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JEE-MAIN 2017 (PEN-PAPER MODE:2-APRIL-2017)HINTS & SOLUTIONS FOR CODE-A
PHYSICS 1. A man grows into a graint such that his linear dimensions increase by a factor of 9. Assuming thta his
density remains same, the streass in the leg will change by a factor of :
(1) 9 (2) 91
(3) 81 (4) 811
Ans. [1]
Sol. Stress = AA
AVg
AMg lg
2. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity
vs time ?
(1) (2) (3) (4)
Ans. [3]
Sol. gtuv –
gut
gtov
3. A body of mass m = 10–2 kg is moving in a medium and expriences a frictional force F = – kv2. Its
initial speed is v0 = 10 ms–1. If, after 10s, its energy is 81
mv02, the value of k will be :
(1) 10–3 kg m–1 (2) 10–3 kg s–1 (3) 10–4 kg m–1 (4) 10–1 kg m–1 s–1
Ans. [3]
Sol. F = – kv2
2–kvdt
mdv
10
02 –
0
kdtvdvm
v
v
]0–10[1–1
0k
vvm
mk
vv101–1
0
10110001
k
v
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20
2 )(81
21 vmmvE
410
)110(10 2
24 k
21104 k4–10k
4. A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work
done by the force during the first 1 sec. will be :
(1) 4.5 J (2) 22 J (3) 9 J (4) 18 J
Ans. [1]
Sol. tdtdvm 6
mdv = 6tdt
1
00
6 tdtdvmv
m.v =
1
0
2
26
t
31133 2
mtv
9121
21 2 mvw
5.4w5. The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I.
What is the ratio l/R such that the moment of inertia is minimum ?
(1) 23
(2) 23
(3) 1 (4) 23
Ans. [1]
Sol.124
22 MlMrI (1)
lrM 2. (2)
422
22
.124 r
MMMrI
522
3 4–12
2.4 r
MrMdrdI
= 0 (3)
solving equation (1) (2) (3)
l/R = 23
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6. A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical
plane (see figure). There is negligible friction at the pivot. The free end is held vertically above thee pivot
and then released. The angular acceleration of the rod when it makes an angle with the vertical is :
(1) sin23
lg
(2) sin32
lg
(3) cos23
lg
(4) cos32
lg
Ans. [1]
Sol. i
3
sin231 2mlmg
sin32
lg
7. The variation of acceleration due to gravity g with distance d from centre of the earth is best represented
by (R = Earth’s radius) :
(1) (2)
(3) (4)
Ans. [4]
Sol. rgm
21r
gout
8. A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100
gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found
the be 750C. T is given by :
Given : room temperature = 300C, specific heat of copper = 0.1 cal/gm0C)
(1) 8000C (2) 8850C (3) 12500C (4) 8250C
Ans. [2]
Sol. Energy release by ball = energy current by water + collision
100 × .1 (T–15) = 100 × .1 × (75–30) + 170 × 1 ( 75–30)
T = 8850C
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9. An external pressure P is applied on a cube at 00C so that it is equally compressed from all sides. K is
the bulk modulus of the material of the cube and is its coefficient of linear expansion. Suppose we want
to bring the cube to its original size by heating. The temperature should be raised by :
(1) KP3 (2) K
P (3) PK
3(4) 3PK
Ans. [1]
Sol. vvkp
vttvv 3)0–(3
by (1) and (2)
kpT
vvtkp
33.
10. Cp and Cv are specific heats at constant pressure and constant volume respectively. It so observed that
Cp –Cv = a for hydrogen gas
Cp –Cv = b for nitrogen gas
The correct relation between a and b is :
(1) ba141
(2) a = b (3) a = 14 b (4) a = 28 b
Ans. [3]
Sol. Cp – CV = a
Cp – Cv = b
If Cp = Cv represend gram
2a = 28 b
a = 14 b
11. The temperature of an open room of volume 30 m3 increases from 170C to 270C due to the sunshine. The
atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the number of molecules in the
room before and after heating, then n f–n i will be :
(1) – 1.61 × 1023 (2) 1.38 × 1023 (3) 2.5 × 1025 (4) –2.5 × 1025
Ans. [4]
Sol. Aif Nnn
3002.83010–
2902.83010–
55
–2.5 × 1025
12. A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of
equilibrium. The kinetic energy – time graph of the particle will look like :
(1) (2)
(3) (4)
Ans. [4]
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Sol. 2]cos[21 wtAwmE
13. An observer is moving with half the speed of light towards a stationary microwave source emitting waves
at frequency 10 GHz. What is the frequency of the microwave measured by the observer ?
(speed of light = 3 × 108 ms–1)
(1) 10.1 GHz (2) 12.1 GHz (3) 17.3 GHz (4) 15.3 GHz
Ans. [3]
Sol.cvcv
ff–1
1'
3.17'f GHz
14. An electric dipole has a fixed dipole moment p , which makes angle with respect to x-axis. When
subjected to an electric field iEE ˆ1
, it experiences a torque kT ˆ1
. When subjected to another electric
field jEE ˆ3 12
it experiences a torque 12 –TT
. The angle is :
(1) 300 (2) 450 (3) 600 (4) 900
Ans. [3]
Sol. sin11 pE (1)
)90sin(22 pE (2)
12 3EE
15. A capacitance of 2 F is required in an electrical circuit across a potential difference of 1.0 kV. A large
number of 1 F capacitors are available which can withstand a potential difference of not more than 300 V..
