
(1) • •(2) 2
(3) •
(4)
••
85. In a triangle ABC, right angled at the vertex
A, if the position vectors of A, B and C are
respectively • • •i j k i j p k
∧ ∧ ∧ ∧ ∧ ∧
• • • • • • • •• • •• • •
and • • •i q j k
∧ ∧ ∧
•• • •• • then the point (p, q) lies
on a line :
(1) parallel to x-axis.
(2) parallel to y-axis.
(3) making an acute angle with the
positive direction of x-axis.
(4) making an obtuse angle with the
positive direction of x-axis.
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86. If the mean deviation of the numbers
1, 1+d, ..., 1+100d from their mean is 255,
then a value of d is :
(1) 10.1
(2) 20.2
(3) 10
(4) 5.05
87. If A and B are any two events such that
P(A)=•• and
•• •• • •• • • •∩ • , then the
conditional probability, P(A(A∪B)),
where A denotes the complement of A, is
equal to :
(1)••
(2)•• •
(3)•• •
(4)• •• •
88. The number of x [0, 2π] for which
• • • •
• • • •• ••• • • •• • • •• • • • •••• • • •• •• • •
=1 is :
(1) 2
(2) 4
(3) 6
(4) 8
86. Å¡ k„¿epAp¡ 1, 1+d, ..., 1+100d _y„ dÝeL$ \udÝeL$-rhQg_ (kf¡fpi rhQg_) 255 lp¡e sp¡ d _uqL„$ds L¡$V$gu \pe ?(1) 10.1
(2) 20.2
(3) 10
(4) 5.05
87. Å¡ OV$_pAp¡ A A_¡ B dpV¡$ P(A)=•• A_¡
•• •• • •• • • •∩ • lp ¡e sp ¡ ifsu k „cph_p
P(A(A∪B)) L¡$V$gu \pe ? Al] A A¡ A _p¡ |fL$v$ip®h¡ R>¡ ?
(1)••
(2)•• •
(3)•• •
(4)• •• •
88. x [0, 2π] A_¡
• • • •
• • • •• ••• • • •• • • •• • • • •••• • • •• •• • •
=1 lp¡e s¡hp L¡$V$gp x dmi¡ ?(1) 2
(2) 4
(3) 6
(4) 8
86. ÿÁŒ ‚¥ÅÿÊ•Ê¥ 1, 1+d, ..., 1+100d ∑§ ◊Êäÿ ‚◊Êäÿ-Áflø‹Ÿ 255 „Ò, ÃÊ d ∑§Ê ∞∑§ ◊ÊŸ „Ò —(1) 10.1
(2) 20.2
(3) 10
(4) 5.05
87. ÿÁŒ A ÃÕÊ B ŒÊ ∞‚Ë ÉÊ≈UŸÊ∞° „Ò¥ Á∑§ P(A)=••
ÃÕÊ •• •• • •• • • •∩ • „Ò, ÃÊ ¬˝ÁÃ’¥ÁœÃ ¬˝ÊÁÿ∑§ÃÊ
P(A(A∪B)), ¡„Ê° A, A ∑§ ¬Í⁄U∑§ ‚◊ÈìÊÿ ∑§ÊÁŸÁŒ¸c≈U ∑§⁄UÃÊ „Ò, ’⁄UÊ’⁄U „Ò —
(1)••
(2)•• •
(3)•• •
(4)• •• •
88. x [0, 2π] ∑§Ë ‚¥ÅÿÊ, Á¡Ÿ∑§ Á‹∞
• • • •
• • • •• ••• • • •• • • •• • • • •••• • • •• •• • •
=1, „Ò —(1) 2
(2) 4
(3) 6
(4) 8
SET - 04 ENGLISH SET - 04 HINDI SET - 04 GUJARATI
Set - 04 46
89. If m and M are the minimum and the
maximum values of
• ••• • • ••• • • •• •• • ••
• •• • x R, then
M−m is equal to :
(1)
• ••
(2)
••
(3)
••
(4)
••
90. Consider the following two statements :
P : If 7 is an odd number, then 7 is
divisible by 2.
Q : If 7 is a prime number, then 7 is an
odd number.
If V1 is the truth value of the contrapositive
of P and V2
is the truth value of
contrapositive of Q, then the ordered pair
(V1, V
2) equals :
(1) (T, T)
(2) (T, F)
(3) (F, T)
(4) (F, F)
- o O o -
89. Å¡ m A_¡ M A¡ • ••• • • ••• • • •• •• • ••
• •• •
x R _p„ Þe|_sd A_¡ dlÑd d|ëep¡ lp¡e sp¡ M−m
_u qL„$ds L¡$V$gu \i¡ ?
(1)
• ••
(2)
••
(3)
••
(4)
••
90. _uQ¡_p b¡ rh^p_p¡ Ap ¡g R>¡ :
P : Å¡ 7 A¡ AeyÁd k„¿ep lp¡e sp¡ 7 _¡ 2 hX¡$r_:i¡j cpNu iL$pe R>¡.
Q : Å¡ 7 A¡ ArhcpÄe k„¿ep lp¡e sp¡ 7 AeyÁdR>¡.
Å¡ V1 A¡ P dp„ kdp_p\ â¡fZ (contrapositive)
_y„ kÐep\®sp d|ëe A_¡ V2 A¡ Q _p kdp_p\ â¡fZ_p„
kÐep\®sp d|ëe lp¡e sp¡ ¾$dey¼s Å¡X$ (V1, V
2) bfpbf:
(1) (T, T)
(2) (T, F)
(3) (F, T)
(4) (F, F)
- o O o -
89. ÿÁŒ m ÃÕÊ M, √ÿ¥¡∑§
• ••• • • ••• • • •• •• • ••
• •• • x R ∑§ ∑ ˝§◊‡Ê—
ãÿÍŸÃ◊ ÃÕÊ •Áœ∑§Ã◊ ◊ÊŸ „Ò¥, ÃÊ M−m ’⁄UÊ’⁄U„Ò —
(1)
• ••
(2)
••
(3)
••
(4)
••
90. ÁŸêŸ ŒÊ ∑§ÕŸÊ¥ ¬⁄U ÁfløÊ⁄U ∑§ËÁ¡∞ —
P : ÿÁŒ 7 ∞∑§ Áfl·◊ ‚¥ÅÿÊ „Ò, ÃÊ 7, 2 ‚ ÷Êíÿ„Ò–
Q : ÿÁŒ 7 ∞∑§ •÷Êíÿ ‚¥ÅÿÊ „Ò, ÃÊ 7 ∞∑§ Áfl·◊‚¥ÅÿÊ „Ò–
ÿÁŒ V1, P ∑§ ¬˝ÁÃœŸÊà◊∑§ ∑§Ê ‚àÿ◊ÊŸ „Ò ÃÕÊ
V2, Q ∑§ ¬˝ÁÃœŸÊà◊∑§ ∑§Ê ‚àÿ◊ÊŸ „Ò, ÃÊ ∑˝§Á◊à ÿÈÇ◊
(V1, V
2) ’⁄UÊ’⁄U „Ò —
(1) (T, T)
(2) (T, F)
(3) (F, T)
(4) (F, F)
- o O o -
Question and Answer Key - April 9 OnlineQuestion No. Answer Key Question No. Answer Key Question No. Answer Key
Q1 3 Q31 3 Q61 2Q2 2 Q32 1 Q62 2Q3 1 Q33 4 Q63 4Q4 2 Q34 3 Q64 4Q5 1 Q35 2 Q65 3Q6 3 Q36 2 Q66 4Q7 3 Q37 4 Q67 1Q8 3 Q38 4 Q68 4Q9 4 Q39 1 Q69 2
Q10 1 Q40 2 Q70 2Q11 3 Q41 3 Q71 3Q12 1 Q42 4 Q72 1Q13 2 Q43 2 Q73 3Q14 2 Q44 4 Q74 2Q15 1 Q45 1 Q75 3Q16 1 Q46 2 Q76 2Q17 2 Q47 4 Q77 4Q18 2 Q48 3 Q78 3Q19 1 Q49 3 Q79 3Q20 2 Q50 3 Q80 2Q21 3 Q51 2 Q81 2Q22 3 Q52 3 Q82 4Q23 2 Q53 2 Q83 3Q24 3 Q54 2 Q84 1Q25 2 Q55 2 Q85 3Q26 2 Q56 2 Q86 1Q27 2 Q57 1 Q87 2Q28 2 Q58 4 Q88 4Q29 4 Q59 4 Q89 2Q30 2 Q60 1 Q90 3
Note:-
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 1
1. A, B, C and D are four different physical
quantities having different dimensions.
None of them is dimensionless. But we
know that the equation AD=C ln(BD)
holds true. Then which of the combination
is not a meaningful quantity ?
(1) A2
− B2C
2
(2)
• • • •• •••
(3) • •• •••
(4)
•• •• • •
• • ••
2. A particle of mass M is moving in a circle
of fixed radius R in such a way that its
centripetal acceleration at time t is given
by n2
R t2 where n is a constant. The power
delivered to the particle by the force acting
on it, is :
(1) M n2
R2
t
(2) M n R
2 t
(3) M n R2
t2
(4)
•• M n
2 R
2 t2
1. A, B, C ÃÕÊ D øÊ⁄U Á÷ÛÊ ◊ÊòÊÊ∞° „Ò¥ Á¡Ÿ∑§Ë Áfl◊Ê∞¥Á÷㟠„Ò¥– ∑§Ê߸ ÷Ë ◊ÊòÊÊ Áfl◊Ê-⁄UÁ„à ◊ÊòÊÊ Ÿ„Ë¥ „Ò¥,‹Á∑§Ÿ AD=C ln(BD) ‚àÿ „Ò– Ã’ ÁŸêŸ ◊¥ ‚∑§ÊÒŸ •Ê‡Êÿ-⁄UÁ„à ◊ÊòÊÊ „Ò?(1) A
2 − B
2C
2
(2)
• • • •• •••
(3) • •• •••
(4)
•• •• • •
• • ••
2. Œ˝√ÿ◊ÊŸ M ∑§Ê ∞∑§ ∑§áÊ ÁŸÁ‡øà ÁòÊíÿÊ R ∑§ flÎûÊËÿ¬Õ ¬⁄U-ß‚ ¬∑§Ê⁄U ø‹ ⁄U„Ê „Ò Á∑§ ‚◊ÿ ‘t’ ¬⁄U •Á÷∑§ãŒËàfl⁄UáÊ n2
R t2 mÊ⁄UÊ ÁŒÿÊ ¡Ê ‚∑§ÃÊ „Ò, ÿ„Ê° ‘n’ •ø⁄U
„Ò– Ã’ ∑§áÊ ¬⁄U ‹ª ⁄U„ ’‹ mÊ⁄UÊ ©‚∑§Ê ŒË ªß¸ ‡ÊÁÄÃ„Ò —(1) M n
2 R
2 t
(2) M n R
2 t
(3) M n R2
t2
(4)
•• M n
2 R
2 t2
1. A, B, C A_¡ D A¡ Qpf Sy>v$p-Sy>v$p `qfdpZ ^fphsuSy>v$u-Sy>v$u cp¥rsL$fpriAp¡ R>¡. s¡dp„_u L$p¡C`Z `qfdpZfrls _\u. Å¡ AD=C ln(BD) kduL$fZ kpQy„ lp¡esp¡ _uQ¡ Ap ¡g ¥L$u L$ey„ k„ep¡S>_ A¡ A\®kcf (kpQu)fpri v$ip®hsy„ _\u ?(1) A
2 − B
2C
2
(2)
• • • •• •••
(3) • •• •••
(4)
•• •• • •
• • ••
2. M v$m ^fphsp¡ L$Z R S>¡V$gu AQm rÓÄep ^fphsphsy®mpL$pf dpN® f A¡hu fus¡ Nrs L$f¡ R>¡ L¡$ t kde¡ s¡_p¡L¡$ÞÖNpdu âh¡N n2
R t2 hX¡$ Ap`u iL$pe; Äep„ n A¡
AQmp„L$ R>¡. sp¡ L$Z `f gpNsp bm hX¡$ L$Z_¡ dmsp¡`phf (L$pe®v$nsp) __________ \i¡.(1) M n
2 R
2 t
(2) M n R
2 t
(3) M n R2
t2
(4)
•• M n
2 R
2 t2
JEE Main 2016 Question Paper 1 Online (April 10, 2016)JEE Main 2016 Question Paper 1 Online (April 10, 2016)JEE Main 2016 Question Paper 1 Online (April 10, 2016)
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 2
3. Concrete mixture is made by mixing
cement, stone and sand in a rotating
cylindrical drum. If the drum rotates too
fast, the ingredients remain stuck to the
wall of the drum and proper mixing of
ingredients does not take place. The
maximum rotational speed of the drum in
revolutions per minute(rpm) to ensure
proper mixing is close to :
(Take the radius of the drum to be 1.25 m
and its axle to be horizontal) :
(1) 0.4
(2) 1.3
(3) 8.0
(4) 27.0
3. ∑¥§∑˝§Ë≈U Á◊Ä‚ø⁄U ’ŸÊŸ ∑§ Á‹ÿ ‚Ë◊¥≈U, ⁄Uà ÃÕÊ ⁄UÊ«∏Ë∑§Ê ∞∑§ ÉÊÍáÊ˸ÿ ’‹ŸÊ∑§Ê⁄U «˛U◊ ◊¥ «UÊ‹Ê ¡ÊÃÊ „Ò– ÿÁŒ«˛U◊ ∑§Ë ÉÊÍáʸŸ-ªÁà ’„Èà á „Ê ÃÊ ‚¥ÉÊ≈U∑§ «˛U◊ ∑§ËŒËflÊ⁄U ‚ Áø¬∑§ ⁄U„à „Ò¥ •ÊÒ⁄U Á◊Ä‚ø⁄U ∆UË∑§ ‚ Ÿ„Ë¥’ŸÃÊ– ÿÁŒ «˛U◊ ∑§Ë ÁòÊíÿÊ 1.25 m „Ò •ÊÒ⁄U ß‚∑§ËœÈ⁄UË ˇÊÒÁá „Ò, Ã’ •ë¿UË Ã⁄U„ Á◊Ä‚ „ÊŸ ∑§ Á‹ÿ¡M§⁄UË •Áœ∑§Ã◊ ÉÊÍáÊ˸ÿ-ªÁà rpm ◊¥ „Ò —(1) 0.4
(2) 1.3
(3) 8.0
(4) 27.0
3. A¡L$ QpL$Nrs L$fsp _mpL$pfue X²$ddp„ rkd¡ÞV$, `Õ\fA_¡ f¡su_¡ c¡Np L$fu_¡ L$p¢q¾$V$ rdîZ s¥epf L$fhpdp„Aph¡ R>¡. lh¡ Å¡ X²$d M|b S> TX$`\u QpL$Nrs L$f¡ sp¡rdîZ L$f¡gp sÒhp¡ X²$d_u qv$hpg_¡ Qp¢V$u Åe R>¡ A_¡s¡d_y„ ep¡Áe rdîZ b_phu iL$psy„ _\u. sp¡ `qfc°dZârs rd_uV$ (rpm) _p `v$dp„, ep¡Áe rdîZ b_hpdpV¡$_u X²$d_u dlÑd L$p¡Zue TX$` __________ _u_ÆL$_u li¡. (X²$d_u rÓÄep 1.25 m A_¡ s¡_u AnkdrnrsS> R>¡ s¡d ^pfp¡).(1) 0.4
(2) 1.3
(3) 8.0
(4) 27.0
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 3
4. Velocity-time graph for a body of mass
10 kg is shown in figure. Work-done on
the body in first two seconds of the motion
is :
(1) 12000 J
(2) −12000 J
(3) −4500 J
(4) −9300 J
5. In the figure shown ABC is a uniform wire.
If centre of mass of wire lies vertically below
point A, then
• •• •
is close to :
(1) 1.85
(2) 1.37
(3) 1.5
(4) 3
4. 10 kg Œ˝√ÿ◊ÊŸ ∑§ Á¬¥«U ∑§ Á‹ÿ flª-‚◊ÿ ª˝Ê»§ ÁøòÊ◊¥ ÁŒÿÊ „Ò– Á¬¥«U ¬⁄U ¬„‹ 2 ‚. ◊¥ Á∑§ÿÊ ªÿÊ ∑§Êÿ¸„Ò —
(1) 12000 J
(2) −12000 J
(3) −4500 J
(4) −9300 J
5. ÁŒÿ ªÿ ÁøòÊ ◊¥ ÃÊ⁄U ABC ∞∑§ ‚◊ÊŸ „Ò– ÿÁŒ‚¥„ÁÃ-∑§ãŒ˝ Á’¥ŒÈ A ∑ ™§äflʸœ⁄U ŸËø ÁSÕà „Ò, Ã’
• •• •
‹ª÷ª „Ò —
(1) 1.85
(2) 1.37
(3) 1.5
(4) 3
4. 10 kg v$m fphsp v$p\® dpV¡$_p¡ h¡N-kde_p¡ Apg¡MApL©$rÑdp„ v$ip®h¡g R>¡. v$p\® f â\d 2 k¡L$ÞX$dp„ \suNrs v$fçep_ \sy„ L$pe® __________\i¡.
(1) 12000 J
(2) −12000 J
(3) −4500 J
(4) −9300 J
5. ApL©$rÑdp„ v$ip®ìep dyS>b ABC A¡ kdp_ spf R>¡. Å¡spf_y„ Öìedp_ L¡$ÞÖ tbvy$ A _u bfp¡bf _uQ¡ Aphsy„
lp¡e sp¡ • •• •
_y„ d|ëe _________ _u _ÆL$_y„ \i¡.
(1) 1.85
(2) 1.37
(3) 1.5
(4) 3
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 4
6. An astronaut of mass m is working on a
satellite orbiting the earth at a distance h
from the earth’s surface. The radius of the
earth is R, while its mass is M. The
gravitational pull FG
on the astronaut is :
(1) Zero since astronaut feels weightless
(2)• •• • •• • ••
• • •••
(3) •• • •• •• •• • • ••
• • • • • ••• • ••
(4)• • •• • • •
•• • ••• •
••
6. ¬ÎâflË ∑§Ë ‚Ä ‚ ‘h’ ŒÍ⁄UË ¬⁄U ÁSÕà ∞∑§ ©¬ª˝„ ¬⁄U∞∑§ ‘m’ Œ˝√ÿ◊ÊŸ ∑§Ê •¥ÃÁ⁄UˇÊ-ÿÊòÊË ∑§Ê◊ ∑§⁄U ⁄U„Ê „Ò–¬ÎâflË ∑§Ê Œ˝√ÿ◊ÊŸ ‘M’ ÃÕÊ ÁòÊíÿÊ ‘R’ „Ò– Ã’ ©‚ÿÊòÊË ¬⁄U ‹ª ⁄U„Ê ªÈL§àflËÿ ’‹ F
G „Ò —
(1) ‡ÊÍãÿ, ÄÿÊ¥Á∑§ fl„ ÿÊòÊË ÷Ê⁄U„ËŸÃÊ ◊„‚Í‚ ∑§⁄UÃÊ„Ò–
(2)• •• • •• • ••
• • •••
(3) •• • •• •• •• • • ••
• • • • • ••• • ••
(4)• • •• • • •
•• • ••• •
••
6. `©Õhu_u k`pV$u\u h KQpC A¡ `qfc°dZ L$fspk¡V¡gpCV$dp„ m v$m ^fphsp¡ MNp¡mipõÓu L$pe® L$f¡ R>¡.©Õhu_u rÓÄep R R>¡ Äepf¡ s¡_y„ v$m M R>¡. MNp¡mipõÓu
`f gpNsy„ NyfyÐhpL$j M¢QpZ FG
__________ li¡.
(1) i|Þe, L$pfZ L¡$ MNp¡mipõÓu hS>_frls[õ\rsdp„ R>¡.
(2)• •• • •• • ••
• • •••
(3) •• • •• •• •• • • ••
• • • • • ••• • ••
(4)• • •• • • •
•• • ••• •
••
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 5
7. A bottle has an opening of radius a and
length b. A cork of length b and radius
(a+∆a) where (∆a<<a) is compressed to fit
into the opening completely (See figure).
If the bulk modulus of cork is B and
frictional coefficient between the bottle and
cork is µ then the force needed to push the
cork into the bottle is :
(1) (πµB b) ∆a
(2) (2πµB b) ∆a
(3) (πµB b) a
(4) (4πµB b) ∆a
7. ∞∑§ ’ÊË ∑§ ◊È°„ ∑§Ë ÁòÊíÿÊ ‘a’ „ÒÒ ÃÕÊ ‹ê’Ê߸ ‘b’ „Ò–∞∑§ ‘b’ ‹ê’Ê߸ •ÊÒ⁄U (a+∆a) ÁòÊíÿÊ (∆a<<a) flÊ‹∑§Ê∑¸§ ∑§Ê ©‚∑§ ◊È°„ ◊¥ ¬Í⁄UË Ã⁄U„ ∆Í°U‚ ÁŒÿÊ ªÿÊ „Ò (ÁøòÊŒÁπÿ)– ÿÁŒ ∑§Ê∑¸§ ∑§Ê •Êÿß ¬˝àÿÊSÕÃÊ ªÈáÊÊ¥∑§ B„Ò ÃÕÊ ’ÊË •ÊÒ⁄U ∑§Ê∑¸§ ∑§ ’Ëø ÉÊ·¸áÊ-ªÈáÊÊ¥∑§ µ „Ò,Ã’ ∑§Ê∑§ ∑§Ê ◊È°„ ◊¥ ÉÊÈ‚ÊŸ ∑§ Á‹ÿ •Êfl‡ÿ∑§ ’‹ „Ò —
(1) (πµB b) ∆a
(2) (2πµB b) ∆a
(3) (πµB b) a
(4) (4πµB b) ∆a
7. A¡L$ bp¡V$g_p Nmp_u rÓÄep a A_¡ g„bpC b R>¡.ApL©$rÑdp„ v$ip®ìep dyS>b, b g„bpC_p A_¡ (a+∆a)
Äep„ (∆a<<a) _u rÓÄep ^fphsp b|Q_¡ v$bpZ |h®L$hX¡$ bp¡V$g_p Nmphpmp cpN_¡ k„ |Z® b„ L$fhpdp„ Aph¡R>¡. Å¡ b|Q_p¡ L$v$ [õ\rsõ\p`L$spA„L$ B lp¡e A_¡bp¡V$g A_¡ b|Q hÃQ¡_p¡ OjZp¯L$ µ lp¡e sp¡ b|Q_¡bp¡V$gdp„ O|kpX$hp S>ê$fu bm ________ \i¡.
(1) (πµB b) ∆a
(2) (2πµB b) ∆a
(3) (πµB b) a
(4) (4πµB b) ∆a
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 6
8. A Carnot freezer takes heat from water at
0C inside it and rejects it to the room at a
temperature of 27C. The latent heat of ice
is 336×103
J kg−1
. If 5 kg of water at 0C is
converted into ice at 0C by the freezer, then
the energy consumed by the freezer is close
to :
(1) 1.67×105
J
(2) 1.68×106
J
(3) 1.51×105
J
(4) 1.71×107
J
8. ∞∑§ ∑§ÊŸÊ≈U »˝§Ë¡⁄U •¬Ÿ •¥Œ⁄U 0C ¬⁄U ⁄Uπ „È∞ ¡‹ ‚™§c◊Ê ‹∑§⁄U ©‚ ∑§◊⁄U ∑§ Ãʬ◊ÊŸ 27C ¬⁄U ÁŸc∑§ÊÁ‚Ã∑§⁄UÃÊ „Ò– ’»¸§ ∑§Ë ªÈåà ™§c◊Ê 336×10
3 J kg
−1 „Ò–ÿÁŒ »˝§Ë¡⁄U ◊¥ ⁄UπÊ 0C ¬⁄U 5 kg ¡‹, 0C ¬⁄U ’»¸§ ◊¥’Œ‹ÃÊ „Ò Ã’ »˝§Ë¡⁄U mÊ⁄UÊ π¬ÊßZ ªß¸ ™§¡Ê¸ ‹ª÷ª „Ò —(1) 1.67×10
5 J
(2) 1.68×106
J
(3) 1.51×105
J
(4) 1.71×107
J
8. A¡L$ L$p_p£V$ f¡qäS>f¡V$f `pZudp„\u 0C sp`dp_¡ Dódpip¡ju 27C sp`dp_¡ fl¡gp hpsphfZdp„ a¢L¡$ R>¡. bfa_ug¡V¡$ÞV$ (Nyá) Dódp 336×10
3 J kg
−1 R>¡. Å¡f¡qäS>f¡V$fdp„ 0C _y„ 5 kg pZu 0C bfadp„ ê$`p„sfusL$fhy„ lp¡e sp¡ f¡qäS>f¡V$f¡ hp`f¡g EÅ® __________ _u_ÆL$_y„ d|ëe ^fphi¡.(1) 1.67×10
5 J
(2) 1.68×106
J
(3) 1.51×105
J
(4) 1.71×107
J
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 7
9. Which of the following shows the correct
relationship between the pressure ‘P’ and
density ρ of an ideal gas at constant
temperature ?
(1)
(2)
(3)
(4)
9. Á∑§‚Ë •ÊŒ‡Ê¸ ªÒ‚ ∑§ Á‹ÿ ÁSÕ⁄U Ãʬ◊ÊŸ ¬⁄U ©‚∑§ ŒÊ’‘P’ ÃÕÊ ÉÊŸàfl ‘ρ’ ∑§ ’Ëø ‚¥’¥œ ∑§ Á‹ÿ ÁŸêŸ ◊¥ ‚∑§ÊÒŸ-‚Ê ÁøòÊ ‚„Ë „Ò?
(1)
(2)
(3)
(4)
9. _uQ¡ Ap ¡g ApL©$rsAp¡ ¥L$u L$C ApL©$rÑ AQm sp`dp_¡fl¡g Apv$i® hpey dpV¡$ v$bpZ ‘P’ A_¡ O_sp ρ hÃQ¡_p¡kpQp¡ k„b„ v$ip®h¡ R>¡.
(1)
(2)
(3)
(4)
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 8
10. In an engine the piston undergoes vertical
simple harmonic motion with amplitude
7 cm. A washer rests on top of the piston
and moves with it. The motor speed is
slowly increased. The frequency of the
piston at which the washer no longer stays
in contact with the piston, is close to :
(1) 0.1 Hz
(2) 1.2 Hz
(3) 0.7 Hz
(4) 1.9 Hz
11. A toy-car, blowing its horn, is moving with
a steady speed of 5 m/s, away from a wall.
An observer, towards whom the toy car is
moving, is able to hear 5 beats per second.
If the velocity of sound in air is 340 m/s,
the frequency of the horn of the toy car is
close to :
(1) 680 Hz
(2) 510 Hz
(3) 340 Hz
(4) 170 Hz
10. ∞∑§ ߥ¡Ÿ ∑§Ê Á¬S≈UŸ 7 cm •ÊÿÊ◊ ∑§Ë ‚⁄U‹-•Êflø-ªÁà ™§äflʸœ⁄U ◊¥ ∑§⁄U ⁄U„Ê „Ò– Á¬S≈UŸ ∑§ ™§¬⁄U ∞∑§flʇÊ⁄U ⁄UπÊ „Ò ¡Ê ©‚∑§ ‚ÊÕ ø‹ÃÊ „Ò– ◊Ê≈U⁄U ∑§Ë ªÁÃœË⁄U-œË⁄U ’…∏Ê߸ ¡ÊÃË „Ò ÃÊ Á¬S≈UŸ ∑§Ë •ÊflÎÁûÊ Á¡‚ ¬⁄UflʇÊ⁄U Á¬S≈UŸ ∑§Ê ‚ÊÕ ¿UÊ«∏ ŒÃÊ „Ò, fl„ ‹ª÷ª „Ò —(1) 0.1 Hz
(2) 1.2 Hz
(3) 0.7 Hz
(4) 1.9 Hz
11. ∞∑§ Áπ‹ÊÒŸÊ ∑§Ê⁄U „ÊŸ¸ ’¡ÊÃË „È߸ 5 m/s ∑§Ë ÁSÕ⁄UªÁà ‚ ∞∑§ ŒËflÊ⁄U ‚ ŒÍ⁄U ÃÕÊ ∞∑§ √ÿÁÄà ∑§Ë •Ê⁄U ¡Ê⁄U„Ë „Ò– ©‚ √ÿÁÄà ∑§Ê 5 ’Ë≈U/‚¥. ‚ÈŸÊ߸ ŒÃË „Ò¥– ÿÁŒ„flÊ ◊¥ äflÁŸ ∑§Ë ªÁà 340 m/s „Ò, Ã’ „ÊŸ¸ ∑§Ë•ÊflÎÁûÊ ‹ª÷ª „Ò —
(1) 680 Hz
(2) 510 Hz
(3) 340 Hz
(4) 170 Hz
10. A¡L$ A¡[ÞS>_dp„ r`õV$_ DÝh® qv$ipdp„ 7 cm _pL„$`rhõspf kp\¡ SHM (k.Ap.N) L$f¡ R>¡. r`õV$__uD`f fl¡g hp¸if s¡_u kp\¡ S> Nrs L$f¡ R>¡. r`õV$__¡Nrs L$fphsu dp¡V$f_u TX$` ud¡\u h^pfhpdp„ Aph¡ R>¡.hp¸if r`õV$__u k`pV$u_¡ R>p¡X$u v$¡ s¡ r`õV$__u S>ê$fuAph©rÑ __________ _u _ÆL$_y„ d|ëe li¡.(1) 0.1 Hz
(2) 1.2 Hz
(3) 0.7 Hz
(4) 1.9 Hz
11. A¡L$ lp¡_® hNpX$su fdL$X$p„_u L$pf qv$hpg \u v|$f sfa5 m/s _u AQm TX$`\u Nrs L$f ¡ R> ¡. A¡L$Ahgp¡L$_L$pf, L¡$ S>¡_u sfa fdL$X$p„_u L$pf Nrs L$f¡ R>¡,A¡L$ k¡L$ÞX$dp„ 5 õ „v$ (beats) kp„cm¡ R>¡. Å¡ lhpdp„Ýhr__p¡ h¡N 340 m/s lp¡e sp¡ L$pf_p lp¡_®_u Aph©qÑ_______ _u _ÆL$_u li¡.(1) 680 Hz
(2) 510 Hz
(3) 340 Hz
(4) 170 Hz
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 9
12. Within a spherical charge distribution of
charge density ρ(r), N equipotential
surfaces of potential V0, V
0+∆V, V
0+2∆V,
.......... V0+N∆V (∆V>0), are drawn and
have increasing radii
r0, r
1, r
2,..........r
N, respectively. If the
difference in the radii of the surfaces is
constant for all values of V0 and ∆V then :
(1) ρ (r) α r
(2) ρ (r) = constant
(3) ( )•• • •••
• •
(4) ( ) ••• • •••
• •
12. •Êfl‡Ê-ÉÊŸàfl ρ (r) ∑§ Á∑§‚Ë ªÊ‹Ëÿ-•Êfl‡Ê-ÁflÃ⁄UáÊ,∑§ •ãŒ⁄U N ‚◊Áfl÷fl-¬Îc∆U, Á¡Ÿ∑§Ë Áfl÷fl „ÒV
0, V
0+∆V, V
0+2∆V,
........ V
0+N∆V
(∆V>0), •Ê⁄UÁπà Á∑§ÿ ªÿ „Ò¥ •ÊÒ⁄U ©Ÿ∑§Ë ÁòÊíÿÊ∞¥∑˝§◊‡Ê— r
0, r
1, r
2,..........r
N „Ò¥– ÿÁŒ ÁòÊíÿÊ•Ê¥ ∑§Ê
•ãÃ⁄UÊ‹, ‚÷Ë V0 ÃÕÊ ∆V ∑§ ◊ÊŸÊ¥ ∑§ Á‹ÿ, ÁSÕ⁄U „Ò
Ã’ —(1) ρ (r) α r
(2) ρ (r)=•ø⁄U
(3) ( )•• • •••
• •
(4) ( ) ••• • •••
• •
12. ρ(r) S>¡V$gu Np¡gue rhÛyscpf rhsfZ dpV¡$, A_y¾$d¡r
0, r
1, r
2,..........r
N rÓÄep ^fphsu A_¡
V0, V
0+∆V, V
0+2∆V, .........., V
0+N∆V
(∆V>0) S>¡V$gy„ [õ\rsdp_ ^fphsu N kd[õ\rsdp_k`pV$u ( ©›$) v$p¡fhpdp„ Aph¡ R>¡. V
0 A_¡ ∆V _p
b^pS> d|ëe dpV¡$ Å¡ k`pV$uAp¡_u rÓÄep hÃQ¡_p¡ saphsAQm fl¡sp¡ lp¡e sp¡ ________.
