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TThhiiss ssoolluuttiioonn wwaass ddoowwnnllooaadd ffrroomm RReessoonnaannccee JJEEEE ((MMAAIINN)) 22002200 SSoolluuttiioonn ppoorrttaall
7340010333
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JJEEEE ((MMaaiinn)) 22002200
COMPUTER BASED TEST (CBT)
Memory Based Questions & Solutions
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TThhiiss ssoolluuttiioonn wwaass ddoowwnnllooaadd ffrroomm RReessoonnaannccee JJEEEE ((MMAAIINN)) 22002200 SSoolluuttiioonn ppoorrttaall
7340010333
| JEE MAIN-2020 | DATE : 07-01-2020 (SHIFT-1) | PAPER-1 | MEMORY BASED | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222 To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
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7340010333
PART : PHYSICS
Single Choice Type (,dy fodYih; izdkj) This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its answer, out of which Only One is correct. bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. A block of mass m is suspended from a pulley in form of a circular disc of mass m & radius R. The system is released from rest, find the angular velocity of disc when block has dropped by height h. (there is no slipping between string & pulley) ,d m nzO;eku dk CykWd f?kjuh ls jLlh ds }kjk tqM+k gqvk gSA f?kjuh pdrh ds :i esa gSA ftldk nzO;eku m rFkk
f=kT;k R gSA ;fn fudk; dks fojke ls NksM+k tkrk gSA rc CykWd }kjk h Å¡pkbZ uhps vkus ij pdrh dk dks.kh; osx
gksxk (f?kjuh rFkk pdrh esa dksbZ fQlyu ugha gS)
m
m
(1) 3gh4
R1 (2)
3gh2
R1 (3)
3gh2R (4)
3gh4R
Ans. (1)
Sol. mgh = 22
21mv
21
v = R (no slipping dksbZ fQlyu ugha)
mgh = 22
22
2mR
21Rm
21
mgh = 22Rm43
3gh4
R1
R3gh4
2
2. Three point masses 1kg, 1.5 kg, 2.5 kg are placed at the vertices of a triangle with sides 3cm, 4cm and
5cm as shown in the figure. The location of centre of mass with respect to 1kg mass is : 2.5 kg
5cm 4 cm
1 kg
3 cm
1.5 kg
(1) 0.6 cm to the right of 1 kg and 2 cm above 1 kg mass (2) 0.9 cm to the right of 1kg and 2 cm above 1 kg mass (3) 0.9 cm to the left of 1kg and 2 cm above 1kg mass (4) 0.9 cm to the right of 1 kg and 1.5 cm above 1kg mass
| JEE MAIN-2020 | DATE : 07-01-2020 (SHIFT-1) | PAPER-1 | MEMORY BASED | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222 To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
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PAGE # 2
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fp=k esa fn[kk, vuqlkj rhu fcUnq nzO;eku ftuds nzO;eku Øe'k% 1kg, 1.5kg rFkk 2.5kg gS] ,d f=kHkqt ds dksuksa ij
j[ks gSA f=kHkqt dh Hkqtk,sa Øe'k% 3 cm, 4cm rFkk 5cm gSA rc nzO;eku dsUnz dh fLFkfr 1kg nzO;eku ds lkis{k gksxh :
2.5 kg
5cm 4 cm
1 kg
3 cm
