JEE (Main) : 2017 C - Pathfinderpathfinder.edu.in/wp-content/uploads/2017/05/JEEMain... · 2017-05-31 · JEE (Main) : 2017 (Question with Solution) Q.1 The freezing point of benzene

JEE (Main) : 2017(Question with Solution)

Q.1 The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene.

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) 94.6% (2) 64.6% (3) 80.4% (4) 74.6%

Solution : [1]

Bf f

A B

w 1000T i K .

w M

0.2 10000.45 i 5.12

20 60

i 0.527

22A A

1 0

12

i 12

1 0.527 0.4732

94.6%

Q.2 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) [Co(H2O)5Cl]Cl2.H2O (2) [Co(H2O)4Cl2]Cl.2H2O

(3) [Co(H2O)3Cl3].3H2O (4) [Co(H2O)6]Cl3

Solution : [1]

100 mL 0.1 M = 10 mmol = 1 × 10–2 mol

2222

23

1.2 101.2 10 ions mol

6.02 10 = 2 × 10–2 mol

Formula [Co(H2O)5Cl]Cl2.H2O

HEAD OFFICE : 96K, S.P. Mukherjee Road, Hazra More, Kolkata-700026, Ph. 24551840 / 24544817 WEBSITE : www.pathfinder.edu.in 1

CDate : 02.04.2017

C

JEE(Main)-2017-SOL

CHEAD OFFICE : 96K, S.P. Mukherjee Road, Hazra More, Kolkata-700026, Ph. 24551840 / 24544817 WEBSITE : www.pathfinder.edu.in 2

Q.3 Which of the following compounds will form significant amount of meta product during mono-nitration

reaction ?

(1)

NHCOCH3

(2)

OH

(3)

OCOCH3

(4)

NH2

Solution : [4]

In presence of nitrating agent (conc. HNO3 + conc. H2SO4) – NH2 becomes protonated.

NH3

+

m-directing group

NH2 NH2 NH2 NH2

NO2

NO2

NO2nitration + +

(51%) (47%) (2%)

Q.4 The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are :

(1) Cl– and 2ClO (2) ClO– and 3ClO (3) 2ClO and 3ClO (4) Cl– and ClO–

Solution : [4]

Cl2 + NaOH NaCl + NaOCl + H2O

(Cold dilute)

Q.5 Both lithium and magnesium display several similar properties due to the diagonal relationship; however,the one which is incorrect, is :

(1) nitrates of both Li and Mg yield NO2 and O2 on heating

(2) both form basic carbonates

(3) both form soluble bicarbonates

(4) both form nitrides

Solution : [2]

Li being monovalent, so Li cannot form basic salt.

Q.6 A water sample has ppm level concentration of following anions

F– = 10; 24SO = 100; 3NO 50

The anion/anions that make/makes the water sample unsuitable for drinking is/are :

(1) only 24SO (2) only 3NO

(3) both 24SO and 3NO (4) only F–

JEE(Main)-2017-SOL


Solution : [4]

More than 2 ppm F– affects bone health and teeth.

NO3– limiting concentration 50 ppm

24SO 500 pm is unsuitable.

Q.7 The formation of which of the following polymers involves hydrolysis reaction ?

(1) Terylene (2) Nylon 6 (3) Bakelite (4) Nylon 6, 6Solution : [2]

hydrolysispolymerisationO

NH

caprolactum

—HN—(CH ) —C—2 5( )n

O

Q.8 The Tyndall 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 wavelength of the light used.

(c) The refractive indices of the dispersed phase and dispersion medium are almost similar inmagnitude.

(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

(1) (b) and (c) (2) (a) and (d) (3) (b) and (d) (4) (a) and (c)

Solution : [3]

Q.9 pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of theirsalt (AB) solution is :

(1) 1.0 (2) 7.2 (3) 6.9 (4) 7.0

Solution : [3]

a b1 1

pH 7 pK pK2 2

1 17 3.2 3.4

2 2= 7 + 1.6 – 1.7 = 6.9

Q.10 The major product obtained in the following reaction is :

O

COOH

ID BAL H

(1)

CHO

CHO (2)

COOH

CHO

OH

(3)

CHO

CHO

OH

(4)

COOH

CHO

JEE(Main)-2017-SOL


Solution : [2]

DIBAL–H

OO

CO H2 CO H2

OH

CHO

Q.11 Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution ?

(1)

OH

OH

HOH2CCH2OCH3

(2) HOH2C

HOOCOCH3

OH

CH2OH

(3) HOH2C

HO

OH

CH2OH (4) HOH2C

HO

OH

CH2OH

OCH3

Solution : [2]

HOH2C

HOOCOCH3

OH

CH2OH

on basic hydrolysis gives hemiacetal, which further in basic medium gives free

aldehyde group which is reducing in nature.

Acetals or ether linkages are stable in basic medium.

Q.12 The correct sequence of reagents for the following conversion will be

O

CHO

HO CH3

HO CH3CH3

(1) [Ag(NH3)2]+OH–, CH3MgBr, H+/CH3OH (2) [Ag(NH3)2]+OH–, H+/CH3OH, CH3MgBr

(3) CH3MgBr, H+/CH3OH, [Ag(NH3)2]+OH– (4) CH3MgBr, [Ag(NH3)2]+OH–, H+/CH3OH

Solution : [2]

[Ag(NH ) ] OH3 2+ –

mild oxidising agent esterification

H /CH OH+3

i) CH MgBr3

(3 mols)

O O O HO

CHO CO H2

CH3

CH3

COOCH3 CH –C–OH 3

ii) H O3+

+ CH OH3

JEE(Main)-2017-SOL


Q.13 Which of the following species is not paramagnetic ?

(1) B2 (2) NO (3) CO (4) O2

Solution : [3]

CO is diamagnetic

Q.14 Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails todecolourize the colour of bromine ?