(1) ρ (r) α r
(2) ρ (r) = AQm
(3) ( )•• • •••
• •
(4) ( ) ••• • •••
• •
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 10
13. Figure shows a network of capacitors where
the numbers indicates capacitances in
micro Farad. The value of capacitance C if
the equivalent capacitance between point
A and B is to be 1 µF is :
(1)
• • • •• •
•
(2)• • • •• •
•
(3)• • • •• •
•
(4)
• • • •• •
•
13. ÁøòÊ ‚¥œÊÁ⁄UòÊÊ¥ ∑§Ê ÁŸ∑§Êÿ Œ‡ÊʸÃÊ „Ò, ¡„Ê° •¥∑§ µF ◊¥œÊÁ⁄UÃÊ Œ‡Êʸà „Ò¥– A fl B ∑§ ’Ëø ¬˝÷ÊflË œÊÁ⁄UÃÊ1 µF „ÊŸ ∑§ Á‹ÿ C ∑§Ë œÊÁ⁄UÃÊ „ÊŸË øÊÁ„ÿ —
(1)
• • • •• •
•
(2)• • • •• •
•
(3)• • • •• •
•
(4)
• • • •• •
•
13. ApL©$rÑdp„ k„OpfL$p¡_p¡ b_¡gy„ _¡V$hL®$ v$ip®h¡ R>¡ Äep„L¡ $`¡kuV$f_u bpSy>dp„ gM¡g _„bf s¡ L¡ $`¡kuV$f_y „dpC¾$p¡a¡fpX$dp„ d|ëe v$ip®h¡ R>¡. Å¡ A A_¡ B hÃQ¡_y„`qfZpdu L¡$`¡kuV$Þk (k„OpfL$sp) 1 µF Å¡Csy„ lp¡e sp¡k„OpfL$ C _y„ d|ëe ________ Å¡Ci¡.
(1)
• • • •• •
•
(2)• • • •• •
•
(3)• • • •• •
•
(4)
• • • •• •
•
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 11
14. The resistance of an electrical toaster has a
temperature dependence given by
R(T)=R0
[1+α(T−T0)] in its range of
operation. At T0=300 K, R=100 Ω and at
T=500 K, R=120 Ω. The toaster is
connected to a voltage source at 200 V and
its temperature is raised at a constant rate
from 300 to 500 K in 30 s. The total work
done in raising the temperature is :
(1)
• •• • • • • ••• •••••
(2)
•• • • • • •••
••
(3)
•• • • • • •••
••
(4) 300 J
14. Á’¡‹Ë ‚ ø‹Ÿ flÊ‹ ≈UÊS≈U⁄U ∑§ ¬˝ÁÃ⁄UÊœ ∑§Ê Ãʬ◊ÊŸ ‚’Œ‹Êfl R(T)=R
0 [1+α(T−T
0)] mÊ⁄UÊ ÁŒÿÊ ªÿÊ
„ Ò– T0=300 K ¬⁄ U R=100 Ω „ Ò ÃÕÊ
T=500 K ¬⁄U R=120 Ω „Ò– ≈UÊS≈U⁄U 200 VU ∑§dÊà ‚ ¡È«∏Ê „Ò, ÃÕÊ ©‚∑§Ê Ãʬ◊ÊŸ 300 K ‚ ∞∑§‚◊ÊŸ Œ⁄U ¬⁄U ’…∏∑§⁄U 30 s ◊¥ 500 K „Ê ¡ÊÃÊ „Ò– Ã’ß‚ ¬˝∑˝§◊ ◊¥ Á∑§ÿÊ ªÿÊ ∑ȧ‹ ∑§Êÿ¸ „Ò —
(1)
• •• • • • • ••• •••••
(2)
•• • • • • •••
••
(3)
•• • • • • •••
••
(4) 300 J
14. A¡L$ Cg¡¼V²$uL$ V$p¡õV$f_p¡ s¡_p D`ep¡N v$fçep_, sp`dp_kp\¡_p¡ k„b„^ R(T)=R
0 [1+α(T−T
0)] hX¡$
Ap`hpdp „ Aph ¡ R> ¡ . T0=300 K dpV ¡ $
R=100 Ω A_¡ T=500 K dpV¡$ R=120 Ω R>¡.V$p¡õV$f_¡ A¡L$ 200 V _p hp¡ëV¡$S> Dv¹$Nd kp\¡ Å¡X$u s¡_y„sp`dp_ AQm v$f¡ 300 \u 500 K ky^u gC S>hpdp„Aph¡ R>¡. s¡ dpV¡$ gpNsp¡ kdeNpmp¡ 30 s R>¡. sp¡sp`dp__p¡ h^pfp ¡ L$fhp dpV¡ $ S>ê$fu L y $g L$pe®__________.
(1)
• •• • • • • ••• •••••
(2)
•• • • • • •••
••
(3)
•• • • • • •••
••
(4) 300 J
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 12
15. Consider a thin metallic sheet
perpendicular to the plane of the paper
moving with speed ‘v’ in a uniform
magnetic field B going into the plane of the
paper (See figure). If charge densities σ1 and
σ2 are induced on the left and right surfaces,
respectively, of the sheet then (ignore
fringe effects) :
(1) σ1=
0 v B, σ
2=−
0 v B
(2)• •
• ••• • •• • • •
• •• ••• •• • −
= =
(3) σ1=σ
2=
0 vB
(4)• •
• •• •• • • • •
• •• ••• •• •−
= =
15. ∞∑§ ¬Ã‹Ë œÊÃÈ ‡ÊË≈U ¬Îc∆U ∑§ ‹ê’flà ⁄UπË „Ò •ÊÒ⁄UÁøòÊ ◊¥ ÁŒπÊß ÁŒ‡ÊÊ ◊¥ flª ‘v’ ‚ ∞∑§ ‚◊ÊŸ øÈê’∑§Ëÿ-ˇÊòÊ B ◊¥ ø‹ ⁄U„Ë „Ò– øÈê’∑§Ëÿ-ˇÊòÊ ß‚ ‚◊Ë ¬Îc∆U◊¥ ¬˝fl‡Ê ∑§⁄U ⁄U„Ê „Ò– ÿÁŒ ß‚ ‡ÊË≈U ∑§Ë ’ÊßZ •ÊÒ⁄U ŒÊßZ‚Äʥ ¬⁄U ∑˝§◊‡Ê— ¬Îc∆U-•Êfl‡Ê-ÉÊŸàfl σ
1 ÃÕÊ σ
2 ¬Á⁄UÃ
„Êà „Ò¥, Ã’ ©¬Ê¥Ã-¬˝÷Êfl ∑§Ê Ÿªáÿ ◊ÊŸÃ „È∞σ
1 ÃÕÊ σ
2 ∑§ ◊ÊŸ „Ê¥ª —
(1) σ1=
0 v B, σ
2=−
0 v B
(2)• •
• ••• • •• • • •
• •• ••• •• • −
= =
(3) σ1=σ
2=
0 vB
(4)• •
• •• •• • • • •
• •• ••• •• •−
= =
15. kdp„Nu Qy„bL$ue n¡Ó B dp„, yõsL$_p ¡S>_¡ g„b A_¡‘v’ S>¡V$gu AQm TX$`\u `¡S>_u A„v$f v$pMg \su^psy_u `psmu s[¼s (`©›$) _¡ Ýep_dp „ gp ¡.(ApL©$rÑ Sy>Ap¡) Å¡ s¡_u X$pbu A_¡ S>dZu k`pV$u `fA_y¾$d¡ σ
1 A_¡ σ
2 S>¡V$gu ©›$rhÛyscpf O_sp Dv¹$ch¡
sp¡ __________ \i¡. (qäÞS> Akf AhNZp¡.)
(1) σ1=
0 v B, σ
2=−
0 v B
(2)• •
• ••• • •• • • •
• •• ••• •• • −
= =
(3) σ1=σ
2=
0 vB
(4)• •
• •• •• • • • •
• •• ••• •• •−
= =
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 13
16. A fighter plane of length 20 m, wing span
(distance from tip of one wing to the tip of
the other wing) of 15 m and height
5 m is flying towards east over Delhi. Its
speed is 240 ms−1
. The earth’s magnetic
field over Delhi is 5×10−5
T with the
declination angle ~0 and dip of θ such that
sin θ =
•• . If the voltage developed is V
B
between the lower and upper side of the
plane and VW
between the tips of the wings
then VB
and V
W are close to :
(1) VB=45 mV; V
W=120 mV with right
side of pilot at higher voltage
(2) VB=45 mV; V
W=120 mV with left
side of pilot at higher voltage
(3) VB=40 mV; V
W=135 mV with right
side of pilot at high voltage
(4) VB=40 mV; V
W=135 mV with left
side of pilot at higher voltage
16. ∞∑§ ‹«∏Ê∑ͧ ¡„Ê¡ ∑§Ë ‹ê’Ê߸ 20 m, ¬¥πÊ¥ ∑§ Á‚⁄UÊ¥ ∑§’Ëø ŒÍ⁄UË 15 m ÃÕÊ ™°§øÊ߸ 5 m „Ò, •ÊÒ⁄U ÿ„ ÁŒÀ‹Ë∑§ ™§¬⁄U ¬Ífl¸-ÁŒ‡ÊÊ ◊¥ 240 ms
−1 ªÁà ‚ ©«∏ ⁄U„Ê „Ò–ÁŒÀ‹Ë ∑§ ™§¬⁄U ¬ÎâflË ∑§Ê øÈê’∑§Ëÿ-ˇÊòÊ 5×10
−5 T
„Ò, Á«UÁÄ‹Ÿ‡ÊŸ ∑§ÊáÊ ~0 „Ò, ÃÕÊ Á«U¬ ∑§ÊáÊ θ ∑§ Á‹ÿ
sin θ =
•• „Ò– ÿÁŒ ¬Á⁄UÃ-Áfl÷fl „Ò¥ — V
B ¡„Ê¡ ∑§
™§¬⁄U fl ŸËø ∑§ ’Ëø ; VW
¬¥πÊ¥ ∑§ Á‚⁄UÊ¥ ∑§ ’Ëø–Ã’ —
(1) VB=45 mV ; V
W=120 mV ŒÊÿÊ¥ ¬¥π-
Á‚⁄UÊ+ve
(2) VB=45 mV ; V
W=120 mV ’ÊÿÊ¥ ¬¥π-
Á‚⁄UÊ−ve
(3) VB=40 mV ; V
W=135 mV ŒÊÿÊ¥ ¬¥π-
Á‚⁄UÊ+ve
(4) VB=40 mV ; V
W=135 mV ’ÊÿÊ¥ ¬¥π-
Á‚⁄UÊ−ve
16. 20 m g„bpC fphsy„, 15 m p„Muep_u lp¡mpC (A¡L$bpSy>_u `p„Muep_p R>¡X$p\u buÆ bpSy>_p `p„rMep_pR>¡X$p ky u) A_¡ 5 m S>¡V$gu KQpC fphsy„ A¡L$ gX$peL$rhdp_ qv$ëlu_p |h® sfa EX$u füy„ R>¡. s¡_u TX$`240 ms
−1 R>¡. qv$ëlu D`f ©Õhu_y„ Qy„bL$ue n¡Ó_y„d|ëe 5×10
−5 T d¡Á_¡V$uL$ X¡$[¼g_¡i_ ~0 A_¡ X$u`
A¡ÞNg θ A¡hp¡ R>¡ L¡$ S>¡\u sin θ =
••
\pe. Å¡
àg¡__p _uQ¡_p A_¡ D`f_p R>¡X$p hÃQ¡ VB S>¡V$gp¡
hp¡ëV¡$S> A_¡ VW
S>¡V$gp¡ hp¡ëV¡$S> b¡ p„Mp¡_p R>¡X$p hÃQ¡DÑ`Þ_ \sp¡ lp¡e sp¡ V
B A_¡ V
W _y „ d|ëe
__________ _ÆL$_y„ li¡..
(1) VB=45 mV A_¡ V
W= `pegp¡V$_u S>dZu
bpSy> KQp hp¡ëV¡$S>¡ lp¡e s¡ fus¡ 120 mV
(2) VB=45 mV; V
W= `pegp¡V$_u X$pbu bpSy>
KQp hp¡ëV¡$S>¡ lp¡e s¡ fus¡ 120 mV
(3) VB=40 mV; V
W= pegp¡V$_u S>dZu bpSy>
KQp hp¡ëV¡$S>¡ lp¡e s¡ fus¡ 135 mV
(4) VB=40 mV; V
W= `pegp¡V$_u X$pbu bpSy>
KQp hp¡ëV¡$S>¡ lp¡e s¡ fus¡ 135 mV
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 14
17. A conducting metal circular-wire-loop of
radius r is placed perpendicular to a
magnetic field which varies with time as
•• ••
• • ••• • , where B
0 and τ are constants,
at time t=0. If the resistance of the loop is
R then the heat generated in the loop after
a long time • •• →• is :
(1)
• • ••
•• ••
π
τ
(2)
• • ••
•• ••
π
τ
(3)
• • •• •• • •π
τ
(4)
• • ••• •
•π
τ
17. ‘r’ ÁòÊíÿÊ ∑§ œÊÃÈ flÎûÊËÿ-ÃÊ⁄U-‹Í¬ ∑§Ê ¬Îc∆U,
•• ••
• • ••• • mÊ⁄UÊ ’Œ‹Ã „È∞ øÈê’∑§Ëÿ-ˇÊòÊ ∑§
‹ê’flà ⁄UπÊ „Ò– ¡„Ê° ‚◊ÿ t=0 ¬⁄U B0 ÃÕÊ τ •ø⁄U
„Ò¥– ÿÁŒ ‹Í¬ ∑§Ê ¬˝ÁÃ⁄UÊœ R „Ò, Ã’ ∑§Ê»§Ë íÿÊŒÊ ‚◊ÿ
• •• →• ªÈ¡⁄UŸ ∑§ ’ÊŒ ©‚ ‹Í¬ ◊¥ ¬ÒŒÊ „È߸ ™§¡Ê¸ „Ò —
(1)
• • ••
•• ••
π
τ
(2)
• • ••
•• ••
π
τ
(3)
• • •• •• • •π
τ
(4)
• • ••• •
•π
τ
17. A¡L$ kyhplL$ ^psy_y„ hsy®mpL$pf N|„Qmy„ L¡$ S>¡_u rÓÄep rlp¡e s¡_¡ Qy„bL$ue n¡Ó_¡ g„bê$ ¡ d|L$hpdp„ Aph¡ R>¡. Ap
Qy„bL$ue n¡Ó kde kp\¡ •• ••
• • ••• • âdpZ¡ bv$gpe
R>¡, Äep„ B0 A_¡ τ t=0 kde¡ AQmp„L$ R>¡. Å¡ N|„Qmp_p¡
Ahfp¡ R lp¡e sp¡ M|b S> gp„bp kde • •• →• _¡
A„s¡ N|„Qmpdp„ DÐ`Þ_ Dódp _______ \i¡.
(1)
• • ••
•• ••
π
τ
(2)
• • ••
•• ••
π
τ
(3)
• • •• •• • •π
τ
(4)
• • ••• •
•π
τ
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 15
18. Consider an electromagnetic wave
propagating in vacuum. Choose the correct
statement :
(1) For an electromagnetic wave
propagating in +x direction the
electric field is
( )•
•• • •• • •• • ••
• • • • • • •• •→ ∧ ∧
= − and the
magnetic field is
( )• • •• • •• • ••
B B x, t y zyz
→ ∧ ∧
•• • •
(2) For an electromagnetic wave
propagating in +x direction the
electric field is
( )•• • • •• • • •• • ••
• • • • •• • • •• •→ ∧ ∧
• • and
the magnetic field is
( )•• • • •• • • •• • ••
• • • • •• • • •• •→ ∧ ∧
• •
(3) For an electromagnetic wave
propagating in +y direction the
electric field is •• • • •• • •••
• • • • • •• •→ ∧
•
and the magnetic field is
•• • • •• • •••
B B x, t zyz
→ ∧
•
18. √ÿÊ◊ ◊¥ ø‹ ⁄U„Ë flÒlÈÃ-ëÊÈê’∑§Ëÿ Ã⁄¥Uª ∑§ Á‹∞ ‚„ËÁfl∑§À¬ øÈÁŸ∞–
(1) +x ÁŒ‡ÊÊ ◊¥ øÊÁ‹Ã flÒlÈÃ-øÈê’∑§Ëÿ Ã⁄¥Uª ∑§
Á‹ÿ ( )•
•• • •• • •• • ••
• • • • • • •• •→ ∧ ∧
= −,
( )• • •• • •• • ••
B B x, t y zyz
→ ∧ ∧
•• • •
(2) +x ÁŒ‡ÊÊ ◊¥ øÊÁ‹Ã flÒlÈÃ-øÈê’∑§Ëÿ Ã⁄¥Uª ∑§
Á‹ÿ ( )•• • • •• • • •• • ••
• • • • •• • • •• •→ ∧ ∧
• • ,
( )•• • • •• • • •• • ••
• • • • •• • • •• •→ ∧ ∧
• •
(3) +y ÁŒ‡ÊÊ ◊¥ ø‹ ⁄U„Ë flÒlÈÃ-øÈê’∑§Ëÿ Ã⁄¥Uª ∑§
Á‹ÿ •• • • •• • •••
• • • • • •• •→ ∧
• ,
•• • • •• • •••
B B x, t zyz
→ ∧
•
18. i|ÞephL$pidp„ âkfsp rhÛysQy„bL$ue sf„N_¡ Ýep_dp„gp¡. _uQ¡dp„\u kpQy„ rh^p_ `k„v$ L$fp¡.
(1) +x -qv$ipdp„ Nrs L$fsp rhÛysQy„bL$ue sf„NdpV¡$ rhÛys n¡Ó
( )•
•• • •• • •• • ••
• • • • • • •• •→ ∧ ∧
= − A_¡
Qy„bL$ue n¡Ó
( )• • •• • •• • ••
B B x, t y zyz
→ ∧ ∧
•• • • \i¡.
(2) +x -qv$ipdp„ Nrs L$fsp rhÛysQy„bL$ue sf„NdpV¡$ rhÛys n¡Ó
( )•• • • •• • • •• • ••
• • • • •• • • •• •→ ∧ ∧
• • A_¡
Qy„bL$ue n¡Ó
( )•• • • •• • • •• • ••
• • • • •• • • •• •→ ∧ ∧
• • \i¡.
(3) +y - qv$ipdp„ Nrs L$fsp rhÛysQy„bL$ue sf„N
dpV¡$ rhÛys n¡Ó •• • • •• • •••
• • • • • •• •→ ∧
•
A_¡ Qy„bL$ue n¡Ó •• • • •• • •••
B B x, t zyz
→ ∧
•
\i¡.
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 16
(4) For an electromagnetic wave
propagating in +y direction the
electric field is •• • • •• • •••
• • • • • •• •→ ∧
•
and the magnetic field is
•• • • •• • •••
B B x, t yz
→ ∧
•
19. A hemispherical glass body of radius
10 cm and refractive index 1.5 is silvered
on its curved surface. A small air bubble is
6 cm below the flat surface inside it along
the axis. The position of the image of the
air bubble made by the mirror is seen :
(1) 14 cm below flat surface
(2) 30 cm below flat surface
(3) 20 cm below flat surface
(4) 16 cm below flat surface
(4) +y ÁŒ‡ÊÊ ◊¥ ø‹ ⁄U„Ë flÒlÈÃ-øÈê’∑§Ëÿ Ã⁄¥Uª ∑§
Á‹ÿ •• • • •• • •••
• • • • • •• •→ ∧
• ,
•• • • •• • •••
B B x, t yz
→ ∧
•
19. ∞∑§ ∑§Ê°ø ∑§ •h¸ªÊ‹Ëÿ ∆UÊ‚ ∑§Ë ÁòÊíÿÊ 10 cm ÃÕÊ•¬fløŸÊ¥∑§ 1.5 „Ò– ©‚∑§Ë fl∑˝§Ëÿ ‚Ä ¬⁄U øÊ°ŒË ∑§Ë¬⁄Uà ø…∏Ê߸ ªß¸ „Ò– ‚◊Ë ¬Îc∆U ∑§ 6 cm ŸËø ÃÕÊ•ˇÊ ¬⁄U, ∞∑§ ‚Í ◊ „flÊ ∑§Ê ’È‹’È‹Ê ÁSÕà „Ò– Ã’fl∑˝§Ëÿ-Œ¬¸áÊ ‚ ’Ÿ ⁄U„ ’È‹’È‹ ∑§Ë ¬˝ÁÃÁ’ê’ ŒÍ⁄UË „Ò —
(1) ‚◊Ë ‚Ä ‚ 14 cm ŸËø
(2) ‚◊Ë ‚Ä ‚ 30 cm ŸËø
(3) ‚◊Ë ‚Ä ‚ 20 cm ŸËø
(4) ‚◊Ë ‚Ä ‚ 16 cm ŸËø
(4) +y -qv$ipdp„ Nrs L$fsp rhÛysQy„bL$ue sf„N
dpV¡ $ rhÛysn¡Ó •• • • •• • •••
• • • • • •• •→ ∧
•
A_¡ Qy„bL$ue n¡Ó •• • • •• • •••
B B x, t yz
→ ∧
•
\i¡.
19. 10 cm _u rÓÄep fphsu A_¡ 1.5 h¾$uch_p„L$ fphspÁgpk_p A¡L$ A ®Np¡mpL$pf cpN_u h¾$ k`pV$u f Qp„v$u_p¡Y$p¡m QY$phhpdp„ Aph¡g R>¡. s¡_u ku^u k`pV$u\u 6 cm
_uQ¡, A ®Np¡mpL$pf_u A„v$f_p cpNdp„, A¡L$ lhp_p¡ _p_p¡`f`p¡V$p¡ s¡_u An `f fl¡g R>¡. Afukp Üpfp lhp_p`f`p¡V$p_p„ ârstbb_y„ õ\p_ _________ v$¡Mpi¡.
(1) ku^u k`pV$u_u _uQ¡ 14 cm A„sf¡
(2) ku^u k`pV$u_u _uQ¡ 30 cm A„sf¡
(3) ku^¡ k`pV$u_u _uQ¡ 20 cm A„sf¡
(4) ku^u k`pV$u_u _uQ¡ 16 cm A„sf¡
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 17
20. Two stars are 10 light years away from the
earth. They are seen through a telescope
of objective diameter 30 cm. The
wavelength of light is 600 nm. To see the
stars just resolved by the telescope, the
minimum distance between them should
be (1 light year=9.46× 1015
m) of the order
of :
(1) 106 km
(2) 108 km
(3) 1011
km
(4) 1010
km
21. A photoelectric surface is illuminated
successively by monochromatic light of
wavelengths λ and ••
. If the maximum
kinetic energy of the emitted
photoelectrons in the second case is
3 times that in the first case, the work
function of the surface is :
(1) •• ••
(2) •• ••
(3)
• ••
(4)
• •• ••
20. ŒÊ ÃÊ⁄U ¬ÎâflË ‚ 10 ¬˝∑§Ê‡Ê-fl·¸ ∑§Ë ŒÍ⁄UË ¬⁄U „Ò¥– ©Ÿ∑§Ê∞∑§ ≈UÁ‹S∑§Ê¬ mÊ⁄UÊ ŒπÊ ¡ÊÃÊ „Ò, Á¡‚∑§Ê •Á÷ŒÎ‡ÿ∑§30 cm √ÿÊ‚ ∑§Ê „Ò– ¬˝∑§Ê‡Ê ∑§Ë Ã⁄¥UªŒÒÉÿ¸ 600 nm
„Ò– (1 ¬˝∑§Ê‡Ê-fl·¸ =9.46× 1015
m) „Ò– ≈UÁ‹S∑§Ê¬•ª⁄U ©Ÿ ÃÊ⁄UÊ¥ ∑§Ê ‹ª÷ª Áfl÷ÁŒÃ Œπ ¬Ê ⁄U„Ê „Ò, Ã’©Ÿ∑§ ’Ëø ∑§Ë ŒÍ⁄UË ∑§Ê order „Ò —(1) 10
6 km
(2) 108 km
(3) 1011
km
(4) 1010
km
21. ∞∑§ ¬˝∑§Ê‡Ê-flÒlÈà ‚Ä ¬⁄U ¬„‹Ë ’Ê⁄U λ ÃÕÊ ŒÍ‚⁄UË
’Ê⁄U ••
Ã⁄¥UªŒÒÉÿ¸ ∑§Ê ¬˝∑§Ê‡Ê «UÊ‹Ê ¡ÊÃÊ „Ò– ÿÁŒ
©à‚Á¡¸Ã ¬˝∑§Ê‡Ê-ß‹Ä≈˛UÊÚŸ ∑§Ë •Áœ∑§Ã◊ ªÁá-™§¡Ê¸ŒÍ‚⁄UË ’Ê⁄U ◊¥ ¬„‹Ë ’Ê⁄U ∑§Ë ÁÃªÈŸË „Ê, Ã’ ©‚ ‚Ä∑§Ê ∑§Êÿ¸-»§‹Ÿ „Ò —
(1) •• ••
(2) •• ••
(3)
• ••
(4)
• •• ••
20. b¡ spfpAp¡ `©Õhu\u 10 âL$pihj® v|$f R>¡. s¡Ap¡_¡30 cm ìepk fphsp Ap¡bS>¡¼¹$V$uh (hõsy-g¡Þk) hpmpV¡$guõL$p¡`\u Å¡hpdp„ Aph¡ R>¡. âL$pi_u sf„Ng„bpC600 nm R>¡. Ap b¡ spfp_¡ just R|>V$p `X¡ $gp(rhc¡qv$s \e¡gp) Å¡hp dpV¡$ s¡d_u hÃQ¡_y„ A„sf_________ _p ¾$d_y„ lp¡hy„ Å¡CA¡.
(1 âL$pi hj®=9.46× 1015
m)
(1) 106 km
(2) 108 km
(3) 1011
km
(4) 1010
km
21. A¡L$ ap¡V$p¡Cg¡¼V²$uL$ k`pV$u_¡ hpfpasu A_y¾$d¡ λ A_¡ ••
sf„Ng„bpC ^fphsp A¡L$f„Nu âL$pi\u âL$pris L$fhpdp„Aph¡ R>¡. buÅ qL$õkpdp„ Å¡ DÐkÅ®sp ap¡V$p¡Cg¡¼V²$p¡__udlÑd Nrs EÅ® â\d qL$õkp L$fsp„ ÓZ NZu dmsulp¡e sp¡ k`pV$u_y„ hL®$a„¼i_ _________ \i¡.
(1) •• ••
(2) •• ••
(3)
• ••
(4)
• •• ••
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 18
22. ‘v’ S>¡V$gu TX$`\u Nrs L$fsp¡ ÞeyV²$p¡_ A¡L$ [õ\f A_¡^fp[õ\rsdp„ fl¡gp lpCX²$p¡S>_ `fdpÏ kp\¡ ku^uA\X$pdZ A_ych¡ R>¡. ÞeyV²$p¡__u A[õ\rsõ\p`L$A\X $ pdZ \pe s ¡ dpV ¡ $_u dlÑd NrsEÅ®__________ \i¡.(1) 10.2 eV
(2) 16.8 eV
(3) 12.1 eV
(4) 20.4 eV
23. ApL©$rÑdp„ v$ip®h¡g qf`\ dpV¡$ ApDV$ yV$ 1 dm¡ s¡ dpV¡$S>ê$fu C_`yV$ __________ \i¡.
(1) a=0, b=1, c=0
(2) a=1, b=0, c=0
(3) a=1, b=0, c=1
(4) a=0, b=0, c=1
22. A neutron moving with a speed ‘v’ makes
a head on collision with a stationary
hydrogen atom in ground state. The
minimum kinetic energy of the neutron for
which inelastic collision will take place is :
(1) 10.2 eV
(2) 16.8 eV
(3) 12.1 eV
(4) 20.4 eV
23. To get an output of 1 from the circuit shown
in figure the input must be :
(1) a=0, b=1, c=0
(2) a=1, b=0, c=0
(3) a=1, b=0, c=1
(4) a=0, b=0, c=1
22. ªÁà ‘v’ ‚ ø‹ÃÊ „È•Ê ∞∑§ ãÿÍ≈˛UÊÚŸ ∞∑§ ÁSÕ⁄U „Ê߸«˛UÊ¡Ÿ¬⁄U◊ÊáÊÈ, ¡Ê •¬ŸË •Êl-•flSÕÊ ◊¥ „Ò, ‚ ‚ê◊Èπ≈UÄ∑§⁄U ∑§⁄UÃÊ „Ò– ãÿÈ≈˛UÊÚŸ ∑§Ë fl„ ãÿÍŸÃ◊ ªÁá ™§¡Ê¸’ÃÊÿ¥ Á¡‚ ∑§ „ÊŸ ¬⁄U ÿ„ ≈UÄ∑§⁄U •¬˝àÿÊSÕ „ÊªË —(1) 10.2 eV
(2) 16.8 eV
(3) 12.1 eV
(4) 20.4 eV
23. ÁŒÿ ªÿ ¬Á⁄U¬Õ ‚ 1 ÁŸª◊ ¬Êåà ∑§⁄UŸ ∑§ Á‹ÿ •Êfl‡ÿ∑§ÁŸfl‡Ê „ÊŸÊ øÊÁ„ÿ —
(1) a=0, b=1, c=0
(2) a=1, b=0, c=0
(3) a=1, b=0, c=1
(4) a=0, b=0, c=1
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 19
24. A modulated signal Cm
(t) has the form
Cm
(t)=30 sin 300πt+10 (cos 200πt
−cos 400πt). The carrier frequency fc, the
modulating frequency (message frequency)
fω
, and the modulation index µ are
respectively given by :
(1) fc=200 Hz ; f
ω=50 Hz ;
•• ••
• •
(2) fc=150 Hz ; f
ω=50 Hz ;
•• ••
• •
(3) fc=150 Hz ; f
ω=30 Hz ;
•• ••
• •
(4) fc=200 Hz ; f
ω=30 Hz ;
•• ••
• •
25. A particle of mass m is acted upon by a force
F given by the empirical law •• • ••• • ••••
• .
If this law is to be tested experimentally by
observing the motion starting from rest, the
best way is to plot :
(1) v(t) against t2
(2) log v(t) against •••
(3) log v(t) against t
(4) log v(t) against
••
24. Cm
(t)=30 sin 300πt+10 (cos 200πt
−cos 400πt) ∞∑§ ◊Ê«ÈUÁ‹Ã Á‚ÇŸ‹ ∑§Ê Œ‡ÊʸÃÊ „Ò–Ã’ flÊ„∑§ •ÊflÎÁûÊ f
c, ◊Ê«ÈU‹∑§ •ÊflÎÁûÊ f
ω ÃÕÊ ◊Ê«ÈU‹∑§
ߟ«UÄ‚ µ ∑˝§◊‡Ê— „Ò¥ —
(1) fc=200 Hz ; f
ω=50 Hz ;
•• ••
• •
(2) fc=150 Hz ; f
ω=50 Hz ;
•• ••
• •
(3) fc=150 Hz ; f
ω=30 Hz ;
•• ••
• •
(4) fc=200 Hz ; f
ω=30 Hz ;
•• ••
• •
25. m Œ˝√ÿ◊ÊŸ ∑§ ∑§áÊ ¬⁄U F ’‹ ‹ª ⁄U„Ê „Ò, •ÊÒ⁄U ©‚∑§
Á‹ÿ •ÊŸÈ÷Áfl∑§ ‚ê’¥œ „Ò •• • ••• • ••••
• ß‚ ‚ê’¥œ
∑§ ‚àÿʬŸ ∑§ Á‹∞ ÁSÕ⁄U •flSÕÊ ‚ ∑§áÊ ∑§Ë ªÁà ∑§Ê¬ ÊáÊ (Observation) ∑§⁄U ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ‚Ê ª˝Ê»§ ‚flÊûÊ◊ „ʪÊ?
(1) t2 ∑§ ÁflL§h v(t)
(2) ••• ∑§ ÁflL§h log v(t)
(3) t ∑§ ÁflL§h log v(t)
(4)
•• ∑§ ÁflL§h log v(t)
24. dp¡X$éygf (Ar^rdrîs) \e¡gp rkÁ_g Cm
(t) _uQ¡dyS>b Ap`u iL$pe R>¡. C
m(t)=30 sin 300πt+10
(cos 200πt−cos 400πt) L¡$fuef Aph©rÑ fc
, k„v$¡ip(dp¡X$éyg¡V$]N) Aph©rÑ f
ω, A_¡ dp¡X$éyg¡i_ A„L$
(index) µ, A_y¾$d¡ ______ hX¡$ Ap`u iL$pe.
(1) fc=200 Hz ; f
ω=50 Hz ;
•• ••
• •
(2) fc=150 Hz ; f
ω=50 Hz ;
•• ••
• •
(3) fc=150 Hz ; f
ω=30 Hz ;
•• ••
• •
(4) fc=200 Hz ; f
ω=30 Hz ;
•• ••
• •
25. m v$m ^fphsp¡ L$Z •• • ••• • ••••
• _¡ A_ykfsp
F bm_u Akf l¡W$m R>¡. Å¡ Ap bm-r_ed_¡ âpep¡rNL$fus¡, L$Z_u Nrs [õ\f [õ\rsdp„\u iê$ \pe s¡ fus¡QL$pkhp¡ lp¡e sp¡ __________ _p¡ N°pa A¡ kp¥\u kpfp¡rhL$ë` \i¡.