1.5 kg
(1) 1kg ds nka;h vksj 0.6 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij (2) 1kg ds nka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij (3) 1kg ds cka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij (4) 1 kg ds nka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 1.5 cm Åij Ans. (2) Sol. Take 1kg mass at origin 1 kg nzO;eku dks ewy fcUnq ysus ij
2.5 kg
5cm 4 cm
1 kg
3 cm
1.5 kg
Y
X
cm1 0 1.5 3 2.5 0X 0.9cm
5
cm1 0 1.5 0 2.5 4Y 2cm
5
3. In a single slit diffraction set up, second minima is observed at an angle of 60°. The expected position of first minima is
,dy fNnz foorZu izfØ;k esa] 2nd fufEu"B (minima) 60° ds dks.k ij fn[kkbZ nsrk gS] rks 1st fufEu"B (minima) dh
visf{kr fLFkfr gksxh & (1) 25° (2) 20° (3) 30° (4) 45° Ans. (1) Sol. For 2nd minima f}rh; fufEu"B ds fy, d sin = 2
sin = 23 (given)
43
d
… (i)
So for 1st minima is blfy, izFke fufEu"B ds fy,
d sin =
sin = 43
d
(from equation (i) (lehdj.k (i) ls
= 25.65° (from sin table) (sin rkfydk ls) 25
| JEE MAIN-2020 | DATE : 07-01-2020 (SHIFT-1) | PAPER-1 | MEMORY BASED | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222 To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
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PAGE # 3
7340010333
4. There are two infinite plane sheets each having uniform surface charge density + C/m2. They are inclined to each other at an angle 30° as shown in the figure. Electric field at any arbitrary point P is:
fp=k esa nks vuUr yEckbZ dh lery ifêdk gS ftudk le:i i"Bh; vkos'k ?kuRo + C/m2 gSA nksuksa ifêdk ,d nwljs
ls 30° dks.k ij >qdh gSA rc fp=kkuqlkj fdlh Hkh ;knfPNd fcUnq P ij fo|qr {ks=k gksxk:
30°+
+
P
Y
X
(1) 0
3 1ˆ ˆ1– y – x2 2 2
(2) 0
3 1ˆ ˆ1 y – x2 2 2
(3) 0
3 1ˆ ˆ1– y x2 2 2
(4) 0
3 1ˆ ˆ1 y x2 2 2
Ans. (1) Sol.
30° +
+
P60°
0 0 0
ˆ ˆE cos60 –x – sin60 y2 2 2
E
= 0
3 1ˆ ˆ1– y – x2 2 2
5. A parallel plate capacitor with plate area A & plate separation d is filled with a dielectric material of
dielectric constant given by k = k0(1 + x). Calculate capacitance of system: (given d << 1)
,d lekUrj iV~V la/kkfj=k ftldh lrg dk {ks=kQy A gS rFkk ifV~Vdkvksa ds e/; dh nwjh d gSA buds e/; ,d
ijkoS|qrkad inkFkZ Hkjk gSA inkFkZ dk ijkoS|qrkad fu;rkad k = k0(1 + x) gSA rc fudk; dh /kkfjrk gksxh:
(fn;k gS d << 1)
x
(1) 2200 d1d
Ak
(2)
2d1
dAk 00 (3) d1
d2Ak 00
(4)
2d1
d2Ak 00
Ans. (2)
| JEE MAIN-2020 | DATE : 07-01-2020 (SHIFT-1) | PAPER-1 | MEMORY BASED | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222 To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
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Sol. Capacitance of element ?kVd dh /kkfjrk = dx
Ak 0
dxx
Capacitance of element ?kVd dh /kkfjrk, C' = dx
A)x1(k 00