(1)

O

Br (2)

O

Br(3)

Br

C H6 5

(4)

O

Br

Solution : [2]

O

C OC H6 5

from (1) from (3) from (4)

all decolourise bromine water.

O O

is unreactive to Br2/H2O

Q.15 Which of the following reactions is an example of a redox reaction ?

(1) XeF6 + 2H2O XeO2F2 + 4HF (2) XeF4 + O2F2 XeF6 + O2

(3) XeF2 + PF5 [XeF]+ PF6– (4) XeF6 + H2O XeOF4+ 2HF

Solution : [2]

XeF + O F XeF + O4 2 2 6 2

+4 +1 –1 +6 0

oxidation

reduction

Q.16 U is equal to

(1) Isothermal work (2) Isochoric work (3) Isobaric work (4) Adiabatic work

Solution : [4]

U = q + W

in adiabatic process q = 0 U = Wadiabatic

Q.17 Which of the following molecules is least resonance stabilized ?

(1)O

(2) (3)O

(4) N

Solution : [1]

O is least resonance stabilised. Rest are aromatic.

JEE(Main)-2017-SOL


Q.18 The increasing order of the reactivity of the following halides for the SN1 reaction is

CH CHCH CH3 2 3

Cl( )I

CH CH CH Cl3 2 2

( )II

p–H CO–C H –CH Cl3 6 4 2

( )III

(1) (II) < (III) < (I) (2) (III) < (II) < (I) (3) (II) < (I) < (III) (4) (I) < (III) < (II)

Solution : [3]

The relative stability of carbocations is —

CH2MeO

+ M effect

> CH –CH–CH –CH > CH –CH –CH3 2 3 3 2 2

+ +

2º 1º

+

Q.19 1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molarmass of M2CO3 in g mol–1 is :

(1) 11.86 (2) 1186 (3) 84.3 (4) 118.6

Solution : [3]

M2CO3 + 2HCl 2MCl + H2O + CO2

0.01186 mol

0.01186 mol 1 g

11 mol g 84.31 g

0.01186

Q.20 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) Na2C2O4 (2) C6H5COONa (3) HCOONa (4) CH3COONa

Solution : [1]

Na C O + Conc.H SO CO + CO2 2 4 2 4 2

CaCl2

MnO4–

Calcium oxalate

CO + Mn22+

H+

(X)

Q.21 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) 10 kg (2) 15 kg (3) 37.5 kg (4) 7.5 kg

Solution : [4]

Mass of a heavy man increase = 75 × 0.1 = 7.5 kg

JEE(Main)-2017-SOL


Q.22 The major product obtained in the following reaction is

BrH

C H6 5

C H6 5

tBuOK

(+)

(1) (–)C6H5CH(OtBu)CH2C6H5 (2) (±)C6H5CH(OtBu)CH2C6H5

(3) C6H5CH=CHC6H5 (4) (+)C6H5CH(OtBu)CH2C6H5

Solution : [3]

tBuOKC H6 5C H6 5

C H6 5C H6 5

Br H

Q.23 Given C(graphite) + O2(g) CO2(g); rHº=–393.5 kJ mol–1

2 2 21

H g O g H O l2

; rHº = –285.8 kJ mol–1

CO2 (g) + 2H2O(l) CH4(g) + 2O2(g); rHº = +890.3 kJ mol–1

Based on the above thermochemial equations, the value of rHº at 298 K for the reaction

C(graphite) + 2H2(g) CH4(g) will be

(1) –144.0 kJ mol–1 (2) +74.8 kJ mol–1 (3) +144.0 kJ mol–1 (4) –74.8 kJ mol–1

Solution : [4]

C (graphite) + 2H2(g) CH4(g)

0rH (i) 2 (ii) (iii)

= (– 393.5) + 2(–285.8) + 890.3 = –74.8 kJmol–1

Q.24 In the following reactions, ZnO is respectively acting as a/an

(a) ZnO + Na2O Na2ZnO2 (b) ZnO + CO2 ZnCO3

(1) acid and base (2) base and acid

(3) base and base (4) acid and acid

Solution : [1]

ZnO + Na2O Na2ZnO2 ; ZnO + CO2 ZnCO3

acid base base acid

Q.25 The radius of the second Bohr orbit for hydrogen atom is

(Planck’s Const. h = 6.6262 × 10–34 Js; mass of electron = 9.1091 × 10–31 kg; charge of electron e =

1.60210 × 10–19 C; permittivity of vacuum 12 1 3 20 8.854185 10 kg m A )

(1) 2.12 Å (2) 1.65 Å (3) 4.76 Å (4) 0.529 Å

Solution : [1]

2

22

r 0.53 Å 2.12Å1

JEE(Main)-2017-SOL


Q.26 Two reactions R1 and R2 have identical pre-exponential factors. Activation energy or R1 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

ln(k2/k1) is equal to :

(R = 8.314 J mol–1 K–1)

(1) 4 (2) 8 (3) 12 (4) 6

Solution : [1]

Ea

RTk Ae

a a1 2

1(E E )

1 RT

2

ke

k

1 23 3

a a2

1

E Ek 10 10 10 10ln 4

k RT RT 8.314 300

Q.27 A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ‘a’, the closestapproach between two atoms in metallic crystal will be :

(1)a

2(2) 2a (3) 2 2 a (4) 2 a

Solution : [1]

a2a 4r 2r

2

Q.28 The group having isoelectronic species is

(1) O–, F–, Na+, Mg2+ (2) O2–, F–, Na+, Mg2+ (3) O–, F–, Na, Mg+ (4) O2–, F–, Na, Mg2+

Solution : [2]