(1) v(t) rhfyÝ^ t2
(2) log v(t) rhfyÝ^ •••
(3) log v(t) rhfyÝ^ t
(4) log v(t) rhfyÝ^ ••
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 20
26. A thin 1 m long rod has a radius of 5 mm.
A force of 50 πkN is applied at one end to
determine its Young’s modulus. Assume
that the force is exactly known. If the least
count in the measurement of all lengths is
0.01 mm, which of the following statements
is false ?
(1)
•••
gets minimum contribution
from the uncertainty in the length.
(2) The figure of merit is the largest for
the length of the rod.
(3) The maximum value of Y that can be
determined is 1014
N/m2.
(4)
•••
gets its maximum contribution
from the uncertainty in strain.
26. 1 m ‹ê’Ë ¬Ã‹Ë ¿U«∏ ∑§Ë ÁòÊíÿÊ 5 mm „Ò– ÿ¥ª◊Ê«U‹‚ ÁŸ∑§Ê‹Ÿ ∑§ Á‹ÿ ß‚ ∑§ Á‚⁄U ¬⁄U 50 πkN
∑§Ê ’‹ ‹ªÊÿÊ ªÿÊ– ◊ÊŸ¥ Á∑§ ’‹ Á’‹∑ȧ‹ ∆UË∑§ ‚ôÊÊà „Ò– ÿÁŒ ‹ê’ÊßÿÊ ¥ ∑§ ◊ʬŸ ∑§ •À¬Ê¥‡Ê0.01 mm „Ò¥– Ã’ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ª‹Ã
„Ò?
(1)
•••
◊¥ ‹ê’Ê߸ ∑§Ë •ÁŸÁ‡øÃÃÊ ∑§Ê ÿʪŒÊŸ
ãÿÍŸÃ◊ „Ò–
(2) ¿U«∏ ∑§Ë ‹ê’Ê߸ ∑§ Á‹ÿ ŒˇÊÃÊ¥∑§ ‚’‚ ’«∏Ê „Ò–
(3) Y ∑§Ê •Áœ∑§Ã◊ ¬˝Ê# „Ê ‚∑§Ÿ flÊ‹Ê ◊ÊŸ10
14 N/m
2.
(4)
•••
◊¥ Áfl∑ΧÁà ∑§Ë •ÁŸÁ‡øÃÃÊ ∑§Ê ÿʪŒÊŸ
•Áœ∑§Ã◊ „Ò–
26. 1 m gp„bp A_¡ 5 mm rÓÄep ^fphsp¡ A¡L$ `psmp¡krmep ¡ R> ¡ . s ¡_p ¡ ×Y $ [õ\rsõ\p`L $sp A„L $(e„N dp¡X$éygk) ip¡^hp dpV¡$ s¡_p A¡L$ R>¡X$p D`f50 πkN S>¡V$gy„ bm gNpX$hpdp„ Aph¡ R>¡. A¡hy„ ^pfp¡ L¡$Ap bm_y„ d|ëe kQp¡V$ fus¡ dpg|d R>¡. Å¡ g„bpC_pb^p S> dp`_dp„ gOyÑd dp` i[¼s 0.01 mm lp¡esp¡ _uQ¡ ¥L$u_y„ L$ey„ rh^p_ Mp¡Vy„$ li¡ ?
(1) g„bpC_u AQp¡½$kpC_¡ L$pfZ¡ •••
dp„ gOyÑd
apmp¡ Aphsp¡ li¡.
(2) krmep_u g„bpC dpV¡$ v$nsp„L$ (figure of
merit) kp¥\u h^pf¡ \i¡.
(3) d¡mhu iL$pe s¡hu Y _u dlÑd qL„ $ds10
14 N/m
2 \i¡.
(4)
•••
dp „ dlÑd apmp ¡ rhL © $ rsdp „ fl¡g
AQp¡½$kpC_¡ L$pfZ¡ li¡.
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 21
27. N¡ëh¡_p¡duV$fdp„ 50 L$p`p R>¡. b¡V$fu_p¡ Ap„sqfL$ Ahfp¡i|Þe R>¡. Äepf¡ =2400 Ω lp¡e R>¡ Ðepf¡ N¡ëh¡_p¡duV$fdp„ 40 L$p`p (divisions) ky u_y„ Aphs®_ dm¡ R>¡.Äepf¡ Ahfp¡ ¡V$udp„\u 4900 Ω S>¡V$gp¡ L$pY$hpdp„ Aph¡R>¡ Ðepf¡ 20 divisions S>¡V$gy„ Aphs®_ \pe R>¡. sp¡Ap`Z¡ spfZ Ap`u iL$uA¡ L¡$ __________.
(1) N¡ëh¡_p¡duV$f_p¡ Ahfp¡ 200 Ω li¡.
(2) |Z®-õL¡$g Aphs®_ dpV¡$ âhpl 2 mA li¡.
(3) N¡ëh ¡_p ¡duV $f_u ` °hpl k „h ¡ qv $sp20 µA/division \i¡.
(4) 10 divisions S>¡V$gy„ Aphs®_ d¡mhhp dpV¡$Ahfp¡ ¡V$udp„ Ahfp¡ _y„ d|ëe 9800 Ω Å¡Ci¡.
28. V²$ph¡g]N dpC¾$p¡õL$p¡`_u dv$v$\u Ágpk_p Qp¡kgp_p¡h¾$uch_p„L$ dp`hp dpV¡$ Ap¡R>pdp„ Ap¡R>p S>ê$fuAhgp¡L$_p¡_u k„¿ep __________ li¡.
(1) b¡
(2) ÓZ
(3) Qpf
(4) `p„Q
27. ∞∑§ ªÀflŸÊ◊Ë≈U⁄U ∑§Ë S∑§‹ 50 ÷ʪʥ ◊¥ ’¥≈UË „Ò– ’Ò≈U⁄UË∑§Ê •Ê ¥ÃÁ⁄ U∑§ ¬ ˝ ÁÃ⁄ U Ê œ ‡Ê Í ãÿ „ Ò– ÿÁŒR=2400 Ω „Ò ÃÊ ÁflˇÊ¬ =40 ÷ʪ „Ò– ÿÁŒR=4900 Ω „Ò ÃÊ ÁflˇÊ¬ =20 ÷ʪ „Ò– Ã’ „◊ÁŸœÊ¸Á⁄Uà ∑§⁄U ‚∑§Ã „Ò¥ Á∑§ —
(1) ªÀflŸÊ◊Ë≈U⁄U ∑§Ê ¬˝ÁÃ⁄UÊœ 200 Ω „Ò–
(2) »È§‹-S∑§‹ ÁflˇÊ¬ ∑§ Á‹ÿ œÊ⁄UÊ 2 mA „Ò–
(3) ªÀflŸÊ◊Ë≈U⁄U ∑§Ë œÊ⁄UÊ-‚¥flŒŸ‡ÊË‹ÃÊ 20 µA
¬˝Áà ÷ʪ „Ò–
(4) ÁflˇÊ¬ =10 ÷ʪ ∑§ Á‹ÿ R=9800 Ω.
28. ∑§Ê°ø ∑§Ë S‹Ò’ ∑§Ê ø‹-◊Ê߸∑˝§ÊS∑§Ê¬ mÊ⁄UÊ •¬fløŸÊ¥∑§ÁŸ∑§Ê‹Ÿ ∑§ Á‹ÿ ¡M§⁄UË ¬Ê∆˜ÿÊ¥∑§Ê¥ ∑§Ë ãÿÍŸÃ◊ ‚¥ÅÿÊ„Ò —
(1) ŒÊ
(2) ÃËŸ
(3) øÊ⁄U
(4) ¬Ê°ø
27. A galvanometer has a 50 division scale.
Battery has no internal resistance. It is
found that there is deflection of 40 divisions
when R=2400 Ω. Deflection becomes
20 divisions when resistance taken from
resistance box is 4900 Ω. Then we can
conclude :
(1) Resistance of galvanometer is 200 Ω.
(2) Full scale deflection current is 2 mA.
(3) Current sensitivity of galvanometer
is 20 µA/division.
(4) Resistance required on R.B. for a
deflection of 10 divisions is 9800 Ω.
28. To determine refractive index of glass slab
using a travelling microscope, minimum
number of readings required are :
(1) Two
(2) Three
(3) Four
(4) Five
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 22
29. A¡L$ V²$p[ÞTõV$f_p common emitter (CE) k„fQ_pdp„C_ yV$ gpnrZL$sp dpV¡$ C_ yV$ Ahfp¡ _p ìeõs_p¡ kp¥\uh^y hpõsrhL$ N°pa __________ \i¡.
(1)
(2)
(3)
29. ∞∑§ ≈˛ U Ê ¥ Á¡S≈U⁄ U ∑§Ê common-emitter (CE)
•Á÷ÁflãÿÊ‚ ◊¥ ÁŸfl‡Ê-•Á÷‹ÊˇÊÁáÊ∑§ ◊ʬŸ Á∑§ÿÊ ªÿÊ–Ã’ ÁŸfl‡Ê-¬˝ÁÃ⁄UÊœ ∑§ √ÿÈà∑˝§◊ ∑§Ê ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚Ê ª˝Ê»§ ©Áøà „Ò?
(1)
(2)
(3)
29. A realistic graph depicting the variation of
the reciprocal of input resistance in an input
characteristics measurement in a common-
emitter transistor configuration is :
(1)
(2)
(3)
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 23
(4)
30. V²$p[ÞTõV$f_u C_`yV$ A_¡ ApDV$`yV$ gpnrZL$sp_pdp`_dp„ ApDV$ yV$ Ahfp¡ r
0 A_¡ C_ yV$ Ahfp¡ r
i
_p¡ NyZp¡Ñf (R) __________ S>¡V$gu f¡ÞS>dp„ lp¡e R>¡.(1) R~10
2−10
3
(2) R~1−10
(3) R~0.1−0.01
(4) R~0.1−1.0
31. 0.1N X$peb¡rTL$ A¡rkX$_y„ L$v$ iy„ li¡ L¡$ S>¡ 1 g b¡CT_psV$õ\uL$fZ L$fhp dpV¡$ ep®á lp¡e L¡$ S>¡_p S>gue ÖphZdp„0.04 dp¡g OH
− Aph¡gp R>¡ ?(1) 200 mL
(2) 400 mL
(3) 600 mL
(4) 800 mL
(4)
30. Á∑§‚Ë ≈˛UÊ¥Á¡S≈U⁄U ∑§Ë ÁŸfl‡Ê-ÁŸª¸◊ •Á÷‹ÊˇÊÁáÊ∑§◊ʬŸ ∑§ Á‹ÿ ¬˝ÿÈÄà ÁŸª¸◊-¬˝ÁÃ⁄UÊ œ (r
0) fl
ÁŸfl‡Ê-¬˝ÁÃ⁄UÊœ (ri) ∑§ •ŸÈ¬Êà (R) ∑§Ê •ÊÿÊ◊
(range) „ʪÊ?(1) R~10
2−10
3
(2) R~1−10
(3) R~0.1−0.01
(4) R~0.1−1.0
31. 0.1N ÁmˇÊÊ⁄UËÿ •ê‹ ∑§Ê •Êÿß ÄÿÊ „ÊªÊ ¡Ê 1 ª˝Ê◊ˇÊÊ⁄U∑§ Á¡‚∑§ ¡‹Ëÿ Áfl‹ÿŸ ◊¥ 0.04 ◊Ê‹ OH
− „Ò∑§Ê ©ŒÊ‚ËŸ ∑§⁄UŸ ∑§ Á‹ÿ ¬ÿʸ# „Ò?(1) 200 mL
(2) 400 mL
(3) 600 mL
(4) 800 mL
(4)
30. The ratio (R) of output resistance r0, and
the input resistance ri in measurements of
input and output characteristics of a
transistor is typically in the range :
(1) R~102−10
3
(2) R~1−10
(3) R~0.1−0.01
(4) R~0.1−1.0
31. The volume of 0.1N dibasic acid sufficient
to neutralize 1 g of a base that furnishes 0.04
mole of OH−
in aqueous solution is :
(1) 200 mL
(2) 400 mL
(3) 600 mL
(4) 800 mL
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 24
32. r_ròs sp`dp_ `f, iê$Apsdp„, N2 AÏAp¡_p¡
kf¡fpi hN®d|m (root mean square) (rms) h¡N u
R>¡. Å¡ Ap sp`dp_ bdÏ„ (Doubled) L$fhpdp„ Aph¡sp¡ b^p S> _pCV²$p¡S>_ AÏAp¡_y„ rhep¡S>_ _pCV²$p¡S>_`fdpÏAp¡dp„ \pe sp¡, R>u _hy„ kf¡fpi hN®d|m (rms)
h¡N iy„ \i¡ ?(1) u/2
(2) 2u
(3) 4u
(4) 14u
33. Cg¡¼V²$p¡r_L$ rhÞepk 1s22s
22p
63s
23p
6 ^fphsp L$epnpf_y„ S>gue ÖphZ Ape_p¡ ^fphsp„ _\u ?(1) NaF
(2) NaCl
(3) KBr
(4) CaI2
34. L$ep k„ep¡S>_dp„ H-X-H b„ M|Zp¡ khp®r^L$ R>¡ ?(1) CH
4
(2) NH3
(3) H2O
(4) PH3
32. ∞∑§ Áfl‡Ê· Ãʬ ¬⁄U ¬˝Ê⁄Uê÷ ◊¥ ŸÊß≈˛UÊ¡Ÿ •áÊÈ•Ê¥ (N2)
∑§Ê flª¸ ◊Êäÿ ◊Í‹ flª u „Ò– ÿÁŒ ß‚ Ãʬ ∑§Ê ŒÈªÈŸÊ∑§⁄U ÁŒÿÊ ¡Êÿ •ÊÒ⁄U ‚÷Ë ŸÊß≈˛UÊ¡Ÿ •áÊÈ ÁflÿÊÁ¡Ã„Ê∑§⁄U ŸÊß≈˛UÊ¡Ÿ ¬⁄U◊ÊáÊÈ ’Ÿ ¡Ê∞ ÃÊ ŸÿÊ flª¸ ◊Êäÿ◊Í‹ flª „ÊªÊ —(1) u/2
(2) 2u
(3) 4u
(4) 14u
33. Á∑§‚ ‹fláÊ ∑§ ¡‹Ëÿ Áfl‹ÿŸ ◊¥ 1s22s
22p
63s
23p
6
ß‹Ä≈˛UÊÚÁŸ∑§ ÁflãÿÊ‚ ∑§ •ÊÿŸ Ÿ„Ë¥ „Ê¥ª?(1) NaF
(2) NaCl
(3) KBr
(4) CaI2
34. Á∑§‚ ÿÊÒÁª∑§ ◊¥ H-X-H •Ê’㜠∑§ÊáÊ ‚flʸÁœ∑§ „Ò?(1) CH
4
(2) NH3
(3) H2O
(4) PH3
32. Initially, the root mean square (rms) velocity
of N2
molecules at certain temperature is u.
If this temperature is doubled and all the
nitrogen molecules dissociate into nitrogen
atoms, then the new rms velocity will be :
(1) u/2
(2) 2u
(3) 4u
(4) 14u
33. Aqueous solution of which salt will not
contain ions with the electronic
configuration 1s22s
22p
63s
23p
6 ?
(1) NaF
(2) NaCl
(3) KBr
(4) CaI2
34. The bond angle H-X-H is the greatest in the
compound :
(1) CH4
(2) NH3
(3) H2O
(4) PH3
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Set - 03 25
35. Å¡ H2O
2 _p 100 dp¡g 1 bpf (bar) A_¡ 300 K f
rhOV$us \pe sp¡, 1 bpf (bar) v$bpZ_p rhfyÝ^ 1
dp¡g O2(g) _p rhõsfZ \hp_¡ gu ¡ \e¡g L$pe® (kJ)
ip¡^p¡.
• • • •• • • ••• • • • •••• •• •• •••
(R = 8.3 J K−1
mol−1
)
(1) 62.25
(2) 124.50
(3) 249.00
(4) 498.00
36. A¡L$ r_ròs sp`dp_ `f, npf MX2 _p S>gue
ÖphZ_p¡ hpÞV$-lp¡a Aheh 2 R>¡. sp¡ npf_p ApÖphZ_p¡ rhep¡S>_A„i ip¡ p¡.(1) 0.33
(2) 0.50
(3) 0.67
(4) 0.80
37. A¡L$ b„ (sealed) r_hp®rss (evacuated) `pÓdp„fpMhpdp„ Aph¡g O_ (solid) XY rhOV$us \C_¡ T
sp`dp_¡ hpeyAp¡_y„ rdîZ X A_¡ Y b_¡ R>¡. Ap pÓdp„k„syg_ v$bpZ 10 bar (bpf) R>¡. Ap âq¾$ep dpV¡$ K
p
iy„ R>¡ ?(1) 5
(2) 10
(3) 25
(4) 100
35. ÿÁŒ H2O
2 ∑§ 100 ◊Ê‹ 1 bar ÃÕÊ 300 K ¬⁄U
ÁflÿÊÁ¡Ã „Ê ÃÊ 1 bar ŒÊ’ ∑§ ÁflL§h 1 ◊Ê‹ •ÊÚÄ‚Ë¡Ÿ∑§ ÁflSÃÊÁ⁄Uà „ÊŸ ¬⁄U Á∑§ÿÊ „È•Ê ∑§Êÿ¸ (kJ ◊¥) „ÊªÊ —
• • • •• • • ••• • • • •••• •• •• •••
(R = 8.3 J K−1
mol−1)
(1) 62.25
(2) 124.50
(3) 249.00
(4) 498.00
36. Á∑§‚Ë Áfl‡Ê· Ãʬ ¬⁄U, ∞∑§ ‹fláÊ MX2
∑§ ¡‹ËÿÁfl‹ÿŸ ∑§Ê flÊã≈U •ÊÚ»§ »Ò§Ä≈U⁄U 2 „Ò– ‹fláÊ ∑§ ß‚Áfl‹ÿŸ ∑§ Á‹∞ ÁflÿÊ¡Ÿ ◊ÊòÊÊ „ÊªË —(1) 0.33
(2) 0.50
(3) 0.67
(4) 0.80
37. ∞∑§ ’¥Œ (‚ËÀ«U) ÁŸflʸÁÃà ¬ÊòÊ ◊¥ ⁄UπÊ ªÿÊ ∆UÊ‚ XY
ÁflÉÊÁ≈Uà „Ê∑§⁄U Ãʬ T ¬⁄U ŒÊ ªÒ‚¥ X ÃÕÊ Y ∑§Ê Á◊üÊáÊ’ŸÊÃÊ „Ò– ß‚ ¬ÊòÊ ◊¥ ‚Êêÿ ŒÊ’ 10 bar „Ò– ß‚•Á÷Á∑˝§ÿÊ ∑§ Á‹ÿ K
p „ÊªÊ —
(1) 5
(2) 10
(3) 25
(4) 100
35. If 100 mole of H2O
2 decompose at 1 bar and
300 K, the work done (kJ) by one mole of
O2(g) as it expands against 1 bar pressure
is :
• • • •• • • ••• • • • •••• •• •• •••(R = 8.3 J K
−1 mol
−1)
(1) 62.25
(2) 124.50
(3) 249.00
(4) 498.00
36. An aqueous solution of a salt MX2
at certain
temperature has a van’t Hoff factor of 2.
The degree of dissociation for this solution
of the salt is :
(1) 0.33
(2) 0.50
(3) 0.67
(4) 0.80
37. A solid XY kept in an evacuated sealed
container undergoes decomposition to
form a mixture of gases X and Y at
temperature T. The equilibrium pressure
is 10 bar in this vessel. Kp
for this reaction
is :
(1) 5
(2) 10
(3) 25
(4) 100
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Set - 03 26
38. k[¼k_¡V$ Ape__p Ap¡[¼kX¡$i_\u Cr\gu_ A_¡ L$pb®_X$pep¡¼kpCX$ hpeyAp¡ DÐ`Þ_ \pe R>¡. `p¡V¡$riedk[¼k_¡V$_p S>gue ÖphZdp„ 0.2 a¡fpX¡$ rhÛysâhpl kpfL$fsp, STP `f (1 atm A_¡ 273 K) A¡ hpey_p¡ _y„Ly$g L$v$ (L¡$\p¡X$ A_¡ A¡_p¡X$ b„_¡ `f) ip¡ p¡.(1) 2.24 L
(2) 4.48 L
(3) 6.72 L
(4) 8.96 L
39. _uQ¡ Ap ¡g âq¾$ep dpV¡$_p¡ h¡N-r_ed k [A][B] ìe„S>L$\u ìe¼s (expression) L$f¡g R>¡.
A+B _u`S>
A _u kp„Ösp 0.1 mole (dp¡g) fpMuA¡ A_¡ Å¡ B _ukp„Ösp 0.1 \u h^pfu_¡ 0.3 dp¡g L$fhpdp„ Aph¡ sp¡h¡NAQmp„L$ iy„ li¡ ?(1) k
(2) k/3
(3) 3k
(4) 9k
40. L¡$V$gpL$ L$rggp¡_p õhZp¯L$ (Np¡ëX$ _„bf) Ap âdpZ¡ R>¡.Æg¡V$u_ (Gelatin) : 0.005 - 0.01, Nd Af¡rbL$(Gum Arabic) : 0.15 - 0.25; Ap ¡ rgA¡V $ $(Oleate) : 0.04 - 1.0; õV$pQ® (Starch) : 15 - 25.
Apdp„\u L$ep¡ kp¥\u h^pfp¡ kpfp¡ frns L$rgg R>¡ ?
(1) Æg¡V$u_ (Gelatin)
(2) Nd Af¡rbL$ (Gum Arabic)
(3) Ap¡rgA¡V$ (Oleate)
(4) õV$pQ® (Starch)
38. ‚ÁÄ‚Ÿ≈U •ÊÿŸ ∑§ •ÊÚÄ‚Ë∑§⁄UáÊ ‚ ∞ÁÕ‹ËŸ ÃÕÊ ∑§Ê’Ÿ«UÊß•ÊÚÄ‚Êß«U ªÒ‚¥ ’ŸÃË „Ò¥– ¬Ê≈Ò˜UÁ‡Êÿ◊ ‚ÁÄ‚Ÿ≈U ∑§¡‹Ëÿ Áfl‹ÿŸ ‚ 0.2 »Ò§⁄UÊ«U ÁfllÈà ¬˝flÊÁ„à ∑§⁄UŸ ¬⁄UªÒ‚Ê¥ ∑§Ê ∑ȧ‹ •Êÿß (∑Ò§ÕÊ«U ÃÕÊ ∞ŸÊ«U ŒÊŸÊ¥ ¬⁄U)STP (1 atm ÃÕÊ 273 K) ¬⁄U „ÊªÊ —(1) 2.24 L
(2) 4.48 L
(3) 6.72 L
(4) 8.96 L
39. ŸËø ŒË ªß¸ •Á÷Á∑˝§ÿÊ ∑§ Á‹∞ Œ⁄U ÁŸÿ◊ k [A][B]
√ÿ¥¡∑§ ‚ √ÿÄà Á∑§ÿÊ ¡ÊÃÊ „Ò
A+B ©à¬ÊŒ
A ∑§Ë ‚ÊãŒ˝ÃÊ ∑§Ê ◊ÊŸ 0.1 ◊Ê‹ ¬⁄U ⁄Uπà „È∞ ÿÁŒ B∑§Ë ‚ÊãŒ˝ÃÊ 0.1 ‚ ’…∏Ê∑§⁄U 0.3 ◊Ê‹ ∑§⁄U ŒË ¡ÊÃË „Ò ÃÊŒ⁄U ÁSÕ⁄UÊ¥∑§ „ÊªÊ —(1) k
(2) k/3
(3) 3k
(4) 9k
40. ∑ȧ¿U ∑§Ê ‹Êß«UÊ ¥ ∑ § SfláÊÊ Z∑§ „ Ò ¥ , Á¡‹Á≈ UŸ —0.005 - 0.01, ª◊ ∞⁄ U Á’∑§ — 0.15 - 0.25;
•ÊÁ‹∞≈U — 0.04 - 1.0; S≈UÊø¸ — 15 - 25, ߟ◊¥ ∑§ÊÒŸ-‚Ê ’„Ã⁄U ⁄UˇÊË ∑§Ê‹Êÿ«U „ʪÊ?
(1) Á¡‹Á≈UŸ
(2) ª◊ ∞⁄UÁ’∑§
(3) •ÊÁ‹∞≈U
(4) S≈UÊø¸
38. Oxidation of succinate ion produces
ethylene and carbon dioxide gases. On
passing 0.2 Faraday electricity through an
aqueous solution of potassium succinate,
the total volume of gases (at both cathode
and anode) at STP (1 atm and 273 K) is :
(1) 2.24 L
(2) 4.48 L
(3) 6.72 L
(4) 8.96 L
39. The rate law for the reaction below is given
by the expression k [A][B]
A+B Product
If the concentration of B is increased from
0.1 to 0.3 mole, keeping the value of A at
0.1 mole, the rate constant will be :
(1) k
(2) k/3
(3) 3k
(4) 9k
40. Gold numbers of some colloids
are : Gelatin : 0.005 - 0.01, Gum
Arabic : 0.15 - 0.25; Oleate : 0.04 - 1.0;
Starch : 15 - 25. Which among these is a
better protective colloid ?
(1) Gelatin
(2) Gum Arabic
(3) Oleate
(4) Starch
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 27
41. _uQ¡_p rh^p_p¡ Aphs®L$p¡ô$dp„_p sÒhp¡_¡ k„b„r^s R>¡._uQ¡_pdp„\u L$ey„ kpQy„ R>¡ ?
(1) kd|l 17 dp„ b^p S> sÒhp¡ hpeyAp¡ R>¡.
(2) kd|l 13 dp„ b^p S> sÒhp¡ ^psyAp¡ R>¡.
(3) Aphs®_¡ A_ygnu_¡ kd|l 15 _p sÒhp¡_usyg_pdp„ kd|l 16 _p sÒhp¡_u â\d Ape_uL$fZA¡Þ\pë`u Ap¡R>u R>¡.
(4) kd|l 15 _p sÒhp¡ dpV¡$, kd|ldp„ _uQ¡ S>CA¡s¡d +5 Ap¡[¼kX¡$i_ Ahõ\p_u [õ\fsp h ¡R>¡.
42. âÖphZ (smelting) Üpfp L$p¡ f_y„ r_óL$j®Z L$fsu hMs¡Dd¡f¡g rkrgL$p_p¡ D`ep¡N _uQ¡_pdp„\u iy„ v|$f L$fhp\pe R>¡ ?(1) Cu
2S
(2) FeO
(3) FeS
(4) Cu2O
43. âq¾$ep Ap¡mMu bsphp¡ L¡$ S>¡dp„ lpCX²$p¡S>_ DÐkS>®_\sp¡ _\u.
(1) S>gue ApëL$gu_u T]L$ kp\¡_u âq¾$ep
(2) Pt Cg¡¼V²$p¡X$p¡_p¡ D`ep¡N L$fu_¡ A¡rkqX$L$ `pZu_y„rhÛysrhcpS>_
(3) âhplu A¡dp¡r_epdp„ kp¡qX$ed_p ÖphZ_¡ fpMu_¡R>p¡X$u v$¡hy„
(4) B2H
6 kp\¡ rgr\ed lpCX²$pCX$_u âq¾$ep
41. ÁŸêŸ ∑§ÕŸ •Êflà ÃÊÁ‹∑§Ê ◊¥ ©¬ÁSÕà ÃàflÊ¥ ‚ ‚ê’¥ÁœÃ„Ò¥– ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚Ê ‚àÿ „Ò?
(1) ªÈ¬ 17 ◊¥ ‚÷Ë Ãàfl ªÒ‚ „Ò¥–
(2) ªÈ¬ 13 ◊¥ ‚÷Ë Ãàfl œÊÃÈ „Ò¥–
(3) ªÈ¬ 15 ∑§ ÃàflÊ¥ ∑§Ë ÃÈ‹ŸÊ ◊¥ ‚¥ªÃ •Êflø ∑§ª˝È¬ 16 ∑§ ÃàflÊ¥ ◊¥ •ÊÿŸŸ ∞ãÕÒÀ¬Ë ∑§Ê ◊ÊŸ∑§◊ ⁄U„ÃÊ „Ò–
(4) ª˝È¬ 15 ∑§ ÃàflÊ¥ ∑§ Á‹∞, ª˝È¬ ◊¥ ŸËø ¡ÊŸ ¬⁄U+5 •ÊÚÄ‚Ë∑§⁄UáÊ •flSÕÊ ∑§Ê SÕÊÁÿàfl ’…∏ÃÊ„Ò–
42. S◊ÁÀ≈¥Uª mÊ⁄UÊ ∑§ÊÚ¬⁄U ∑§ ÁŸc∑§·¸áÊ ◊¥ Á‚Á‹∑§Ê ÿÊíÿ ∑§M§¬ ◊¥ ÁŸêŸ ◊¥ ‚ Á∑§‚∑§Ê „≈UÊŸ ∑§ Á‹∞ ∑§Ë ¡ÊÃË „Ò ?(1) Cu
2S
(2) FeO
(3) FeS
(4) Cu2O
43. ©‚ •Á÷Á∑˝§ÿÊ ∑§Ê ’ÃÊß∞ Á¡‚◊¥ „Êß«˛UÊ¡Ÿ ©à‚Á¡¸ÃŸ„Ë¥ „ÊÃË „Ò —
(1) ¡‹Ëÿ ˇÊÊ⁄U ∑§ ‚ÊÕ Á¡¥∑§ ∑§Ë •Á÷Á∑˝§ÿÊ
(2) å‹Á≈UŸ◊ ß‹Ä≈˛UÊ«UÊ¥ ∑§Ê ¬˝ÿʪ ∑§⁄U∑§ •ê‹Ë∑Χá‹ ∑§Ê ÁfllÈà •¬ÉÊ≈UŸ
(3) Œ˝fl •◊ÊÁŸÿÊ ◊¥ ‚ÊÁ«Uÿ◊ ∑§ Áfl‹ÿŸ ∑§Ê ÁSÕ⁄U„ÊŸ ∑§ Á‹∞ ¿UÊ«∏ ŒŸÊ
(4) B2H
6 ∑§ ‚ÊÕ ‹ËÁÕÿ◊ „Êß«˛ UÊß«U ∑§Ë
•Á÷Á∑˝§ÿÊ
41. The following statements concern elements
in the periodic table. Which of the
following is true ?
(1) All the elements in Group 17 are
gases.
(2) The Group 13 elements are all metals.
(3) Elements of Group 16 have lower
ionization enthalpy values compared
to those of Group 15 in the
corresponding periods.
(4) For Group 15 elements, the stability
of +5 oxidation state increases down
the group.
42. Extraction of copper by smelting uses silica
as an additive to remove :
(1) Cu2S
(2) FeO
(3) FeS
(4) Cu2O
43. Identify the reaction which does not
liberate hydrogen :
(1) Reaction of zinc with aqueous alkali.
(2) Electrolysis of acidified water using
Pt electrodes.
(3) Allowing a solution of sodium in
liquid ammonia to stand.
(4) Reaction of lithium hydride with
B2H
6.
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 28
44. L¡$ëiued Ap¡¼kpCX$_y„ ìep`pfu ^p¡fZ¡ _pd Ap`p¡.
(1) v|$r^ep¡ Q|_p¡ (rdëL$ Ap¡a gpCd) (Milk of
lime)
(2) ap¡X¡$gp¡ Q|_p¡ (Slaked lime)
(3) Q|_p_p¡ `Õ\f (Limestone)
(4) L$mu Q|_p¡ (Quick lime)
45. L$\_ : L$pb®__p blºê$`p¡dp„, lufp¡ A¡ AhplL$ R>¡Äepf¡ N°¡apCV$ A¡ rhÛys_p kyhplL$ R>¡.
L$pfZ : lufpdp„ A_¡ N°¡apCV$dp„ L$pb®__y„ k„L$fZA_y¾d¡ sp
3 A_¡ sp2 R>¡.
(1) L$\_ A_¡ L$pfZ b„_¡ kpQp R>¡, s\p L$pfZ A¡L$\_ dpV¡$_u kpQu kdS|>su R>¡.
(2) L$\_ A_¡ L$pfZ b„_¡ kpQp R>¡, `Z L$pfZ A¡L$\_ dpV¡$_u kpQu kdS|>su _\u.
(3) L$\_ A¡ AkÐe rh^p_ R>¡. `Z L$pfZ kÐeR>¡.
(4) L$\_ A_¡ L$pfZ b„_¡ AkÐe R>¡.
46. Mp¡Vy„$ rh^p_ ip¡ u bsphp¡.
(1) S2 A¡ Ap¡[¼kS>__u dpaL$ A_yQy„bL$ue R>¡.
(2) f¹lp¡[çbL$ A_¡ dp¡_p¡[¼gr_L$ këafdp„ S8
AÏAp¡ R>¡.
(3) S8 hge_p¡ ApL$pf dyNV$ (¾$pD_) S>¡hp¡ R>¡.
(4) S8 A_¡ S
6 hgep¡dp„ S-S-S b„^ M|ZpAp¡
A¡L$kfMp R>¡.