The minimum number of capacitors required to achieve this is :
(1) 2 (2) 16 (3) 24 (4) 32
Ans. [4]
Sol.
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16. In the given circuit diagram when the current reaches steady state in the circuit, the change on the capacitor
of capacitance C will be :
(1) CE (2) )( 2
1
rrrCE (3) )( 2
2
rrrCE (4) )( 1
1
rrrCE
Ans. [3]
Sol. At steasty state current in capacitor
brach is zero
so 2rr
Ei
potential on r2 = 2
2
rrEr
Q = Cv for capacitor
2
2
rrrCEQ
17.
In the above circuit the current in each resistance is :(1) 1A (2) 0.25 A (3) 0.5 A (4) 0 A
Ans. [4]
Sol. applying kvl by taking all all loops
in any clase loop E = 0
so i = 0
18. A magnetic needle of magnetic moment 6.7 × 10–2 Am2 and moment of inertia 7.5 × 10–6 kg m2 is performing
simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is :
(1) 6.65 s (2) 8.89 s (3) 6.98 s (4) 8.76 s
Ans. [1]
Sol.MB
I2
19. When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 , it shows fullscale deflection. The value of the resistance to be put in series with the galvanometer to convert i t into avoltmeter of range 0 – 10 V is :
(1) 1.985 × 103 (2) 2.045 × 103 (3) 2.535 × 103 (4) 4.005 × 103
Ans. [1]
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Sol. GgvR –
1.985 × 103
20. In a coil of resistance 100 , a current is induced by changing the magnetic flux through it as shown in
the figure. The magnitude of change in flux through the coil is :
(1) 200 Wb (2) 225 Wb (3) 250 Wb (4) 275 Wb
Ans. [3]
Sol. dtde
dedtRidtdt
= 25.10100
= 250 Wb
21. An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It
produces continuous as well as characteristic X-rays. If min with log V is correctly represented in :
(1) (2)
(3) (4)
Ans. [1]
Sol. evhc
min
take by both side
vehc log–loglog min
22. A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converginglengs of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The finalimage formed is :(1) Real and at a distance of 40 cm from convergent lens.(2) Virtual and at a distance of 40 cm from convergent lens.
(3) Real and at a distance of 40 cm from the divergent lens
(4) Real and at a distance of 6 cm from the convergent lens
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Ans. [1]
Sol. For first lens v = – 25
for second lens u = –40
using fuu11–1
23. In a Young’s double slit experiment, slits are separated by 0.5mm, and the screen is placed 150 cm away.
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain inteference fringes
on the screen. The least distance from the common central maximum to the point where the bright fringes
due to both the wavelengths coincide is :
(1) 1.56 mm (2) 7.8 mm (3) 9.75 mm (4) 15.6 mm
Ans. [2]
Sol. 2211 nn
1
5520650
n
41 n
dDy 14
mmy 8.7
24. A particle A of mass m and initial velocity v collides with a particle B of mass 2m
which is at reast. The
collision is heat on, and elastic. The ratio of the de-Broglie wavelengths A to B after the collision is :
(1) 31
B
A
(2) 2B
A
(3) 32
B
A
(4) 21
B
A
Ans. [2]
Sol. from momentum conservation
mv = mv1 + m/2 v2 (i)
vovvl
–– 21 (ii)
from (i) and (ii)
34,
3 21vvvv
11 p
hl 2
2 phl
1
2
2
1
pp
ll
22
1 ll
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25. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths 21 / r , is givenby :
(1) 34
r (2) 32
r (3) 43
r (4) 31
r
Ans. [4]
Sol. –E – (–4/E/3) = 2
hc
E = 2
hc (i)
–E –(–2E) = 1
hc
E = 1
hc (ii)
(i) & (ii)
31
2
1
26. A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no mucleus B. At
sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by :
(1) 3.1log2log
2Tt (2) 2log
3.1logTt (3) )3.1log(Tt (4) )3.1log(tt
Ans. [2]
Sol. At time t number of mole for ‘A’ = N = teN –0
at time ‘t’ number of mole for ‘B’ = N0 = N
according 3.0–0 N
NN
TB
NN 2
0 log,3.01–
2log3.1logTt
27. In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between th input and
the output voltages will be :
(1) 450 (2) 900 (3) 1350 (4) 1800
Ans. [4]
Sol. input and output are in opp. phage
phage difference = 1800
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28. In amplitude modulation, sinusoidal carrier frequency used is denoted by c and the signal frequency is
denoted by m . The bandwidth )( m of the signal is such that cm . Which of the following
frequencies is not contained in the modulated wave ?
(1) m (2) c (3) cm (4) mc –
Ans. [1]
Sol.
29. Which of the following statements is false ?
(1) Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude.
(2) In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is
disturbed.
(3) A rheostat can be used as a potential divider
(4) Kirchhoff’s second law represents energy conservation
Ans. [2]
Sol.
30. The following observations were taken for determining surface tension T of water by capillary method .
diameter of capillary, D = 1.25 × 10–2 m
rise of water, h = 1.45 × 10–2 m.
