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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


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


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


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]


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.


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]


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


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


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


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


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


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


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


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.


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]


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]


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


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


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


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


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


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


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


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


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


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



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


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


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


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]


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)


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


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)


Now cos2x = 2cos2 x–1 = 2 113 3

cos4x = 2cos2 2x–1 = 12 19

= 7

9


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

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