jee main online examination - 2017 - daily careers · 08-04-2017 · jee main online examination -...

JEE Online Paper _ Date [8-4-2017] (Page # 1)

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[MATHEMATICS]

1. If ( 27)999 is divided by 7, then the remainder is :(1) 3 (2) 1 (3) 6 (4) 2

Sol. 3

99928 17

= 28 1

7

28 7 1 1

7

= 7 4 1 6

7

Rem = 6

2. If 0 cos x sin x

S x 0, 2 : sin x 0 cos x 0cos x sin x 0

, Then x S

tan x3

(1) 4 2 3 (2) 2 3 (3) 2 3 (4) 4 2 3 Sol. 2

0(0 – cosx) – cosx(0 – cos2x) – sinx (sin2x – 0) = 0cos3x – sin3x = 0tan3 = 1 tanx = 1

3 tan x1 3

3 1 1 31 3 1 3

1 3 2 32

=

24 2 32 3

2 3

3. Let A be any 3 × 3 invertible matrix. Then which one of the folliwing is not Always true ?(1) adj(adj(1)) = |A|2 . (adj(1))–1 (2) adj(adj(1)) = |A|. (adj(1))–1

(3) ajd(adj(1)) = |A|.A (4) adj(1) = |A|.A–1

Sol. 3See theorey

4. The value of 2 2

1

2 2

1 x 1 xtan1 x 1 x

, 1x2

, x 0 , is equal to :

(1) 1 21 cos x4 2

(2) 1 2cos x

4

(3) 1 21 cos x4 2

(4) 1 2cos x

4

Sol. 1

x2 = cos2 ; = 21 cos x2

1 1 cos 2 1 cos 2tan1 cos 2 1 cos 2

1 1 tantan1 tan

= 1tan tan

4

= 21 cos x4 2

JEE Online Paper _ Date [8-4-2017](Page # 2)

Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 4 April. 2017

5. The area (in sq. units) of the parallelogram whosed diagonals are along the vectors ˆ ˆ8i 6j and

ˆ ˆ ˆ3i 4 j 12k , is :

(1) 20 (2) 65 (3) 52 (4) 26Sol. 2

d1 × d2 =

i j k8 6 03 4 12

= ˆ ˆ ˆ72i 96 j 50k

= 5178 + 7056 + 2500

1 2d d = 16900 = 130

A = 1 21 d d2

= 1 1302

= 65

6. The integral

4

3

12`

8cos 2xtan x cot x

dx equals :

(1) 13256 (2)

1564 (3)

1332 (4)

1 51 2 8

Sol. 4

4

3

12

cos 2x1

sin 2x

=

42

12

cos 2x sin 2x.sin 2x dx

= 1 sin 4x. 1 cos 4x dx4

= 4

12

1 sin 4x4

–

4

12

sin 8x

7. Let z C , the set of complex numbers. Then the equation, 2|z + 3i| – |z – i| = 0 a circle with

radius 83 .

(1) a cirlce with radius 83 (2) an ellipse with length of minor axis

169

(3) an ellipse with length of major axis 163 (4) a circle with diameter

103

Sol. 1

2 x i y 3 = x i y 1



= 222 x y 3 = 22x y 1

= 4x2 + 4( y + 3)2 = x2 + ( y – 1)2

= 3x2 = y2 – 2y + 1 – 4y2 - 24y – 36= 3x2 + 3y2 + 26y + 35 = 0

= x2 + y2 + 26 y3 +

353 = 0

= r = 169 350

9 3

= 649

= 83

8. If the um of the first n terms of the series 3 + 75 + 243 + 507 + ...... is 435 3 , then nequals :(1) 13 (2) 15 (3) 29 (4) 18

Sol. 2

3 1 25 81 69 ...... = 435 3

3 [ 1 + 5 + 9 + 13 +.....Tn] = 435 3

= n3 2 n 1 42

= 435 3

2n + 4n2 – 4n = 870= 4n2 – 2n – 870 = 0= 2n2 – n – 435 = 0

n = 1 1 4 2 435

4

= 1 59

4

= 1 59

4

= 4; 1 59

4

9. The tangent at the point (2,–2) to the curve x2y2 – 2x = 4(1 – y) does not pass through thepoint :

(1) (–2,–7) (2) (8,5) (3) (–4,–9) (4) 14,3

Sol. 1x2y2 –2x = 4 – 4y

2xy2 + 2y.x2. dydx –2 = – 4.