Using g = 9.80 m/s2 and the simplified relation 3102
rhgT N/m, the possible error in surface tension is
closest to :
(1) 0.15% (2) 1.5% (3) 2.4% (4) 10%
Ans. [2]
Sol. dhrt %%%% = .8 + .6 + .1%T = 1.5 %
CHEMISTRY 31. Given
2 2graphite0 1
r
C O g CO g ;H 393.5kJmol
2 2 2
0 1r
1H g O g H O 1 ;2
H 285.8kJ mol
2 2 4 2CO g 2H O 1 CH g 2O g ; 0 1
rH 890.3kJ mol Based on the above thermochemical equations, the value of 0rD H at 298 K for the
reaction
2 4graphiteC 2H g CH g will be :
(1) 174.8kJ mol (2) 1144kJ mol (3) 174.8kJ mol (4) 1144.0kJ mol
Ans. [1]
Sol. 0R n R e ac tan t P r oud uctH H com b. H com b.
2 4g rap hite H C H
H com b. 2 H com b. H com b.
393.5 2 285.8 890.3
=-74.8 kJ/mol–1
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32. 1 gram of a carbonate 2 3M CO on treatment with excess HCl produces 0.01186 mole of CO2. The molar
mass of M2CO2 in g mol-1 is :
(1) 118.6 (2) 11.86 (3) 1186 (4) 84.3
Ans. [4]
Sol. 2 3 2 2M CO 2HCl 2MCl CO H O 1 mol of CO2 is obtained by -1 mol of m2CO30.01186 mol of CO2 is obtained by = 0.01186 mol of M2CO3
WnM
wMn
10.01186
M = 84.31
33. U is equal to :
(1) Adiabatic work (2) Isothermal work (3) Isochoric work (4) Isobaric work
Ans. [1]
Sol. For adiabatic process q 0
from Ist law of thermodynamics . U q w
q 0
U w
34. The Tynadall effect is observed only when following conditions are satisfied:
(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used.
(b) The diameter of the dispersed particle is not much smaller than the wavelengths of the light used.
(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude
(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.
(1) (a) and (b) (2) (b) and (c) (3) (a) and (d) (4) (b) and (d)
Ans. [4]
Sol. (Fact based)
35. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ‘a’, the closest
approach between two atoms in metallic crystal will be
(1) 2a (2) 12 (3) 2a (4) 2 2a
Ans. [2]
Sol. In Fcc unit cell -
A B
C
a
a
Closest distance - 2r
2 2AC a a
4r 2a
2a2r2
a2r2
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36. Given 3 /Cr2
0 0Cl /Cl Cr
E 1.36V,E 0.74V
2 3 2r 7 4
0 0C O /Cr MnO /Mn
E 1.33V,E 1.51V
Among the following, the strongest reducing agent is :
(1) Cr3+ (2) Cl- (3) Cr (4) Mn2+
Ans. [3]
Sol. Cr is the strongest reducing reagent.
37. The freezing point of benzene decreases by 0.450 C when 0.2 g of acetic acid is added to 20 g of benze.
If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will
be: (Kf for benzene = 5.12 K kg mol -1)
(1) 74.6 % (2) 94.6 % (3) 64.6% (4) 80.4 %
Ans. [2]
Sol. f fT i K m
.2 10000.45 i 5.1260 20
i 0.52for Association of CH3COOH in benzene 2 CH3COOH (CH3COOH)2 n = 2
i 111n
0.52 1112
0.946
94.6%
38. The radius of the second Bohr orbit for hydrogen atom is :
(Plank’s Const. h = 6.6262×10-34 Js; mass of electron = 9.1091×10-31 kg; Charge of electron e = 1.620210×10-
19 C; permittivity of vacuum 12 1 3 20 8.854185 10 kg m A )
(1) 0.529Å (2) 2.12Å (3) 1.65 Å (4) 4.76Å
Ans. [2]
Sol. r2 = 0.529×2n
Zfor hydrogen - Z = 1
n = 22
22r 0.5291
2r 2.116
2r 2.12Å
39. Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R 1 exceeds that of R2
by 10 kJ mol-1 . If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then In(k2/
k1) is equal to : (R = 8.314 J mol-1 K-1)
(1) 6 (2) 4 (3) 8 (4) 12
Ans. [2]
Sol. Given- 1 2E E 10
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1 2E E 10kJ / mol
1 2E E 10 1000 J / mol
2 1 2
1
K E ElnK RT RT
2 1 2
1
K E ElnK RT
2
1
K 10 1000lnK 8.314 300
2
1
Kln 4K
40. pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their
salt (AB) solution is :
(1) 7.0 (2) 1.0 (3) 7.2 (4) 6.9
Ans. [4]
Sol. pH of salt made by weak acid & weak base.
k k1ph 7 P a P b2
1pH 7 3.2 3.42
1pH 7 0.22
pH 7 0.1
pH 6.9
41. Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the
one which is incorrect, s :
(1) both form nitrides (2) nitrates of both Li and Mg yield NO2
(3) both form basic carbonates (4) both form soluble bicarbonates
Ans. [2]
Sol. Fact based.
42. Which of the following species is not paramagnetic?
(1) O2 (2) B2 (3) NO (4) CO
Ans. [4]
Sol. CO molecule do not have any unpaired
So that it is diamagnetic.
43. Which of the following reactions is an example of a redox reaction?
(1) 6 2 4XeF H O XeOF 2HF (2) 6 2 2 2XeF 2H O XeO F 4HF
(3) 4 2 2 6 2XeF O F XeF O (4) 2 5 6XeF PF XeF PF
Ans. [3]
Sol.