Q.29 Given

32

o oCl /Cl Cr /Cr

E 1.36 V, E 0.74 V , 2 3 22 7 4

o oCr O /Cr MnO /Mn

E 1.33 V, E 1.51V

Among the following, the strongest reducing agent is

(1) Cl– (2) Cr (3) Mn2+ (4) Cr 3+

Solution : [2]

32

0 0Cl /Cl Cr/Cr

E 1.36V E 0.74V

2 3 24 2 7

0 0Mn /MnO Cr /Cr O

E 1.51V E 1.33

Q.30 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. Thenumber of possible stereoisomers for the product is

(1) Four (2) Six (3) Zero (4) Two

Solution : [1]

HBrH O2 2

Br

4 stereoisomers

JEE(Main)-2017-SOL


Q.31 The integral

3

4

4

dx

1 cos x is equal to :

(1) 4 (2) –1 (3) –2 (4) 2

Solution : [4] ; 2

I

3

4

4

dx

1 cos x

I

3

4

4

dx

1 cos x

I =

3

4

4

1 12 dx

1 cosx 1 cosx

=

3

4

2

4

2dx

sin x

3 /4

/42 cot x

= 2 [1 + 1] I = 2

Q.32 Let I nn tan xdx, n 1 . If I4 + I6 = a tan5 x + bx5 + C, where C is a constant of integration, then the

ordered pair (a, b) is equal to :

(1)1

, 15

(2)1

,05

(3)1

,15

(4)1

,05

Solution : [4] ; 1

,05

Inntan xdx n 1

I4 + I6 = (a tan5 x + bx5 + c)

I4 + I6 = 4 2tan x sec x dx

55tan x

0x c5

So, 1

a,b ,05

JEE(Main)-2017-SOL


Q.33 The area (in sq. units) of the region 2{ x,y : x 0,x y 3,x 4y and y 1 x } is :

(1)7

3(2)

5

2(3)

59

12(4)

3

2

Solution : [2] ; 5

2

y 1 x

y 1 x

x = (y – 1)2

and x2 4y

x + y = 3

x2 = 4 (3 – x)

x2 + 4x – 12 = 0

x2 + 6x – 2x – 12 = 0

x(x + 6) – 2(x + 6) = 0

(x + 6) (x – 2) = 0

x = – 6, 2

Area 1 22 2

0 1

x x1 x dx 3 x dx

4 4

1323 2 32

10

2x x x xx 3x

3 12 2 12

2 1 2 1 11 6 2 3

3 12 3 2 12

1 1 1 1 1 54 1 3 1 1 2

2 12 12 2 2 2

Q.34 A box contains 15 green and 10 yellow balls, If 10 balls are randomly drawn, one-by-one, withreplacement, then the variance of the number of green balls drawn is :

(1) 4 (2)6

25(3)

12

5(4) 6

JEE(Main)-2017-SOL


Solution : [3] ; 12

5

15 G10 Y

Probability of getting a green ball,

151

251

C 15 3p

25 5C

n = 10

3 2p , q

5 5

variance = npq 3 2

105 5

60 12

25 5

Q.35 If (2 + sin x) dy

y 1 cos x 0dx

and y(0) = 1, then y2

is equal to :

(1)1

3(2)

4

3(3)

1

3(4)

2

3

Solution : [3] ; 1

3

dy2 sin x y 1 cos x 0

dx

y 1 cos xdy

dx 2 sinx

dy cos x dx

y 1 2 sinx

ln (y + 1) = – ln(2 + sinx) + ln C

Cy 1

2 sinx

C2 C 4

2 [ y(0) = 1]

4y 1

2 sinx

4 1y 1

2 3 3

Q.36 Let be a complex number such that 2 + 1 = z where z 3 .

If 2 2

2 7

1 1 1

1 1 3k

1

, then k is equal to :

(1) – 1 (2) 1 (3) – z (4) z

JEE(Main)-2017-SOL


Solution : [3] ; – z

2 + 1 = z Where z 3 3i

1 3 1 3 i

2 2

2 2

2 7

1 1 1

1 1 3k

1

; 2 2

2

1 1 1

0 2 1

0 1 1

= 22 21 2 1 1

= –1(3 + 2 – 2 – 2) – 4 + 22 – 1

= –1 – 2 + 2 + 2 – + 22 –1

2 1 3 i 1 3 i 2 3 i3 3 3 3 3 3 i 3( z) 3k k z

2 2

Q.37 Let a 2i j 2k and b 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) 5 (2)1

8(3)

25

8(4) 2

Solution : [4] ; 2

i j k

a b 2 1 2

1 1 0

2 i 2 j k

a b 4 4 1 3

a b c 3

a b . c sin30 n 3

1

3. c . .n 32

1

3 c . 3 c 22

Again, c a 3

2

c a 9

2 2

c a 2 c . a 9

4 (4 1 4) 2 a . c 9 c 2

a . c 4

a . c 2

JEE(Main)-2017-SOL


Q.38 The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y x

is :

(1) 4 2 1 (2) 4 2 1

(3) 2 2 1 (4) 2 2 1

Solution : [1] ; 4 2 – 1

x – y = 0

Y

X

y = x

y = – x

(0,

)

| 0 |r

2

2 r

Equation of the circle is 2 2 2x (y 2r) r

2 2 2x y 2 2ry r 0 , which passes through (0, 4)

216 8 2 r r 0

2r 8 2 r 16 0

min8 2 128 64

r r2

8 2 8

4( 2 1)2

Q.39 If for 1

x 0,4

, the derivative of 13

6x xtan

1 9x is x g x , then g(x) equals :

(1) 3

3x

1 9x(2) 3

3

1 9x(3) 3

9

1 9x(4) 3

3x x

1 9x

Solution : [3] ; 3

9

1+ 9x

1x 0,

4

1 1 13 2

6x x 2.3x xy tan tan 2tan (3x x )

1 9x 1 (3x x )