44. ∑Ò§ÁÀ‡Êÿ◊ •ÊÚÄ‚Êß«U ∑§Ê √ÿÊfl‚ÊÁÿ∑§ ŸÊ◊ „Ò —
(1) Á◊À∑§ •ÊÚ»§ ‹Êß◊
(2) S‹ÒÄ«U ‹Êß◊
(3) ‹Êß◊S≈UÊŸ
(4) ÁÄfl∑§ ‹Êß◊
45. ∑§ÕŸ — ∑§Ê’Ÿ ∑§ •¬⁄UM§¬Ê¥ ◊¥, «UÊÿ◊¥«U ∑ȧøÊ‹∑§„Ò ¡’ Á∑§ ª˝»§Êß≈U ∞∑§ ÁfllÈà ‚ÈøÊ‹∑§„Ò–
∑§Ê⁄UáÊ — «UÊÿ◊ã«U ÃÕÊ ª˝»§Êß≈U ◊¥ ∑§Ê’¸Ÿ ∑§Ê‚¥∑§⁄UáÊ ∑˝§◊‡Ê— sp
3 ÃÕÊ sp
2 „Ò–
(1) ∑§ÕŸ ÃÕÊ ∑§Ê⁄UáÊ ŒÊŸÊ¥ ‚„Ë „Ò¥ ÃÕÊ ∑§Ê⁄UáÊ ∑§ÕŸ∑§Ë ‚„Ë √ÿÊÅÿÊ „Ò–
(2) ∑§ÕŸ ÃÕÊ ∑§Ê⁄UáÊ ŒÊŸÊ¥ ‚„Ë „Ò¥ ¬⁄UãÃÈ ∑§Ê⁄UáÊ,∑§ÕŸ ∑§Ë ‚„Ë √ÿÊÅÿÊ Ÿ„Ë¥ „Ò–
(3) ∑§ÕŸ •‚àÿ „Ò ¬⁄UãÃÈ ∑§Ê⁄UáÊ ‚àÿ „Ò–
(4) ∑§ÕŸ ÃÕÊ ∑§Ê⁄UáÊ ŒÊŸÊ¥ „Ë •‚àÿ „Ò¥–
46. •‚àÿ ∑§ÕŸ ∑§Ê ¬„øÊÁŸ∞ —
(1) •ÊÚÄ‚Ë¡Ÿ ∑§Ë Ã⁄U„ S2 •ŸÈøÈê’∑§Ëÿ „Ò–
(2) ⁄UÊÁê’∑§ (Áfl·◊‹¥’ÊˇÊ) ÃÕÊ ◊ÊŸÊÄ‹ËÁŸ∑§‚À»§⁄U ◊¥ S
8 •áÊÈ „Ò¥–
(3) S8 fl‹ÿ ∑§Ê •Ê∑§Ê⁄U ◊È∑ȧ≈U ∑§Ë Ã⁄U„ „Ò–
(4) S8 ÃÕÊ S
6 fl‹ÿÊ¥ ◊¥ S-S-S •Ê’㜠∑§ÊáÊ ∞∑§
¡Ò‚ „Ò¥–
44. The commercial name for calcium oxide
is :
(1) Milk of lime
(2) Slaked lime
(3) Limestone
(4) Quick lime
45. Assertion : Among the carbon
allotropes, diamond is an
insulator, whereas,
graphite is a good
conductor of electricity.
Reason : Hybridization of carbon
in diamond and graphite
are sp3 and sp
2,
respectively.
(1) Both assertion and reason are correct,
and the reason is the correct
explanation for the assertion.
(2) Both assertion and reason are correct,
but the reason is not the correct
explanation for the assertion.
(3) Assertion is incorrect statement, but
the reason is correct.
(4) Both assertion and reason are
incorrect.
46. Identify the incorrect statement :
(1) S2 is paramagnetic like oxygen.
(2) Rhombic and monoclinic sulphur
have S8 molecules.
(3) S8 ring has a crown shape.
(4) The S-S-S bond angles in the S8 and
S6 rings are the same.
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 29
47. kpQy„ rh^p_ ip¡ p¡.
(1) Ape_® (gp¡M„X$) Ap¡[¼kS>_-dy¼s `pZudp„npqfs \pe R>¡.
(2) npfhpmp `pZudp„ Ape_® (gp¡M„X$) M|b S>TX$`\u npqfs \pe R> ¡ L $pfZ L ¡ $ s ¡_p ¡rhÛysfpkperZL$ `p¡V¡$[Þieg KQp¡ R>¡.
(3) Ape_® (gp¡M„X$) _y„ npfZ, Äepf¡ s¡_p¡ buÆKQp fuX$n_ `p¡V¡$[Þieg ^fphsu ^psy kp\¡k„`L®$ L$fsp Ap¡Ry>„ L$fu iL$pe R>¡.
(4) Ape_® (gp¡M„X$) _y„ npfZ, s¡_u k`pV$u `fA`pfNçe Ahfp¡ b_phu_¡ Ap¡Ry>„ L$fu iL$peR>¡.
48. _uQ¡ Ap` ¡gpAp ¡dp „\u L $e y „ A ¡L $ lp ¡dp ¡g ¡àV $ uL $(homoleptic) kdp_uL©$s k„L$uZ®_y„ Dv$plfZ R>¡ ?(1) [Co(NH
3)6]Cl
3
(2) [Pt(NH3)2Cl
2]
(3) [Co(NH3)4Cl
2]
(4) [Co(NH3)5Cl]Cl
2
49. dpZ¡L$ (Ruby) A_¡ `Þ_p (emerald) dp„ f„Np¡ dpV¡$S>hpbv$pf k„¾$p„rs ^psy Ape_p¡ A_y¾$d¡ _uQ¡_pdp„\uip¡^p¡.
(1) Cr3+
A_¡ Co3+
(2) Co3+
A_¡ Cr3+
(3) Co3+
A_¡ Co3+
(4) Cr3+
A_¡ Cr3+
47. ‚„Ë ∑§ÕŸ ∑§Ê ¬„øÊÁŸÿ —
(1) •Êÿ⁄UŸ •ÊÚÄ‚Ë¡Ÿ-◊ÈÄà ¡‹ ◊¥ ‚¥ˇÊÊÁ⁄Uà „ÊÃÊ„Ò–
(2) ‹fláÊËÿ ¡‹ ◊¥ •Êÿ⁄UŸ ¡ÀŒË ‚ ‚¥ˇÊÊÁ⁄Uà „ÊÃÊ„Ò ÄÿÊ¥Á∑§ ß‚∑§Ê ÁfllÈà ⁄UÊ‚ÊÿÁŸ∑§ Áfl÷fl ©ìÊ„Ò–
(3) •Êÿ⁄UŸ ∑§Ê ‚¥ ÊÊ⁄UáÊ ß‚∑§Ê ©ìÊ •¬øÿŸ Áfl÷flflÊ‹ œÊÃÈ ∑§ ‚ê¬∑¸§ ◊¥ ‹ÊŸ ¬⁄U ∑§◊ Á∑§ÿÊ ¡Ê‚∑§ÃÊ „Ò–
(4) •Êÿ⁄UŸ ∑§Ê ‚¥ˇÊÊ⁄UáÊ ß‚∑§ ‚Ä ¬⁄U •¬Ê⁄Uªêÿ•fl⁄UÊœ ’ŸÊ∑§⁄U ∑§◊ Á∑§ÿÊ ¡Ê ‚∑§ÃÊ „Ò–
48. ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ „Ê◊Ê‹Áå≈U∑§ (homoleptic) ‚¥∑ȧ‹∑§Ê ∞∑§ ©ŒÊ„⁄UáÊ „Ò?(1) [Co(NH
3)6]Cl
3
(2) [Pt(NH3)2Cl
2]
(3) [Co(NH3)4Cl
2]
(4) [Co(NH3)5Cl]Cl
2
49. M§’Ë ∞fl¥ ß◊⁄UÊÀ«U ◊¥ Á¡Ÿ ‚¥∑˝§◊áÊ œÊÃÈ•Ê¥ ∑§ •ÊÿŸÊ¥∑§Ë ©¬ÁSÕÁà ∑§ ∑§Ê⁄UáÊ ⁄¥Uª „ÊÃÊ „Ò, fl ∑˝§◊‡Ê— „Ò¥ —
(1) Cr3+
ÃÕÊ Co3+
(2) Co3+
ÃÕÊ Cr3+
(3) Co3+
ÃÕÊ Co3+
(4) Cr3+
ÃÕÊ Cr3+
47. Identify the correct statement :
(1) Iron corrodes in oxygen-free water.
(2) Iron corrodes more rapidly in salt
water because its electrochemical
potential is higher.
(3) Corrosion of iron can be minimized
by forming a contact with another
metal with a higher reduction
potential.
(4) Corrosion of iron can be minimized
by forming an impermeable barrier
at its surface.
48. Which of the following is an example of
homoleptic complex ?
(1) [Co(NH3)6]Cl
3
(2) [Pt(NH3)2Cl
2]
(3) [Co(NH3)4Cl
2]
(4) [Co(NH3)5Cl]Cl
2
49. The transition metal ions responsible
for color in ruby and emerald are,
respectively :
(1) Cr3+
and Co3+
(2) Co3+
and Cr3+
(3) Co3+
and Co3+
(4) Cr3+
and Cr3+
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 30
50. X²$pe¼gu_]Ndp„ _uQ¡ Ap ¡gp `v$p\p£dp„\u L$ep¡ A¡L$_p¡D`ep¡N hpsphfZue âv|$jZ_¡ r_e„ÓZ L$fhpdp„ h yL$pe®nd R>¡ ?
(1) V¡$V²$p¼gp¡fp¡Cr\rg_
(2) L$pb®_ X$pep¡¼kpCX
(3) këaf X$pep¡¼kpCX$
(4) _pCV²$p¡S>_ X$pep¡¼kpCX$
51. l¡gp¡S>_p¡_u L$kp¡V$u L$fsp l¡gp kp¡qX$ed AL®$ (extract)
_¡ kp„Ö HNO3 _u kp\¡ Nfd L$fhpdp„ Aph¡ R>¡ L$pfZ
L¡$...
(1) rkëhf l¡gpCX$p¡ _pCV²$uL$ A¡rkXdp„ k„`|Z® fus¡AÖpìe R>¡.
(2) A¡rkqX$L$ dpÝeddp„ Ag2S A_¡ AgCN Öpìe
\pe R>¡.
(3) Å¡ S2− A_¡ CN
− lpS>f lp¡e sp¡ kp„ÖHNO
3 \u rhOV$_ \pe R>¡. s¡\u `funZdp„
(L$kp¡V$udp„) v$MgNufu L$fsp„ _\u.
(4) A¡rkqX$L$ dpÝeddp„ Ag _u âq¾$ep l¡gpCX$p¡ kp\¡TX$`u \pe R>¡.
50. «˛UÊ߸ċËÁŸ¥ª ◊¥ ¬˝ÿÈÄà ÁŸêŸ ¬ŒÊÕÊZ ◊¥ ‚ Á∑§‚∑§Ê ¬˝ÿʪflÊÃÊfl⁄UáÊ ¬˝ŒÍ·áÊ ∑§ ÁŸÿ¥òÊáÊ ∑§Ë ’„Ã⁄U ∑§Êÿ¸ ŸËÁà „Ò?
(1) ≈U≈˛UÊÄ‹Ê⁄UÊ∞ÁÕ‹ËŸ
(2) ∑§Ê’¸Ÿ «UÊß•ÊÚÄ‚Êß«U
(3) ‚À»§⁄U «UÊß•ÊÚÄ‚Êß«U
(4) ŸÊß≈˛UÊ¡Ÿ «UÊß•ÊÚÄ‚Êß«U
51. „Ò‹Ê¡Ÿ˜‚ ∑§Ë ¡Ê°ø ∑§ ¬„‹ ‚ÊÁ«Uÿ◊ ∞ÄS≈˛ÒUÄ≈U ∑§Ê‚ÊãŒ˝ HNO
3 ∑§ ‚ÊÕ ª◊¸ Á∑§ÿÊ ¡ÊÃÊ „Ò ÄÿÊ¥Á∑§ —
(1) Á‚Àfl⁄U „Ò‹Êß«U ŸÊßÁ≈˛U∑§ •ê‹ ◊¥ ¬ÍáʸM§¬áÊ•ÉÊÈ‹Ÿ‡ÊË‹ „Ò¥–
(2) •ê‹Ëÿ ◊Êäÿ◊ ◊ ¥ Ag2S ÃÕÊ AgCN
ÉÊÈ‹Ÿ‡ÊË‹ „Ò¥–
(3) ÿÁŒ S2−
ÃÕÊ CN−
©¬ÁSÕà „Ò¥ ÃÊ ‚ÊãŒ˝HNO
3 ‚ ÁflÉÊÁ≈Uà „Ê ¡Êà „Ò¥ ß‚Á‹ÿ ¬⁄UˡÊáÊ
◊¥ „SÃˇÊ¬ Ÿ„Ë¥ ∑§⁄UÖ
(4) •ê‹Ëÿ ◊Êäÿ◊ ◊¥ Á‚Àfl⁄U, „Ò‹Êß«UÊ¥ ∑§ ‚ÊÕá •Á÷Á∑˝§ÿÊ ∑§⁄UÃÊ „Ò–
50. Which one of the following substances used
in dry cleaning is a better strategy to control
environmental pollution ?
(1) Tetrachloroethylene
(2) Carbon dioxide
(3) Sulphur dioxide
(4) Nitrogen dioxide
51. Sodium extract is heated with concentrated
HNO3 before testing for halogens
because :
(1) Silver halides are totally insoluble in
nitric acid.
(2) Ag2S and AgCN are soluble in acidic
medium.
(3) S2−
and CN−
, if present, are
decomposed by conc. HNO3 and
hence do not interfere in the test.
(4) Ag reacts faster with halides in acidic
medium.
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 31
52. kpe¼gp ¡l ¡¼Tu__y „ b ° p ¡ rd_ ¡i_ Ap`¡g _uQ¡_u`qf[õ\rsAp¡dp„ L$C _u`S> Ap`i¡ ?
• •• • • •
(1)
• •
• •
(2)
• •
(3)
• •
• •
(4)
• •
• •
52. ŸËø ÁŒÿ ªÿ ¬ÁÃ’ãœÊ¥ ◊¥ ‚ÊßÄ‹Ê„Ä‚ËŸ ∑§Ê ’Ê◊ËŸ‡ÊŸŒÃÊ „Ò —
• •• • • •
(1)
• •
• •
(2)
• •
(3)
• •
• •
(4)
• •
• •
52. Bromination of cyclohexene under
conditions given below yields :
• •• • • •
(1)
• •
• •
(2)
• •
(3)
• •
• •
(4)
• •
• •
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 32
53. _uQ¡ Ap`¡g âq¾$ep_p¡ ¾$d Ýep_dp„ gp¡.
• • • •
X iy„ R>¡ ?
(1)
• •
• •
• •• •
(2)
• •
• •
• • • •
(3)
• • • •
53. ŸËø ŒË ªß¸ •Á÷Á∑˝§ÿÊ ∑˝§◊ ¬⁄U ÁfløÊ⁄U ∑§ËÁ¡∞ —
• • • •
X
„Ò —
(1)
• •
• •
• •• •
(2)
• •
• •
• • • •
(3)
• • • •
53. Consider the reaction sequence below :
• • • •
X
is :
(1)
• •
• •
• •• •
(2)
• •
• •
• • • •
(3)
• • • •
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 33
(4)
• •• •
54. _uQ¡ Ap`¡gpdp„\u L$ep¡ âq¾$eL$ A¡ rhgp¡`_ âq¾$ep dpV¡$kyk„Ns _\u ?
• •
(1) NaOH/H2O
(2) NaOEt/EtOH
(3) NaOH/H2O-EtOH
(4) NaI
55. PETN _u b_phV $dp „ h`fpsp ¡ Cfu\ ° uV $ p ¡g(C(CH
2OH)
4) _p k„ïg¡jZ dpV¡$ L$ey„ rh^p_ kpQy„
R>¡ ?
(1) k„ïg¡jZ dpV¡$ rd\¡_p¡g A_¡ C\¡_p¡g_u hÃQ¡Qpf ApëX$p¡g k„O___u AphíeL$sp lp¡e R>¡.
(2) k„ïg¡jZdp„ b¡ ApëX$p¡g k„O__ A_¡ b¡ L¡$r_Tpfp¡âq¾$ep_u AphíeL$sp R>¡.
(3) k„ïg¡jZdp„ ÓZ ApëX$p¡g k„O__ A_¡ A¡L$L¡$r_Tpfp¡ âq¾$ep_u AphíeL$sp R>¡.
(4) Ap âq¾$epdp„ C\¡_p¡g_p¡ Apëap lpCX²$p¡S>_ A_¡rd\¡_p¡g cpN g¡ R>¡.
(4)
• •• •
54. Áfl‹Ê¬Ÿ •Á÷Á∑˝§ÿÊ ∑§ Á‹∞ ߟ •Á÷∑§Ê⁄U∑§Ê¥ ◊¥ ‚∑§ÊÒŸ-‚Ê ©¬ÿÈÄà Ÿ„Ë¥ „Ò?
• •
(1) NaOH/H2O
(2) NaOEt/EtOH
(3) NaOH/H2O-EtOH
(4) NaI
55. PETN ∑§ ’ŸÊŸ ◊¥ ¬˝ÿÈÄà ß⁄UËÁÕ˝≈UÊÚ‹ (C(CH2OH)
4)
∑§ ‚¥‡‹·áÊ ∑§ ‚ê’㜠◊¥ ‚„Ë ∑§ÕŸ „Ò —
(1) ‚¥‡‹·áÊ ◊¥ ◊ÕŸÊÚ‹ ÃÕÊ ∞ÕŸÊÚ‹ ∑§ ’Ëø øÊ⁄U∞À«UÊ‹ ‚¥ÉÊŸŸ ∑§Ë •Êfl‡ÿ∑§ÃÊ „ÊÃË „Ò–
(2) ‚¥‡‹·áÊ ◊¥ ŒÊ ∞À«UÊ‹ ‚¥ÉÊŸŸ ÃÕÊ ŒÊ ∑Ò§ÁŸ¡Ê⁄UÊ•Á÷Á∑˝§ÿÊ ∑§Ë ¡M§⁄Uà „ÊÃË „Ò–
(3) ‚¥‡‹·áÊ ◊¥ ÃËŸ ∞À«UÊ‹ ‚¥ÉÊŸŸ ÃÕÊ ∞∑§∑Ò§ÁŸ¡Ê⁄UÊ •Á÷Á∑˝§ÿÊ ∑§Ë •Êfl‡ÿ∑§ÃÊ „ÊÃË „Ò–
(4) ß‚ •Á÷Á∑˝§ÿÊ ◊¥ ∞ÕŸÊÚ‹ ∑§ •À»§Ê „Êß«˛UÊ¡ŸÃÕÊ ◊ÕŸÊÚ‹ ÷ʪ ‹Ã „Ò¥–
(4)
• •• •
54. Which one of the following reagents is not
suitable for the elimination reaction ?
• •
(1) NaOH/H2O
(2) NaOEt/EtOH
(3) NaOH/H2O-EtOH
(4) NaI
55. The correct statement about the synthesis
of erythritol (C(CH2OH)
4) used in the
preparation of PETN is :
(1) The synthesis requires four aldol
condensations between methanol
and ethanol.
(2) The synthesis requires two aldol
condensations and two Cannizzaro
reactions.
(3) The synthesis requires three aldol
condensations and one Cannizzaro
reaction.
(4) Alpha hydrogens of ethanol and
methanol are involved in this
reaction.
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 34
56. L$p¡C A¡fp¡d¡qV$L$ hge_y„ ãgp¡qf_¡i_ M|b S> kl¡gpC\uk„ch lp¡e R>¡, Äepf¡ s¡_p X$peTp¡_ued npf_¡ HBF
4
kp\¡ âq¾$ep L$fhpdp„ Aph¡ R>¡. D`f_u °q¾$ep dpV¡$ _uQ¡_pAp`¡gpdp„\u L$C `qf[õ\rs kpQu R>¡ ?
(1) a¼s Dódp(2) NaNO
2/Cu
(3) Cu2O/H
2O
(4) NaF/Cu
57. _uQ¡ Ap ¡gpdp„\u L$ep blºgL$_y„ k„ïg¡jZ A¡ dy¼s dygL$blºguL$fZ sL$_uL$_p¡ D`ep¡N L$fu_¡ L$fhpdp„ Aph¡ R>¡?
(1) V¡$agp¡_
(2) V¡$qfrg_
(3) d¡g¡dpC_ blºgL$
(4) _pegp¡_ 6,6
58. “N” L¡$ S>¡ Ap ¡g k„ep¡S>__u b¡rTL$spdp„ cpN g¡sp¡_\u s¡ bsphp¡.
•
• •
•
•
••
••
••
•• •
(1) N 7
(2) N 9
(3) N 1
(4) N 3
56. Á∑§‚Ë ∞⁄UÊÒ◊ÒÁ≈U∑§ fl‹ÿ ∑§Ê ç‹È•Ê⁄UË∑§⁄UáÊ •Ê‚ÊŸË ‚‚¥÷fl „ÊÃÊ „Ò ÿÁŒ ©‚∑§ «UÊß∞¡ÊÁŸÿ◊ ‹fláÊ ∑§ÊHBF
4 ∑§ ‚ÊÕ ©¬øÊÁ⁄Uà Á∑§ÿÊ ¡Êÿ– ß‚ •Á÷Á∑˝§ÿÊ
∑§ ‚ê’㜠◊¥ ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ‚Ë ¬Á⁄UÁSÕÁé¬ÿÈÄà „Ò?
(1) ∑§fl‹ ™§c◊Ê(2) NaNO
2/Cu
(3) Cu2O/H
2O
(4) NaF/Cu
57. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ-‚Ê ’„È‹∑§ ◊ÈÄà ◊Í‹∑§’„È‹∑§Ë∑§⁄UáÊ ÁflÁœ mÊ⁄UÊ ‚¥‡‹Á·Ã Á∑§ÿÊ ¡ÊÃÊ „Ò?
(1) ≈Uç‹ÊÚŸ
(2) ≈ÒU⁄UË‹ËŸ
(3) ◊Ò‹Ò◊Êߟ ’„È‹∑§
(4) ŸÊÿ‹ÊÚŸ 6,6
58. fl„ “N” ¡Ê ÁŸêŸ ÿÊÒÁª∑§ ∑§Ë ÊÊ⁄UËÿ ¬˝flÎÁûÊ ◊¥ ÿʪŒÊŸŸ„Ë¥ ŒÃÊ „Ò, fl„ „Ò —
•
• •
•
•
••
••
••
•• •
(1) N 7
(2) N 9
(3) N 1
(4) N 3
56. Fluorination of an aromatic ring is easily
accomplished by treating a diazonium salt
with HBF4. Which of the following
conditions is correct about this reaction ?
(1) Only heat
(2) NaNO2/Cu
(3) Cu2O/H
2O
(4) NaF/Cu
57. Which of the following polymers is
synthesized using a free radical
polymerization technique ?
(1) Teflon
(2) Terylene
(3) Melamine polymer
(4) Nylon 6,6
58. The “N” which does not contribute to the
basicity for the compound is :
•
• •
•
•
••
••
••
•• •
(1) N 7
(2) N 9
(3) N 1
(4) N 3
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 35
59. _uQ¡_pdp„\u L$ey„ A¡L$ b¡¼V¡$qfep_piL$ ârsÆhuAp¡(antibiotic) R>¡ ?
(1) Cqf\°p¡dpeku_
(2) V¡$V²$pkpe¼gu_
(3) ¼gp¡fA¡ça¡r_L$p¡g
(4) Ap¡ãgp¡¼k¡rk_
60. ""fyldp_ Å„byX$uep¡''(` ®g) (Rhumann’s purple)
Ahgp¡L$_ A¡ r_Zp®eL$ L$kp¡V$u _uQ¡_pdp„\u L$p¡_u lpS>fuk|Qh¡ R>¡ ?
(1) fuX$éyk]N iL®$fp
(2) ¼eyâuL$ Ape_
(3) âp¡V$u_
(4) õV$pQ®
61. ^pfp ¡ L ¡ $ P : sin cos 2 cos • • • •= − =
A_¡ Q : sin cos 2 sin • • • •••= + = b¡
NZp¡ R>¡. sp¡ :
(1) P ⊂ Q A_¡ Q−P ≠ φ
(2) Q ⊄ P
(3) P ⊄ Q
(4) P = Q
59. ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚Ê ’ÒÄ≈UËÁ⁄UÿʟʇÊË ¬˝ÁáÒÁfl∑§ „Ò?
(1) ∞Á⁄UÕ˝Ê◊Êß‚ËŸ
(2) ≈U≈˛UÊ‚ÊÿÄ‹ËŸ
(3) Ä‹Ê⁄U∞ê»Ò§ÁŸ∑§Ê‹
(4) •ÊÚ»˜§‹ÊÄ‚Ò‚ËŸ
60. L§„◊ÒŸ ŸË‹ ‹ÊÁ„à (¬¬‹) ∑§Ê ¬∑§≈U „ÊŸÊ ÁŸêŸÁ‹ÁπÃ◊¥ ‚ Á∑§‚∑§Ê ‚¥¬ÈÁc≈U ¬⁄UˡÊáÊ „Ò?
(1) •¬øÊÿ∑§ ‡Ê∑¸§⁄UÊ
(2) ÄÿͬÁ⁄U∑§ •ÊÿŸ
(3) ¬˝Ê≈UËŸ
(4) S≈UÊø¸ (◊¥«U)
61. ◊ÊŸÊ P : sin cos 2 cos • • • •= − = ÃÕÊ
Q : sin cos 2 sin • • • •••= + = ŒÊ‚◊Èëøÿ „Ò¥, ÃÊ —
(1) P ⊂ Q ÃÕÊ Q−P ≠ φ
(2) Q ⊄ P
(3) P ⊄ Q
(4) P = Q
59. Which of the following is a bactericidal
antibiotic ?
(1) Erythromycin
(2) Tetracycline
(3) Chloramphenicol
(4) Ofloxacin
60. Observation of “Rhumann’s purple” is a
confirmatory test for the presence of :
(1) Reducing sugar
(2) Cupric ion
(3) Protein
(4) Starch
61. Let P : sin cos 2 cos • • • •= − = and
Q : sin cos 2 sin • • • •••= + = be two
sets. Then :
(1) P ⊂ Q and Q−P ≠ φ
(2) Q ⊄ P
(3) P ⊄ Q
(4) P = Q
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
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62. Å¡ kduL$fZ
•• • • • • • • • • •• •• • ••
• • •
+ − − = _p¡ A¡L$
DL¡$g x R>¡, sp¡ •• • •• • • =__________ \pe.
(1)
••
(2)
••
(3) 2
(4) • •
63. ^pfp¡ L¡$ z=1+ai, a > 0, A¡L$ A¡hu k„L$f k„¿ep R>¡ L¡$S> ¡\u z
3 hpõsrhL$ k„¿ep \pe. sp ¡ kfhpmp ¡
1+z+z2+.....+z
11 =__________ \pe.
(1) • • • • • • ••
(2) • • • • • • •
(3) • • • • • • •
(4) • • • • • • ••
62. ÿÁŒ ‚◊Ë∑§⁄UáÊ
•• • • • • • • • • •• •• • • ••
• • •
+ − − = ∑§Ê x
∞∑§ „‹ „Ò, ÃÊ •• • •• • • ’⁄UÊ’⁄U „Ò —
(1)
••
(2)
••
(3) 2
(4) • •
63. ◊ÊŸÊ z=1+ai, a > 0 ∞∑§ ∞‚Ë ‚Áê◊üÊ ‚¥ÅÿÊ „Ò,Á∑§ z
3 ∞∑§ flÊSÃÁfl∑§ ‚ ¥ÅÿÊ „ Ò , ÃÊ ÿÊ ª
1+z+z2+.....+z
11 ’⁄UÊ’⁄U „Ò —
(1) • • • • • • ••
(2) • • • • • • •
(3) • • • • • • •
(4) • • • • • • ••
62. If x is a solution of the equation,
•
1
2 1 2 1 1, , then
2
4 1 is equal to :
• • •
•
+ − − =
−
(1)
••
(2)
••
(3) 2
(4) • •
63. Let z=1+ai be a complex number, a > 0,
such that z3 is a real number. Then the sum
1+z+z2+.....+z
11 is equal to :
(1) • • • • • • ••
(2) • • • • • • •
(3) • • • • • • •
(4) • • • • • • ••
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
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64. ^pfp ¡ L ¡ $ A A¡L $ 3×3 î¡ rZL $ R> ¡ L ¡ $ S > ¡\uA
2−5A+7I=O.
rh^p_ - I :• •• ••• •• • ••
••
= −
rh^p_ - II : blº`v $ u A3−2A
2−3A+I _¡
5(A−4I) dp„ ê$`p„sqfs L$fu iL$pe R>¡.sp¡ __________.
(1) rh^p_-I kÐe R>¡, f„sy rh^p_-II AkÐe R>¡.
(2) rh^p_-I AkÐe R>¡, f„sy rh^p_-II kÐe R>¡.
(3) b„_¡ rh^p_p¡ kÐe R>¡.
(4) b„_¡ rh^p_p¡ AkÐe R>¡.
65. Å¡ • •
• •• •
• •• •• •
• •• , sp ¡ î ¡ rZL $
(A2016
−2A2015
−A2014
) _p ¡ r_òpeL $__________ R>¡.(1) 2014
(2) −175
(3) 2016
(4) −25
64. ◊ÊŸÊ A, 3×3 ∑§Ê ∞∑§ ∞‚Ê •Ê√ÿÍ„ „ Ò Á∑§A
2−5A+7I=O „Ò–
∑§ÕŸ - I : • •• ••• •• • ••
••
= −
∑§ÕŸ - II : ’„ȬŒ A3−2A
2−3A+I ∑§Ê
5(A−4I) ◊¥ ¬Á⁄UflÁÃà Á∑§ÿÊ ¡Ê ‚∑§ÃÊ„Ò–
ÃÊ,
(1) ∑§ÕŸ - I ‚àÿ „Ò ‹Á∑§Ÿ ∑§ÕŸ - II •‚àÿ „Ò–
(2) ∑§ÕŸ - I •‚àÿ „Ò ‹Á∑§Ÿ ∑§ÕŸ - II ‚àÿ „Ò–
(3) ŒÊŸÊ¥ ∑§ÕŸ ‚àÿ „Ò¥–
(4) ŒÊŸÊ¥ ∑§ÕŸ •‚àÿ „Ò¥–
65. ÿÁŒ • •
• •• •
• •• •• •
− −
= „ Ò , ÃÊ •Ê√ÿ Í„
(A2016
−2A2015
−A2014
) ∑§Ê ‚Ê⁄UÁáÊ∑§ „Ò —(1) 2014
(2) −175
(3) 2016
(4) −25
64. Let A be a 3×3 matrix such that
A2−5A+7I=O.
Statement - I :
• •• ••• •• • •••
• = −
Statement - II : The polynomial
A3−2A
2−3A+I can be
reduced to 5(A−4I).
Then :
(1) Statement-I is true, but Statement-II
is false.
(2) Statement-I is false, but Statement-II
is true.
(3) Both the statements are true.
(4) Both the statements are false.
65. If
• •• •
• •• •• •• •
− −
= , then the determinant of
the matrix (A2016
−2A2015
−A2014
) is :
(1) 2014
(2) −175
(3) 2016
(4) −25
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
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66. Å¡ •
••
•
• • •• • ••
•
•
•
• • sp¡ n _uQ¡_pdp„\u L$ey„ kduL$fZ
k„`p¡j¡?(1) n
2+3n−108=0
(2) n2+5n−84=0
(3) n2+2n−80=0
(4) n2+n−110=0
67. Å¡ ••
••
• ••
••
•
•• • , (x > 0) _p rhõsfZdp„ x−2
A_¡ x−4 _p klNyZL$p¡ A_y¾$d¡ m A_¡ n lp¡e sp¡,
•• =__________ \pe.
(1) 182
(2)
••
(3)
••
(4) 27
68. ^pfp¡ L¡$ a1, a
2, a
3, ......, a
n, ..... kdp„sf î¡Zudp„ R>¡.