dydx

2dy 2y.x 4dx

= 2– 2x.y2

2, 2

dydx

= 2 2 2 4

2 2 4 4 =

1412

= 76

(y + 2) = 7 x 26

7x – 6y = 26



10. Let f(x) = 210dx + 1 and g(x) = 310x – 1. If (fog)(x) = x, then x is equal to :

(1) 10

10 10

2 12 3

(2) 10

10 10

1 23 2

(3) 10

10 10

3 13 2

(4) 10

10 10

1 32 3

Sol. 2f(g(x)) = xf(310x – 1) = 210(310.x – 1) = x

= 10 110

13 2

210(310x –1) + 1 = xx(610 –1) = 210 – 1

x = 10

10

2 16 1

= 10

10 10

1 23 2

11. The proposition (~p)V(P^~q) is equalvalent to :(1) P ~q (2) P^~q (3) q p (4) p v~ q

Sol. 2(~p) V ( P ^ ~q)

p q ~ p ~ q p^ ~ q ~ p v p^ ~ qT T F F F FT F F T T TF T T F F FF F T T F F

12. If all the words, with or without meaning, are written using the letters of the word QUEEN andare arranged as in English dictionary, then the position of the word QUEEN is :(1) 47th (2) 45th (3) 46th (4) 44th

Sol. 3E,E, N, Q,U(i) E......... = 24

(ii) N ......... = 4! 122

(iii) Q E........ = 3! = 6

(iv) Q N ......= 3!2! = 3

(v) Q U E E N = 1Total = (i) + (ii) + (iii) + (iv) + (v) = 46th

13. The curve ssatisfying the differential equation, ydx – (x + 3y2)dy = 9 and passing through thepoint (1,1), also passes through the point :

(1) 1 1,4 2

(2) 1 1,3 3

(3) 1 1,4 2

(4) 1 1,3 3

Sol. 2ydx – xdy – 3y2 dy = 0

dxdy =

x 3yy

dx xdy y

= 3y



I.f. = 1 dyye

= e–lnyy = 1y

solution is

x 13y. dyy y

xy = 3y + c

pass through (1,1) 1 = 3 + c ; c = –2x = 3y2 – 2y

(i) 1 1,4 2

= 1 3 14 4

(ii) 1 1 23 3 3

= 13

14. An unbiased coin is tossed eight times. The probabiity of obtaining at least one head and atleast one tail is :

(1) 127128 (2)

6364 (3)

255256 (4)

12

Sol. 1

1 P All Head P All Tail

8 8

1 112 2

= 7

112

= 11

128 =

127128

15. If the arithmetic mean of two nubers and b,a, > b > 0. is five times their geometric mean,

then a ba b

is equal to :

(1) 7 312

(2) 3 2

4(3)

62

(4) 5 612

Sol. 4

a b 5 ab2

a bab

= 10

ab =

10 9610 96

= 10 4 610 4 6

Use C and D

a b 20a b 8 6

=

52 6 =

5 612



16.x 3

3x 3lim2x 4 2

is equal to :

(1) 12 (2)

12 2 (3)

32

(4) 3

Sol. 1

x 3

3x 3lim2x 4 2

Rationalize

x 3

3x 9 2x 4 2lim

2x 4 2 3x 3

= x 3

3 x 3 2x 4 2lim2 x 3 3x 3

= 3 2 22 6 =

12

17. The locus of the point of intersection of the straigh lines,tx – 2y – 3t = 0

x – 2ty + 3 = 0 t R , is :

(1) A hyperbola with the length of conjugate axis 3

(2) a hyperbola with eccentricity 5(3) an ellipse with the length of major axis 6

(4) an ellipse with eccentricity 25

Sol. 1tx – 2y – 3t = 0x – 2ty + 3 = 0

tx – 2y – 3t = 0tx – 2t y + 3t = 02

+ –y(2t –2)2 = 6t

t x – 2ty – 3t = 0x – 2ty + 3 = 0

2 2

– –(t -1)x= (3t + 1)2 2+

2

6ty2t 2

= 2

3tt 1 x = – 3sec2

2y = 3(–tan2)sec22 – tan22 = 1

2 2x y9 9 / 4 = 1

a2 = 9 ; b2 = 9/4

(T.A) = 6 ; 2 9 / 4e 19

= 114

; e 5e

2



18. If 15

2y x x 1 + 15

2y x x 1 , then (x2 – 1)2

2

d y dyxdxdx

is equal to :