4 6
2 02 2
Xe Xe oxidationO O reduction
Hence (3) is redox reaction
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44. A water sample has ppm level concentration of following anions 24 3F 10; SO 100; NO 50
The anion/anions that make/makes the water sample unsuitable for drinking is/are:
(1) only F– (2) only 24SO (3) only 3NO (4) both 2
4SO and 3NO
Ans. [1]
Sol. level of concentration should between .7 to 1.2 ppm
45. The group having isoelectronic species is :
(1) 2 2O ,F ,Na,Mg (2) 2O ,F , Na ,Mg (3) 2 2O ,F ,Na ,Mg (4) O ,F ,Na,Mg
Ans. [3]
Sol. 2 2O ,F , Na ,Mg all have 10 electrons.
46. The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are :
(1) Cl– and ClO– (2) Cl and 2ClO (3) ClO and 3ClO (4) 2ClO and 3ClO
Ans. [1]
Sol. Cold
2 2dil.Cl NaOH NaCl NaOCl H O
47. In the following reactions, ZnO is respectively acting as a/an :
(a) 2 2 2ZnO Na O Na ZnO (b) 2 3ZnO CO ZnCO
(1) acid and acid (2) acid and base (3) base and acid (4) base and base
Ans. [2]
Sol. (a) since Na2O is more basic in nature so ZnO act as acid.
(b) CO2 is acidic and ZnO act as basic.
48. Sodium salt of an organic acid ‘X’ produces effervescence with conc. H2SO4 . ‘X’ reacts with the acidified
aqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of KMnO4. ‘X’ is
(1) CH3COONa (2) Na2C2O4 (3) C6H5COONa (4) HCOONa
Ans. [2]
Sol. Oxalates reacts with conc. H2SO4 to produce CO2 while produces effervescence. Further it reacts with aq.
CaCl2 (acidified) to form CaC2O4 (white ppt). CaC2O4 reacts with acidified KMNO4 which decolourises pink
violet colour to colourless.
49. The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%); Carbon
(22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would gain if all 1H
atoms are replaced by 2H atoms is :
(1) 7.5 kg (2) 10 kg (3) 15 kg (4) 37.5 kg
Ans. [1]
Sol. If 1 2H H
then 10% H has weight of 7.5 kg in 75 kg if it replace by 2 H it becomes 15 kg. so difference - 7.5
kg.
50. On treatment of 100 mL of 0.1 M solution of CoCl3.6H2O with excess AgNO3; 1.2×1022 ions are precipitated.
the complex is
(1) 2 36Co H O Cl (2) 2 2 25Co H O Cl Cl .H O
(3) 2 2 24Co H O Cl Cl.2H O (4) 2 3 23Co H O Cl 3H O
Ans. [2]
Sol. Mole of comp.
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nmv
n = 0.1×0.1n = 0.01
moles of ions precipitated = 22
23
1.2 106.02 10
= 0.02 moles
Only Cl– ions are precipitated.0.1 moles of compound gives = 0.02 moles of Cl–
1 moles of compound gives = 2 moles of Cl–
For the formation of 2 moles of Cl– formula of compound must be- 2 2 25Co H O Cl Cl .H O
51. Which of the following compound will form significant amount of meta product during mono-nitration of
meta product during mono-nitration reaction?
(1)
NH2
(2)
NHCOCH6
(3)
OH
(4)
OCOCH3
Ans. [1]
Sol. Aniline form Anelinium
NH3
ion which is meta evreeting.
52. Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to
decolourize the colour bromine?
(1)
O
Br(2)
O
Br(3)
Br
O
(4) Br
C H6 5
Ans. [3]
53. The formation of which of the following polymers involves hydrolysis reaction?
(1) Nylon 6,6 (2) Terylene (3) Nylon 6 (4) Bakelite
Ans. [3] Nylon-6
Sol. n
H C2
H C2
H C2 CH2
CH2
C = 0
H|
N
Caprolactum
0
2
260 270 CH O
2 5 n
O||C NH CH
Nylon 6
54. Which of the following molecules is least resonance stabilized?
(1) N
(2) O
(3) (4) O
Ans. [2]
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Sol.OAbove structure can’t not stabilized by resonance.
55. The increasing order of the reactivity of the following halides for the SN1 reaction is :
3 2 3CH CHCH CH|
Cl(I)
3 2 2CH CH CH Cl
(II)
3 6 4 2p H CO C H CH Cl
(III)
(1) (I) < (III) < (II) (2) (II) < (III < (I) (3) (III) < (II) < (I)(4) (II) < (I) < (III)
Ans. [4] (II) < (I) < (III)
I)
Sol.3 2 3CH CH CH CH
|Cl(I)
3 2 3CH CH CH CH
(I)
3 2 2CH CH CH Cl
(II)
3 2 2CH CH CH
(II)
3 6 4 2P H CO C H CH Cl
CH2
OCH3
(II
Stability of carbocation III > I > IISN1 order - III > I > II
56. The major product obtained in the following reaction is :
Br
H
C H6 5
C H6 5
(+)
t BuOK
(1) t6 5 2 6 5C H CH O Bu CH C H (2) t
6 5 2 6 5C H CH O Bu CH C H
(3) t6 5 2 6 5C H CH O Bu CH C H (4) 6 5 6 5C H CH CHC H
Ans. [4]
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Sol.
Br
H
C H6 5
C H6 5
tBuOK(Elimination)
C H -CH=CH-C H6 5 6 5
57. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?
(1)
OHOH C2 CH OH2
OCH3
OH
HO(2)
HOH C2
O CH OCH2 3
OHOH
OH
(3)
OHOH C2 CH OH2
OCOCH3
OH
HO(4)
OHOH C2 CH OH2
OH
HO
Ans. [3]
Sol. Fact based on reactants.
Because it has free ketone group, so it act as reducing agent.