Å¡ a3+a
7+a
11+a
15=72, sp¡ s¡_p â\d 17 v$p¡_p¡
kfhpmp¡ __________ \pe.(1) 306
(2) 153
(3) 612
(4) 204
66. ÿÁŒ •
••
•
• • •• • ••
•
•
•
• • „Ò, ÃÊ n ÁŸêŸ ◊¥ ‚ Á∑§‚
‚◊Ë∑§⁄UáÊ ∑§Ê ‚¥ÃÈc≈U ∑§⁄UÃÊ „Ò?(1) n
2+3n−108=0
(2) n2+5n−84=0
(3) n2+2n−80=0
(4) n2+n−110=0
67. ÿÁŒ ••
••
• ••
••
•
•• • , (x > 0), ∑§ ¬˝‚Ê⁄U ◊¥ x−2
ÃÕÊ x−4 ∑§ ªÈáÊÊ¥∑§ ∑˝§◊‡Ê— m ÃÕÊ n „Ò¥, ÃÊ
••
’⁄UÊ’⁄U „Ò —(1) 182
(2)
••
(3)
••
(4) 27
68. ◊ÊŸÊ a1, a
2, a
3, ......, a
n, ..... ∞∑§ ‚◊Ê¥Ã⁄U üÊ…∏Ë ◊¥ „Ò¥–
ÿÁŒ a3+a
7+a
11+a
15=72 „Ò, ÃÊ ©‚∑§ ¬˝Õ◊ 17
¬ŒÊ¥ ∑§Ê ÿʪ ’⁄UÊ’⁄U „Ò —(1) 306
(2) 153
(3) 612
(4) 204
66. If
••
••
• • •• • ••
•
•
•
• • then n satisfies the
equation :
(1) n2+3n−108=0
(2) n2+5n−84=0
(3) n2+2n−80=0
(4) n2+n−110=0
67. If the coefficients of x−2
and x−4
in the
expansion of
••
••
• ••
••
•
•• • , (x > 0), are
m and n respectively, then
•• is equal to :
(1) 182
(2)
••
(3)
••
(4) 27
68. Let a1, a
2, a
3, ......, a
n, ..... be in A.P. If
a3+a
7+a
11+a
15=72, then the sum of its
first 17 terms is equal to :
(1) 306
(2) 153
(3) 612
(4) 204
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69.( )
• ••
• •• • •• ••
•• ••
•• • • =__________.
(1) (11)!
(2) 10×(11!)
(3) 101×(10!)
(4) 11×(11!)
70.
( )•
•
• • •• • •••• •
• ••• • • ••• • ••
•• • • •→
• •• •
=__________.
(1) −2
(2)
••
• •
(3)
••
(4) 2
69. ÿʪ»§‹ ( )• •
•
• •• • •• ••
•• ••
•• • • ’⁄UÊ’⁄U „Ò —
(1) (11)!
(2) 10×(11!)
(3) 101×(10!)
(4) 11×(11!)
70.
( )•
•
• • •• • •••• •
• ••• • • ••• • ••
•• • • •→
• •• •
’⁄UÊ’⁄U „Ò —
(1) −2
(2)
••
• •
(3)
••
(4) 2
69. The sum ( )• •
•
• •• • •• ••
•• ••
•• • • is equal to :
(1) (11)!
(2) 10×(11!)
(3) 101×(10!)
(4) 11×(11!)
70.
( )•
•
• • •• • •••• •
• ••• • • ••• • ••
•• • • •→
• •• •
is :
(1) −2
(2)
••
• •
(3)
••
(4) 2
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 40
71. ^pfp¡ L¡$ a, b R, (a ≠ 0).
Å¡ rh^¡e
•
•
•
• • ••• •• • • • •• ••
• •• •••• •• • • • •• • • •
• • ••••• • • • •• •
• ••
• • • •
• • ••
••••••• ∞••
≤
= ≤
−
≤
A„sfpg [0, ∞) dp„ kss lp¡e, sp¡ ¾$dey¼s Å¡X$(a, b) = __________ R>¡.
(1) ( )• ••• • •• •
(2) ( )• ••• • •• • • •
(3) ( )• •• • • •• • •
(4) ( )• ••• • •• • • •
72. ^pfp¡ L¡$ f(x)=sin4x+cos
4x. sp¡ _uQ¡_pdp„\u L$ep
A„sfpgdp„ f h^sy„ rh ¡e R>¡ ?
(1) • •••
• •• •• •
π
(2) ••• •
• •• •• •π π
(3)
•••• •
• •• •• •π π
(4)
• •••• •
• •• •• •π π
71. ◊ÊŸÊ a, b R, (a ≠ 0)– ÿÁŒ »§‹Ÿ f ¡Ê, ÁŸêŸ mÊ⁄UʬÁ⁄U÷ÊÁ·Ã „Ò —
•
•
•
• • ••• •• • • • •• ••
• •• •••• •• • • • •• • • •
• • ••••• • • • •• •
• ••
• • • •
• • ••
••••••• ∞••
≤
= ≤
−
≤
•¥Ã⁄UÊ‹ [0, ∞) ◊¥ ‚Ãà „Ò, ÃÊ ∞∑§ ∑˝§Á◊à ÿÈÇ◊(a, b) „Ò —
(1) ( )• ••• • •• •
(2) ( )• ••• • •• • • •
(3) ( )• •• • • •• • •
(4) ( )• ••• • •• • • •
72. ◊ÊŸÊ f(x)=sin4x+cos
4x „Ò, ÃÊ ÁŸêŸ ◊¥ ‚ Á∑§‚
•¥Ã⁄UÊ‹ ◊¥ f ∞∑§ flœ¸◊ÊŸ »§‹Ÿ „Ò?
(1) • •••
• •• •• •
π
(2) ••• •
• •• •• •π π
(3)
•••• •
• •• •• •π π
(4)
• •••• •
• •• •• •π π
71. Let a, b R, (a ≠ 0). If the function f defined
as
•
•
•
• • ••• •• • • • •• ••
• •• •••• •• • • • •• • • •
• • ••••• • • • •• •
• ••
• • • •
• • ••
••••••• ∞••
≤
= ≤
−
≤
is continuous in the interval [0, ∞), then an
ordered pair (a, b) is :
(1) ( )• ••• • •• •
(2) ( )• ••• • •• • • •
(3) ( )• •• • • •• • •
(4) ( )• ••• • •• • • •
72. Let f(x)=sin4x+cos
4x. Then f is an
increasing function in the interval :
(1) • •••
• •• •• •
π
(2) ••• •
• •• •• •π π
(3)
•••• •
• •• •• •π π
(4)
• •••• •
• •• •• •π π
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
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73. ^pfp¡ L¡$ A¡L$ h¾$ C A¡ •• •• • • • • • •• • •• • • • • •
••
• > Üpfp v$ip®h¡g R>¡. Å¡ C `f A¡L$ A¡hy„ tbvy$ P R>¡
L¡$ S>¡\u tbvy$ P ApNm_p õ`i®L$_p¡ Y$pm ••
\pe, sp¡ P
ApNm_p¡ Arcg„b __________ tbvy$dp„\u kpf \peR>¡.(1) (2, 3)
(2) (4, −3)
(3) (1, 7)
(4) (3, −4)
74.
( ) •• • • •
• •
• • •• • • •=__________.
(Äep„ C k„L$g__p¡ AQmp„L$ R>¡.)
(1)
• •• • •• •
••
• •• •• •• •
(2)
• •• • •• •
••
• •• •• •• •
(3)
• •• •• •
••
• •• •• •• •
(4)
• •• • •• •
••
• • •• •• •
73. ◊ÊŸÊ C ∞∑§ fl∑˝§ „Ò ¡Ê •• •• • • • • •• • •• • • • • • ,
••
• > mÊ⁄UÊ ¬˝ŒûÊ „Ò– ÿÁŒ C ¬⁄U P ∞∑§ ∞‚Ê Á’¥ŒÈ „Ò
Á∑§ P ¬⁄U πË¥øË ªß¸ S¬‡Ê¸ ⁄UπÊ ∑§Ë …Ê‹ ••
„Ò, ÃÊ fl„
Á’¥ŒÈ Á¡‚‚ P ¬⁄U πË¥øÊ ªÿÊ •Á÷‹¥’ ªÈ$¡⁄UÃÊ „Ò, „Ò —(1) (2, 3)
(2) (4, −3)
(3) (1, 7)
(4) (3, −4)
74. ‚◊Ê∑§‹ ( ) •• • • •
• •
• • •• • • • ’⁄UÊ’⁄U „Ò —
(¡„Ê° C ∞∑§ ‚◊Ê∑§‹Ÿ •ø⁄U „Ò–)
(1)
• •• • •• •
••
• •• •• •• •
(2)
• •• • •• •
••
• •• •• •• •
(3)
• •• •• •
••
• •• •• •• •
(4)
• •• • •• •
••
• • •• •• •
73. Let C be a curve given by
•• •• • • • • • •• • •= + −
••
• > . If P is a point
on C, such that the tangent at P has slope
••
, then a point through which the normal
at P passes, is :
(1) (2, 3)
(2) (4, −3)
(3) (1, 7)
(4) (3, −4)
74. The integral
( ) •• • • •
• •
• • •• • • • is equal
to :
(where C is a constant of integration.)
(1)
• •• • •• •
••
• •• •• •• •
(2)
• •• • •• •
••
• •• •• •• •
(3)
• • •• •
••
• •• •• •• •
(4)
• •• • •• •
••
• • •• •• •
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75.
•
• •
•
• • • • • • •
• • •
• • •
• •• •
• • • •• • • •• • • • • •
=_________.
Äep„ [x] A¡ x \u _p_p A\hp x _¡ kdp_ sdpd|Zp¯L$p¡dp„ kp¥\u dp¡V$p¡ |Zp¯L$ v$ip®h¡ R>¡.
(1) 6
(2) 3
(3) 7
(4)
••
76. Å¡ x R, x ≠ 0 dpV¡$, y(x) A¡L$ rhL$g_ue rh^¡e R>¡ L¡$
S > ¡\u • ••• •• • • • • •• • •• ••
• •• • • • • • • • • • •• •• • • • , sp ¡
y(x)=__________.
(Äep„ C AQm R>¡.)
(1)
•• • •••
• •
(2)
•
•• •••
• •
(3)
•
•• • •••
• •
(4)
••• • •• •
75. ‚◊Ê∑§‹ •
• •
•
• • • • • • •
• • •
• • •
• •• •
• • • •• • • •• • • • • •
, ¡„Ê°
[x], x ‚ ∑§◊ ÿÊ x ∑§ ’⁄UÊ’⁄U ◊„ûÊ◊ ¬ÍáÊÊZ∑§ „Ò, ∑§Ê◊ÊŸ „Ò —(1) 6
(2) 3
(3) 7
(4)
••
76. x R, x ≠ 0, ∑§ Á‹∞, ÿÁŒ y(x) ∞∑§ ∞‚Ê •fl∑§‹ŸËÿ»§‹Ÿ „Ò Á∑§
• ••• •• • • • • •• • •• ••
• •• • • • • • • • • • •• •• • • • „Ò, ÃÊ y (x)
’⁄UÊ’⁄U „Ò —
(¡„Ê° C ∞∑§ •ø⁄U „Ò–)
(1)
•• • •••
• •
(2)
•
•• • •••
• •
(3)
•
•• • •••
• •
(4)
••• • •• •
75. The value of the integral
•
• •
•
• • • • • • •
• • •
• • •
• •• •
• • • •• • • •• • • • • •
, where [x]
denotes the greatest integer less than or
equal to x, is :
(1) 6
(2) 3
(3) 7
(4)
••
76. For x R, x ≠ 0, if y(x) is a differentiable
function such that
• ••• •• • • • • •• • •• ••
• •• • • • • • • • • • •• •• • • • , then y(x)
equals :
(where C is a constant.)
(1)
•• • •••
• •
(2)
•
•• • •••
• •
(3)
•
•• • •••
• •
(4)
••• • •• •
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77. rhL$g kduL$fZ •• •••• • • • •
• • •• • • ••• • ••• • • Äep„
• • • •• ••
• π• , A_¡ y(0)=1, _p¡ DL¡$g _________
R>¡.
(1) • • ••• • • •• •
••• •
• • • •• •
(2)• • • •
•• • • •• •••
• •• • • •
• •
(3)• • • •
•• • • •• •••
• •• • • •
• •
(4) • • ••• • • •• •
••• •
• • • •• •
78. âL$pi_y„ A¡L$ qL$fZ f¡Mp 7x−y+1=0 `f Ap`ps\pe R>¡ S>¡ tbv$y$ (0, 1) ApNm dm¡ R>¡. Ðepfbpv$ ApqL$fZ Ap tbvy$dp„\u `fphrs®s \C f¡Mp y+2x=1
`f fl¡ R>¡. sp¡ Ap`ps qL$fZ_y„ kduL$fZ __________
R>¡.(1) 41x−38y+38=0
(2) 41x+25y−25=0
(3) 41x+38y−38=0
(4) 41x−25y+25=0
77. •fl∑§‹ ‚◊Ë∑§⁄UáÊ •• •••• • • • •
• • •• • • ••• • ••• • • ¡„Ê°
• • • •• ••
• π• „Ò ÃÕÊ y(0)=1 „Ò, ∑§Ê „‹ „Ò —
(1) • • ••• • • •• •
••• •
• • • •• •
(2)• • • •
•• • • •• •••
• •• • • •
• •
(3)• • • •
•• • • •• •••
• •• • • •
• •
(4) • • ••• • • •• •
••• •
• • • •• •
78. ¬˝∑§Ê‡Ê ∑§Ë ∞∑§ Á∑§⁄UáÊ ∞∑§ ⁄UπÊ ∑§Ë ÁŒ‡ÊÊ ◊¥ •Ê¬ÁÃÃ„Ò ¡Ê ∞∑§ •ãÿ ⁄UπÊ 7x−y+1=0 ∑§Ê Á’¥ŒÈ(0, 1) ¬⁄U Á◊‹ÃË „Ò– fl„ Á∑§⁄UáÊ Á»§⁄U ß‚ Á’¥ŒÈ ‚ ⁄UπÊy+2x=1 ∑§Ë ÁŒ‡ÊÊ ◊¥ ¬Á⁄UflÁøà „ÊÃË „Ò, ÃÊ •Ê¬ÁÃì˝∑§Ê‡Ê ∑§Ë Á∑§⁄UáÊ ∑§Ê ‚◊Ë∑§⁄UáÊ „Ò —(1) 41x−38y+38=0
(2) 41x+25y−25=0
(3) 41x+38y−38=0
(4) 41x−25y+25=0
77. The solution of the differential equation
•• •••• • • • •• • •
• • • ••• • ••• • • where • • • •• •
•• π• ,
and y(0)=1, is given by :
(1) • • ••• • • •• •
••• •
• • • •• •
(2)• • • •
•• • • •• •••
• •= +
+
(3)• • • •
•• • • •• •••
• •= −
+
(4) • • ••• • • •• •
••• •
• • • •• •
78. A ray of light is incident along a line which
meets another line, 7x−y+1=0, at the
point (0, 1). The ray is then reflected from
this point along the line, y+2x=1. Then
the equation of the line of incidence of the
ray of light is :
(1) 41x−38y+38=0
(2) 41x+25y−25=0
(3) 41x+38y−38=0
(4) 41x−25y+25=0
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 44
79. ENdtbvy$ O dp„\u `kpf \su A¡L$ f¡Mp, f¡MpAp¡3y=10−4x A_¡ 8x+6y+5=0 _¡ A_y¾$d¡tbvy$Ap¡ A A_¡ B dp„ dm¡ R>¡. sp¡ O A¡ f¡MpM„X$ AB _¡__________ NyZp¡Ñfdp„ rhcpS>_ L$f¡ R>¡.(1) 2 : 3
(2) 1 : 2
(3) 4 : 1
(4) 3 : 4
80. S>¡_y„ L¡$ÞÖ f¡MpAp¡ x−y=1 A_¡ 2x+y=3 _y„ R>¡v$tbvy$lp¡e s¡hp hsy®m_¡ tbvy$ (1, −1) ApNm_p õ`i®L$_y„kduL$fZ __________ R>¡.(1) 4x+y−3=0
(2) x+4y+3=0
(3) 3x−y−4=0
(4) x−3y−4=0
81. P A_¡ Q `fhge y2=4x `f Aph¡gp b¡ rcÞ_
tbv$y$Ap¡ R>¡ S>¡_p âpQgp¡ A_y¾$d¡ t A_¡ t1 R>¡. Å¡ P
ApNm_p¡ Arcg„b Q dp„\u `kpf \sp¡ lp¡e, sp¡ •••
_y„ Þe|_sd d|ëe __________ R>¡.(1) 2
(2) 4
(3) 6
(4) 8
79. ◊Í‹ Á’¥ŒÈ O ‚ „Ê∑§⁄U ¡ÊŸ flÊ‹Ë ∞∑§ ‚⁄U‹ ⁄UπÊ ⁄UπÊ•Ê¥3y=10−4x ÃÕÊ 8x+6y+5=0 ∑§Ê ∑˝§◊‡Ê—Á’¥ŒÈ•Ê¥ A ÃÕÊ B ¬⁄U Á◊‹ÃË „Ò¥, ÃÊ Á’¥ŒÈ O ⁄UπÊπ¥«UAB ∑§Ê Á¡‚ •ŸÈ¬Êà ◊¥ Áfl÷ÊÁ¡Ã ∑§⁄UÃÊ „Ò, fl„ „Ò —(1) 2 : 3
(2) 1 : 2
(3) 4 : 1
(4) 3 : 4
80. ©‚ flÎûÊ Á¡‚∑§Ê ∑§ãŒ˝ ‚⁄U‹ ⁄UπÊ•Ê¥ x−y=1 ÃÕÊ2x+y=3 ∑§Ê ¬˝ÁÃë¿UŒ Á’¥ŒÈ „Ò, ∑§ Á’¥ŒÈ (1, −1) ¬⁄UπË¥øË ªß¸ S¬‡Ê¸ ⁄πÊ ∑§Ê ‚◊Ë∑§⁄UáÊ „Ò —(1) 4x+y−3=0
(2) x+4y+3=0
(3) 3x−y−4=0
(4) x−3y−4=0
81. P ÃÕÊ Q ¬⁄Ufl‹ÿ y2=4x ¬⁄U ÁSÕà ŒÊ Á÷㟠Á’¥ŒÈ „Ò
Á¡Ÿ∑§ ¬˝Êø‹ ∑˝§◊‡Ê— t ÃÕÊ t1 „Ò¥– ÿÁŒ P ¬⁄U πË¥øÊ
ªÿÊ •Á÷‹¥’ Q ‚ „Ê∑§⁄U ¡ÊÃÊ „Ò, ÃÊ ••• ∑§Ê ãÿÍŸÃ◊
◊ÊŸ „Ò —(1) 2
(2) 4
(3) 6
(4) 8
79. A straight line through origin O meets the
lines 3y=10−4x and 8x+6y+5=0 at
points A and B respectively. Then O
divides the segment AB in the ratio :
(1) 2 : 3
(2) 1 : 2
(3) 4 : 1
(4) 3 : 4
80. Equation of the tangent to the circle, at the
point (1, −1), whose centre is the point of
intersection of the straight lines x−y=1
and 2x+y=3 is :
(1) 4x+y−3=0
(2) x+4y+3=0
(3) 3x−y−4=0
(4) x−3y−4=0
81. P and Q are two distinct points on the
parabola, y2=4x, with parameters t and t
1
respectively. If the normal at P passes
through Q, then the minimum value of •••
is :
(1) 2
(2) 4
(3) 6
(4) 8
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 45
82. A¡L $ Arshge S> ¡_p ¡ d y¿e An ip „L $h••
•• •
•• •• • •• • _p â^p_ An `f R>¡ s\p s¡_p
rifp¡tbvy$Ap¡ Ap ip„L$h_u _prcAp¡ `f R>¡. Å¡
Arshge_u DÐL¡$ÞÖsp •• lp¡e, sp¡ _uQ¡_pdp„\u L$ey„
tbvy$ s¡_p `f Aph¡gy„ _l] lp¡e ?(1) (0, 2)
(2) ( )• ••• •
(3) ( )• • ••• •
(4) ( )• ••• •
83. A¡L$ kdsgdp„ A¡L$ rÓL$p¡Z ABC R>¡. S>¡_p rifp¡tbvy$Ap¡A(2, 3, 5), B(−1, 3, 2) A_¡ C(λ, 5, µ) R>¡. Å¡ Adp„\u _uL$msu dÝeNp epdpnp kp\¡ kdp_ fus¡ Y$m¡gR>¡, sp¡ (λ3
+µ3+5) _u qL„$ds __________ R>¡.
(1) 1130
(2) 1348
(3) 676
(4) 1077
82. ∞∑§ •Áì⁄Ufl‹ÿ, Á¡‚∑§Ê •ŸÈ¬˝SÕ •ˇÊ ‡ÊÊ¥∑§fl••
•• •
•• •• • •• • ∑§ ŒËÉʸ •ˇÊ ∑§Ë ÁŒ‡ÊÊ ◊¥ „Ò ÃÕÊ
Á¡‚∑§ ‡ÊË·¸ ß‚ ‡ÊÊ¥∑§fl ∑§Ë ŸÊÁ÷ÿÊ¥ ¬⁄U „Ò– ÿÁŒ
•Áì⁄Ufl‹ÿ ∑§Ë ©à∑§ãŒ˝ÃÊ •• „Ò, ÃÊ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ
‚Ê Á’¥ŒÈ ß‚ ¬⁄U ÁSÕà Ÿ„Ë¥ „Ò?(1) (0, 2)
(2) ( )• ••• •
(3) ( )• • ••• •
(4) ( )• ••• •
83. ∞∑§ ‚◊Ë ◊¥ ∞∑§ ÁòÊ÷È¡ ABC „Ò Á¡‚∑§ ‡ÊË·¸A(2, 3, 5), B(−1, 3, 2) ÃÕÊ C(λ, 5, µ) „Ò¥– ÿÁŒA ‚ „Ê∑§⁄U ¡ÊÃË ◊ÊÁäÿ∑§Ê ÁŸŒ¸‡ÊÊ¥∑§ •ˇÊÊ¥ ¬⁄U ‚◊ÊŸM§¬ ‚ ! ÊÈ∑§Ë „Ò, ÃÊ (λ3
+µ3+5) ∑§Ê ◊ÊŸ „Ò —
(1) 1130
(2) 1348
(3) 676
(4) 1077
82. A hyperbola whose transverse axis is along
the major axis of the conic,
•••
• ••• •• • •• •
and has vertices at the foci of this conic. If
the eccentricity of the hyperbola is
•• , then
which of the following points does NOT
lie on it ?
(1) (0, 2)
(2) ( )• ••• •
(3) ( )• • ••• •
(4) ( )• ••• •
83. ABC is a triangle in a plane with vertices
A(2, 3, 5), B(−1, 3, 2) and C(λ, 5, µ). If the
median through A is equally inclined to the
coordinate axes, then the value of
(λ3+µ
3+5) is :
(1) 1130
(2) 1348
(3) 676
(4) 1077
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 46
84. S>¡_p dpV¡$ f¡MpAp¡ •• •• • • •
• ••• •
•− − +
= =
A_¡ •
• •• • • ••• •
•• ••− − −
= = kdsgue \pe
s¡hu λ _u rcÞ_ hpõsrhL$ qL „ $dsp ¡_u k„¿ep__________ R>¡.(1) 4
(2) 1
(3) 2
(4) 3
85. ^pfp¡ L¡$ ABC A¡L$ rÓL$p¡Z R>¡ S>¡_y„ `qfL¡$ÞÖ P ApNmR>¡. Å¡ A, B, C A_¡ P _p õ\p_kqv$ip¡ A_y¾$d¡
• •• • • • •→ → →
A_¡ • • • ••
• • •→ → →
• • lp¡e, sp¡ Ap
rÓL$p¡Z_p g„bL¡$ÞÖ_p¡ õ\p_kqv$i __________ R>¡.
(1) • • • •• • •→ → →
• •
(2)• • • ••
•• • •→ → →
• ••
(3) •→
(4) • • • •
•
• • •→ → →
• •
84. λ ∑§ fl„ Á÷㟠flÊSÃÁfl∑§ ◊ÊŸÊ¥ ∑§Ë ‚¥ÅÿÊ Á¡Ÿ∑§
Á‹∞ ⁄ πÊ∞° •• •• • • ••
• ••• •
•− − +
= = ÃÕÊ
•• •• • • ••
• ••• ••− − −
= = ‚◊ËËÿ „Ò¥, „Ò —
(1) 4
(2) 1
(3) 2
(4) 3
85. ◊ÊŸÊ ABC ∞∑§ ÁòÊ÷È¡ „Ò Á¡‚∑§Ê ¬Á⁄U∑§ãŒ˝ P ¬⁄U „Ò–ÿÁŒ Á’¥ŒÈ•Ê¥ A, B, C ÃÕÊ P ∑§ ÁSÕÁà ‚ÁŒ‡Ê ∑˝§◊‡Ê—
• •• • • • •→ → →
ÃÕÊ • • • ••
• • •→ → →
• • „Ò¥, ÃÊ ß‚ ÁòÊ÷È¡
∑§ ‹¥’ - ∑§ãŒ˝ ∑§Ê ÁSÕÁà ‚ÁŒ‡Ê „Ò —
(1) • • • •• • •→ → →
• •
(2)• • • ••
•• • •→ → →
• ••
(3) •→
(4) • • • •
•
• • •→ → →
• •
84. The number of distinct real values of λ for
which the lines •• •• • • ••
• ••• •
•− − +
= =
and •
• •• • • ••• •
•• ••− − −
= = are
coplanar is :
(1) 4
(2) 1
(3) 2
(4) 3
85. Let ABC be a triangle whose circumcentre
is at P. If the position vectors of A, B, C
and P are • •• • • • •→ → →
and • • • •
•• • •→ → →
• •
respectively, then the position vector of the
orthocentre of this triangle, is :
(1) • • • •• • •→ → →
• •
(2)• • • ••
•• • •→ → →
• ••
(3) •→
(4) • • • •
•
• • •→ → →
• •
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 47
86. 5 Ahgp¡L$_p¡_p¡ dÝeL$ 5 A_¡ rhQfZ 124 R>¡. Å¡s¡dp„\u ÓZ Ahgp¡L$_p¡ 1, 2 A_¡ 6 lp¡e, sp¡ dprlsu_y„dÝeL$\u kf¡fpi rhQg_ __________ \pe.(1) 2.4
(2) 2.8
(3) 2.5
(4) 2.6
87. A¡L$ âep¡N S>¡V$gp hMs Akam \pe R>¡ s¡_p\u b¡NZp¡ kam \pe R>¡. Ap âep¡N_p R> âeÐ_p¡dp„\uAp¡R>pdp„ Ap¡R>u 5 kamsp dmhp_u k„cph_p__________ R>¡.
(1)
• • •• • •
(2)
• • •• • •
(3)
• • •• • •
(4)
• • •• • •
86. 5 ¬˝ˇÊáÊÊ¥ ∑§Ê ◊Êäÿ 5 „Ò ÃÕÊ ©Ÿ∑§Ê ¬˝‚⁄UáÊ 124 „Ò–ÿÁŒ ©Ÿ◊¥ ‚ ÃËŸ ¬˝ˇÊáÊ 1, 2 ÃÕÊ 6 „Ò¥, ÃÊ ßŸ •Ê°∑§«∏Ê¥∑§Ê ◊Êäÿ ‚ ◊Êäÿ Áflø‹Ÿ „Ò —(1) 2.4
(2) 2.8
(3) 2.5
(4) 2.6
87. ∞∑§ ¬˝ÿʪ ∑§ ‚»§‹ „ÊŸ ∑§Ê ‚¥ÿʪ ©‚∑§ Áfl»§‹„ÊŸ ∑§ ‚¥ÿʪ ∑§Ê ŒÈªÈŸÊ „Ò– ß‚ ¬˝ÿʪ ∑§ 6 ¬⁄UˡÊáÊÊ¥◊¥ ‚ ∑§◊ ‚ ∑§◊ ¬Ê°ø ∑§ ‚»§‹ „ÊŸ ∑§Ë ¬˝ÊÁÿ∑§ÃÊ„Ò —
(1)
• • •• • •
(2)
• • •• • •
(3)
• • •• • •
(4)
• • •• • •
86. The mean of 5 observations is 5 and their
variance is 124. If three of the observations
are 1, 2 and 6 ; then the mean deviation
from the mean of the data is :
(1) 2.4
(2) 2.8
(3) 2.5
(4) 2.6
87. An experiment succeeds twice as often as
it fails. The probability of at least 5
successes in the six trials of this experiment
is :
(1)
• • •• • •
(2)
• • •• • •
(3)
• • •• • •
(4)
• • •• • •
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 48
88. If A>0, B>0 and • • •• • ••π• • , then the
minimum value of tanA+tanB is :
(1) • • • ••
(2) • • • ••
(3) • • •• ••
(4)••
89. The angle of elevation of the top of a
vertical tower from a point A, due east of it
is 45. The angle of elevation of the top of
the same tower from a point B, due south
of A is 30. If the distance between A and
B is • • • •• , then the height of the tower
(in metres), is :
(1) • • •(2) 54
(3) • • •(4) 108
88. ÿÁŒ A>0, B>0 ÃÕÊ • • •• • ••π• • „ Ò , ÃÊ
tanA+tanB ∑§Ê ãÿÍŸÃ◊ ◊ÊŸ „Ò —
(1) • • • ••
(2) • • • ••
(3) • • •• ••
(4)••
89. Á’¥ŒÈ A ‚, ¡Ê ∞∑§ ™§äflʸœ⁄U ◊ËŸÊ⁄U ∑§ ¬Ífl¸ ∑§Ë •Ê⁄U„Ò, ◊ËŸÊ⁄U ∑§ ‡ÊË·¸ ∑§Ê ©ãŸÿŸ ∑§ÊáÊ 45 „Ò– Á’¥ŒÈ B,
¡Ê Á’¥ŒÈ A ∑§ ŒÁˇÊáÊ ◊¥ „Ò, ‚ ©‚Ë ◊ËŸÊ⁄U ∑§ ‡ÊË·¸ ∑§Ê©ãŸÿŸ ∑§ÊáÊ 30 „Ò– ÿÁŒ A ÃÕÊ B ∑§ ’Ëø ∑§Ë ŒÍ⁄UË
• • • ◊Ë. „Ò, ÃÊ ◊ËŸÊ⁄U ∑§Ë ™§°øÊ߸ (◊Ë. ◊¥) „Ò —
(1) • • •(2) 54
(3) • • •(4) 108
88. Å¡ A>0, B>0 A_¡ • • •• • ••π• • lp ¡e, sp ¡
tanA+tanB _y„ Þe|_sd d|ëe __________ R>¡.
(1) • • • ••
(2) • • • ••
(3) • • •• ••
(4)••
89. A¡L$ rifp¡g„b V$phf_u |h® sfa_p tbvy$ A \u V$phf_uV$p¡Q_p¡ DÐk¡ L$p¡Z 45 R>¡. tbvy$ A _u v$rnZ¡ Aph¡gtbvy$ B \u V$phf_u V$p¡Q_p¡ DÐk¡ L$p¡Z 30 R>¡. Å¡ A
A_¡ B hÃQ¡_y„ A„sf • • • •• lp¡e, sp¡ V$phf_uKQpC (duV$fdp„) __________ R>¡.
(1) • • •(2) 54
(3) • • •(4) 108
SET - 03 ENGLISH SET - 03 HINDI SET - 03 GUJARATI
Set - 03 49
90. The contrapositive of the following
statement,
“If the side of a square doubles, then its area
increases four times”, is :
(1) If the side of a square is not doubled,
then its area does not increase four
times.
(2) If the area of a square increases four
times, then its side is doubled.
(3) If the area of a square increases four
times, then its side is not doubled.
(4) If the area of a square does not
increase four times, then its side is not
doubled.
- o O o -
90. ÁŸêŸ ∑§ÕŸ ∑§Ê ¬˝ÁÃœŸÊà◊∑§ (contrapositive)
„Ò —
““ÿÁŒ Á∑§‚Ë flª¸ ∑§Ë ÷È¡Ê ŒÈªÈŸË „Ê ¡Ê∞, ÃÊ ©‚∑§ÊˇÊòÊ»§‹ øÊ⁄U ªÈŸÊ ’…∏ ¡ÊÃÊ „Ò”” —
(1) ÿÁŒ ∞∑§ flª¸ ∑§Ë ÷È¡Ê ŒÈªÈŸË Ÿ ∑§Ë ¡Ê∞, ÃÊ©‚∑§Ê ˇÊòÊ»§‹ øÊ⁄U ªÈŸÊ Ÿ„Ë¥ ’…∏ÃÊ–
(2) ÿÁŒ Á∑§‚Ë flª¸ ∑§Ê ÊòÊ»§‹ øÊ⁄U ªÈŸÊ ’…∏ ¡Ê∞,ÃÊ ©‚∑§Ë ÷È¡Ê ŒÈªÈŸË „Ê ¡ÊÃË „Ò–
(3) ÿÁŒ Á∑§‚Ë flª¸ ∑§Ê ˇÊòÊ»§‹ øÊ⁄U ªÈŸÊ ’…∏ ¡ÊÃÊ„Ò, ÃÊ ©‚∑§Ë ÷È¡Ê ŒÈªÈŸË Ÿ„Ë¥ „ÊÃË–
(4) ÿÁŒ Á∑§‚Ë flª¸ ∑§Ê ÊòÊ»§‹ øÊ⁄U ªÈŸÊ Ÿ„Ë¥ ’…∏ÃÊ,ÃÊ ©‚∑§Ë ÷È¡Ê ŒÈªÈŸË Ÿ„Ë¥ „ÊÃË–
- o O o -
90. _uQ¡_p rh^p__y„ kdp_p\ â¡fZ L$ey„ R>¡ ?
""Å¡ L$p¡C Qp¡fk_u bpSy> bdZu L$fhpdp„ Aph¡, sp¡ s¡_y„n¡Óam QpfNÏ„ h ¡.''