d
0 00 )x1(Akdx
'C1
)d1(nAk
1C1
00
Given fn;k gS& d << 1
2dd
Ak1
C1 22
00
2d1
Akd
C1
00
2d1
dAk
C 00
6. A long solenoid of radius R carries a time dependent current I = I0 t(1 – t). A ring of radius 2R is placed
coaxially near its centre. During the time interval 0 t 1, the induced current R and the induced emf VR in the ring vary as:
(1) current will change its direction and its emf will be zero at t = 0.25sec. (2) current will not change its direction & emf will be maximum at t = 0.5sec (3) current will not change direction and emf will be zero at 0.25sec. (4) current will change its direction and its emf will be zero at t = 0.5sec. ,d R f=kT;k dh ijhuyhdk ftlesa ,d le; fuHkZj /kkjk I = I0 t(1 – t) izokfgr gks jgh gSA ,d 2R f=kT;k dh oy;
ifjuyhdk ds lek{kh; :i ls mlds dsUnz ds ikl fLFkr gSA le; vUrjky 0 t 1, ds nkSjku oy; esa izsjhr /kkjk R
vkSj izsfjr fo+|qr okgd cy VR ds laxr lgh fodYi gksxk
(1) /kkjk viuh fn'kk cnysxh vkSj fo|qr okgd cy le; t = 0.25sec ij 'kwU; gksxkA (2) /kkjk viuh fn'kk ugha cnysxh vkSj fo|qr okgd cy le; t = 0.5sec ij vf/kdre gksxkA (3) /kkjk viuh fn'kk ugha cnysxh vkSj fo|qr okgd cy le; t = 0.25sec ij 'kwU; gksxkA (4) /kkjk viuh fn'kk cnysxh vkSj fo|qr okgd cy t = 0.5sec ij 'kwU; gksxkA Ans. (4) Sol. I = I0t – I0t2 = BA = 0nIA
VR = dtd– = –0nAI0 (1 – 2t)
VR = 0 at t = 21
| JEE MAIN-2020 | DATE : 07-01-2020 (SHIFT-1) | PAPER-1 | MEMORY BASED | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222 To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
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PAGE # 5
7340010333
and rFkk R = loopofcetansisRe
VR sèk dk izfrjkyqi
RV
e
1/2 t
7. If 10% of intensity is passed from analyser, then, the angle by which analyser should be rotated such
that transmitted intensity becomes zero. (Assume no absorption by analyser and polarizer). ;fn ,d èkqzod ls 10% rhozrk xqtjrh gks rks èkqzod dks fdrus dks.k ls ?kqek;s tk;s fd fuxZr rhozrk 'kwU; gks tk;sA
(eku fyft,s dh /kzqod ls vo'kks"k.k 'kwU; gS) (1) 60° (2) 18.4° (3) 45° (4) 71.6°
Ans. (B) Sol. I = I0 cos2
20
0 cos10
II
cos = 101 = 0.31 <
21 which is 0.707
So > 45° and rFkk 90 – < 45º so only one option is correct blfy, dsoy ,d fodYi lgh gksxkA
i.e. vFkkZr 18.4º angle rotated should be ?kqek;k gqvk dks.k = 90° – 71.6° = 18.4°
8. Three moles of ideal gas A with 34
CC
V
P is mixed with two moles of another ideal gas B with 35
CC
V
P .
The v
P
CC of mixture is (Assuming temperature is constant)
vkn'kZ xSl A ds 3 eksy ftlds fy, fd 34
CC
V
P gS] dks nwljh vkn'kZ xSl B ds 2 eksy ftlds fy, 35
CC
V
P gS] ds
lkFk feyk;k tkrk gS rks feJ.k dh v
P
CC dk eku crkb, (rkieku dks ,d&leku eku ekfu;sA)
(1) 1.5 (2) 1.42 (3) 1.7 (4) 1.3 Ans. (2)
Sol.