1

23

dy 1 32. 3 x

dx 21 9x

3

9 x

1 9x

3

9g(x)

1 9x

Q.40 If two different numbers are taken from the set {0, 1, 2, 3, ......, 10}; then the probability that their sumas well as absolute difference are both multiple of 4,is :

(1)14

45(2)

7

55(3)

6

55(4)

12

55

JEE(Main)-2017-SOL


Solution : [3] ; 6

55

Let ‘A’ be the event that “sum of their is multiple of 4”

‘B’ be the event that “difference of their is multiple of 4”

A = {(0, 4), (0, 8), (1, 3), (1, 7), (2, 6), (2, 10), (3, 5), (3, 9), (4, 8), (5, 7), (6,10), (7, 9)}

B = {(0, 4), (0, 8), (1, 5), (1, 9), (2, 6), (2, 10), (3, 7), (4, 8), (6,10)}

A B = {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}

P(A B) = 112

n(A B) 6 6 611.10n(S) 55C

2

Q.41 3x

2

cot x cos xlim

2x equals :

(1)1

8(2)

1

4(3)

1

24(4)

1

16

Solution : [4] ; 1

16

3 3x x

2 2

cot x cosx 1 cot x cos xlim lim .

8( 2x)x

2

x t2

3t 0

cot t cos t1 2 2

. lim8 t

3t 0

1 tant sint. lim

8 t

2t 0 t 0

1 sin t sec t 1lim lim

8 t t

2 2

t 0 t 0 t 0 t 0

1 cos t 1 cos t1 1 cos t 1 1 1. lim lim . lim . lim

8 cos t 8 1 cos tt t

2

2

1 sin t 1 1. .

8 2 16t

Q.42 The value of

(21C1 – 10C1) + (21C2 – 10C2) + (21C3 – 10C3) + (21C4 – 10C4) + ..... + (21C10 – 10C10) is :

(1) 220 – 29 (2) 220 – 210 (3) 221 – 211 (4) 221 – 210

JEE(Main)-2017-SOL


Solution : [2] ; 220 – 210

21 10 21 10 21 10 21 101 1 2 2 3 3 10 10C C C C C C ...... C C

21 21 21 21 10 10 10 101 2 3 10 1 2 3 10C C C ...... C C C C .... C

21 21 21 21 10 10 100 1 2 10 0 1 10C C C ...... C 1 C C .... C 1

2110 20 102

2 2 22

Q.43 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 C or A occurs)1

4 and P(All the three events occur simultaneously)

1

16

Then the probability that at least one of the events occurs, is :

(1)7

64(2)

3

16(3)

7

32(4)

7

16

Solution : [4] ; 7

16

1P A B B A P B C C B P C A A C

4

1P A B C

16

Now, 1P A B P B A

4

1P(A) P(A B) P(B) P(A B)

4

1P(A) P(B) 2P(A B) ......(i)

4

Similarly, P(B) + P(C) – 2P(BC) 1

4 ........ (ii)

and P(C) + P(A) – 2P(C A) 1

4 ........ (iii)

Adding (i), (ii) and (iii), we get

2[P(A) + P(B) + P(C) – P(A B) – P(B C) – P(C A)]3

4

P(A) + P(B) + P(C) – P(A B) – P(B C) – P(C A)]3

8

P(ABC)3 1 6 1 7

8 16 16 16

JEE(Main)-2017-SOL


Q.44 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 than is equal to :

(1)2

9(2)

4

9(3)

6

7(4)

1

4

Solution : [1] ; 2

9

AB 1tan

AP 2

P A

C

B

AC 1tan( )

AP 4

tan tantan( )

1 tan tan

1tan1 2

14 1 tan2

1 1 2tan

4 2 tan

2 + tan = 4 – 8tan

9 tan = 2

2tan

9

Q.45 The eccentricity of an ellipse whose centre is at the origin is 1

2. If one of its directrices is x = – 4, then

the equation of the normal to it at 3

1,2

is :

(1) 4x + 2y = 7 (2) x + 2y = 4 (3) 2y – x = 2 (4) 4x – 2y = 1

Solution : [4] ; 4x – 2y = 1

ax

e

Y

X

x=

–4

31,

2

a

4e

a = 4e

1a 4

2 = 2

22

2

be 1

a

21 b1

4 4

2b 3

4 4 b2 = 3 b 3

2 2x y1

4 3

2x 2y dy. 0

4 3 dx

JEE(Main)-2017-SOL


y dy x dy 3x

3 dx 4 dx 4y

31,

2

dy 3.1 3 13dx 6 24.2

Normal : 3

y 2(x 1)2

2y – 3 = 4x – 4

4x – 2y – 1 = 0 4x – 2y = 1

Q.46 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) 10 (2) 11 (3) 12 (4) 9

Solution : [2] ; 11

x(x + 1) + (x + 1)(x + 2) + ........... + (x + n – 1) (x + n) = 10n

nx2 + x(1 + 3 + 5 + ....... to n terms) + 1.2 + 2.3 + ...... to (n – 1) terms = 10n

n 12 2

n 1

nx x.n n(n 1) 10n

2 2 (n 1)n(2n 1) (n 1)nnx n x 10n

6 2

2 n 1

x nx (2n 1 3) 106

2 (n 1)(n 1)x nx 10

3

If n = 11, then

2 10 12x 11x 10

3

x2 + 11x + 30 = 0

(x + 5)(x + 6) = 0

x = – 5, – 6

So, n = 11

Q.47 The following statement

(pq)[(~pq)q] is :

(1) equivalent to p~q (2) a fallacy

(3) a tautology (4) equivalent to ~pq

Solution : [3] ; a tautology

(P q) [(~P q) q]

p q p q ~ p (~ p q) (~ p q) q Final

T T T F T T T

T F F F T F T

F T T T T T T

F F T T F T T

Given statement be a tautology.