(1) Å¡ L$p¡C Qp¡fk_u bpSy> bdZu L$fhpdp„ _ Aph¡,sp¡ s¡_y„ n¡Óam QpfNÏ„ h^i¡ _rl.
(2) Å¡ L$p¡C Qp¡fk_y„ n¡Óam QpfNÏ„ h^pfhpdp„ Aph¡,sp¡ s¡_u bpSy> bdZu \pe.
(3) Å¡ L$p¡C Qp¡fk_y„ n¡Óam QpfNÏ„ h^pfhpdp„Aph¡, sp¡ s¡_u bpSy> bdZu _ \pe.
(4) Å¡ L$p¡C Qp¡fk_y„ n¡Óam QpfNÏ„ h^pfhpdp„ _Aph¡, sp¡ s¡_u bpSy> bdZu _ \pe.
- o O o -
Question and Answer Key - April 10 OnlineQuestion No. Answer Key Question No. Answer Key Question No. Answer Key
Q1 2 Q31 2 Q61 4Q2 1 Q32 2 Q62 1Q3 4 Q33 1 Q63 4Q4 3 Q34 1 Q64 3Q5 2 Q35 2 Q65 4Q6 4 Q36 2 Q66 1Q7 4 Q37 3 Q67 1Q8 1 Q38 4 Q68 1Q9 4 Q39 2 Q69 2
Q10 4 Q40 1 Q70 1Q11 4 Q41 3 Q71 1Q12 3 Q42 2 Q72 2Q13 2 Q43 4 Q73 3Q14 1 Q44 4 Q74 2Q15 2 Q45 2 Q75 2Q16 2 Q46 4 Q76 3Q17 2 Q47 4 Q77 3Q18 1 Q48 1 Q78 1Q19 3 Q49 4 Q79 3Q20 2 Q50 2 Q80 2Q21 2 Q51 2 Q81 4Q22 4 Q52 2 Q82 4Q23 3 Q53 4 Q83 2Q24 2 Q54 4 Q84 4Q25 4 Q55 3 Q85 4Q26 3 Q56 1 Q86 2Q27 3 Q57 1 Q87 3Q28 2 Q58 2 Q88 3Q29 2 Q59 4 Q89 2Q30 1 Q60 3 Q90 4
Note:-
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Page 1 PHYSICS : English & Hindi 091. x and y displacements of a particle aregiven as x(t)5a sinvt and y(t)5a sin2vt.Its trajectory will look like :
(1)
(2)
(3)
(4)
1. ØçÎ ç·¤âè ·¤æ ·ð¤ x ÌÍæ y çßSÍæÂÙæð´ ·¤æð ·ý¤×àæÑ,x(t)5a sinvt ÌÍæ y(t)5a sin2vt, âð çÙM¤çÂÌç·¤Øæ ÁæÌæ ãñ Ìæð, §â·¤æ Âýÿæð - ÂÍ, çÙÙæ´ç·¤Ì ×ð´ âðç·¤â·ð¤ Áñâæ çιæ§ü Îð»æ?
(1)
(2)
(3)
(4)
Page 2 PHYSICS : English & Hindi 092. If a body moving in a circular pathmaintains constant speed of 10 ms21, thenwhich of the following correctly describesrelation between acceleration andradius ?
(1)
(2)
(3)
(4)
2. ØçÎ ßëææ·¤æÚU ÂÍ ×ð´ »çÌ ·¤ÚUÌð ãé° ç·¤âè ç´ÇU (ßSÌé)·¤è ¿æÜ 10 ms21 ãñ ¥æñÚU Øã ¥¿ÚU ÕÙè ÚUãÌè ãñ Ìæð,çÙÙæ´ç·¤Ì ×ð´ âð ·¤æñÙâæ ¥æÜð¹, ßÚUæ ÌÍæ çæØæ ·ð¤Õè¿ âÕÏ ·¤æ ÆUè·¤ (âãè) ç¿ææ ·¤ÚUÌæ ãñ?
(1)
(2)
(3)
(4)
Page 3 PHYSICS : English & Hindi 093. A block of mass m510 kg rests on ahorizontal table. The coefficient of frictionbetween the block and the table is 0.05.When hit by a bullet of mass 50 g movingwith speed v, that gets embedded in it, theblock moves and comes to stop aftermoving a distance of 2 m on the table.
If a freely falling object were to acquirespeed
10
v after being dropped from heightH, then neglecting energy losses andtaking g510 ms22, the value of H is closeto :(1) 0.2 km(2) 0.3 km(3) 0.4 km(4) 0.5 km
4. A block of mass m50.1 kg is connected toa spring of unknown spring constant k. Itis compressed to a distance x from itsequilibrium position and released from rest.After approaching half the distance
2
x
from equilibrium position, it hits anotherblock and comes to rest momentarily, whilethe other block moves with a velocity3 ms21. The total initial energy of thespring is :(1) 1.5 J(2) 0.6 J(3) 0.3 J(4) 0.8 J
5. A uniform solid cylindrical roller of massm is being pulled on a horizontal surfacewith force F parallel to the surface andapplied at its centre. If the acceleration ofthe cylinder is a and it is rolling withoutslipping then the value of F is :(1) ma(2) 2 ma(3) 3
ma2
(4) 5 ma
3
3. ç·¤âè Üæò·¤ (»éÅU·ð¤) ·¤æ ÎýÃØ×æÙ m510 kg ãñÐØã °·¤ ÿæñçÌÁ ×ðÁ ÂÚU ÚU¹æ ãñÐ §Ù ÎæðÙæð´ ·ð¤ Õè¿æáüæ »éææ´·¤50.05 ãñÐ §â Üæò·¤ ÂÚU 50 g ÎýÃØ×æÙ·¤è °·¤ »æðÜè v ¿æÜ âð ÅU·¤ÚUæÌè ¥æñÚU §â×ð´ Ï´âÁæÌè ãñÐ §ââð Øã Üæò·¤, ×ðÁ ÂÚU 2 m çßSÍæçÂÌãæð·¤ÚU L¤·¤ ÁæÌæ ãñÐØçÎ, H ª¡¤¿æ§ü âð ×éÌ M¤Â âð ç»ÚUæÙð ·ð¤ Âà¿æÌ÷ ·¤æð§üßSÌé
10
v ¿æÜ ÂýæÌ ·¤ÚU ÜðÌè ãñ Ìæð, ª¤Áæü-ÿæØ ·¤æðÙ»Ø ×æÙÌð ãé°, H ·¤æ âçÙ·¤ÅU ×æÙ ãæð»æ Ñ(g510 ms22)(1) 0.2 km(2) 0.3 km(3) 0.4 km(4) 0.5 km
4. ç·¤âè Üæò·¤ ·¤æ ÎýÃØ×æÙ m50.1 kg ãñÐ Øã °·¤°ðâè ·¤×æÙè (çSÂý») âð ÁéÇæ ãñ çÁâ·¤æ ·¤×æÙè çSÍÚUæ·¤k ãñÐ §â·¤æð §â·¤è âæØæßSÍæ âð x ÎêÚUè Ì·¤ ÎÕæ·¤ÚUçßÚUæ×æßSÍæ âð ÀUæðǸ çÎØæ ÁæÌæ ãñÐ âæØæßSÍæ âð
2
x
ÎêÚUè ÂÚU ¥æÙð ÂÚU Øã °·¤ ¥Ø Üæò·¤ âð ÅU·¤ÚUæÌæãñ ¥æñÚU ÿæçæ·¤ M¤Â âð L¤·¤ ÁæÌæ ãñ, ÁÕç·¤, ÎêâÚUæÜæò·¤ 3 ms21 ·ð¤ ßð» âð »çÌ ·¤ÚUÙð Ü»Ìæ ãñÐ Ìæð,·¤×æÙè ·¤è ·é¤Ü ÂýæÚ´UçÖ·¤ ª¤Áæü ãñ Ñ(1) 1.5 J(2) 0.6 J(3) 0.3 J(4) 0.8 J
5. m ÎýÃØ×æÙ ·ð¤ ç·¤âè °·¤â×æÙ ÆUæðâ çâçÜÇUÚU ·ð¤·ð¤Îý ÂÚU °·¤ ÕÜ F ܻ淤ÚU, ©âð ç·¤âè â×ÌÜâÌã ÂÚU, ©â·ð¤ â×æÌÚU ¹è´¿æ Áæ ÚUãæ ãñÐ ØçÎ ØãçâçÜÇUÚU Õ»ñÚU (çÕÙæ) çȤâÜð a ßÚUæ âð Üéɸ·¤ÚUãæ ãñ Ìæð, F ·¤æ ×æÙ ãæð»æ Ñ(1) ma(2) 2 ma(3) 3
ma2
(4) 5 ma
3
Page 4 PHYSICS : English & Hindi 096. Consider a thin uniform square sheet madeof a rigid material. If its side is a, mass mand moment of inertia I about one of itsdiagonals, then :
(1) 2maI >
12
(2) 2 2ma ma< I <
24 12
(3) 2maI
125
(4) 2maI
245
7. A very long (length L) cylindrical galaxyis made of uniformly distributed mass andhas radius R (R<<L). A star outside thegalaxy is orbiting the galaxy in a planeperpendicular to the galaxy and passingthrough its centre. If the time period of staris T and its distance from the galaxys axisis r, then :(1) T2 ; r3(2) T ; r2(3) T ; r(4) T r;
8. If it takes 5 minutes to fill a 15 litre bucketfrom a water tap of diameter 2
cmp
thenthe Reynolds number for the flow is(density of water5103 kg/m3 andviscosity of water51023 Pa.s) close to :(1) 5500(2) 11,000(3) 550(4) 1100
6. °·¤ ÂÌÜè, °·¤â×æÙ, ß»æü·¤æÚU, ¿æÎÚU (àæèÅU) ç·¤âèÎëɸ ÂÎæÍü ·¤è ÕÙè ãñÐ ØçÎ §â·¤è °·¤ ÖéÁæ a,ÎýÃØ×æÙ m ÌÍæ ç·¤âè °·¤ çß·¤æü ·ð¤ ÂçÚUÌÑ, §â·¤æÁÇ¸ß ¥ææêæü I ãñ Ìæð Ñ(1) 2ma
I > 12
(2) 2 2ma ma< I <
24 12
(3) 2maI
125
(4) 2maI
245
7. °·¤ ÕãéÌ ÜÕè »ñÜðâè (×´Îæç·¤Ùè) (ÜÕæ§ü L)°·¤â×æÙ çßÌçÚUÌ ÎýÃØ ·¤è ÕÙè ãñ, §â·¤è çæØæR (R<<L) ãñÐ §â »ñÜðâè ·ð¤ ÕæãÚ °·¤ ÌæÚUæ,»ñÜðâè ·¤è ÂçÚU·ý¤×æ ·¤ÚU ÚUãæ ãñÐ §â·¤è ÂçÚU·ý¤×æ ·¤æâ×ÌÜ »ñÜðâè ·ð¤ â×ÌÜ ·ð¤ ÜÕßÌ÷ ãñ ÌÍæ §â·ð¤·ð¤Îý âð ãæð·¤ÚU »é$ÁÚUÌæ ãñÐ ØçÎ, ÌæÚðU ·¤è »ñÜðâè ·¤è¥ÿæ âð ÎêÚUè r ãñ ¥æñÚU ÌæÚðU ·¤æ ¥æßÌü ·¤æÜ T ãñ Ìæð ÑU(1) T2 ; r3(2) T ; r2(3) T ; r(4) T r;
8. ÂæÙè ·ð¤ ÙÜ ·¤è °·¤ ÅUæð´ÅUè ·¤æ ÃØæâ 2 cmp
ãñЧââð 15 çÜÅUÚU ·¤è °·¤ ÕæËÅUè ·¤æð ÖÚUÙð ×ð´ 5 ç×ÙÅU·¤æ â×Ø Ü»Ìæ ã ñÐ (ØçÎ, ÁÜ ·¤ææÙß5103 kg/m3 ÌÍæ ÁÜ ·¤è àØæÙÌæ51023Pa.s ãñ) Ìæð, §â Âýßæã ·ð¤ çÜØð ÚðUÙËÇ÷Uâ â´Øæãæð»è Ñ(1) 5500(2) 11,000(3) 550(4) 1100
Page 5 PHYSICS : English & Hindi 09
9. ØçÎ ·¤æ¡¿ ·¤è Îæð ÜðÅUæð´ ·ð¤ Õè¿ ×ð´ ÁÜ ·¤è °·¤ÂÌÜè ÂÚUÌ ãæð (¥æÚðU¹ Îðç¹Øð) Ìæð ©Ù ÜðÅUæð´ ·¤æð¹è´¿·¤ÚU ¥Ü» ·¤ÚUÙæ ÕãéÌ ·¤çÆUÙ ãæðÌæ ãñÐ
§â·¤æ ·¤æÚUæ Øã ãñ ç·¤, ÁÜ, ç·¤ÙæÚUæð´ ÂÚU çâçÜ´ÇUÚUè(ÕðÜÙæ·¤æÚU) âÌãð´ ÕÙæ ÎðÌæ ãñ çÁââð ßæØé×´ÇUÜ ·¤èÌéÜÙæ ×ð´ ßãæ¡ ÎæÕ ·¤× ãæð ÁæÌæ ãñÐ ØçÎ §â çâçÜ´ÇUÚUèâÌã (ÂëcÆU) ·¤è çæØæ R ãñ ÌÍæ ÁÜ ·¤æ ÂëcÆU ÌÙæßT ãñ Ìæð, Îæð ÜðÅUæð´ ·ð¤ Õè¿ ÁÜ ×ð´ ÎæÕ ç·¤ÌÙæ ·¤×ãæð»æ?(1) T
4R
(2) T
2R
(3) 4T
R
(4) 2T
R
9. If two glass plates have water betweenthem and are separated by very smalldistance (see figure), it is very difficult topull them apart. It is because the water inbetween forms cylindrical surface on theside that gives rise to lower pressure in thewater in comparison to atmosphere. If theradius of the cylindrical surface is R andsurface tension of water is T then thepressure in water between the plates islower by :
(1) T
4R
(2) T
2R
(3) 4T
R
(4) 2T
R
Page 6 PHYSICS : English & Hindi 0910. An ideal gas goes through a reversible
cycle a®b®c®d has the V - T diagramshown below. Process d®a and b®c areadiabatic.
The corresponding P - V diagram for theprocess is (all figures are schematic andnot drawn to scale) :
(1)
(2)
(3)
(4)
10. °·¤ ¥æÎàæü »ñ⠷𤠩·ý¤×æèØ ¿·ý¤ a®b®c®d ,·ð¤ çÜØð V - T ¥æÚðU¹ Øãæ¡ ÎàææüØæ »Øæ ãñÐ Âý·ý¤×d®a ÌÍæ b®c L¤Î÷Ïæðc× ãñ´Ð
Ìæð, §â Âý·ý¤× ·ð¤ çÜØð, â´»Ì P - V ¥æÚðU¹ ãæð»æ(âÖè ¥æÚðU¹ ÃØßSÍæ ¥æÚðU¹ ãñ ¥æñÚU S·ð¤Ü ·ð¤ ¥ÙéâæÚUÙãè´ ãñ´) Ñ
(1)
(2)
(3)
(4)
Page 7 PHYSICS : English & Hindi 0911. In an ideal gas at temperature T, theaverage force that a molecule applies onthe walls of a closed container depends onT as Tq. A good estimate for q is :
(1) 2(2) 1(3) 1
2
(4) 1
4
12. A simple harmonic oscillator of angularfrequency 2 rad s21 is acted upon by anexternal force F5sint N. If the oscillatoris at rest in its equilibrium position at t50,its position at later times is proportionalto :(1) 1
sin sin 22
t t1
(2) 1sin cos 2
2t t1
(3) 1cos sin 2
2t t2
(4) 1sin sin 2
2t t2
13. A bat moving at 10 ms21 towards a wallsends a sound signal of 8000 Hz towardsit. On reflection it hears a sound offrequency f. The value of f in Hz is close to(speed of sound5320 ms21)(1) 8258(2) 8516(3) 8000(4) 8424
11. ç·¤âè ¥æÎàæü »ñâ ×ð´, ç·¤âè ¥æé mæÚUæ »ñâ ·ð¤ ÕÎÂææ ·¤è ÎèßæÚUæð´ ÂÚU Ü»æØæ »Øæ ¥æñâÌ ÕÜ, »ñâ ·ð¤Ìæ T ÂÚU, Tq ·ð¤ ¥ÙéâæÚU çÙÖüÚU ·¤ÚUÌæ ãñÐ Ìæð, q ·¤æâçÙ·¤ÅU ×æÙ ãñ Ñ(1) 2(2) 1(3) 1
2
(4) 1
4
12. ç·¤âè âÚUÜ ¥æßÌü ÎæðçÜæ ·¤è ·¤æðæèØ ¥æßëçæ2 rad s21 ãñÐ §â ÂÚU °·¤ Õæs ÕÜ, F5sintØêÅUÙ (N) Ü»Ìæ ãñÐ ØçÎ â×Ø t50 ÂÚU, Øã ÎæðçÜæ,¥ÂÙè âæØæßSÍæ ×ð´ çßÚUæ× çSÍçÌ ×ð´ ãñ Ìæð, §â·ð¤Âà¿æÌ÷ ·ð¤ ç·¤âè â×Ø ×ð´, §â·¤è çSÍçÌ çÙÙæ´ç·¤Ì×ð´ ç·¤â·ð¤ â×æÙéÂæÌè ãæð»è?(1) 1
sin sin 22
t t1
(2) 1sin cos 2
2t t1
(3) 1cos sin 2
2t t2
(4) 1sin sin 2
2t t2
13. 10 ms21 ·¤è ¿æÜ âð ç·¤âè ÎèßæÚU ·¤è ¥æðÚU ÁæÌæãé¥æ °·¤ ¿×»æÎǸ, ÎèßæÚU ·¤è ¥æðÚU 8000 Hz ·¤æßçÙ â´·ð¤Ì (çâÙÜ) ÂýðçáÌ ·¤ÚUÌæ (ÖðÁÌæ) ãñÐÎèßæÚU âð ÂÚUæßÌüÙ ·ð¤ Âà¿æÌ÷ ßã f ¥æßëçæ ·¤èßçÙ, âéÙÌæ ãñÐ ØçÎ, ßçÙ ·¤è ¿æÜ 320 ms21 ãñÌæð, Hz ×ð´ f ·¤æ âçÙ·¤ÅU ×æÙ ãæð»æ Ñ(1) 8258(2) 8516(3) 8000(4) 8424
Page 8 PHYSICS : English & Hindi 0914. A thin disc of radius b52a has a concentrichole of radius a in it (see figure). It carriesuniform surface charge s on it. If theelectric field on its axis at height h (h<<a)from its centre is given as Ch then valueof C is :
(1)0a
s
e
(2)0
2a
se
(3)04a
s
e
(4)08a
s
e
14. °·¤ ÂÌÜè çÇUS·¤ (¿ç·ý¤·¤æ) ·¤è çæØæ b ãñÐ §â×ð´ÕÙð °·¤ â´·ð¤Îýè çÀUÎý (ÀðUÎ) ·¤è çæØæ a ãñÐ(b52a) Ð çÇS·¤ ÂÚU °·¤â×æÙ ÂëcÆU ¥æßðàæ s ãñÐØçÎ §â·¤è ¥ÿæ ÂÚU ÌÍæ §â·ð¤ ·ð¤Îý âð h ª¡¤¿æ§üÂÚU, (h<<a), çßléÌ ÿæðæ Ch ãæð Ìæð, C ·¤æ ×æÙãñ Ñ
(1)0a
s
e
(2)0
2a
se
(3)04a
s
e
(4)08a
s
e
Page 9 PHYSICS : English & Hindi 0915. Shown in the figure are two point charges
1Q and 2Q inside the cavity of aspherical shell. The charges are kept nearthe surface of the cavity on opposite sidesof the centre of the shell. If s1 is the surfacecharge on the inner surface and Q1 netcharge on it and s2 the surface charge onthe outer surface and Q2 net charge on itthen :
(1) s1 ¹ 0, Q1 ¹ 0s2 ¹ 0, Q2 ¹ 0
(2) s1 ¹ 0, Q1 5 0s2 ¹ 0, Q2 5 0
(3) s1 ¹ 0, Q1 5 0s2 5 0, Q2 5 0
(4) s1 5 0, Q1 5 0s2 5 0, Q2 5 0
16. If the capacitance of a nanocapacitor ismeasured in terms of a unit u made bycombining the electronic charge e, Bohrradius a0, Plancks constant h and speedof light c then :(1) 2
0
e c
hau 5
(2) 2
0
e h
cau 5
(3) 20e a
hc
u 5
(4) 20
hc
e au 5
15. Øãæ¡ ¥æÚðU¹ ×ð´, ç·¤âè »æðÜæ·¤æÚU ·¤æðàæ (àæñÜ) ·ð¤ ·¤æðÅUÚU·ð¤ ÖèÌÚU Îæð çÕÎé-¥æßðàæ 1Q ÌÍæ 2Q ÎàææüØð »Øðãñ´Ð Øð ¥æßðàæ ·¤æðÅUÚU ·¤è âÌã ·ð¤ çÙ·¤Å U§â Âý·¤æÚUÚU¹ð »Øð ãñ´ ç·¤, °·¤ ¥æßðàæ ·¤æðàæ ·ð¤ ·ð¤Îý ·¤è °·¤¥æðÚU ãñ ¥æñÚU ÎêâÚUæ ·ð¤Îý ·ð¤ çßÂÚUèÌ ÎêâÚUè ¥æðÚUÐ ØçÎ,ÖèÌÚUè ÌÍæ ÕæãÚUè âÌãæð´ (ÂëcÆUæð´) ÂÚU, ÂëcÆU ¥æßðàæ·ý¤×àæÑ s1 ÌÍæ s2 ¥æñÚU ÙðÅU ¥æßðàæ ·ý¤×àæÑ Q1 ÌÍæQ2 ãæð Ìæð Ñ
(1) s1 ¹ 0, Q1 ¹ 0s2 ¹ 0, Q2 ¹ 0
(2) s1 ¹ 0, Q1 5 0s2 ¹ 0, Q2 5 0
(3) s1 ¹ 0, Q1 5 0s2 5 0, Q2 5 0
(4) s1 5 0, Q1 5 0s2 5 0, Q2 5 0
16. ØçÎ ç·¤âè ÙñÙæð â´ÏæçÚUæ ·¤è ÏæçÚUÌæ, °·¤ °ðâð ×ææ·¤u ×ð´ ×æÂè ÁæØ, Áæð §ÜðÅþUæòÙ ¥æßðàæ e , ÕæðÚU-çæØæa0, Üæ´·¤ çSÍÚUæ´·¤ h ÌÍæ Âý·¤æàæ ·¤è ¿æÜ c ·ð¤â´ØæðÁÙ âð ÕÙæ ãñ Ìæð Ñ(1) 2
0
e c
hau 5
(2) 2
0
e h
cau 5
(3) 20e a
hc
u 5
(4) 20
hc
e au 5
Page 10 PHYSICS : English & Hindi 0917. Suppose the drift velocity vd in a materialvaried with the applied electric field E as
d Ev ; . Then V - I graph for a wiremade of such a material is best given by :
(1)
(2)
(3)
(4)
17. ØçÎ, ç·¤âè ÂÎæÍü ×ð ¥Âßæã ßð», vd ·¤æ ×æÙ, Ü»æØð»Øð çßléÌ ÿæðæ E ÂÚU §â Âý·¤æÚU çÙÖüÚU ·¤ÚUÌæ ãñ, ç·¤
d Ev ; Ð Ìæð çÙÙæ´ç·¤Ì ×ð´ âð ·¤æñÙ-âæ »ýæȤ(¥æÜð¹), §â ÂÎæÍü âð ÕÙð ÌæÚU ·ð¤ çÜØð, âçÙ·¤ÅUV - I »ýæȤ ãæð»æ?
(1)
(2)
(3)
(4)
Page 11 PHYSICS : English & Hindi 09
18. °·¤ ßæðËÅU×èÅUÚU âð â×æÌÚU ·ý¤× ×ð´, Îæð ÕñÅUçÚUØæ¡, ÁæðǸ軧ü ãñÐ ÂãÜè, 10V ÌÍæ 1V ¥æÌçÚU·¤ ÂýçÌÚUæðÏ ·¤è¥æñÚU ÎêâÚUè, 15V ÌÍæ 0.6V ¥æÌçÚU·¤ ÂýçÌÚUæðÏ ·¤è(¥æÚðU¹ Îðç¹Øð) Ìæð, ßæðËÅU×èÅUÚU ·ð¤ ÂÆUÙ (ÚUèçÇ´U»)·¤æ âçÙ·¤ÅU ×æÙ ãæð»æ Ñ
(1) 11.9 V(2) 12.5 V(3) 13.1 V(4) 24.5 V
19. çßÖßæÌÚ V mæÚUæ ßçÚUÌ, °·¤ ÂýæðÅUæòÙ (ÎýÃØ×æÙ m),ç·¤âè ¥ÙéÂýSÍ °·¤â×æÙ ¿éÕ·¤èØ ÿæðæ B âð ãæð·¤ÚUÌèßý ¿æÜ âð »é$ÁÚUÌæ ãñÐ Øã ¿éÕ·¤èØ ÿæðæ d ¿æñǸæ§üÌ·¤ çßSÌçÚUÌ ãñÐ ØçÎ, Øã ÂýæðÅUæòÙ, ¿éÕ·¤èØ ÿæððæ ·ð¤·¤æÚUæ ¥ÂÙè »çÌ ·¤è ÂýæÚ´UçÖ·¤ çÎàææ âð a ·¤æðæ âðçß¿çÜÌ ãæ ð ÁæÌæ ã ñ (¥æÚ ð U¹ Î ð ç¹Øð) Ìæ ð ,sin a ·¤æ ×æÙ ãæð»æ Ñ
(1) qdB
2 mV
(2) qB
d 2mV
(3) qBd
2mV
(4) BdqV
2m
18. A 10V battery with internal resistance1V and a 15V battery with internalresistance 0.6V are connected in parallelto a voltmeter (see figure). The reading inthe voltmeter will be close to :
(1) 11.9 V(2) 12.5 V(3) 13.1 V(4) 24.5 V
19. A proton (mass m) accelerated by apotential difference V flies through auniform transverse magnetic field B. Thefield occupies a region of space by widthd. If a be the angle of deviation of protonfrom initial direction of motion (seefigure), the value of sin a will be :
(1) qdB
2 mV
(2) qB
d 2mV
(3) qBd
2mV
(4) BdqV
2m
Page 12 PHYSICS : English & Hindi 09
20. ç·¤âè ÂçÚUÙæçÜ·¤æ ·¤è ÜÕæ§ü 25 cm ÌÍæ çæØæ2 cm ãñ ¥æñÚU §â×ð´ ÌæÚU ·ð¤ ·é¤Ü 500 Èð¤ÚðU ÜÂðÅðU »Øðãñ´Ð §ââð 15 A ·¤è ÏæÚUæ ÂýßæçãÌ ãæð ÚUãè ãñÐ ØãÂçÚUÙæçÜ·¤æ, §âè âæ§Á ÌÍæ ¿éÕ·¤Ù M
→ (¿éÕ·¤èØ¥ææêæü/¥æØÌÙ), ·ð¤ ÌéËØ ãñ Ìæð , M
→? ? ãñ Ñ(1) 3p Am21(2) 30000 Am21(3) 300 Am21(4) 30000p Am21
21. ç·¤âè ·é´¤ÇUÜè âð ÂýßæçãÌ çßléÌ ÏæÚUæ ·¤æ ×æÙ, 0.1 s×ð´, 5 A âð 2 A ãæð ÁæÌæ ãñ çÁââð, 50 V ·¤è ¥æñâÌßæðËÅUÌæ ©ÂÙ ãæðÌè ãñÐ Ìæð, §â ·é¤ÇUÜè ·¤æ SßÂýðÚU·¤ßãñ Ñ(1) 0.67 H(2) 1.67 H(3) 3 H(4) 6 H
20. A 25 cm long solenoid has radius 2 cmand 500 total number of turns. It carries acurrent of 15 A. If it is equivalent to amagnet of the same size and magnetizationM→ (magnetic moment/volume), then M
→? ?is :(1) 3p Am21(2) 30000 Am21(3) 300 Am21(4) 30000p Am21
21. When current in a coil changes from 5 Ato 2 A in 0.1 s, an average voltage of 50 Vis produced. The self - inductance of thecoil is :(1) 0.67 H(2) 1.67 H(3) 3 H(4) 6 H
Page 13 PHYSICS : English & Hindi 09
22. Îæð ÂçÚUÂÍæð´ (a) ÌÍæ (b) ×ð çSß¿ S1 ÌÍæ S2, t50ÂÚU ÕÎ ç·¤Øð ÁæÌð ãñ´ ¥æñÚU ÂØæüÌ ÜÕð â×Ø Ì·¤ÕÎ ÚU¹ð ÁæÌð ãñ´, Ìæð t ³ 0 ·ð¤ çÜØð, Îæð ÂçÚUÂÍæð´ ×ð´çßléÌ ÏæÚUæ¥æð´ ·ð¤ çß¿ÜÙ (ÂçÚUßÌüÙ) ·¤æð, ·¤æñÙâæ»ýæȤ çÙ·¤ÅUÌ× Îàææ üÌæ ã ñ? (¥æÚ ð U¹ ·ð¤ßÜÃØßSÍæ×·¤ ãñ´ ¥æñÚU S·ð¤Ü ·ð¤ ¥ÙéâæÚU Ùãè´ ãñ´)
(1)
(2)
(3)
(4)
22. In the circuits (a) and (b) switches S1 andS2 are closed at t50 and are kept closedfor a long time. The variation of currentsin the two circuits for t ³ 0 are roughlyshown by (figures are schematic and notdrawn to scale) :
(1)
(2)
(3)
(4)
Page 14 PHYSICS : English & Hindi 09
23. x-çÎàææ ×ð´ ¿ÜÌè ãé§ü ç·¤âè çßléÌ ¿éÕ·¤èØ ÌÚ´U» ·¤è¥æßëçæ 231014 Hz ãñ ÌÍæ §â·¤æ çßléÌ ÿæðæ27 Vm21 ãñÐ Ìæð, çÎØð »Øð çÙÙæ´ç·¤Ì çß·¤ËÂæð´ ×ð´âð ·¤æñÙ âæ çß·¤ËÂ, §â ÌÚ´U» ·ð¤ ¿éÕ·¤èØ ÿæðæ ·¤æðÂý·¤ÅU ·¤ÚUÌæ ãñ?(1) 8B ( , t) (3 10 T)x j
→ ∧25 3
8 14sin 2 (1.5 10 2 10 t)x
2p 3 23
(2) 8B ( , t) (9 10 T)x k→ ∧25 3
6 14sin 2 (1.5 10 2 10 t)x
2p 3 23
(3) 8B ( , t) (9 10 T)x i→ ∧25 3
8 14sin 2 (1.5 10 2 10 t)x
2p 3 23
(4) 8B ( , t) (9 10 T)x j→ ∧25 3
6 14sin 1.5 10 2 10 tx
23 23
24. ¥æ·¤æð °·¤ àæðçß´» ÎÂüæ ÕÙæÙð ·¤æð ·¤ãæ ÁæÌæ ãñÐØçÎ ·¤æð§ü ÃØçÌ §â ÎÂüæ ·¤æð ¥ÂÙð ¿ðãÚðU âð 10 cmÎêÚU ÚU¹Ìæ ãñ ¥æñÚU ¿ðãÚðU ·ð¤ ¥æßçÏüÌ ÂýçÌçÕÕ ·¤æð,âéçßÏæÁÙ·¤-ÎàæüÙ ·¤è çÙ·¤ÅUÌ× ÎêÚUè, 25 cm ÂÚUÎð¹Ìæ ãñ Ìæð, ÎÂüæ ·¤è ß·ý¤Ìæ çæØæ ãæð»è Ñ(1) 30 cm(2) 24 cm(3) 60 cm(4) 224 cm
23. An electromagnetic wave travelling in thex-direction has frequency of 231014 Hzand electric field amplitude of 27 Vm21.From the options given below, which onedescribes the magnetic field for thiswave ?(1) 8B ( , t) (3 10 T)x j
→ ∧25 3
8 14sin 2 (1.5 10 2 10 t)x
2p 3 23
(2) 8B ( , t) (9 10 T)x k→ ∧25 3
6 14sin 2 (1.5 10 2 10 t)x
2p 3 23
(3) 8B ( , t) (9 10 T)x i→ ∧25 3
8 14sin 2 (1.5 10 2 10 t)x
2p 3 23
(4) 8B ( , t) (9 10 T)x j→ ∧25 3
6 14sin 1.5 10 2 10 tx
23 23
24. You are asked to design a shaving mirrorassuming that a person keeps it 10 cm fromhis face and views the magnified image ofthe face at the closest comfortable distanceof 25 cm. The radius of curvature of themirror would then be :(1) 30 cm(2) 24 cm(3) 60 cm(4) 224 cm
Page 15 PHYSICS : English & Hindi 09
25. x-çÎàææ ×ð´ »çÌ ·¤ÚUÌæ ãé¥æ °·¤ â×æÌÚU §ÜðÅþUæòÙ Âé´Ád ¿æñÇæ§ü ·¤è çÛæÚUè ÂÚU ¥æÂçÌÌ ãæðÌæ ãñ (¥æÚðU¹ Îðç¹Øð)ÐØçÎ §â çÛæÚUè âð çÙ·¤ÜÙð ·ð¤ Âà¿æÌ÷ §ÜðÅþUæòÙ,y-çÎàææ ×ð´, py â´ßð» ÂýæÌ ·¤ÚU ÜðÌð ãñ´ Ìæð, çSÜÅU âð»éÁÚUÙð ßæÜð ¥çÏ·¤æ´àæ §ÜðÅþUæòÙæð´ ·ð¤ çÜØð (ØçÎ hÜæ´·¤ çÙØÌæ´·¤ ãñ) Ñ
(1) ?py?d; h(2) ?py?d>h(3) ?py?d<h(4) ?py?d>>h
26. ç·¤âè ÎêÚUÎàæü·¤ ·ð¤ ¥çÖÎëàØ·¤ ÌÍæ Ùðçæ·¤æ ·¤è Ȥæð·¤âÎêçÚUØæ¡ ·ý¤×àæÑ 150 cm ÌÍæ 5 cm ãñÐ ØçÎ 1 kmÎêÚU çSÍÌ ç·¤âè 50 m ª¡¤¿ð ÅUæßÚU (×èÙæÚU) ·¤æð, âæ×æØçßØæâ ×ð´, ÎêÚUÎàæü·¤ âð Îð¹Ùð ÂÚU, ÅUæßÚU ·ð¤ ÂýçÌçÕÕmæÚUæ ÕÙæØæ »Øæ ·¤æðæ, u ãæð Ìæð, u ·¤æ ×æÙ ãæð»æֻܻ Ñ(1) 18(2) 158(3) 308(4) 608
27. 50 V ßæðËÅUÌæ mæÚUæ ßçÚUÌ §ÜðÅþUæòÙ ·¤è Îð-ÕýæòÜèÌÚ´U»ÎñØü ·¤æ çÙ·¤ÅUÌ× ×æÙ ãæð»æ Ñ(?e?51.6310219 C, me59.1310231 kg,h56.6310234 Js)(1) 0.5 Å(2) 1.2 Å(3) 1.7 Å(4) 2.4 Å
25. A parallel beam of electrons travelling inx-direction falls on a slit of width d (seefigure). If after passing the slit, an electronacquires momentum py in the y-directionthen for a majority of electrons passingthrough the slit (h is Plancks constant) :
(1) ?py?d; h(2) ?py?d>h(3) ?py?d<h(4) ?py?d>>h
26. A telescope has an objective lens of focallength 150 cm and an eyepiece of focallength 5 cm. If a 50 m tall tower at adistance of 1 km is observed through thistelescope in normal setting, the angleformed by the image of the tower is u, thenu is close to :(1) 18(2) 158(3) 308(4) 608
27. de-Broglie wavelength of an electronaccelerated by a voltage of 50 V is close to(?e?51.6310219 C, me59.1310231 kg,h56.6310234 Js) :(1) 0.5 Å(2) 1.2 Å(3) 1.7 Å(4) 2.4 Å
Page 16 PHYSICS : English & Hindi 09
28. ØçÎ ¿éÕ·¤èØ ÿæðæ B ·ð¤ ÂýÖæß ×ð´, â×ÌÜ ×ð´ »çÌ·¤ÚUÌð ãé°, m ÎýÃØ×æÙ ÌÍæ q ¥æßðàæ ·ð¤ ·¤æ ÂÚU,ÕæðÚU ×æòÇUÜ Üæ»ê ç·¤Øæ (ÂýØæð» ×ð´ ÜæØæ) ÁæØ Ìæð,¥æßðçàæÌ ·¤æ ·¤è n ßè´ SÌÚU ×ð´ ª¤Áæü ãæð»è Ñ(1) hqB
n2 m
p
(2) hqBn
4 m
p
(3) hqBn
8 m
p
(4) hqBn
m
p
29. ç·¤âè ¥-ÕæØçâÌ n-p â´çÏ ×ð´ §ÜðÅþUæòÙ n - ÿæðæ âðp - ÿæðæ ·¤æð çßâçÚUÌ ãæðÌð ãñ´ Øæð´ç·¤ Ñ(1) p - ÿæðæ ×ð´ çÀUÎý (ãæðÜ) ©ã𴠥淤çáüÌ ·¤ÚUÌð
ãñÐ(2) §ÜðÅþUæòÙ çßÖßæÌÚU ·ð¤ ·¤æÚUæ â´çÏ ·ð¤ ÂæÚU
¿Üð ÁæÌð ãñ´Ð(3) n - ÿæðæ ×ð´ §ÜðÅþUæòÙæð´ ·¤è âæ´ÎýÌæ p - ÿæðæ âð
¥çÏ·¤ ãæðÌè ãñÐ(4) §ÜðÅþUæòÙ ·ð¤ßÜ n âð p ÿæðæ ·¤æð ÁæÌð ã´ñ §â·ð¤
çßÂÚèUÌ (p âð n ·¤æð ) Ùãè´Ð
28. If one were to apply Bohr model to aparticle of mass m and charge q movingin a plane under the influence of amagnetic field B, the energy of thecharged particle in the nth level will be :(1) hqB
n2 m
p
(2) hqBn
4 m
p
(3) hqBn
8 m
p
(4) hqBn
m
p
29. In an unbiased n-p junction electronsdiffuse from n - region to p - regionbecause :(1) holes in p - region attract them(2) electrons travel across the junctiondue to potential difference(3) electron concentration in n - regionis more as compared to that inp - region(4) only electrons move from n to pregion and not the vice - versa
Page 17 PHYSICS : English & Hindi 0930. Diameter of a steel ball is measured usinga Vernier callipers which has divisions of0.1 cm on its main scale (MS) and 10divisions of its vernier scale (VS) match 9divisions on the main scale. Three suchmeasurements for a ball are given as :
S.No. MS (cm) VS divisions
1. 0.5 8
2. 0.5 4
3. 0.5 6
If the zero error is 20.03 cm, then meancorrected diameter is :(1) 0.56 cm(2) 0.59 cm(3) 0.53 cm(4) 0.52 cm
- o 0 o -
30. SÅUèÜ ·¤è °·¤ »æðÜè ·¤æ ÃØæâ °·¤ °ðâð ßçÙüØÚU·ñ¤ÜèÂâü âð ÙæÂæ ÁæÌæ ãñ çÁâ·ð¤ ×éØ Âñ×æÙð ·¤æ °·¤Öæ» (MSD) 0.1 cm ãñ, ÌÍæ §â×ð´ ßçÙüØÚ Âñ×æÙðU(VS) ·ð¤ 10 Öæ», ×éØ Âñ×æÙð ·ð¤ 9 Öæ»æð´ ·ð¤ ÕÚUæÕÚUãñ´Ð »æðÜè ·ð¤ ÃØæâ ·ð¤ çÜØð ÌèÙ ÂæÆ÷UØæ´·¤ (ÚUèçÇU´»)Øãæ¡ çÎØð »Øð ãñ´ Ñ
â¼Ë ¼Î½ §Ö¼Ë¾Õ Í ¼Ë§ cm
Ä̾὿U §Ö¼Ë¾Õ Õ »Ë
1. 0.5 8
2. 0.5 4
3. 0.5 6
ØçÎ ßçÙüØÚU ·ñ¤ÜèÂâü ·¤è àæêØ æéçÅU 20.03 cm, ãñÌæð, ÃØæâ ·¤æ ×æØ â´àææðçÏÌ ×æÙ ãæð»æ Ñ(1) 0.56 cm(2) 0.59 cm(3) 0.53 cm(4) 0.52 cm
- o 0 o -
Page 1 Chemistry : English & Hindi 091. A sample of a hydrate of barium chlorideweighing 61 g was heated until all thewater of hydration is removed. The driedsample weighed 52 g. The formula of thehydrated salt is : (atomic mass,Ba5137 amu, Cl535.5 amu)
(1) BaCl2.H2O(2) BaCl2.2H2O(3) BaCl2.3H2O(4) BaCl2.4H2O
2. Which of the following is not anassumption of the kinetic theory of gases ?(1) A gas consists of many identicalparticles which are in continualmotion.(2) Gas particles have negligible volume.(3) At high pressure, gas particles aredifficult to compress.(4) Collisions of gas particles areperfectly elastic.