58. 3-Methyl-pen-2-ene on reaction with HBr in presence of peroxide forms an addition product. the number
of possible stereoisomers for the product is :
(1) Two (2) Four (3) Six (4) Zero
Ans. [2]
Sol. 3 2 3
3
CH CH C CH CH|CH
HBrPeroxide
* *
3 2 3
3
CH CH CH CH CH| |Br CH
Chiral Carbon - 2 (n)Sterioisomers = 2n
= 22
= 4
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59. The correct sequence of reagents for the following conversion will be:
O
CHO
HO CH3
CH3
CH3
HO
(1) 3 3 32CH MgBr, Ag NH OH ,H / CH OH
(2) 3 3 32Ag NH OH ,CH MgBr,H / CH OH
(3) 3 3 32Ag NH OH ,H / CH OH,CH MgBr,
(4) 3 3 3 2CH MgBr,H / CH OH, Ag NH OH
Ans. [3]
Sol. Fact based on reactants.
60. The major product obtained in the following reaction is :
OO
COOH
DIBAL H
(1)COOH
CHO(2)
CHO
CHO(3)
COOH
CHO
OH
(4)
CHO
CHO
OH
Ans. [4]
Sol. Fact based.
MATHEMATICS
61. The function f : R 1 1,2 2
defined as f(x) =
1 1,2 2
is :
(1) Injective but not surjective (2) surjective but not injective
(3) neither injective nor surjective (4) invertible
Ans. [2]
Sol. For one one
f(x1) = f(x2)
1 22 2
1 2
x x1 x 1 x
2 21 2 2 1 21x x x x x x
1 2 1 2 2 1(x x ) x x (x x ) 0
Page No. 19 ||www.gyankuteer.com||
62. If, for a positive integer n, the quadratic equation
x(x+1) + (x+1) (x+2) + ....
+ (x n 1) (x+n) = 10n
has two consecutive integral solutions, then n is equal to :
(1) 9 (2) 10 (3) 11 (4) 12
Ans. [3]
Sol. Given equation is
x(x+1) + (x+1) (x+2) + ....................... (x+n–1) (x+n) = 10n
(x2+x+0) + (x2+3x+2) + .................. + x2 + (2n–1)x + n(n–1) = 10n
nx2 + x (1+3+ ...... + 2n–1) + (0+2+6+12+ .... + upto n terms) 10n
x2 + nx + (n 1)(2n 1)
6
n(n 1)
2
= 10
x2 + nx + n 1
2
2n 1 1
3
= 10
x2 + nx + 2n 13
= 10
3x2 + 3nx + n2–1 = 30 3x2 + 3nx + n2–31 = 0Now D = 9n2–12 (n2–31)D = 372 – 3n2
for n=11 only D is perfect square.and where n=11, equ. (i) becomes3x2 + 33x + 90 = 0
or x2 + 11x + 30 = 0 which has roots –5, –6 and are consecutive
Hence (3) is correct.
63. Let be a complex number such that 2 1 z where z = 3 . If
2 2
2 7
1 1 11 11
= 3k,
then k is equal to :
(1) z (2) –1 (3) 1 (4) -z
Ans. [4]
Sol.2 2
2
1 1 11 w 1 w1 w w
= 3k
3 4 2 2 21 ( w w) w 1 w w 1 w w 1 3k
2 21 w w w w 1 2w 3k –1 – 3w + w2 + 1 + 2w2 = 3k–32 + 3w2 = 3k– w + w2 = k–w – w + w + w2 = kk = –z
Page No. 20 ||www.gyankuteer.com||
64. If A = 2 34 1
, then adj (3A2 + 12A) is equal to :
(1)2 34 1
(2) 51 8463 72
(3) 72 6384 51
(4) 72 8463 51
Ans. [1]
Sol. A = 2 34 1
A2 = 16 912 13
3A2 = 48 2736 39
12A = 24 3648 12
3A2 + 12A = 72 6384 51
Adj (3A2 + 12A) = 51 6384 72
65. If S is the set distinct values of ‘b’ for which the following system of linear equations
x + y + z = 1
x + xy + z = 1
ax + by + z = 0
has no solution then S is :
(1) an infinite set (2) a finite set containing two or more elements
(3) a singleton (4) an empty set
Ans. [3]
Sol. Given equations are
x+y+x=1
x+ay+z=1
ax+by+z=0
for no solution must be zero
1 1 11 a 1a b 1
= 0
(1) (a–b) –1(1–a)+1 (b-a2) = 0 a–b–1+a+b–a2=0 –a2+2a–1=0 a2–2a+1=0 (a–1)2 = 0 a=1
When a=1, equations becomesx+y+z=1x+y+z=1& x+by+z=0They represent two coincident planes & third plane as x+by+x=0
Page No. 21 ||www.gyankuteer.com||
for no solution, no interiection should be there. Normals of x+y+z=1 & x+by+z=0 should be parallel.
1111
11
bb
Hence for single value of ‘b’ system of equations, has no solution.
3 is correct.