JEE(Main)-2017-SOL


Q.48 The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passesthrough the point :

(1)1 1

,2 3 (2)

1 1,

2 3(3)

1 1,

2 2(4)

1 1,

2 2

Solution : [4] ; 1 1

,2 2

When x = 0, then y = 1,

Point on y-axis is (0, 1)

2

x 6y

x 5x 6

2

2

dy x 5x 6 (x 6)(2x 5)

dx (x 5x 6)

2 2

2 2

x 5x 6 2x 7x 30

(x 5x 6)

2

2 2

x 12x 36

(x 5x 6)

(0,1)

dy 361

dx 36

Normal : y – 1 = – 1.x

x + y – 1 = 0

Here, 1 1

1 02 2

1 1,

2 2

satisfies the above equation.

Q.49 For any three positive real numbers a, b and c, 9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c).

Then :

(1) a, b and c are in A.P. (2) a, b and c are in G.P.

(3) b, c and a are in G.P. (4) b, c and a are in A.P.

Solution : [4] ; b, c and a are in A.P.

9 × 25a2 + 9b2 + 25c2 – 75ac = 45ab + 15bc

(15a)2 + (3b)2 + (5c)2 – 15a.5c – 15a.3b – 3b.5c = 0

x2 + y2 + z2 – xy – yz – zx = 0

2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx = 0

(x – y)2 + (y – z)2 + (z – x)2 = 0

x = y = z

15a = 3b = 5c

b ca

5 3

c ba k

3 5

a = k, c = 3k, b = 5k

JEE(Main)-2017-SOL


Q.50 If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line,

x y z

1 4 5 is Q, then PQ is equal to :

(1) 42 (2) 6 5 (3) 3 5 (4) 2 42

Solution : [4] ; 2 42

x 1 y 2 z 3r

1 4 5

x = (r + 1), y = (4r – 2), z = (5r + 3)

2(r + 1) + 3(4r – 2) – 4(5r + 3) + 22 = 0

2r + 12r – 20r = – 22 + 12 + 6 – 2

– 6r = – 6 r = 1

x = 2, y = 2, z = 8

2 2 2PQ 2 (1 2) ( 2 2) (3 8)

2 1 16 25 2 42 2 42

Q.51 If 5(tan2x – cos2x) = 2 cos 2x + 9, then the value of cos 4x is :

(1)2

9(2)

7

9(3)

3

5(4)

1

3

Solution : [2] ; –7

9

5(tan2x – cos2x) = 2cos2x + 9

1 cos2x 55. (1 cos2x) 2cos2x 9

1 cos2x 2

10(1 – cos2x) – 5(1 + cos2x)2 = 2(1 + cos2x)(2cos2x + 9)

10 – 10cos2x – 5(1 + 2 cos2x + cos22x) = 2[2cos2x + 9 + 2cos22x + 9cos2x]

9cos22x + (10 + 10 + 4 + 18)cos2x + 18 + 5 – 10 = 0

9cos22x + 42cos2x + 13 = 0

(3cos2x + 1)(3cos2x + 13) = 0

1 13cos2x cos2x

3 3

2 1 2 9 7cos4x 2cos 2x 1 2. 1

9 9 9

Q.52 Let a, b, c R. If f(x) = ax2 + bx + c is such that a + b + c = 3 and

f(x + y) = f(x) + f(y) + xy, x,y, R , then 10

n 1

f n is equal to :

(1) 190 (2) 255 (3) 330 (4) 165

JEE(Main)-2017-SOL


Solution : [3] ; 330

f(x) = ax2 + bx + c

f(1) = (a + b + c) = 3

f(x + y) = f(x) + f(y) + xy

f(2) = f(1) + f(1) + 1 = (3 + 3 + 1) = 7

f(3) = 7 + 3 + 2 = 12

f(4) = 12 + 3 + 3 = 18

S = 3 + 7 + 12 +18 + ...........+ tn

S = 3 + 7 + 12 + ...........+ tn – 1 + tn

– – – – – –

0 = (3 + 4 + 5 + 6 +........to n terms) –tntn = 3 + 4 + 5 + 6 + ....... to n terms

2n n n 5[2.3 (n 1).1] [n 5] n

2 2 2 2

10 10 102

10 nn 1 n 1 n 1

1 5S t n n

2 2

1 10.11.21 5 10.11. .

2 6 2 2 5.11.7 5.5.11 55

(7 5)2 2 2

5512

2 = 55 × 6 = 330

Q.53 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)5

83(2)

10

74(3)

20

74(4)

10

83

Solution : [4] ; 10

83

a(x – 1) + b(y + 1) + c(z + 1) = 0

a – 2b + 3c = 0

2a – b – c = 0

a b c

5 7 3

5(x – 1) + 7 (y + 1) + 3(z + 1) = 0

5x + 7y + 3z + 5 = 0

Required distance 5 1 7 3 3 ( 7) 5 10

83 83

JEE(Main)-2017-SOL


Q.54 If S is the set of distinct values of ‘b’ for which the following system of linear equations

x + y + z = 1 ; x + ay + z = 1 ; ax + by + z = 0

has no solution, then S is :

(1) a finite set containing two or more elements (2) a singleton

(3) an empty set (4) an infinite set

Solution : [3] ; an empty set

b

1 1 1 1

A 1 a 1 1 ,

a b 1 0

2 2 1

3 3 1

R R R

R R aR

1 1 1 1

0 a 1 0 0

0 b a 1 a a

2

1 1 1 1

0 a 1 0 0R (a 1)R (b a)R3 3 2 0 0 (a 1) a(a 1)

There is no value of b such that (A) (Ab)

S =

Q.55 If 2 3

A4 1

, then adj (3A2 + 12A) is equal to :

(1)51 84

63 72(2)

72 63

84 51(3)

72 84

63 51(4)