1–Rn
1–Rn
1–Rn
1–Rn
CnCnCnCn
2
2
1
1
2
22
1
11
V2V1
P2P1mixture
21
21
on rearranging we get iqu O;ofLFkr djus ij,
1–n
1–n
1–nn
2
2
1
1
mix
21
; 3/2
23/1
31–
5mix
1–
5mix
= 9 + 3 = 12 1251
1217
mixure ; mix = 1.42
| JEE MAIN-2020 | DATE : 07-01-2020 (SHIFT-1) | PAPER-1 | MEMORY BASED | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222 To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
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PAGE # 6
7340010333
9. Given magnetic field equation is B = 3 × 10–8 sin(t + kx + ) j , then appropriate equation for electric field (E) will be :
fn;s x;s pqEcdh; {kS=k dh lehdj.k B = 3 × 10–8 sin(t + kx + ) j gS] rc fo|qr {kS=k (E) dh mfpr lehdj.k gksxhA
(1) 20 × 10–9 sin (t + kx + ) k (2) 9 sin (t + kx + ) k (3) 16 × 10–9 sin (t + kx + ) k (4) 3 × 10–9 sin (t + kx + ) k Ans. (2)
Sol. 0
0
EB
= C (speed of light in vacuum fuokZr esa izdk'k dh pky)
E0 = B0C = 3 × 10–8 × 3 × 108
= 9 N/C So blfy, E = 9 sin (t + kx + ) 10. There is a LCR circuit, If it is compared with a damped oscillation of mass m oscillating with force constant
k and damping coefficient 'b'. Compare the terms of damped oscillation with the devices in LCR circuit. ,d LCR ifjiFk fn;k tkrk gSA bldh ,d voefUnr nksyu ds lkFk rqyuk dh tkrh gS ftldk æO;eku m, nksyu dk
cy fu;rkad 'k' gS vkSj voefUnr fu;rkad b gS] rc voefUnr nksyu dh bdkbZ;ksa dks LCR ifjiFk ls rqyuk djksA
(1) L m , C 1k
, R b (2) L m , C k, R b
(3) L k , C b, R m (4) L 1m
, C 1k
, R 1b
Ans. (1) Sol. In damped oscillation voefUnr nksyu esa ma + bv + kx = 0
2
2
d x dxm b kxdtdt
= 0 …(i)
In the circuit ifjiFk esa
– iR – diLdt
– qc
= 0
2
2
d q dq 1L R .qdt cdt
= 0 …(ii)
Comparing equation (i) and (ii) rqyuk djus ij
m = L, b = R, k = 1c
11. A lift can hold 2000kg, friction is 4000N and power provided is 60HP. (1 HP = 746W) Find the maximum speed with which lift can move up.
,d fy¶V 2000kg nzO;eku mBk ldrh gS] ;fn ?k"kZ.k 4000N gS rFkk nh xbZ 'kfDr 60HP gSA (1 HP = 746W) rc fy¶V dh vf/kdre pky D;k gksxh] ftlls og ÆÅij tk ldsA
| JEE MAIN-2020 | DATE : 07-01-2020 (SHIFT-1) | PAPER-1 | MEMORY BASED | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222 To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
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PAGE # 7
7340010333
(1) 1.9 m/s (2) 1.7 m/s (3) 2 m/s (4) 1.5 m/s Ans. (1) Sol. 4000 × V + mg × V = P
20000400074660
= V
V = 1.86 m/s. 9.1 m/s. 12. A H–atom in ground state has time period T = 1.6 × 10–16 sec. find the frequency of electron in first excited
state ,d H–ijek.kq ds bysDVªkWu dk fuEure voLFkk esa vkorZdky T = 1.6 × 10–16 sec. gSA rc bysDVªkWu dh izFke mÙksftr
voLFkk esa vkorhZ gksxh & (1) 7.8 × 1014 (2) 7.8 × 1016 (3) 3.7 × 1014 (4) 3.7 × 1016 Ans. (1)
Sol. T 2
32
zn
zn
zn
vr
81
nn
TT
32
31
2
1
T2 = 8T1 = 8 × 1.6 × 10–16 = 12.8 × 10–16
f2 = 16–108.121
7.8 × 1014