66. A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them
are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in
which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X
and Y are in this party, is :
(1) 468 (2) 469 (3) 484 (4) 485
Ans. [4]
Sol. Total ways = X1L2m Y2L1m + X2L1m Y1L2m + X3L0m Y0L3m+ X0L3m Y3L0m
= 1 2 2 1 2 1 1 2 3 3 3 3C C C C C C C C C C C C4 3 . 3 4 4 3 3 4 4 4 3 .3
4 3 3 4 6 3 3 6 4 4 1
= 144 324 16 1 = 485
67. The value of :21 10 21 10
1 1 2 2( C C ) ( C C ) 21 10 21 10
3 3 4 4( C C ) ( C C ) .... 21 10
10 10( C C ) is :
(1) 221 – 210 (2) 220 – 29 (3) 220 – 210 (4) 221 – 211
Ans. [3]
Sol. 1 2 3 10 1 2 3 10C C C C C C C CS 21 21 21 ...... 21 10 10 10 ...... 10
0 1 2 3 21
21 2 3 21C C C C C1 x 21 21 x 21 x 21 x ......... 21 x
put x 1 0 1 2 21
21C C C C21 21 21 ......... 21 2
21 0C C21 21 , 20 1C C21 21 ,
19 2C C21 21
then 1 2 21
20C C C21 21 ..... 21 2 1
and
0 1 2 3 10
10 2 3 10C C C C C1 x 10 10 x 10 x 10 x ......... 10 x
put x 1 0 1 2 10
10C C C C10 10 10 ......... 10 2
1 2 10
10C C C10 10 ......... 10 2 1
Now S = 20 102 1 2 1 = 20 102 2
68. For any three positive real numbers a, b and c,
9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c)
(1) b, c and a are in A.P. (2) a, b and c are in A.P.
(3) a, b and c in G.P. (4) b, c and a are in G.P.
Ans. [1]
Sol. Given
9(25a2+b2) + 25(C2–3ac) = 15b(3a+c)
225a2 + 9b2 + 25c2 – 75ac – 45ab – 15bc = 0
Page No. 22 ||www.gyankuteer.com||
450a2 + 18b2 + 50c2 – 150ac – 90ab – 30bc = 0
(15a–3b)2 + (3b–5c)2 + (5c–15a)2 = 0
15a–3b=0, 3b–5c=0 and 5c–15a=0
b=5a & 3bc5
and c=3a
Hence a,b,c are equivalent to 1, 5a, 3a or 2c = a+bHence (1) is correct
69. Let a,b,c R. If f(x) = ax2 + bx + c is such that a + b + c = 3 andf(x+y) = f(x) + f(y) + xy, x, y R,
then 10
n 1 f(n) is equal to :
(1) 165 (2) 190 (3) 255 (4) 330
Ans. [4]
Sol. Given f(x) = ax2+bx+c
f(1) = a+b+c=3Also f(x+y) = f=(x) + f(y) + xyPutting x=1, y=1f(3) = f(2) + f(1) + 2×1Sinilarly f(4) = f(3) + f(1) + 3i.e. +(4) = 12+3+3 = 18Hence series is3, 7, 12, 18, ...............Let Sn = 3+7+12+18+ ...... + Tn ....... (i) Sn = 3+7+12+ ....... + Tn-1 +Tn .........(ii)
Tn = 3+[4+5+6+ .............. upto ‘n–1’terms]
Tn = n 13
2
[8 (n 2)(1)]
i.e. Tn = (n 1)3
2
(n+6)
i.e. Tn = 2n 5n
2 2
Sn = n
nn 1
(T ) =
12
n(n 1)(2n 1)6
+
52
n(n 1)2
10
n 1f (n)
= f(1) + f(2) + .... f(10) = S10 =
10 11 2112
+
5 10 114
= 330
Hence (4) is correct.
70. 3x
2
cot x cos xlim2x
equals :
(1) 1
16 (2) 18 (3)
14 (4)
124
Ans. [1]
Sol.
3h o
cot( / 2 h) cos / 2 hlim
2 h2
Page No. 23 ||www.gyankuteer.com||
3h o
1 tanh sinhlim8 h
by DL rule2
2h o
1 sec h coshlim8 3h
by DL rule2
h o
1 2sec h tanh sinhlim8 6h
So 1 2 1 18 6 6 16
71. If for 1x 0,4
, the derivative of 1
3
6x xtan1 9x
is x . g(x), then g(x) equals :
(1) 3
3x x1 9x
(2) 3
3x1 9x
(3) 3
31 9x
(4) 3
91 9x
Ans. [4]
Sol.3/2
13
d 6xtan xdx 1 9x
g(x)
Puting 3/ 23x tan
1d tan tan 2 xdx
g(x)
1 3/2d2 tan (3x ) xdx
g(x)
1/23/2 2
1 32 3 x x1 (3x ) 2
g(x)
So 3
9g(x)1 9x
72. The normal to the curve y(x–2)(x–3) = x + 6 at the point where the curve intersects the y-axis passes
through the point :
(1) 1 1,2 2
(2) 1 1,2 3
(3) 1 1,2 3
(4) 1 1,2 2
Ans. [1]
Sol. clearly normal intersects the y-axis at (0,1)
given equation is y(x–2)(x–3) = x + 6
there fore 1)1,0(
dxdy
there fore slope of normal = –1hence equation of normal is )0–(1–1– xy
that is 01– yx
clearly 1 1,2 2
satisfy the given equation
Page No. 24 ||www.gyankuteer.com||
73. Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the
maximum area (in sq. m) of the flower-bed is :
(1) 10 (2) 25 (3) 30 (4) 12.5
Ans. [2]
Sol. 20 mThis wire is to be used for fencing circular sector (flower bed).Hence l + 2r = 20 ........... (i) l = 20 – 2r
Now, A = 12 lr
A = 12 (20 – 2r) r
A = (10 – r)r r r
l
A = 10 r - r2
Now dAdr = 10 – 2r
for A to be max./min.
dA odr
r = 5 l = 10 from (i)
Now 2
2
d Adr
= –2 < 0
Hence A is max.