(1) 224y2 (2) 125y (3) 225y (4) 225 y2

Sol. 3

y = 15152 2x x 1 x x 1

dydx = 14

215 x x 1 142

2

x15 x x 1 1x 1

2

dy 15 .ydx x 1

...(i)

2 dyx 1.dx

= 15y

2

x dy.dxx 1

+ 2

22

d yx 1dx

= dy15dx

2

22

dy d yx x 1dx dx

= 2

2

1515 x 1. .yx 1

= 225y

19. The area (in sq. units) of the smaller portion enclosed between the curves, x2 + y2 = 4 andy2 =3x, is :

(1) 1 4

33

(2) 1 2

33

(3) 1 _

32 3

(4) 1 2

32 3

Sol. 1x2 + 3x – 4 = 0(x + 4) ( x – 1) = 0x = – 4, x = 1

Area = 1

0

3. xdx +

1 1 22

0 0 1

3. xdx 4 x .dx 2

=

1 23/ 22 1

10

x x x3 4 x 2sin 23/ 2 2 2

= 2 33 2. 23 2 2 3

x + y =

4

2

2

1 2

y = 3x2

2 3 2 22 33

= 1 2 2

32 3

= 1 4

33

d

20. If the common tangents to the parabola, x2 + 4y and the circle, x2 + y2 = 4 intersect at thepoint P, then the distance of P from the origin, is :

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



Sol. 4tangent to x2 + y2 = 4

y = mx ± 22 1 m2x 4y

x2 = 4mx + 28 1 m

x2 = 4mx – 28 1 m = 0

D = 0

16m2 + 24.8 1 m = 0

2 2m 2 1 m 0

or 2 2m 1 m m4 = 4 + 4m2

m4 – 4m2 – 4 = 0

m2 = 4 16 16

2

P

y

x

= 4 4 2

2

m2 = 2 2 2

21. The line of intersection of the planes

ˆ ˆ ˆr. 3i j k 1

and

ˆ ˆ ˆr. i 4j 2k 2

, is :

(1)

6 5x y z13 132 7 13

(2)

4 5x zy7 72 7 13

(3)

6 5x y z13 132 7 13

(4)

4 5x zy7 72 7 13

Sol. 3

1 2n n n

ˆ ˆ ˆi j k3 1 11 4 2

= ˆ ˆ ˆi( 2) j( 7) k(13)

n = ˆ ˆ ˆ2i 7 j 13k Now3x – y + z = 1x + 4y – 2z = 2but z = 03x – y = 1×4x + 4y = 213x = 6 x = 6/13



y = 5/13.... is

x 6 /13 y 5 /13 z 02 7 13

or

x 6 /13 y 5 /13 z2 7 13

22. If a point P has co - ordinates ( 0,–2) and Q ia any point on the circle, x2 +y2 – 5x – y + 5 = 0,then the maximum value of (PQ)2 is :

(1) 8 5 3 (2) 47 10 6

2

(3) 14 5 3 (4) 25 6

2

Sol. 3

(x –5/2)2 254

+(y -1/2)2 – 1/4 + 5 = 0

= (x – 5/2)2 + (y –1/2)2 = 3/2

on circle Q = 5/2 + 3/ 2 cos Q, 1 3/ 2 sin Q2

PQ2 = 25 3/ 2 cosQ

2

+25 3/ 2 sin Q

2

PQ2 = 252

+32

+ 5 3/ 2 (cos Q + tan Q)

= 14 5 3/ 2 (cos Q sin Q)

manmr = 14 5 3/ 2 ( 2)

= 14 5 3

23. The mean age of 25 teaches in a school is 40 years. A teacher retires at the age of 60 yearsand a new teacher is appointed in his place. if now the mean age of the teachers in this schoolis 39 years, then the age ( in years ) of the newly appointed teachers is :(1) 35 (2) 40 (3) 25 (4) 30

Sol. 1

1 2 25x x ... x25

= x 40

x1 +x2 + ... + x25 = 1000x2 + x2 + ...+ x25 – 60 + A = x 251000 - 60 + A = 39 × 25 = 975A = 975 – 940 = 35

24. Let p(x) be a quadratic polynomial such that p(0) = 1. if p(x) leaves remainder 4 when dividedby x – 1 and it leaves remainder 6 when divided by x + 1, then :(1) p(–2) = 19 (2) p(2) = 19 (3) p(–2) = 11 (4) p(2) = 11