13. Magnification of compound microscope is 375. Length of tube is 150mm. Given that focal length of
objective lens is 5mm, then value of focal length of eyepiece is: ,d la;qDr lq{en'khZ dh vko/kZu {kerk 375 gSA uyh dh yEckbZ 150 mm gSA fn;k x;k gS vfHkn`';d ySUl dh Qksdl
nwjh fo = 5mm rc vfHkus=k ysal dh Qksdl nwjh fe dk eku gksxk ? (1) 2mm (2) 22mm (3) 12mm (4) 33mm Ans. (2) Sol. Case-I
If final image is at least distance of clear vision ;fn vfUre izfrfcEc fudVre Li"B nf"V ij gks rks
M.P. =
e0 fD1
fL ; 375 =
ef251
5150
ef251
30375
ef25
30345
fe = 345750 = 2.17 cm; fe 22 mm
Case-II If final image is at infinity ;fn vfUre izfrfcEc vuUr ij gks
M.P. =
e0 fD
fL
= 375
fe 22 mm
| JEE MAIN-2020 | DATE : 07-01-2020 (SHIFT-1) | PAPER-1 | MEMORY BASED | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222 To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
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7340010333
14. 1 litre of a gas at STP is expanded adiabatically to 3 litre. Find work done by the gas. Given = 1.40 and
31.4= 4.65 ,d xSl ds 1 yhVj dks STP ij 3 yhVj rd izlkfjr fd;k tkrk gSA xSl }kjk fd;k x;k dk;Z gksxkA fn;k gS =
1.40 rFkk 31.4= 4.65 (1) 100.8J (2) 90.5J (3) 45J (4) 18J Ans. (2) Sol. P1 = 1atm, T1 = 273K 1 1 2 2P V P V
P2 = P1 1
2
VV
= 1atm 1.41
3
now work done vc fd;k x;k dk;Z = 1 1 2 2P V – P V– 1
= 88.7 J
Closest ans is 90.5 J
15. A string of length 60 cm, mass 6gm and area of cross section 1mm2 and velocity of wave 90m/s. Given young's modulus is Y = 1.6 × 1011 N/m2. Find extension in string
,d 60 cm yEch jLlh] ftldk nzO;eku 6gm vkSj vuqizLFk dkV dk {ks=kQy 1mm2 vkSj rjax dk osx 90m/s gSA (fn;k gS ;ax izR;kLFkrk xq.kkad Y = 1.6 × 1011 N/m2) jLlh esa izlkj crkb,
(1) 0.3 mm (2) 0.2 mm (3) 0.1 mm (4) 0.4 mm Ans. (1)
Sol. Tv
T = v2
2v Y
A
= 2v
AY
after substituting value of v,,A and Y we get v,,A rFkk Y dk eku j[kus ij
= 0.3 mm
16. Which of the following gate is reversible buesa ls dkSulk }kj mRØe.kh; gSA
(1)
(2)
(3)
(4)
Ans. (1) Sol. A logic gate is reversible if we can recover input data from the output eg. NOT gate ,d rdZ }kj mRØe.kh; gksxk ;fn fuxZr ladsr ds vk/kkj ij fuos'kh ladsr izkIr fd;k tk ldsA eg. NOT }kj
| JEE MAIN-2020 | DATE : 07-01-2020 (SHIFT-1) | PAPER-1 | MEMORY BASED | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222 To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
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PAGE # 9
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17. A thin uniform rod is of mass M and length L. Find the radius of gyration for rotation about an axis passing
through a point at a distance of L4
from centre and perpendicular to rod.
,d iryh ,dleku NM+ ftldk nzO;eku M vkSj yEckbZ L gSA rks NM+ ds dsUnz ls L4
nwjh vkSj NM+ ds yEcor v{k ds
lkis{k ?kq.kZu f=kT;k Kkr djsA
(1) 748
L (2) 548
L (3) 724
L (4) 1924
L
Ans. (1)
Sol.