Now Amax = 12 (10) (5) = 25
74. Let nnI tan x dx,(n 1) . If I4 + I6 = tan5 x + bx5 + C, where C is a constant of integration, then the
ordered pair (a,b) is equal to :
(1) 1 ,05
(2) 1 , 15
(3) 1 ,05
(4) 1 ,15
Ans. [1]
Sol. 4 24 6I I tan x 1 tan x dx
4 24 6I I tan x sec x dx
put tanx = tsec2x dx = dt
44 6I I t dt
5
4 6tI I C5
55tan x 0.x c
5
so (a,b) = 1 ,05
Page No. 25 ||www.gyankuteer.com||
75. The integral
34
4
dx1 cos x
is equal to :
(1) 2 (2) 4 (3) –1 (4) –2
Ans. [1]
Sol.34
4
dxI1 cos x
Using property b b
a af x dx f a b x dx
34
4
dxI1 cos x
34
4
22I dx1 cos x 1 cos x
3 3 4
24 42
4 44
1dxI cosec xdx cotxsin x
1 1 2
76. The area (in sq. units) of region 2{(x, y) : x 0, x y 3, x 4y and
y 1 x} is :
(1) 32 (2)
73 (3)
52 (4)
5912
Ans. [3]
Sol. Given region is
2(x, y): x o, x y 3, x 4y and y 1 x
Area =
1 22 2
0 1
x x(1 x ) dx (3x) dx4 4
=
13/2 3
0
2x xx3 12
+
22 3
1
x x3x2 12
(0,3)(0,1)C
(0,0) 0 3,0
(2,0)A
B(1,2) y 1 x
x = 4y2y
= 2 1 8 1 11 6 – 2 33 12 12 2 12
= 5 1 8 1 14 33 12 12 2 12
= 5 1 2 513 2 3 2
sq. units
Hence (3) is correct.
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77. If (2 + sinx) dydx + (y + 1)cos x = 0 and y(0) = 1, then y
2
is equal to :
(1) 23
(2) 13
(3) 43 (4)
13
Ans. [4]
Sol. Given (2+sinx) dydx + (y+1) cosx = 0
(2+sinx) dy
y 1 = (cos x)dx2 sinx
dy
y 1 = cosx dx2 sin x
log|y+1| = – log |2 + sinx|+log |c|
|y+1| = | c |
| 2 sinx |Also y(o) 1
Z = | c |2 |c| = 4
|y+1| = 4
2 sin x as 2 + sinx > 0 where xR
Putting x2
|y+1| = 43
y+1= 43 or y+1 =
43
y = 13 or y =
73
Hence y = 13
(4) is correct
78. Let k be an integer such that the triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units.
Then the orthocentre of this triangle is at the point :
(1) 31,4
(2) 31,4
(3) 12 ,2
(4) 12,2
Ans. [3]
Sol.
k 3k 15 k 1 56k 2 1
on solving we get k = 2 then
Page No. 27 ||www.gyankuteer.com||
coordinate of vertices of triangle are (-2,2), (5,2) and (2,-6) O
(5,2)(-2,2)
(2,-6)
then orthocentre will be 12,2
79. The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y =
|x| is :
(1) 2 2 1 (2) 2 2 1 (3) 4 2 1 (4) 2 2 1
Ans. [2]
Sol. Clearly r = 4 – k – (i)
also length of from (o,k) to the liney = x will be equal to radius r.
(0,4) y = x
y = 4 - x 2
y = -x
(0,k)
r = | 0 k |
2
k > 0
r = k2 ....... (ii)
k = 4 2 - 2 k
2 1 k = 4 2
k = 4 2 2 1
k = 8 – 4 2 r = 4 – (8–4 2 )
r = 4 2 – 4
r = 4( 2 1)
Hence (2) is correct
80. The eccentricity of an ellipse whose centre is a the origin is 12 . If one of its directrices is x = –4, then
the equation of the normal to it at 31,2
is :
(1) 4x – 2y = 1 (2) 4x + 2y = 7 (3) x + 2y = 4 (4) 2y – x = 2
Ans. [1]
Sol. Given that
43
43
41–1
2
2
2
2
2
2
abab
ab
Page No. 28 ||www.gyankuteer.com||
and 4ea
2a = 4 a = 2 3b
there fore equation of ellipse is 134
22
yx
differentiating w.r.t x
21–
43–
03
22
)2/3,1(
dxdy
yx
dxdy
dxdyyx
there fore slope of normal = 2there fore equation of normal
)1–(22/3– xy
that is 4x – 2y = 1
81. A hyperbola passes through the point P( 2, 3) and has foci at ( 2, 0) . Then the tangent to this hyperbola
at P also passes through the point :
(1) (2 2, 3 3) (2) ( 3, 2) (3) ( 2, 3) (4) (3 2,2 3)
Ans. [1]
Sol. 2ae and
2222
2
22 1
baeaabe
there fore a2 = 4 – 2b2 (1)equation of hyperbola is
1– 2
2
2
2
by
ax
it passes though P( 2, 3)
there fore 13–222
ba (2)
solving (1) & (2) we get
32 b and 4–2 b (not possible)
there fore 12 athere fore equation of hyperbola is
13
–1
22
yx
there fore yx
dxdy 3
that is 6)3,2(
dxdy
there fore equation of tangent at P is
03––6
)2–(63–
yx
xy
clearly point (2 2, 3 3) satisfy the given equation
Page No. 29 ||www.gyankuteer.com||
82. The distance of the point (1,3, –7) from the plane passing through the point (1, –1, –1), having normal
perpendicular to both the lines x 1 y 2 z 4
1 2 3
and x 2 y 1 z 7
2 1 1
is :
(1) 1083 (2)
583 (3)
1074 (4)
2074
Ans. [1]
Sol. Eqn. of plane passing through (1, –1, –1) is
0)1()1()1–( zcybxa
normal perpendicular to lines x 1 y 2 z 4
1 2 3
and x 2 y 1 z 7
2 1 1
there fore 032– cba and 0––2 cba . solving these two we get3b = 7cthere fore 3:7:5:::: cbathere fore equation of plane will be
0)1(3)1(7)1–(5 zyx
that is 05375 zyxdistance of plane from (1,3,–7) is
8310
92549521–215
83. If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the l ine,
x y z1 4 5 is O, then PQ is equal to :
(1) 2 42 (2) 42 (3) 6 5 (4) 3 5Ans. [1]
Sol. Equation of line parellel to x y z1 4 5 passes through (1,–2,3)
will be x 1 y 2 z 3 r
1 4 5
point on line (r+1, 4r–2, 5r+3)satisfy the plane2r+2+12r–6–20r–12+22=0r=1R(2,2,8)R is the mid point of P & Q(a, b, c)
then a 1 b 22, 2
2 2
and c 3 8
2
Q(3,6,13)
So PQ 4 64 100 2 42
84. Let ˆ ˆ ˆa 2i j 2k and ˆ ˆb i j
.