51 63

84 72

Solution : [4] ; 51 63

84 72

2 3A

4 1

2 2 3 2 3 16 9A

4 1 4 1 12 13

2 48 273A

36 39

2 48 27 24 36 72 633A 12A

36 39 48 12 84 51

Adj 2 51 63(3A 12A)

84 72

JEE(Main)-2017-SOL


Q.56 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) 3, 2 (2) 2 , 3 (3) 3 2, 2 3 (4) 2 2 ,3 3

Solution : [4] ; 2 2 ,3 3

ae = 2

2 2

2 2

x y1

a b 2 2

2 31

a b

2 2 2

2 31

a a (e 1)

22

32 a

e 1

22

32 a

21

a

[ ae = 2]

22

2

3a2 a

4 a

(2 – a2)(4 – a2) = 3a2

(a2 – 2)(a2 – 4) = 3a2

a4 – 6a2 + 8 = 3a2 a4 – 9a2 + 8 = 0

(a2 – 1)(a2 – 8) = 0

a = 1, 2 2

a = 1

b2 = a2(e2 – 1)

b2 = a2e2 – a2 = 22 – a2 = 22 – 1 = 3

2 2

2 2

x y1

a b

2 2x y1

1 3

Trangent : y. 3

x. 2 13

y

x. 2 13

2 33 2. 2

3 = 6 – 2

3 32 2. 2

3 = 4 – 3 = 1

2 2,3 3 satisfies the equation of the tangent

Q.57 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)3

1,4

(2)1

2,2

(3)1

2,2

(4)3

1,4

JEE(Main)-2017-SOL


Solution : [2] ; 1

2,2

k 3k 11

5 k 1 282

k 2 1

(–2, 2) (5, 2)

(2, –6)

(2, )k.(k – 2) – (3k)(–k –5) + 1.(10 + k2) = 56

(k2 – 2k + 3k2 + 15k + 10 + k2) = 56

5k2 + 13k – 46 = 0

5k2 + 23k – 10k – 46 = 0

k(5k + 23) – 2(5k + 23) = 0 (k – 2)(5k + 23) = 0

k = 2 and the vertices are (2, –6), (5, 2), (–2, 2)

Let (2, ) be the orthocentre

2 2 6. 1

2 2 5 2

2 8. 1

4 3

3 3 12 2

2 2 2

Orthocentre of the triangle is 1

2,2

Q.58 Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then themaximum area (in sq. m) of the flower-bed, is :

(1) 25 (2) 30 (3) 12.5 (4) 10

Solution : [1] ; 25

2r + r = 20

r

r r

20 2r

r

2 21 1 20 2rA .r r

2 2 r

1r(20 2r)

2

= 10r – r2

dA

(10 2r)dr

= 0

r = 5 = (4 – 2) = 2 Radian

2

2

d A2 0

dr

[A]max = 10 × 5 – 52 = 50 – 25 = 25 sq.m

JEE(Main)-2017-SOL


Q.59 The function 1 1

f : R ,2 2

defined as 2

xf x

1 x, is :

(1) surjective but not injective (2) neither injective nor surjective

(3) invertible (4) injective but not surjective

Solution : [1]; surjective but not injective

1 1f : R ,

2 2

2

xf(x)

1 x

2

2 2

(1 x ) x(2x)f (x)

(1 x )

y

y

x x

2 2

2 2

1 x 2xf (x) 0

(1 x )

x = ± 1

2

2 2 2 2

(x 1) (x 1)(x 1)f (x) 0

(x 1) (x 1)

when x [–1, 1] and f(x) 0 when x (– , – 1][1, ]

f(x) is surjective but not injective

Q.60 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 areladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in whichX and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y arein this party, is :

(1) 469 (2) 484 (3) 485 (4) 468

Solution : [3] ; 485

X Y

Ladies(4)

Men(3)

Ladies(3)

Men(4)

0

1

2

3

3

2

1

0

3

2

1

0

0

1

2

3

4 3 3 40 3 3 0C C C C 1

4 3 3 41 2 2 0C C C C 4.3.3.4 144

4 3 3 42 1 1 2

4.3 4.3C C C C .3.3. 6.9.6 324

2 2

4 3 3 43 0 0 3C C C C 4.1.1.4 16

Total = 324 + 144 + 16 + 1 = 485

Total number of ways = 485

JEE(Main)-2017-SOL


ALL THE GRAPHS / DIAGRAMS GIVEN ARE SCHEMATIC AND NOT DRAWN TO SCALE.

Q.61 An observer is moving with half the speed of light towards a stationary microwave source emitting wavesat frequency 10 GHz. What is the frequency of the microwave measured by the observer?

(speed of light = 3 × 108 ms–1)

(1) 12.1 GHz (2) 17.3 GHz (3) 15.3 GHz (4) 10.1 GHz

Solution : [2] ; 17.3 GHz

0

0

C Vf f f 3

C V 0C

V2

Q.62 The following observations were taken for determining surface tension T of water by capilary 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 3rhgT 10 N / m

2, the possible error in surface tension is closest to :

(1) 1.5% (2) 2.4% (3) 10% (4) 0.15%

Solution : [1] ; 1.5%

3 3rhg DhgT 10 10

2 4

T D h g

T D h g

0.01 0.01 0.01

1.25 1.45 9.80

1 1100%

125 145

100 100%

125 145

= 1.5 %

Q.63 Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = 1/2, is givenby :

–E

4E

3

–2 E

–3 E

2

1

(1)2

r3

(2)3

r4

(3)1

r3

(4)4

r3

JEE(Main)-2017-SOL


Solution : [3] ; 1

r =3

1 1

2

2

hcE

1

hc E 3

3

Q.64 A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial

speed is v0 = 10 ms–1, If, after 10 s, its energy is 20

1mv

8, the value of k will be :