L/4
22ML LM12 4
= MK2
2 2L L
12 16 = K2
7K48
L
18. A satellite of mass 'M' is projected radially from surface of earth with speed 'u'. When it reaches a height
equal to radius of earth, it ejects a rocket of mass M10
and it itself starts orbiting the earth in circular path
of radius 2R, find the kinetic energy of rocket. ,d nzO;eku 'M' ds mixzg dks i`Foh dh lrg ls 'u' pky ds lkFk f=kT; fn'kk eas iz{ksfir fd;k tkrk gSA tc og i`Foh
dh f=kT;k ds cjkcj Å¡pkbZ rd igq¡p tkrk gS] rc jkWdsV M10
nzO;eku dks fu"dkflr dj nsrk gS vkSj ;g Loa; 2R
f=kT;k ds oÙkh; iFk esa ifjØek djuk 'kq: dj nsrk gS] rks jkWdsV dh xfrt ÅtkZ gksxh
(1) 2 e119GM5M u
200R
(2) 2 e113GM5M u
200R
(3) 2 e119GMM u20 100R
(4) 2 e113GMM u20 200R
Ans. (1)
Sol. 2 2e eGM M GM M1 1Mu MvR 2 2R 2
M
u
R
R
M
R
v
Me
Me
v = 2 eGMu
R
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PAGE # 10
7340010333
M/10
Me
Vr
9M/10
VT
eGMV
2R
VT Transverse velocity of rocket VR Radial velocity of rocket 2R
eT
GMM 9MV10 10 2R
2 er
GMM V M u10 R
Kinetic energy xfrt ÅtkZ = 2 2 2e eT r
GM GM1 M MV V 81 100u 1002 10 20 2R R
= 2 e119GMM 100u20 2R
= 2 e119GM5M u
200R
19. The current 'i' in the given circuit is fn;s x;s ifjiFk esa /kkjk 'i' gS &
1V
2
1
1 i 2
(1) 0.2A (2) 0.3A (3) 0.5A (4) 0.25A Ans. (1)
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7340010333
Sol.
1V
2
1
1 i 2
I
I = 5.2
1 = 0.4A
i = 2I = 0.2A
20. A current carrying circular loop is placed in an infinite plane if i is the magnetic flux through the inner
region and o is magnitude of magnetic flux through the outer region, then ,d /kkjkokgh oÙkkdkj ywi vuUr ry esa fLFkr gS] ;fn vkUrfjd {kS=k ls lEcaf/kr pqEcdh; ¶yDl i rFkk ckg~; {kS=k ls
lEcaf/kr pqEcdh; ¶yDl o gks rks (1) i > o (2) i < o
(3) i = – o (4) i = o Ans. 3 Sol. As magnetic field lines always form a closed loop, hence every magnetic field line creating magnetic flux
in the inner region must be passing through the outer region. Since flux in two regions are in opposite direction,
i = – 0 Hindi. pawfd pqEcdh; {kS=k js[kk,saa cUn ywi dk fuekZ.k djrh gSA vr% izR;sd pqEcdh; {kS=k js[kk }kjk vkUrfjd {kS=k esa mRiUu
fd;k x;k pqEcdh; ¶yDl ckg~; {kS=k ls xqtjrk gSA vr% bu nksuksa {kS=kks esa mifLFkr ¶yDl foijhr fn'kk esa gksrk gSA i = – 0
Numerical Value Type (la[;kRed izdkj) This section contains 5 Numerical value type questions.