Let c be a vector such that c a = 3, (a b) c
=3 and the angle between c and a × b
be 30°.
Then a . c is equal to :
(1) 2 (2) 5 (3) 18 (4)
258
Ans. [1]
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Sol. | (a b) c | 3
| a b | | c | sin 30 3
| a b | | c | 6
Now, | c a | 3
squaring both side2 2c a 2c.a 9
........... (i)a2 = 9
& | a b | 3
| c | 2
putting the value in ...... (i)So a.c 2
85. A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement,
then the variance of the number of green balls drawn is :
(1) 6 (2) 4 (3) 625 (4)
125
Ans. [4]
Sol. Here n = 10
p = 15 325 5
(p = Probability)
and 10 2q25 5
Now, variance = npq
variance = 3 2105 5
= 125
86. For three events A, B and C,
P(Exactly one of A or B occurs)
=P(Exactly one of B or C occurs)
= P(Exactly one of the C or A occurs) = 14 and P(All the three events occur simultaneously)
116
(1) 7
16 (2) 764 (3)
316 (4)
732
Ans. [1]
Sol. Given (PA) + P(B) – 2P A B = 14 ......(i)
P(B) + P(C) – 2P B C = 14 ......(ii)
and P(C) + P(A) – 2P C A = 14 ......(iii)
Page No. 31 ||www.gyankuteer.com||
32[P(A) P(B) P(C) – P(A B) – P(B C) P(C A) ]8
Also given P(A B C) 1
16 ............. (iv)
P(A B C) P(A) P(B) P(C) P(A B) P(B C) P(C A) P(A B C)
P(A B C) = 3 18 16
= 6 1 716 16
Hence (1) is correct.
87. If two different numbers are taken from the set {0, 1, 2, 3, ......, 10} ; then the probability that their
sum as well as absolute difference are both multiple of 4, is :
(1) 1255 (2)
1445 (3)
755 (4)
655
Ans. [4]
Sol. Tw o different numbers can be selected in 11c2 ways i.e 55 ways
and favourable out comes are
(0,4), (0,8), (2,6), (2,10), (4,8) and (10,6)
Probability = 6
55Hence (4) is correct.
88. If 5(tan2 x – cos2 x) = 2cos 2x + 9, then the value of cos 4x is :
(1) 13 (2)
29 (3)
79
(4) 35
Ans. [3]
Sol. Given 5 (tan2x – cos2x) = 2cos 2x+9
2
22
sin x5 cos xcos x
= 22(2cos x 1) 9
2
22
1 cos x5 cos xcos x
= 22(2cos x 1) 9
Let cos2 x = t
1 t5 t 2(2 t 1) 9
t
21 t t5 4t 7
t
9t2 + 15t–3t–5=0
1 5t , t3 3
i.e. cos2x = 13 , cos2x
53
(not possible)
Page No. 32 ||www.gyankuteer.com||
Now cos2x = 2cos2 x–1 = 2 113 3
cos4x = 2cos2 2x–1 = 12 19
= 7
9
Hence (3) is correct.
89. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a
point on the ground such that AP = 2AB. If <BPC = , then tan is equal to :
(1) 14 (2)
29 (3)
49 (4)
67
Ans. [2]
Sol. Let BC = x,
AC = xand AP = 4xLet CPA
Now tanx = 14
A
C
B
P 4x
x
x
and tan ( ) = 12
tan tan 1
1 tan tan 2
1 tan 141 21 tan4
1 4 tan 14 tan 2
2 8 tan 4 tan
9 tan 2
2tan9
Hence (2) is correct.90. The following statement :
(p q) [(~ p q) q] is :
(1) equivalent to ~ p q (2) equivalent to p ~ q
(3) a fallacy (4) a tautology
Ans. [4]
Sol.
P q pq ~P ~P q (~Pq)q (P q) [(~Pq) q T T F F
T F T F
T F T T
F F T T
T T T F
T F T T
T T T T