3. If the principal quantum number n56, thecorrect sequence of filling of electrons willbe :(1) ns®np®(n21)d®(n22)f(2) ns®(n22)f®(n21)d®np(3) ns®(n21)d®(n22)f®np(4) ns®(n22)f®np®(n21)d
1. 61 g ÕðçÚUØ× ÜæðÚUæ§ÇU ·ð¤ ãæ§ÇþðUÅU ·ð¤ °·¤ Ù×êÙð ·¤æð»ÚU× ·¤ÚU·ð¤ âé¹æØæ »ØæÐ âê¹ð Ù×êÙð ·¤æ ßÁÙ 52 gÍæÐ ãæ§ÇþðUÅU Üßæ ·¤æ âêæ ãñ Ñ (ÂÚU×ææé ÎýÃØ×æÙ,Ba5137 amu, Cl535.5 amu)(1) BaCl2.H2O(2) BaCl2.2H2O(3) BaCl2.3H2O(4) BaCl2.4H2O
2. çÙÙ ×ð´ âð ·¤æñÙ âè »ñâæ𴠷𤠥æé»çÌ·¤ çâhæ´Ì ·¤è¥ßÏæÚUææ Ùãè´ ãñ?(1) °·¤ »ñâ ÕãéÌ âæÚðU â×M¤Â ·¤ææð´ âð ÕÙÌè ãñ
Áæð Ü»æÌæÚU »çÌ·¤ ¥ßSÍæ ×ð´ ÚUãÌð ãñ´Ð(2) »ñ⠷𤠷¤ææð´ ·¤æ ¥æØÌÙ Ù»Ø ãñÐ(3) ©æ ÎæÕ ÂÚU »ñâ ·¤ææð ·¤æ â´ÂèÇUÙ ·¤çÆUÙ ãñÐ(4) »ñ⠷𤠷¤ææð´ ·ð¤ ×Ø â´æ^ ÂêæüÌÑ ÂýØæSÍ
ãæðÌð ãñÐ
3. ×éØ ßæ´ÅU× â´Øæ n56 ·ð¤ çÜ° §ÜðÅþUæòÙæð´ ·ð¤ ÖÚUÙð·¤æ âãè ·ý¤× ãñ Ñ(1) ns®np®(n21)d®(n22)f(2) ns®(n22)f®(n21)d®np(3) ns®(n21)d®(n22)f®np(4) ns®(n22)f®np®(n21)d
Page 2 Chemistry : English & Hindi 094. After understanding the assertion andreason, choose the correct option.
Assertion : In the bonding molecularorbital (MO) of H2, electrondensity is increased betweenthe nuclei.Reason : The bonding MO is cA1cB,which shows destructiveinterference of the combiningelectron waves.(1) Assertion and reason are correct andreason is the correct explanation forthe assertion.(2) Assertion and reason are correct, butreason is not the correct explanationfor the assertion.(3) Assertion is correct, reason isincorrect.(4) Assertion is incorrect, reason iscorrect.
5. The heat of atomization of methane andethane are 360 kJ/mol and 620 kJ/mol,respectively. The longest wavelength oflight capable of breaking the C - C bond is(Avogadro number56.0231023,h56.62310234 J s) :(1) 1.493103 nm(2) 2.483103 nm(3) 2.483104 nm(4) 1.493104 nm
6. A solution at 208C is composed of 1.5 molof benzene and 3.5 mol of toluene. If thevapour pressure of pure benzene and puretoluene at this temperature are 74.7 torrand 22.3 torr, respectively, then the totalvapour pressure of the solution and thebenzene mole fraction in equilibrium withit will be, respectively :(1) 35.0 torr and 0.480(2) 38.0 torr and 0.589(3) 30.5 torr and 0.389(4) 35.8 torr and 0.280
4. ¥çÖ·¤ÍÙ ¥æñÚU Ì·ü¤ ·¤æð â×Ûæ·¤ÚU âãè çß·¤Ë ¿éçÙ°Ð¥çÖ·¤ÍÙ : ãæ§ÇþUæðÁÙ ·ð¤ ¥æÕ´Ïè ¥æçß·¤ ·¤ÿæ·¤
×ð´ §ÜðÅþUæòÙ æÙß ÙæçÖ·¤æð´ ·ð¤ Õè¿Õɸæ ãé¥æ ãæðÌæ ãñÐ
Ì·ü¤ : ¥æÕ´Ïè ¥æçß·¤ ·¤ÿæ·¤ cA1cB ãñÁæð â´ØæðÁè §ÜðÅþUæòÙ ÌÚ´U»æð´ ·¤æ çßÙæàæèÃØçÌ·¤ÚUæ ÎàææüÌæ ãñÐ
(1) ¥çÖ·¤ÍÙ ß Ì·ü¤ ÎæðÙæð´ âãè ãñ´ ¥æñÚU Ì·ü¤¥çÖ·¤ÍÙ ·¤æ âãè SÂcÅUè·¤ÚUæ ãñÐ
(2) ¥çÖ·¤ÍÙ ß Ì·ü¤ ÎæðÙæð´ âãè ãñ´ ×»ÚU Ì·ü¤¥çÖ·¤ÍÙ ·¤æ âãè SÂcÅUè·¤ÚUæ Ùãè´ ãñÐ
(3) ¥çÖ·¤ÍÙ âãè ãñ ×»ÚU Ì·ü¤ »ÜÌ ãñÐ(4) ¥çÖ·¤ÍÙ »ÜÌ ãñ ¥æñÚU Ì·ü¤ âãè ãñÐ
5. ç×ÍðÙ ÌÍæ §ÍðÙ ·ð¤ ·¤æè·¤ÚUæ ª¤c×æ ·ý¤×àæÑ360 kJ/mol ÌÍæ 620 kJ/mol ãñ´Ð C - C ¥æÕ´Ï·¤æð ÌæðǸÙð ·¤è ÿæ×Ìæ ÚU¹Ùð ßæÜè Âý·¤æàæ ·¤è ÎèæüÌ×ÌÚ´U»ÎñØü ãæð»è (¥æßæð»æÎýæð â´Øæ56.0231023,h56.62310234 J s) :(1) 1.493103 nm(2) 2.483103 nm(3) 2.483104 nm(4) 1.493104 nm
6. 208C ÂÚU °·¤ çßÜØÙ ×ð´ 1.5 ×æðÜ ÕðÁèÙ ¥æñÚU3.5 ×æðÜ ÅUæðÜé§Ù ãñ´Ð ¥»ÚU §â Ìæ ÂÚU àæéh ÕðÁèÙ¥æñÚU àæéh ÅUæðÜé§Ù ·ð¤ ßæc ÎæÕ ·ý¤×àæÑ 74.7 torr¥æñÚU 22.3 torr ãñ, ÌÕ çßÜØÙ ·¤æ ·é¤Ü ßæc ÎæÕ¥æñÚU ÕðÁèÙ ·¤æ ×æðÜ ¥´àæ §â·ð¤ âæØ ×ð´ ·ý¤×àæÑ ãñ´ Ñ(1) 35.0 torr ¥æñÚU 0.480(2) 38.0 torr ¥æñÚU 0.589(3) 30.5 torr ¥æñÚU 0.389(4) 35.8 torr ¥æñÚU 0.280
Page 3 Chemistry : English & Hindi 097. Gaseous N2O4 dissociates into gaseousNO2 according to the reaction
N2O4(g)ì 2NO2(g)At 300 K and 1 atm pressure, the degreeof dissociation of N2O4 is 0.2. If one moleof N2O4 gas is contained in a vessel, thenthe density of the equilibrium mixture is :(1) 1.56 g/L(2) 3.11 g/L(3) 4.56 g/L(4) 6.22 g/L
8. A variable, opposite external potential(Eext) is applied to the cellZn|Zn21 (1 M) ?? Cu21 (1 M) | Cu, ofpotential 1.1 V. When Eext < 1.1 V andEext > 1.1 V, respectively electrons flowfrom :(1) anode to cathode and cathode toanode(2) cathode to anode and anode tocathode(3) cathode to anode in both cases(4) anode to cathode in both cases
9. The reaction2N2O5(g) ® 4NO2(g)1O2(g)follows first order kinetics. The pressureof a vessel containing only N2O5 wasfound to increase from 50 mm Hg to87.5 mm Hg in 30 min. The pressureexerted by the gases after 60 min. will be(Assume temperature remains constant) :(1) 106.25 mm Hg(2) 116.25 mm Hg(3) 125 mm Hg(4) 150 mm Hg
7. »ñâèØ N2O4 »ñâèØ NO2 ×ð´ çÙÙ ¥çÖç·ý¤Øæ ·ð¤¥ÙéâæÚU çßØæðçÁÌ ãæðÌæ ãñ ÑN2O4(g)ì 2NO2(g)300 K ¥æñÚU 1 atm ÎæÕ ÂÚU N2O4 ·¤è çßØæðÁÙ×æææ 0.2 ãñÐ ¥»ÚU °·¤ ×æðÜ N2O4 »ñâ ·¤æð °·¤ Âææ×ð´ çÜØæ Áæ° ÌÕ âæØ ç×ææ ·¤æ æÙß ãñ Ñ(1) 1.56 g/L(2) 3.11 g/L(3) 4.56 g/L(4) 6.22 g/L
8. 1.1 V çßÖß ·ð¤ âðÜZn|Zn21 (1 M) ?? Cu21 (1 M) | Cu, ×ð´ °·¤ÂçÚUßÌèü çßÂÚUèÌ Õæs çßÖß (Eext) Ü»æØæ »ØæÐÁÕ Eext < 1.1 V ÌÍæ Eext > 1.1 V, ãæð ÌÕ§ÜðÅþUæòÙæð´ ·¤æ Âýßæã ãæð»æ Ñ(1) °ÙæðÇU âð ·ñ¤ÍæðÇU ÌÍæ ·ñ¤ÍæðÇU âð °ÙæðÇU(2) ·ñ¤ÍæðÇU âð °ÙæðÇU ÌÍæ °ÙæðÇU âð ·ñ¤ÍæðÇU(3) ÎæðÙæð´ çSÍçÌ ×ð´ ·ñ¤ÍæðÇU âð °ÙæðÇU(4) ÎæðÙæð´ çSÍçÌ ×ð´ °ÙæðÇU âð ·ñ¤ÍæðÇU
9. çÙÙ ¥çÖç·ý¤Øæ °·¤ ÂýÍ× ·¤æðÅUè ÕÜ»çÌ·¤è ãñ Ñ2N2O5(g) ® 4NO2(g)1O2(g)30 min ×ð´, °·¤ Âææ çÁâ×ð´ ·ð¤ßÜ N2O5 Íæ, ©â·¤æÎæÕ 50 mm Hg âð 87.5 mm Hg Õɸ »ØæÐ60 min ·ð¤ ÕæÎ Øã ÎæÕ ãæð»æ (×æÙ ÜèçÁ° ·¤è ÌæÂçÙØÌ ãñ) :(1) 106.25 mm Hg(2) 116.25 mm Hg(3) 125 mm Hg(4) 150 mm Hg
Page 4 Chemistry : English & Hindi 0910. The following statements relate to theadsorption of gases on a solid surface.Identify the incorrect statement amongthem :
(1) Enthalpy of adsorption is negative(2) Entropy of adsorption is negative(3) On adsorption, the residual forces onthe surface are increased(4) On adsorption decrease in surfaceenergy appears as heat
11. In the long form of the periodic table, thevalence shell electronic configuration of5s2 5p4 corresponds to the element presentin :(1) Group 16 and period 6(2) Group 17 and period 5(3) Group 16 and period 5(4) Group 17 and period 6
12. In the isolation of metals, calcinationprocess usually results in :(1) metal carbonate(2) metal oxide(3) metal sulphide(4) metal hydroxide
13. Permanent hardness in water cannot becured by :(1) Boiling(2) Ion exchange method(3) Calgons method(4) Treatment with washing soda
10. çÙÙ ·¤ÍÙ ÆUæðâ ÂëcÆU ÂÚU »ñâèØ ¥çÏàææðáæ ·ð¤ â´ÎÖü×ð´ ãñ´Ð §Ù×ð´ âð »ÜÌ ·¤ÍÙ ãñ Ñ(1) ¥çÏàææðáæ ·¤è °ÍñËÂè «¤ææ×·¤ ãñÐ(2) ¥çÏàææðáæ ·¤è °ÅþUæòÂè «¤ææ×·¤ ãñÐ(3) ¥çÏàææðáæ ÂÚU, ÂëcÆU ÂÚU ¥ßçàæcÅU ÕÜ ÕɸÌð
ãñÐ(4) ¥çÏàææðáæ ÂÚU, ÂëcÆU ª¤Áæü ·¤æ Oæâ ª¤c×æ ·ð¤
M¤Â ×ð´ Âý·¤ÅU ãæðÌæ ãñÐ
11. ¥æßÌü âæÚUæè ·ð¤ Îèæü SßM¤Â ×ð´, ¥»ÚU â´ØæðÁè ·¤æðàæ§ÜðÅþUæòÙ çßØæâ 5s2 5p4 ãñ ÌÕ ßã Ìß ©ÂçSÍÌãñ Ñ(1) ß»ü 16 ¥æñÚU ¥æßÌü 6 ×ð´(2) ß»ü 17 ¥æñÚU ¥æßÌü 5 ×ð´(3) ß»ü 16 ¥æñÚU ¥æßÌü 5 ×ð´(4) ß»ü 17 ¥æñÚU ¥æßÌü 6 ×ð´
12. ÏæÌé¥æð´ ·ð¤ çÙc·¤áüæ ×ð´, çÙSÌæÂÙ âð ¥·¤âÚU ÕÙÌðãñ´ Ñ(1) ÏæÌé ·¤æÕæðüÙðÅU(2) ÏæÌé ¥æòâæ§ÇU(3) ÏæÌé âËȤæ§ÇU(4) ÏæÌé ãæ§ÇþUæòâæ§ÇU
13. ÁÜ ·¤è SÍæØè ·¤ÆUæðÚUÌæ ·¤æð §â Âýç·ý¤Øæ âð ÆUè·¤ Ùãè´ç·¤Øæ Áæ â·¤Ìæ ãñ Ñ(1) ©ÕæÜÙæ(2) ¥æØÙ çßçÙ×Ø çßçÏ(3) ·ð¤Ü»æòÙ çßçÏ(4) ÏæßÙ âæðÇUæ ·ð¤ ©Â¿æÚU âð
Page 5 Chemistry : English & Hindi 09
14. ãæ§ÇþUæòâæ§ÇUæð´ ·¤æ ÌæÂèØ SÍæçØß ·¤æ âãè ·ý¤× ãñ Ñ(1) Ba(OH)2 < Sr(OH)2 < Ca(OH)2 <Mg(OH)2(2) Ba(OH)2 < Ca(OH)2 < Sr(OH)2 <Mg(OH)2(3) Mg(OH)2 < Ca(OH)2 < Sr(OH)2 <Ba(OH)2(4) Mg(OH)2 < Sr(OH)2 < Ca(OH)2 <Ba(OH)2
15. âÕâð ·¤× â´Øæ ·ð¤ ¥æâè¥Ü ÕÙæÌæ ãñ Ñ(1) Ùæ§ÅþUæðÁÙ(2) âËȤÚU(3) Üé¥æðÚUèÙ(4) ÜæðÚUèÙ
16. ßè.°â.§ü.Âè.¥æÚU. (VSEPR) çâhæ´Ì ·ð¤ ¥ÙéâæÚU,XeOF4 ·¤è Øæç×çÌ ãñ Ñ(1) çæ·¤æðæèØ çmçÂÚUæç×ÇUè(2) ß»ü çÂÚUæç×ÇUè(3) ¥cÅUȤܷ¤èØ(4) ´¿·¤æðæèØ â×ÌÜèØ
17. Üßæ X ·¤æ ÁÜèØ çßÜØÙ SCN2 ·ð¤ âæÍ ¹êÙèÜæÜ Ú´U» ¥æñÚU K4[Fe(CN)6] ·ð¤ âæÍ ÙèÜæ Ú´U» ÎðÌæãñÐ X °·¤ â·¤æÚUæ×·¤ ·ý¤æðç×Ü ÜæðÚUæ§ÇU ÂÚUèÿææ ÖèÎðÌæ ãñÐ Üßæ X ãñ Ñ(1) CuCl2(2) FeCl3(3) Cu(NO3)2(4) Fe(NO3)3
18. çÙÙ ×ð âð ·¤æñÙ âæ ¥æé/¥æØÙ â´·é¤Ü Øæñç»·¤æð´ ×ð´´çÜ»ñÇU Ùãè´ ãæð â·¤Ìæ ãñ?(1) CO(2) CN2
(3) CH4(4) Br2
14. The correct order of thermal stability ofhydroxides is :(1) Ba(OH)2 < Sr(OH)2 < Ca(OH)2 <Mg(OH)2(2) Ba(OH)2 < Ca(OH)2 < Sr(OH)2 <Mg(OH)2(3) Mg(OH)2 < Ca(OH)2 < Sr(OH)2 <Ba(OH)2(4) Mg(OH)2 < Sr(OH)2 < Ca(OH)2 <Ba(OH)2
15. The least number of oxyacids are formedby :(1) Nitrogen(2) Sulphur(3) Fluorine(4) Chlorine
16. The geometry of XeOF4 by VSEPR theoryis :(1) trigonal bipyramidal(2) square pyramidal(3) octahedral(4) pentagonal planar
17. An aqueous solution of a salt X turns bloodred on treatment with SCN2 and blue ontreatment with K4[Fe(CN)6]. X also givesa positive chromyl chloride test. The saltX is :(1) CuCl2(2) FeCl3(3) Cu(NO3)2(4) Fe(NO3)3
18. Which molecule/ion among the followingcannot act as a ligand in complexcompounds ?(1) CO(2) CN2
(3) CH4(4) Br2
Page 6 Chemistry : English & Hindi 0919. The correct statement on the isomerismassociated with the following complexions,
(a) [Ni(H2O)5NH3]21,(b) [Ni(H2O)4(NH3)2]21 and(c) [Ni(H2O)3(NH3)3]21 is :(1) (a) and (b) show only geometricalisomerism(2) (a) and (b) show geometrical andoptical isomerism(3) (b) and (c) show geometrical andoptical isomerism(4) (b) and (c) show only geometricalisomerism
20. Photochemical smog consists of excessiveamount of X, in addition to aldehydes,ketones, peroxy acetyl nitrile (PAN), andso forth. X is :(1) CH4(2) CO(3) CO2(4) O3
21. 1.4 g of an organic compound was digestedaccording to Kjeldahls method and theammonia evolved was absorbed in 60 mLof M/10 H2SO4 solution. The excesssulphuric acid required 20 mL of M/10NaOH solution for neutralization. Thepercentage of nitrogen in the compoundis :(1) 3(2) 5(3) 10(4) 24
22. The optically inactive compound from thefollowing is :(1) 2 - chloropropanal(2) 2 - chloropentane(3) 2 - chlorobutane(4) 2 - chloro - 2 - methylbutane
19. çÙÙ â´·é¤Ü ¥æØÙæð´ âð âÕçÏÌ â×æßØßÌæ ÂÚUâãè ·¤ÍÙ ãñ´(a) [Ni(H2O)5NH3]21,(b) [Ni(H2O)4(NH3)2]21 ¥æñÚU(c) [Ni(H2O)3(NH3)3]21 :(1) (a) ¥æñÚU (b) ·ð¤ßÜ Øæç×ÌèØ â×æßØßÌæ
ÎàææüÌð ãñ´Ð(2) (a) ¥æñÚU (b) Øæç×ÌèØ ¥æñÚU Ïýéßæ â×æßØßÌæ
ÎàææüÌð ãñ´Ð(3) (b) ¥æñÚU (c) Øæç×ÌèØ ¥æñÚU Ïýéßæ â×æßØßÌæ
ÎàææüÌð ãñ´Ð(4) (b) ¥æñÚU (c) ·ð¤ßÜ Øæç×ÌèØ â×æßØßÌæ
ÎàææüÌð ãñ´Ð
20. Âý·¤æàæ ÚUæâæØçÙ·¤ Ïê× ·¤æðãÚðU ×ð´ °ðçËÇUãæ§ÇU, ·¤èÅUæðÙ,ÂðÚUæðâè °çâÅUæ§Ü Ùæ§ÅþUæ§Ü (PAN) §ØæçÎ ·ð¤ ¥Üæßæ¥çÏ·¤ ×æææ ×ð´ X Öè ãæðÌæ ãñ, X ãñ Ñ(1) CH4(2) CO(3) CO2(4) O3
21. 1.4 g ·¤æÕüçÙ·¤ Øæñç»·¤ ·¤æð ·ñ¤ËÇUæÜ çßçÏ ·ð¤ ¥ÙéâæÚUÂæç¿Ì ç·¤Øæ »Øæ ÌÍæ çÙ·¤Üð ¥×æðçÙØæ ·¤æð 60 mL,M/10, H2SO4 çßÜØÙ ×ð ¥ßàææðçáÌ ç·¤Øæ »ØæÐ¥çÌçÚUÌ âËØêçÚU·¤ ¥Ü ·¤æð ©ÎæâèÙ ·¤ÚUÙð ·ð¤çÜ°, 20 mL, M/10, NaOH Ü»æÐ §â Øæñç»·¤×ð´ Ùæ§ÅþUæðÁÙ ·¤è ÂýçÌàæÌÌæ ãñ Ñ(1) 3(2) 5(3) 10(4) 24
22. çÙÙ ×ð´ âð Ïýéßæ ¥æêæü·¤ Øæñç»·¤ ãñ Ñ(1) 2 - ÜæðÚUæðÂýæðÂðÙÜ(2) 2 - ÜæðÚUæðÂðÅðUÙ(3) 2 - ÜæðÚUæðØêÅðUÙ(4) 2 - ÜæðÚUæð - 2 - ×ðçÍÜØêÅðUÙ
Page 7 Chemistry : English & Hindi 09
23.
A is :(1)
(2)
(3)
(4)
24. A compound A with molecular formulaC10H13Cl gives a white precipitate onadding silver nitrate solution. A onreacting with alcoholic KOH givescompound B as the main product. B onozonolysis gives C and D. C givesCannizaro reaction but not aldolcondensation. D gives aldol condensationbut not Cannizaro reaction. A is :(1)
(2)
(3) C6H52CH22CH22CH22CH22Cl
(4)
23.