bl [k.M esa 5 l[;kRed izdkj ds iz'u gSaA
21. Consider a loop ABCDEFA with coordinates A (0, 0, 0), B(5, 0, 0), C(5, 5, 0), D(0, 5, 0) E(0, 5, 5) and
F(0, 0, 5). Find magnetic flux through loop due to magnetic field ˆ ˆB 3i 4k
ywi ABCDEFA dh dYiuk dhft, ftlds funsZ'kkad Øe'k% A (0, 0, 0), B(5, 0, 0), C(5, 5, 0), D(0, 5, 0)
E(0, 5, 5) rFkk F(0, 0, 5) gSA pqEcdh; {kS=k ˆ ˆB 3i 4k
ds dkj.k ywi ls lEcaf/kr pqEcdh; ¶yDl Kkr djksA Ans. 175
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PAGE # 12
7340010333
Sol. = B.A
= ˆ ˆ(3i 4k) . ˆ ˆ(25i 25k)
= (3 × 25) + (4 × 25) = 175 weber
22. A Carnot's engine operates between two reservoirs of temperature 900K and 300K. The engine performs
1200 J of work per cycle. The heat energy delivered by the engine to the low temperature reservoir in a cycle is:
,d dkuksZ±V batu nks rkih; fudk; ds e/; dk;Z djrk gS ftuds rkieku Øe'k% 900K rFkk 300K gSA batu izR;sd
pØ esa 1200J dk;Z djrk gSA ,d pØ esa batu fuEu rki okys fudk; dks fdruh Å"ek LFkkukUrjhr djrk gSA Ans. 600
Sol. = hQ
W = 9003001 =
32
Qh = 23 W = 1800 J
QL = Qh – W = 600 J
23. A non-isotropic solid metal cube has coefficients of linear expansion as 5 × 10–5 /°C along the x-axis and
5 × 10–6 /°C along y-axis and z-axis. If coefficient of volume expansion of the solid is C × 10–6 /°C then the value of C is
vlekax Bksl /kkfRod ?ku dk rkih; js[kh; izlkj xq.kkad x–v{k ds vuqfn'k 5 × 10–5 /°C rFkk y ,oa z–v{k ds vuqfn'k
5 × 10–6 /°C gSA ;fn Bksl dk vk;ru izlkj xq.kkad C × 10–6 /°C gks] rks C dk eku Kkr djksA Ans. 60 Sol. V = 22 + 1 = 10 × 10–6 + 5 × 10–5 = 60 × 10–6 /°C
24. A particle is released at point A. Find its kinetic energy at point P. (Given m = 1 kg and all surfaces are frictionless)
,d d.k dks fcUnq A ls NksM+k tkrk gSA fcUnq P ij xfrt ÅtkZ Kkr djksA fn;k gS m = 1 kg rFkk lHkh lrg ?k"kZ.kjfgr
gSA A
2m
B
C
1m
P
Ans. 10 Sol. KE = PE1 – PE2 = mgh1 – mgh2 = 1 × 10 × 2 – 1 × 10 × 1 = 10J
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Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222 To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
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PAGE # 13
7340010333
25. On a photosensitive metal of area 1 cm2 and work function 2eV, light of intensity 6.4 × 10–5 W/cm2 and wavelength 310 nm is incident normally. If 1 out of every 103 photons are successful, then number of photoelectrons emitted in one second is 10x. Find x
,d izdk'k laosnh /kkrq dk {kS=kQy 1 cm2 rFkk dk;ZQyu 2eV gSA bl ij 6.4 × 10–5 W/cm2 rhozrk ;qDr 310 nm rajxnS/;Z dk izdk'k yEcor~ vkifrr gSA ;fn 103 QksVksu esa ls dsoy ,d QksVksu] ,d izdk'k bysDVªkWu mRltZu dj
ldrk gks rks ,d lsd.M esa mRlftZr QksVks bysDVªkWu dh la[;k 10x gS] rc x Kkr djksA Ans. 11
Sol. Energy of photon. E = 1240310
= 4eV > 2eV (so photoelectric effect will take place)
= 4 × 1.6 × 10–19 = 6.4 × 10–19 Joule No. of photons falling per second
= 5
1419
6.4 10 1 106.4 10
No. of photoelectron emitted per second
= 14
113
10 1010
Hindi. QksVksu dh ÅtkZ E = 1240310
= 4eV > 2eV (vr% izdk'k fo|qr izHkko mRiUu gksxk)
= 4 × 1.6 × 10–19 = 6.4 × 10–19 Joule izfrlsd.M vkifrr QksVksu
= 5
1419
6.4 10 1 106.4 10
izfrlsd.M mRlftZr QksVksbysDVªkWu
= 14
113
10 1010
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