(1) 10–3 kg s–1 (2) 10–4 kg m–1 (3) 10–1 kg m–1 s–1 (4) 10–3 kg m–1

Solution : [2] ; 10–4 kg m–1

2 2f 0

1 1mv mv

2 8 0

fv

v2

2dvm kv

dt

0

0

v102

2v 0

dv kdt

mv

0

0

v

2

v

1 k.10

v m

0 0

2 1 k.10

v v m

24

0

m 1 10k 10

v 10 100

Q.65 Cp and Cv are specific heats at constant pressure and constant volume respectively. It is observed that

Cp – Cv = a for hydrogen gas

Cp – Cv = b for nitrogen gas

The correct relation between a and b is :

(1) a = b (2) a = 14 b (3) a = 28 b (4)1

a b14

Solution : [2] ; a = 14 b

CP – Cv = R for all gases

Q.66 The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector isI. What is the ratio l/R such that the moment of inertia is minimum?

(1)3

2(2) 1 (3)

3

2(4)

3

2

JEE(Main)-2017-SOL


Solution : [4] ; 3

2

dm = r2dx

x=

2 dxr .l

l

mdx

l

I22

2

2

dm rdm x

4

l

l

222

0

m r m2 .dx. .dx x

4

l

l l

2 3m r m2 . . 2 .

4 2 3 8

l ll l

2 2rm

12 4

l2

2

r m

mr

l

l2 mm

12 4

ll

I2

d 2 mm

d 12 4

ll l

2

m

6 4

ll

3 26. r

4l l

2 26r

4l

3

r 2

l

Q.67 A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. Atsometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by :

(1)log1.3

t Tlog2

(2) t = T log (1.3) (3)T

tlog 1.3

(4)T log2

t2 log1.3

JEE(Main)-2017-SOL


Solution : [1] ; log1.3

t = Tlog2

t0 0

t0

N N e0.3

N e

te 1 0.3

t = ln 1.3

1t ln 1.3

Tln 1.3

ln 2

Q.68 Which of the following statements is false?

(1) In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is

disturbed.

(2) A rheostat can be used as a potential divider.

(3) Kirchhoff’s second law represents energy conservation.

(4) Wheatstone bridge is the most sensitive when all the four resistances are of the same order of

magnitude.

Solution : [1] ; In a balanced wheatstone bridge if the cell and the galvanometer are exchanged,

the null point is disturbed.

is false

Q.69 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) 16 (2) 24 (3) 32 (4) 2

Solution : [3] ; 32

n

m

eqvC

m Cn

1.m 2

n

m 2

n 1

1000300

n

JEE(Main)-2017-SOL


1000n

300

n = 4, 5, ......

For minimum values :

n 4mn 32

m 8

Q.70 In the given circuit diagram when the current reaches steady state in the circuit, the charge on thecapacitor of capacitance C will be :

E r

C

r1

r2

(1) 1

2

rCE

r r(2)

2

2

rCE

r r(3)

1

1

rCE

r r(4) CE

Solution : [2] ; 2

2

rCE

r +r

2 22

Ev i r .r

r r

2

2

rQ CV CE

r r

Q.71 2V 2V 2V

2V 2V 2V

1 1 1

In the above circuit the current in each resistance is :

(1) 0.25 A (2) 0.5 A (3) 0 A (4) 1 A

Solution : [3] ; 0 A

[from symmetry]

Q.72 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 m << c. Which of the following

frequencies is not contained in the modulated wave?

(1) c (2) m + c (3) c – m (4) m

JEE(Main)-2017-SOL


Solution : [4] ; m

Am = (AC + Am sin mt) sin c t

= Ac sin ct + Am

2 cos m

c m c nA

t cos2

c, c – m, c + m

Q.73 In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input

and the output voltages will be :

(1) 90° (2) 135° (3) 180° (4) 45°

Solution : [3] ; 180°

ln CE mode, Ic and Ib are in opposite phase

Q.74 A copper ball of mass 100 gm is at a temperature T. It is dropped in copper calorimeter of mass 100 gm,

filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found

to be 75°C. T is given by : (Given : room temperature = 30°C, specific heat of copper = 0.1 cal/gm°C)

(1) 885°C (2) 1250°C (3) 825°C (4) 800°C

Solution : [1] ; 885°C

100 × 0.1 (T – 75) = (100 × .1 + 170) (75 – 30)

180 45T 75

10

T – 75 = 810

T = 885 ºC

Q.75 In a Young’s double slit experiment, slits are separated by 0.5 mm, 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 interference

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) 7.8 mm (2) 9.75 mm

(3) 15.6 mm (4) 1.56 mm

Solution : [1] ; 7.8 mm

n1 1 = n2 2

1 2

2 1

n 520 4

n 650 5

n1 = 4

Distance = 4 1 = 1D4

d

1504 650 nm

0.05

= 120 × 102 × 650 nm

= 7.8 mm

JEE(Main)-2017-SOL


Q.76 An electric dipole has a fixed dipole moment p

, which makes angle with respect to x-axis. When

subjected to an electric field 1E Ei

, it experiences a torque 1T k

. When subjected to another

electric field 2 1E 3 E j

it experiences a torque 2 1T T

. The angle is :

(1) 45° (2) 60° (3) 90° (4) 30°

Solution : [2] ; 60°

P Pcos i Psin j

1T P E P E i PEsin K

2T P E P 3 E j 3 PEcos K

PE sin = 3 PE cos

tan 3, 60º

Q.77 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 pivot. The free end is held vertically above the pivot and then

released. The angular acceleration of the rod when it makes an angle with the vertical is :

z

x

(1)2g

sin3

l

(2)3g

cos2

l

(3)2g

cos3

l

(4)3g

sin2

l

Solution : [4] ; 3g

sin2l

21M Mg sin

3 2

ll

3gsin

2

l

Q.78 An external pressure P is applied on a cube at 0ºC so that it is equally compressed from all sides. K isthe bulk modulus of the material of the cube and is its coefficient of linear expansion. Suppose we wantto bring the cube to its original size by heating. The temperature should be raised by :

(1)P

K(2)

3

PK

(3) 3PK (4)

P

3 K

JEE(Main)-2017-SOL


Solution : [4] ; P

3 K

Thermal expansion

v = lbh

v + v = l bh (1 + )3

= v(1 + 3)

v3

v

Elasticity

v Pk P

v 3

P

3 k

Q.79 A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converginglens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final imageformed is :

(1) virtual and at a distance of 40 cm from convergent lens.