A ãñ Ñ(1)
(2)
(3)
(4)
24. Øæñç»·¤ A çÁâ·¤æ ¥æéâêæ C10H13Cl ãñ, çâËßÚUÙæ§ÅþðUÅU çßÜØÙ ç×ÜæÙð ÂÚU àßðÌ ¥ßÿæð ÎðÌæ ãñÐA °ðË·¤æðãæòçÜ·¤ KOH ·ð¤ âæÍ ¥çÖç·ý¤Øæ ·¤ÚUÙðÂÚU ×éØ M¤Â âð Øæ ñç»·¤ B ÎðÌæ ãñÐ B ·¤æ¥æðÁæðÙ-¥ÂæÅUÙ ·¤ÚUÙð ÂÚU Øæñç»·¤ C ÌÍæ D ÂýæÌãæðÌð ãñ´Ð C ·ñ¤çÙÁæÚUæð ¥çÖç·ý¤Øæ ÎðÌæ ãñ, ÂÚUÌé °ðËÇUæÜâ´æÙÙ Ùãè´ ÎðÌæÐ D °ðËÇUæÜ â´æÙÙ ÎðÌæ ãñ, ÂÚUÌé·ñ¤çÙÁæÚUæð ¥çÖç·ý¤Øæ Ùãè´ ÎðÌæÐ A ãñ Ñ(1)
(2)
(3) C6H52CH22CH22CH22CH22Cl
(4)
Page 8 Chemistry : English & Hindi 0925. In the presence of a small amount ofphosphorous, aliphatic carboxylic acidsreact with chlorine or bromine to yield acompound in which a - hydrogen has beenreplaced by halogen. This reaction isknown as :
(1) Wolff - Kischner reaction(2) Etard reaction(3) Hell - Volhard - Zelinsky reaction(4) Rosenmund reaction
26. Arrange the following amines in the orderof increasing basicity.
(1)
(2)
(3)
(4)
25. ȤæòSȤæðÚUâ ·¤è ·¤× ×æææ ·¤è ©ÂçSÍçÌ ×ð´ °ÜèÈð¤çÅU·¤·¤æÕæðüçâçÜ·¤ ¥Ü ÜæðÚUèÙ ¥æñÚU Õýæð×èÙ ·ð¤ âæÍ¥çÖç·ý¤Øæ ·¤ÚUÌð ãé° ¥ÂÙð a - ãæ§ÇþUæðÁÙ ·¤æð ãñÜæðÁÙ×ð´ ÂçÚUßçÌüÌ ·¤ÚUÌð ãñ´Ð §â ¥çÖç·ý¤Øæ ·¤æ Ùæ× ãñ Ñ(1) ßæðËȤ-ç·¤àÙÚU ¥çÖç·ý¤Øæ(2) §üÅUæÇüU ¥çÖç·ý¤Øæ(3) ãðÜ-ȤæðÜæÇüU-ÁðçÜ´S·¤è ¥çÖç·ý¤Øæ(4) ÚUæðÁðÙ×é´ÇU ¥çÖç·ý¤Øæ
26. çÙÙ ¥×èÙæð´ ·¤æð ÿææÚU·¤Ìæ ·ð¤ ÕɸÌð ·ý¤× ×ð´ ܻ槰Ð
(1)
(2)
(3)
(4)
Page 9 Chemistry : English & Hindi 0927. Match the polymers in column-A withtheir main uses in column-B and choosethe correct answer :
Column - A Column - B
(A) Polystyrene (i) Paints and
lacquers
(B) Glyptal (ii) Rain coats
(C) Polyvinyl
Chloride
(iii) Manufacture of
toys
(D) Bakelite (iv) Computer discs
(1) (A) - (ii), (B) - (i), (C) - (iii), (D) - (iv)(2) (A) - (iii), (B) - (i), (C) - (ii), (D) - (iv)(3) (A) - (ii), (B) - (iv), (C) - (iii), (D) - (i)(4) (A) - (iii), (B) - (iv), (C) - (ii), (D) - (i)
28. Complete hydrolysis of starch gives :(1) glucose and fructose in equimolaramounts(2) galactose and fructose in equimolaramounts(3) glucose only(4) glucose and galactose in equimolaramounts
29. is used as :
(1) Insecticide(2) Antihistamine(3) Analgesic(4) Antacid
30. The cation that will not be precipitated byH2S in the presence of dil HCl is :(1) Cu21
(2) Pb21
(3) As31
(4) Co21
- o O o -
27. ·¤æòÜ×-A ×ð´ çΰ »° ÕãéÜ·¤æð´ ·¤æð ·¤æòÜ×-B ×ð´ ©Ù·ð¤Âý×é¹ ©ÂØæð» ·ð¤ âæÍ âé×ðçÜÌ ·¤Úð´U ÌÍæ âãè çß·¤Ë¿éÙð´ Ñ
Ë×Á¼ - A Ë×Á¼ - B
(A) §ËÁÍSªUË¿U; (i) §âÁÕ§ ËÖ¿U §âÁËäË º¾Ë¾Õ ¼Õ
(B) ÌóÁåªUÁ (ii) º¿UÇË̱½Ë º¾Ë¾Õ ¼Õ(C) §Ë×ÁÍÄË̾Á
þÁËÕ¿UˬU(iii) ÌÁËÖ¾Õ º¾Ë¾Õ ¼Õ
(D) ºÖÕÁ˪U (iv) å½ÏªU¿U ̬US º¾Ë¾Õ ¼Õ
(1) (A) - (ii), (B) - (i), (C) - (iii), (D) - (iv)(2) (A) - (iii), (B) - (i), (C) - (ii), (D) - (iv)(3) (A) - (ii), (B) - (iv), (C) - (iii), (D) - (i)(4) (A) - (iii), (B) - (iv), (C) - (ii), (D) - (i)
28. SÅUæ¿ü ·ð¤ Âêæü ÁÜ ¥ÂæÅUÙ âð ç×ÜÌæ ãñ Ñ(1) Üê·¤æðâ ¥æñÚU Èýé¤ÅUæðâ ·¤è â××æðÜ ×æææ(2) »ñÜðÅUæðâ ¥æñÚU Èýé¤ÅUæðâ ·¤è â××æðÜ ×æææ(3) ·ð¤ßÜ Üê·¤æðâ(4) Üê·¤æðâ ¥æñÚU »ñÜðÅUæðâ ·¤è â××æðÜ ×æææ
29. ·¤æ ÂýØæð» ç·¤â M¤Â ×ð´ ãæðÌæ ãñ?
(1) ·¤èÅUÙæàæ·¤(2) ÂýçÌçãSÅñUç×Ù(3) ÂèǸæãæÚUè(4) ÂýçÌ¥Ü
30. ßã ÏÙæØÙ Áæð ÌÙé HCl ·ð¤ ©ÂçSÍçÌ ×ð´ H2S âð¥ßÿæðçÂÌ Ùãè´ ãæðÌæ ãñ, ßã ãñ Ñ(1) Cu21(2) Pb21(3) As31(4) Co21
- o O o -
Page 1 MATHEMATICS : English & Hindi 091. In a certain town, 25% of the families owna phone and 15% own a car ; 65% familiesown neither a phone nor a car and 2,000families own both a car and a phone.Consider the following three statements :
(a) 5% families own both a car and aphone.(b) 35% families own either a car or aphone.(c) 40,000 families live in the town.Then,(1) Only (a) and (b) are correct.(2) Only (a) and (c) are correct.(3) Only (b) and (c) are correct.(4) All (a), (b) and (c) are correct.
2. The largest value of r for which the regionrepresented by the set v e C/?v242i?[ris contained in the region represented bythe set z e C/?z21?[?z1i?, is equal to :(1) 17
(2) 2 2
(3) 32
2
(4) 52
2
3. If 213i is one of the roots of the equation2x329x21kx21350, k e R, then the realroot of this equation :(1) does not exist.(2) exists and is equal to 1
2.
(3) exists and is equal to 1
22 .
(4) exists and is equal to 1.
1. ç·¤âè àæãÚU ×ð´, 25% ÂçÚUßæÚUæð´ ·ð¤ Âæâ ȤæðÙ ãñ ÌÍæ15% ·ð¤ Âæâ ·¤æÚU ãñ ; 65% ÂçÚUßæÚUæð´ ·ð¤ Âæâ Ù ÌæðȤæðÙ ãñ ¥æñÚU Ù ãè ·¤æÚU ãñ, ÌÍæ 2,000 ÂçÚUßæÚUæð´ ·ð¤ ÂæâȤæðÙ ÌÍæ ·¤æÚU ÎæðÙæð´ ãñ´Ð çÙÙ ÌèÙ ·¤ÍÙæð´ ÂÚU çß¿æÚU·¤èçÁ° Ñ(a) 5% ÂçÚUßæÚUæð´ ·ð¤ Âæâ ·¤æÚU ÌÍæ ȤæðÙ ÎæðÙæð´ ãñ´Ð(b) 35% ÂçÚUßæÚUæð´ ·ð¤ Âæâ Øæ Ìæð ·¤æÚU ãñ Øæ ȤæðÙ
ãñÐ(c) àæãÚU ×ð´ 40,000 ÂçÚUßæÚU ÚUãÌð ãñ´ÐÌæð,(1) ·ð¤ßÜ (a) ÌÍæ (b) âãè ãñ´Ð(2) ·ð¤ßÜ (a) ÌÍæ (c) âãè ãñ´Ð(3) ·ð¤ßÜ (b) ÌÍæ (c) âãè ãñ´Ð(4) (a), (b) ÌÍæ (c) âÖè âãè ãñÐ
2. r ·¤æ ßã ¥çÏ·¤Ì× ×æÙ çÁâ·ð¤ çÜ° â×éæØv e C/?v242i?[r mæÚUæ çÙÏæüçÚUÌ ÿæðæ, â×éæØz e C/?z21?[?z1i? mæÚUæ çÙÏæüçÚUÌ ÿæðæ ×ð ´âç×çÜÌ ãñ, ãñ Ñ(1) 17
(2) 2 2
(3) 32
2
(4) 52
2
3. ØçÎ 213i, â×è·¤ÚUæ 2x329x21kx21350,k e R ·¤æ °·¤ ×êÜ ãñ, Ìæð §â â×è·¤ÚUæ ·¤æ ßæSÌçß·¤×êÜ Ñ(1) çßl×æÙ Ùãè´ ãñÐ(2) çßl×æÙ ãñ ÌÍæ 1
2 ·ð¤ ÕÚUæÕÚU ãñÐ
(3) çßl×æÙ ãñ ÌÍæ 1
22 ·ð¤ ÕÚUæÕÚU ãñÐ
(4) çßl×æÙ ãñ ÌÍæ 1 ·ð¤ ÕÚUæÕÚU ãñÐ
Page 2 MATHEMATICS : English & Hindi 094. The least value of the product xyz for
which the determinant 1 1
1 1
1 1
x
y
z
is
non-negative, is :(1) 2 22
(2) 16 22
(3) 28(4) 21
5. If 0 1A
1 0
25 , then which one of the
following statements is not correct ?(1) A42I5A21I(2) A32I5A(A2I)(3) A21I5A(A22I)(4) A31I5A(A32I)
6. The number of ways of selecting 15 teamsfrom 15 men and 15 women, such thateach team consists of a man and a woman,is :(1) 1120(2) 1240(3) 1880(4) 1960
7. Let the sum of the first three terms of anA.P. be 39 and the sum of its last four termsbe 178. If the first term of this A.P. is 10,then the median of the A.P. is :(1) 26.5(2) 28(3) 29.5(4) 31
4. »éæÙÈ¤Ü xyz ·¤æ ßã ØêÙÌ× ×êËØ çÁâ·ð¤ çÜ°
âæÚUçæ·¤1 1
1 1
1 1
x
y
z
«¤æðÌÚU ãñ, ãñ Ñ
(1) 2 22
(2) 16 22
(3) 28(4) 21
5. ØçÎ 0 1A
1 0
25 ãñ, Ìæð çÙÙ ×ð´ âð ·¤æñÙ-âæ °·¤
·¤ÍÙ âãè Ùãè´ ãñ?(1) A42I5A21I(2) A32I5A(A2I)(3) A21I5A(A22I)(4) A31I5A(A32I)
6. 15 ÂéL¤áæð´ ÌÍæ 15 ×çãÜæ¥æð´ ×ð´ âð °ðâè 15 ÅUè×ð´,çÁÙ×ð´ ÂýØð·¤ ×ð´ °·¤ ÂéL¤á ÌÍæ °·¤ ×çãÜæ ãæð, ¿éÙÙð·ð¤ ÌÚUè·¤æð´ ·¤è â´Øæ ãñ Ñ(1) 1120(2) 1240(3) 1880(4) 1960
7. ×æÙæ °·¤ â×æ´ÌÚU æðÉ¸è ·ð¤ ÂýÍ× ÌèÙ ÂÎæð´ ·¤æ Øæð» 39ãñ ÌÍæ §â·ð¤ ¥´çÌ× ¿æÚU ÂÎæð´ ·¤æ Øæð» 178 ãñÐ ØçΧâ â×æ´ÌÚU æðÉ¸è ·¤æ ÂýÍ× ÂÎ 10 ãñ, Ìæð §â â×æ´ÌÚUæðÉ¸è ·¤æ ×æØ·¤ ãñ Ñ(1) 26.5(2) 28(3) 29.5(4) 31
Page 3 MATHEMATICS : English & Hindi 098. If the coefficients of the three successiveterms in the binomial expansion of (11x)n
are in the ratio 1 : 7 : 42, then the first ofthese terms in the expansion is :(1) 6th(2) 7th(3) 8th(4) 9th
9. The value of 30
r 16
(r 2)(r 3)∑5
1 2 is equal to :(1) 7785(2) 7780(3) 7775(4) 7770
10.2
20
e cos
sin
x
x
xlim
x→
2 is equal to :(1) 3(2) 3
2
(3) 5
4
(4) 211. The distance, from the origin, of thenormal to the curve, x52 cost 1 2t sint,
y 5 2 sint22t cost at 4
tp
5 , is :(1) 4(2) 2 2
(3) 2(4) 2
8. ØçÎ (11x)n ·ð¤ çmÂÎ çßSÌæÚU ×ð´ ÌèÙ ·ý¤ç×·¤ ÂÎæð´·ð¤ »éææ´·¤æð´ ×ð´ 1 : 7 : 42 ·¤æ ¥ÙéÂæÌ ãñ, Ìæð §Ù ×ð´ âðçßSÌæÚU ×ð ÂãÜæ ÂÎ ãñ Ñ(1) ÀUÆUæ(2) âæÌßæ´(3) ¥æÆUßæ´(4) Ùæñßæ´
9.30
r 16
(r 2)(r 3)∑5
1 2 ·¤æ ×æÙ ÕÚUæÕÚU ãñ Ñ(1) 7785(2) 7780(3) 7775(4) 7770
10.2
20
e cos
sin
x
x
xlim
x→
2 ÕÚUæÕÚU ãñ Ñ(1) 3(2) 3
2
(3) 5
4
(4) 2
11. ß·ý¤ x52 cost12t sint, y52 sint22t cost ÂÚU
4t
p5 ÂÚU ¹è´¿ð »° ¥çÖÜ´Õ ·¤è ×êÜ çÕ´Îé âð
ÎêÚUè ãñ Ñ(1) 4(2) 2 2
(3) 2(4) 2
Page 4 MATHEMATICS : English & Hindi 0912. If Rolles theorem holds for the function
f (x)52x31bx21cx, x e [21, 1], at thepoint 1
2
x 5 , then 2b1c equals :(1) 1(2) 21(3) 2(4) 23
13. Let the tangents drawn to the circle,x21y2516 from the point P(0, h) meet thex-axis at points A and B. If the area ofDAPB is minimum, then h is equal to :(1) 4 3
(2) 3 3
(3) 3 2
(4) 4 2
14. The integral 3 54 4
d
( 1) ( 2)
x
x x∫
1 2 is equal
to :
(1)14 1
4 C 2
x
x
1 12
(2)14 2
4 C 1
x
x
2 11
(3)144 1
C3 2
x
x
12 12
(4)144 2
C3 1
x
x
22 11
12. ØçΠȤÜÙ f (x)52x31bx21cx, x e [21, 1]·ð¤ çÜ° çÕ´Îé 1
2
x 5 ÂÚU ÚUæðÜð ·¤æ Âý×ðØ Üæ»ê ãæðÌæ ãñ,Ìæð 2b1c ÕÚUæÕÚU ãñ Ñ(1) 1(2) 21(3) 2(4) 23
13. ×æÙæ çÕ´Îé P(0, h) âð ßëæ x21y2516 ÂÚU ¹è´¿è»§ü SÂàæü ÚðU¹æ°¡ x-¥ÿæ ·¤æð çÕ´Îé¥æð´ A ÌÍæ B ÂÚUç×ÜÌè ãñÐ ØçÎ DAPB ·¤æ ÿæðæÈ¤Ü ØêÙÌ× ãñ, Ìæðh ÕÚUæÕÚU ãñ Ñ(1) 4 3
(2) 3 3
(3) 3 2
(4) 4 2
14. â×æ·¤Ü 3 54 4
d
( 1) ( 2)
x
x x∫
1 2 ÕÚUæÕÚU ãñ Ñ
(1)14 1
4 C 2
x
x
1 12
(2)14 2
4 C 1
x
x
2 11
(3)144 1
C3 2
x
x
12 12
(4)144 2
C3 1
x
x
22 11
Page 5 MATHEMATICS : English & Hindi 09
15. x > 0 ·ð¤ çÜ° ×æÙæ 1
log t( ) dt
1 t
x
f x ∫51
ãñ, Ìæð1
( ) f x fx
1 ÕÚUæÕÚU ãñ Ñ
(1) 21 (log )
4x
(2) 21 (log )
2x
(3) log x(4) 21
log 4
x
16. ß·ý¤æð´ y12x250 ÌÍæ y13x251 mæÚUæ ÂçÚUÕh ÿæðæ·¤æ ÿæðæÈ¤Ü (ß»ü §·¤æ§Øæð´ ×ð´) ÕÚUæÕÚU ãñ Ñ(1) 3
5
(2) 3
4
(3) 1
3
(4) 4
3
17. ØçÎ y(x), ¥ß·¤Ü â×è·¤ÚUæ
2d( 2) 4 9
d
yx x x
x1 5 1 2 , x ¹ 22 ¥æñÚU
y(0)50, ·¤æ ãÜ ãñ, Ìæð y(24) ÕÚUæÕÚU ãñ Ñ(1) 0(2) 1(3) 21(4) 2
15. For x > 0, let 1
log t( ) dt
1 t
x
f x ∫51
. Then1
( ) f x fx
1 is equal to :
(1) 21 (log )
4x
(2) 21 (log )
2x
(3) log x(4) 21
log 4
x
16. The area (in square units) of the regionbounded by the curves y12x250 andy13x251, is equal to :(1) 3
5
(2) 3
4
(3) 1
3
(4) 4
3
17. If y(x) is the solution of the differentialequation
2d( 2) 4 9
d
yx x x
x1 5 1 2 , x ¹ 22
and y(0)50, then y(24) is equal to :(1) 0(2) 1(3) 21(4) 2
Page 6 MATHEMATICS : English & Hindi 09
18. The points 80,
3
, (1, 3) and (82, 30) :(1) form an obtuse angled triangle.(2) form an acute angled triangle.(3) form a right angled triangle.(4) lie on a straight line.
19. Let L be the line passing through the pointP(1, 2) such that its intercepted segmentbetween the co-ordinate axes is bisectedat P. If L1 is the line perpendicular to Land passing through the point (22, 1),then the point of intersection of L and L1is :(1) 4 12
, 5 5
(2) 11 29,
20 10
(3) 3 17,
10 5
(4) 3 23,
5 10
20. If y13x50 is the equation of a chord ofthe circle, x21y2230x50, then theequation of the circle with this chord asdiameter is :(1) x21y213x19y50(2) x21y223x19y50(3) x21y223x29y50(4) x21y213x29y50
18. çÕ´Îé 80,
3
, (1, 3) ÌÍæ (82, 30) Ñ(1) °·¤ ¥çÏ·¤·¤æðæ çæÖéÁ ÕÙæÌð ãñ´Ð(2) °·¤ ØêÙ·¤æðæ çæÖéÁ ÕÙæÌð ãñ´Ð(3) °·¤ â×·¤æðæ çæÖéÁ ÕÙæÌð ãñ´Ð(4) °·¤ âÚUÜ ÚðU¹æ ÂÚU çSÍÌ ãñ´Ð
19. ×æÙæ L, çÕ´Îé P(1, 2) âð ãæð·¤ÚU ÁæÙð ßæÜè ßã ÚðU¹æ ãñçÁâ·¤æ çÙÎðüàææ´·¤ ¥ÿææð´ ·ð¤ Õè¿ ·¤ÅUæ ÚðU¹æ¹ÇU P ÂÚUâ×çmÖæçÁÌ ãæðÌæ ãñÐ ×æÙæ L1 ßã ÚðU¹æ ãñ Áæð L ÂÚUÜ´ÕßÌ ãñ ÌÍæ çÕ´Îé (22, 1) âð ãæð·¤ÚU ÁæÌè ãñ, ÌæðL ÌÍæ L1 ·¤æ ÂýçÌÀðUÎÙ çÕ´Îé ãñ Ñ(1) 4 12
, 5 5
(2) 11 29,
20 10
(3) 3 17,
10 5
(4) 3 23,
5 10
20. ØçÎ y13x50, ßëæ x21y2230x50 ·¤è °·¤Áèßæ ·¤æ â×è·¤ÚUæ ãñ, Ìæð ©â ßëæ, çÁâ·¤æ ÃØæâ,Øã Áèßæ ãñ, ·¤æ â×è·¤ÚUæ ãñ Ñ(1) x21y213x19y50(2) x21y223x19y50(3) x21y223x29y50(4) x21y213x29y50
Page 7 MATHEMATICS : English & Hindi 0921. If the tangent to the conic, y265x2 at(2, 10) touches the circle,
x21y218x22y5k (for some fixed k) at apoint (a, b) ; then (a, b) is :(1) 6 10
, 17 17
2
(2) 8 2,
17 17
2
(3) 4 1,
17 17
2
(4) 7 6,
17 17
2
22. An ellipse passes through the foci of thehyperbola, 9x224y2536 and its majorand minor axes lie along the transverse andconjugate axes of the hyperbolarespectively. If the product of eccentricitiesof the two conics is 1
2, then which of the
following points does not lie on theellipse ?(1) ( )13 , 0
(2) 39, 3
2
(3) 1 313 ,
2 2
(4) 13, 6
2
23. If the points (1, 1, l) and (23, 0, 1) areequidistant from the plane,3x14y212z11350, then l satisfies theequation :(1) 3x2210x1750(2) 3x2110x1750(3) 3x2110x21350(4) 3x2210x12150
21. ØçÎ àææ´·¤ß y265x2 ·ð¤ çÕ´Îé (2, 10) ÂÚU ¹è´¿è»§ü SÂàæü Úð U¹æ ßëæ x21y218x22y5k ·¤æ ð(ç·¤âè çÙçà¿Ì k ·ð¤ çÜ°) çÕ´Îé (a, b) ÂÚU SÂàæü·¤ÚUÌè ãñ, Ìæð (a, b) ãñ Ñ(1) 6 10
, 17 17
2
(2) 8 2,
17 17
2
(3) 4 1,
17 17
2
(4) 7 6,
17 17
2
22. °·¤ Îèæ üßëæ, ¥çÌÂÚUßÜØ 9x224y2536 ·ð¤ÙæçÖ·ð´¤Îýæð´ âð ãæð·¤ÚU ÁæÌæ ãñ ÌÍæ §â·ð¤ Îèæü ÌÍæ Üæé¥ÿæ ·ý¤×àæÑ ¥çÌÂÚUßÜØ ·ð¤ ¥ÙéÂýSÍ ÌÍæ â´Øé×è¥ÿææð ·ð¤ ¥ÙéçÎàæ ãñÐ ØçÎ §Ù Îæð àææ·¤ßæð ·¤è ©·ð¤ÎýÌæ¥æð·¤æ »éæÙÈ¤Ü 1
2 ãñ, Ìæð çÙÙ ×ð´ âð ·¤æñÙ-âæ çÕ´Îé
Îèæüßëæ ÂÚU çSÍÌ Ùãè´ ãñ?(1) ( )13 , 0
(2) 39, 3
2
(3) 1 313 ,
2 2
(4) 13, 6
2
23. ØçÎ çÕ´Î é (1, 1, l) ÌÍæ (23, 0, 1) â×ÌÜ3x14y212z11350 âð â×ÎêÚUSÍ ãñ´, Ìæð l, çÙÙâ×è·¤ÚUæ ·¤æð â´ÌécÅU ·¤ÚUÌæ ãñ Ñ(1) 3x2210x1750(2) 3x2110x1750(3) 3x2110x21350(4) 3x2210x12150
Page 8 MATHEMATICS : English & Hindi 0924. If the shortest distance between the lines
1 1
1 1
yx z125 5
a 2, (a ¹21) and
x1y1z115052x2y1z13 is 1
3, then
a value of a is :(1) 16
192
(2) 19
162
(3) 32
19
(4) 19
32
25. Let a→ and b
→ be two unit vectors suchthat 5 3 . If
( )c a 2 b 3 a b→→ → →→
5 1 1 3 , then isequal to :(1) 55
(2) 51
(3) 43
(4) 37
26. Let X be a set containing 10 elements andP(X) be its power set. If A and B are pickedup at random from P(X), withreplacement, then the probability that Aand B have equal number of elements, is :(1) 20
10102
C
(2) ( )10
202 1
2
2
(3) ( )10
102 1
2
2
(4) 2010
202
C
24. ØçÎ ÚðU¹æ¥æð´ 1 1
1 1
yx z125 5
a 2, (a ¹21)
ÌÍæ x1y1z115052x2y1z13 ·ð¤ Õè¿·¤è ØêÙÌ× ÎêÚUè 1
3 ãñ, Ìæð a ·¤æ °·¤ ×æÙ ãñ Ñ
(1) 16
192
(2) 19
162
(3) 32
19
(4) 19
32
25. ×æÙæ a→ ÌÍæ b
→ °ðâð ×ææ·¤ âçÎàæ ã ñ ´ ç·¤5 3 ãñÐ ØçÎ
( )c a 2 b 3 a b→→ → →→
5 1 1 3 ãñ, Ìæð ÕÚUæÕÚUãñ Ñ(1) 55
(2) 51
(3) 43
(4) 37
26. ×æÙæ X °·¤ â×éæØ ãñ çÁâ×ð´ 10 ¥ßØß ãñ´ ÌÍæP(X) §â·¤æ ææÌ â×éæØ ãñÐ ØçÎ P(X) âð A ÌÍæB ØæÎëÀUØæ, ÂýçÌSÍæÂÙæ âçãÌ, çÜ° »° ãñ´, ÌæðA ÌÍæ B ×ð´ ÕÚUæÕÚU ¥ßØßæð´ ·ð¤ ãæðÙð ·¤è ÂýæçØ·¤Ìæ ãñ Ñ(1) 20
10102
C
(2) ( )10
202 1
2
2
(3) ( )10
102 1
2
2
(4) 2010
202
C
Page 9 MATHEMATICS : English & Hindi 0927. A factory is operating in two shifts, dayand night, with 70 and 30 workersrespectively. If per day mean wage of theday shift workers is ` 54 and per day meanwage of all the workers is ` 60, then perday mean wage of the night shift workers(in `) is :
(1) 66(2) 69(3) 74(4) 75
28. In a DABC, a 2 3
b5 1 and ÐC5608.
Then the ordered pair (ÐA, ÐB) is equalto :(1) (158, 1058)(2) (1058, 158)(3) (458, 758)(4) (758, 458)
29. If 1 12
2( ) 2tan sin
1
xf x x
x
2 25 11
, x > 1,then f (5) is equal to :(1)
2
p
(2) p(3) 4 tan21(5)(4) 1 65
tan156
2
30. The contrapositive of the statementIf it is raining, then I will not come, is :(1) If I will come, then it is not raining.(2) If I will not come, then it is raining.(3) If I will not come, then it is notraining.(4) If I will come, then it is raining.
- o 0 o -
27. °·¤ Èñ¤ÅþUè Îæð ÂæçÚUØæð´, çÎÙ ÌÍæ ÚUæÌ, ×ð´ ¿ÜÌè ãñçÁÙ×ð´ ·ý¤×àæÑ 70 ÌÍæ 30 ·¤æ×»æÚU ·¤æØü ·¤ÚUÌð ãñ´ÐØçÎ çÎÙ ·¤è ÂæÚUè ·ð¤ ·¤æ×»æÚUæð´ ·¤æ ×æØ ÂýçÌçÎÙßðÌÙ ` 54 ãñ ÌÍæ âÖè ·¤æ×»æÚUæð´ ·¤æ ×æØ ÂýçÌçÎÙßðÌÙ ` 60 ãñ, Ìæð ÚUæÌ ×ð´ ·¤æØü ·¤ÚUÙð ßæÜð ·¤æ×»æÚUæð´ ·¤æ×æØ ÂýçÌçÎÙ ßðÌÙ (` ×ð´) ãñ Ñ(1) 66(2) 69(3) 74(4) 75
28. °·¤ çæÖéÁ ABC ×ð´, a 2 3
b5 1 ÌÍæ ÐC5608
ãñ, Ìæð ·ý¤ç×Ì Øé× (ÐA, ÐB) ÕÚUæÕÚU ãñ Ñ(1) (158, 1058)(2) (1058, 158)(3) (458, 758)(4) (758, 458)
29. ØçÎ 1 12
2( ) 2tan sin
1
xf x x
x
2 25 11
, x > 1ãñ, Ìæð f (5) ÕÚUæÕÚU ãñ Ñ(1)
2
p
(2) p(3) 4 tan21(5)(4) 1 65
tan156
2
30. ·¤ÍÙ ÑÒÒØçÎ ßáæü ãæð ÚUãè ãñ, Ìæð ×ñ´ Ùãè´ ¥æª´¤»æÓÓ·¤æ ÂýçÌÏÙæ×·¤ ·¤ÍÙ ãñ Ñ(1) ØçÎ ×ñ´ ¥æª´¤»æ, Ìæð ßáæü Ùãè´ ãæð ÚUãè ãñÐ(2) ØçÎ ×ñ´ Ùãè´ ¥æª´¤»æ, Ìæð ßáæü ãæð ÚUãè ãñÐ(3) ØçÎ ×ñ´ Ùãè´ ¥æª´¤»æ, Ìæð ßáæü Ùãè´ ãæð ÚUãè ãñÐ(4) ØçÎ ×ñ´ ¥æª´¤»æ, Ìæð ßáæü ãæð ÚUãè ãñÐ
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ANSWER KEY PAPER - 1 JEE (MAIN) - 2014
QUESTION
NO.
06/04/2014
SET E
06/04/2014
SET F
06/04/2014
SET G
06/04/2014
Book H
09/04/2014 11/04/2014 12/04/2014 19/04/2014
1. 1 4 2 3 2 3 2 1
2. 3 4 2 4 3 3 3 1
3. 2 4 3 1 2 1 4 4
4. 1 1 3 3 3 2 2 4
5. 3 3 1 3 3 2 4 3
6. 3 1 1 2 4 4 1 2
7. 4 4 4 3 3 1 2 *
8. 1 1 4 4 3 1 3 3
9. 3 2 1 2 3 2 2 1
10. 1 2 1 1 1 1 2 4
11. 3 1 3 2 1 2 2 3
12. 4 4 3 1 1 3 2 1
13. 1 3 2 3 2 2 1 4
14. 4 3 3 1 1 1 4 3
15. 3 2 2 1 1 4 2 2
16. 3 4 1 4 3 2 2 1
17. 1 1 3 2 4 3 3 4
18. 3 4 3 1 3 3 2 3
19. 2 1 3 4 4 1 3 2
20. 3 3 4 1 1 2 2 2
21. 3 3 3 1 3 1 2 1
22. 3 1 3 1 4 3 2 3
23. 2 2 3 4 4 4 1 3
24. 2 2 2 1 2 4 3 3
25. 4 4 2 4 4 1 2 3
26. 2 4 4 3 4 3 2 2
27. 3 1 4 1 2 2 2 3
28. 1 4 4 3 1 3 4 1
29. 4 3 3 2 2 2 2 2
30. 2 1 4 1 1 4 4 2
31. 1 1 4 1 4 3 4 4
32. 2 2 4 4 1 4 1 2
33. 3 2 4 1 3 4 4 2
34. 2 1 2 4 4 4 2 3
35. 1 3 4 2 2 2 2 2
36. 1 2 2 4 3 4 3 4
37. 3 3 4 4 3 2 2 2
38. 1 2 3 1 3 1 2 1
ANSWER KEY PAPER - 1 JEE (MAIN) - 2014
QUESTION
NO.
06/04/2014
SET E
06/04/2014
SET F
06/04/2014
SET G
06/04/2014
Book H
09/04/2014 11/04/2014 12/04/2014 19/04/2014
39. 2 4 2 2 4 2 3 3
40. 4 2 4 2 3 1 3 4
41. 3 1 1 2 2 3 4 3
42. 2 1 4 4 3 3 3 2
43. 2 3 2 3 2 4 4 2
44. 1 3 2 4 3 2 2 3
45. 4 2 4 4 1 4 4 2
46. 2 1 3 3 2 2 4 2
47. 2 1 1 4 2 3 1 1
48. 1 1 3 1 2 2 2 4
49. 4 4 4 4 3 3 1 4
50. 2 4 3 1 1 2 3 4
51. 2 1 1 3 4 4 1 4
52. 4 1 1 2 3 3 2 4
53. 4 1 3 4 2 1 3 2
54. 3 4 2 4 1 3 1 3
55. 1 4 4 3 2 2 4 2
56. 1 1 2 4 2 3 4 3
57. 4 4 4 3 4 3 3 4
58. 1 2 4 1 2 3 1 4
59. 1 3 3 1 3 3 4 3
60. 3 1 1 3 2 3 4 2
61. 2 4 2 3 4 3 2 4
62. 4 2 2 4 4 1 3 2
63. 3 1 2 4 2 2 3 1
64. 2 2 2 3 1 1 3 2
65. 1 2 2 1 2 4 1 4
66. 4 1 3 3 1 4 2 2
67. 2 1 4 2 3 4 2 2
68. 1 4 3 4 2 3 1 2
69. 2 2 3 4 3 2 3 4
70. 2 1 4 4 1 4 2 4
71. 2 3 2 1 2 3 3 2
72. 2 4 2 4 4 3 2 4
73. 1 2 1 1 2 1 1 3
74. 4 3 4 3 2 3 2 2
75. 2 4 1 2 2 1 4 4
76. 3 3 1 4 3 3 1 1
ANSWER KEY PAPER - 1 JEE (MAIN) - 2014
QUESTION
NO.
06/04/2014
SET E
06/04/2014
SET F
06/04/2014
SET G
06/04/2014
Book H
09/04/2014 11/04/2014 12/04/2014 19/04/2014
77. 3 2 1 4 2 4 4 3
78. 4 2 3 2 3 2 2 4
79. 1 3 3 2 3 2 3 2
80. 1 2 3 2 4 2 4 1
81. 2 1 4 3 4 3 4 1
82. 3 4 3 3 1 4 4 1
83. 3 4 3 3 3 3 2 2
84. 3 3 1 1 3 2 3 3
85. 2 4 2 3 3 4 2 3
86. 1 1 4 4 1 2 2 1
87. 4 2 2 3 4 2 1 2
88. 2 2 2 1 3 2 4 3
89. 2 2 3 3 2 1 3 3
90. 3 2 1 2 2 3 2 3
Note :- * denotes that the question is dropped.