(2) real and at a distance of 40 cm from the divergent lens.

(3) real and at a distance of 6 cm from the convergent lens.

(4) real and at a distance of 40 cm from convergent lens.

Solution : [4] ; real and at a distance of 40 cm from convergent lens.

25

25

40 = 2f2

Image distance = 2f2 = 40 cm (real image)

Q.80 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 is the smallest possible wavelength ofX-ray in the spectrum, the variation of log min with log V is correctly represented in :

(1) log min

log V

(2) log min

log V

(3) log min

log V

(4) log min

log V

JEE(Main)-2017-SOL


Solution : [4] ; log min

log V

ev hc

for ordinary x-ray..

V = constant.

In V + In = constant

x + y = constant

Q.81 The temperature of an open room of volume 30 m3 increases from 17ºC to 27ºC due to the sunshine.The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the number of moleculesin the room before and after heating, then nf – ni will be :

(1) 1.38 × 1023 (2) 2.5 × 1025 (3) –2.5 × 1025 (4) –1.61 × 1023

Solution : [3] ; –2.5 × 1025

Zn = PV

RT

Zf – Zi = PV 1 1

R 300 290

nf – ni = 5

231010 306.023 10

8.31 300 290

= – 2.5 × 1025

Q.82 In a coil of resistance 100 , a current is induced by changing the magnetic flux through it as shown inthe figure. The magnitude of change in flux through the coil is :

0.5 secTime

Current(amp.)

(1) 225 Wb (2) 250 Wb (3) 275 Wb (4) 200 Wb

Solution : [2] ; 250 Wb

dV Ri

dt

d R i dt

= R × area = 1

100 0.5 102

= 250 Wb

Q.83 When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 , it showsfull scale deflection. The value of the resistance to be put in series with the galvanometer to convert itinto a voltmeter of range 0 – 10 V is :

(1) 2.045 × 103 (2) 2.535 × 103 (3) 4.005 × 103 (4) 1.985 × 103

JEE(Main)-2017-SOL


Solution : [4] ; 1.985 × 103 Vg = 15 × 5 mV

V = 10,000 mV

n = 10,000

75

R = (n – 1) Rg = 400

1 15 19853

Q.84 A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the workdone by the force during the first 1 sec. will be :

(1) 22 J (2) 9 J (3) 18 J (4) 4.5 J

Solution : [4] ; 4.5 J

12

0mv F dt 3t 3

2P 9E 4.5J

2m 2 1

Q.85 A magnetic needle of magnetic moment 6.7 × 10–2 Am2 and moment of inertia 7.5 ×10–6 kg m2 isperforming simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 completeoscillations is

(1) 8.89 s (2) 6.98 s (3) 8.76 s (4) 6.65 s

Solution : [4] ; 6.65 s

IT 2

MB

6

2 2

7.5 102

6.7 10 10

2 7.5

10 6.7

6.28 7.5

10 6.7

10T 6.65 sec

Q.86 The variation of acceleration due to gravity g with distance d from centre of the earth is best representedby (R = Earths radius) :

(1)

g

O Rd

(2)

g

O Rd

(3)

g

O Rd

(4)

g

Od

JEE(Main)-2017-SOL


Solution : [3] ;

g

O Rd

g = x

gR

(inside)

= 2

2

gR

x (outside)

Q.87 A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocityvs time ?

(1) t

v

(2)t

v

(3) t

v

(4)

t

v

Solution : [2] ; t

v

Q.88 A particle A of mass m and initial velocity v collides with a particle B of mass m

2 which is at rest. The

collision is head on,and elastic. The ratio of the de-Broglie wavelengths A to B after the collision is :

(1) A

B

2

(2) A

B

2

3

(3) A

B

1

2

(4) A

B

1

3

Solution : [1] ; A

B= 2

1 2 11 1

1 2

m m uV u

m m 3

1 1 12

1 2

2m u 4uV

m m 3

1 2 2 2 2

2 1 1 1 1

m v m v 14 2

m v m v 2

JEE(Main)-2017-SOL


Q.89 A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position ofequilibrium. The kinetic energy - time graph of the particle will look like :

(1)

KE

0 T t(2)

KE

0 T tT

2

(3)

KE

0 tT

2

TT

4

(4)

KE

0 tTT

2

T

Solution : [3] ;

KE

0 tT

2

TT

4

x = a sin t = a sin2

tT

KE = 2 2 21mw a x

2

2 2 2 21m a a sin t

2 2A 1 sin t

1A 1 1 cos 2 t

2

1 1 2A cos 2. t

2 2 T

Time period of KE is T

2.

Q.90 A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that hisdensity remains same, the stress in the leg will change by a factor of

(1)1

9(2) 81 (3)

1

81(4) 9

Solution : [4] ; 9

mgP

A

Vdg

A

2

32 2 1

1 1 2

P V A 19 9

P V A 9

Documents

JEE (Main) : 2017 C - Pathfinderpathfinder.edu.in/wp-content/uploads/2017/05/JEEMain... · 2017-05-31 · JEE (Main) : 2017 (Question with Solution) Q.1 The freezing point of benzene