Sol. 1p(x) = ax2 + bx + cp (0) = 1 = c = 1

p(1) 4p( 1) 6



a b c 4 a 4a b c 6 b 1

p(x) = 4x2 – x + 1p(–2) = 16 + 2 + 1 = 19

25. consider an ellipse, whose centre is at the origin and its major axis is along the x - axis. if its

eccentricity is 35 and the distance between its foci is 6, then the area ( in sq. units ) of the

quadrilateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is :(1) 32 (2) 80 (3) 40 (4) 8

Sol. 3e = 3/5, 2ae = 6, a(5) a = 5

b a

b2 = a2 (1–e2)b2 = 25 (1–9/25)b = 4

area = 4 (1/2 ab)= 2ab = 40

26. If two parallel chords of the a circle, having diameter 4 units, lie on the opposite sides of the

centre and subtend angles 1 1cos

7

and sec–1(7) at the centre respectively, then the distance

between these chords, is :

(1) 87 (2)

167 (3)

47 (4)

87

Sol. 1cos 2Q = 1/7 = 2cos2Q – 1 = 1/7

= 2 cos2Q = 8/7 cos2Q = 4/7

= 2cp

4=

47

= Cp = 47

sec2Q = 7 = 21 7

2 cos Q 1

= 2

2Cp 12 12 7

= 2

2Cp 822 7

r

2

Q

p1

p2

Q1 Q1

2

2

= 24Cp7

4 47 7 =

87



27. The number of real values of for which the system of linear equations2x + 4y – z = 04x + y + 2z = 0x + 2y + 2z = 0has infinitely many solutions, is :(1) 3 (2) 1 (3) 2 (4) 0

Sol. 20

2 4 2 44 z 4 0

z 2 z

= – (32 + 8 – 3) = 0= 3 + 4 – 40 = 0=

28. The coodinates of the foot of the perpendicular from the point (1, – 2,1) on the plane contain-ing the lines,

x 1 y 1 z 36 7 8

and

x 1 y 2 z 33 5 7

, is :

(1) (2, – 4, 2) (2) (1, 1, 1) (3) (0,0,0) (4) (–1, 2, –`=1)Sol. 3

1 2n n n

=

ˆ ˆ ˆi j k6 7 83 5 7

= (9, –18, 9)= (1, –2, 1)

1(x+1)–2 (y–1) + (2–3) = 0

= x 2y z 0

foot to z

x 11

= y 2

2

=z 1

1

=[1 4 1]

6

x 0, y 0, z 0

10,000

29. Three persong P, Q and R independently try to hit a target. If the probabilities of their hitting

the target are 34

, 12

and 58 respectively, then the probability that the target is hit by P or Q

but not by R is :

(1) 3964 (2)

2164 (3)

964 (4)

1564



Sol. 2

34

12

38

+1 1 3 3 1 34 2 8 4 2 8

= 12 9

64

= 2764

30. The integral 1 2cot x cos ecx cot x dx 0 x2

is equal to :

(where C is a constant of integration )

(1) x2log sin C2

(2) x4log sin C2

(3) x4 log cos C2

(4)x2 log cos C2

Sol. 1

22cot x cos ecx cos ec x cot x dx cos | x cot x | dx(cosec cot x)dxcosec dx

22log(log(x ) c



[PHYSICS]

1. A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously,such that its temeprature increass by T. The net change in its length is zero. Let l be the lengthof the rod, A its area of cross-section, Y its Young's modulus, and its coefficient of linearexpansion. Then, F is equal to -

(1) lA Y T (2) A Y T (3) T

AY

(4) l2 Y T

Sol. 3Net change in length = 0Thermal Exp. = l t

y = l/lA/F

ll = Ay

F

l = AyFl

AyFl

= l t

F = Ay t

2. A potentiometer PQ is set up to comapre two resistances as shown in the figure. The ammeter Ain the circuit reads 1.0 A when two way key K3 is open. The balance point is at a length l1 cm fromP when two way key K3 is plugged in between 2 and 1, while the balance point is at a length l2 cm

from P when key K3 is plugged in between 3 and 1. The ratio of the two resistance 2

1

RR

, is found

to be -

K2Rh2E2

A

R12 3R2

1K3

G

Q

K1Rh1E1

P

(1) 21

1

lll (2)

12

2

lll (3)

21

1

lll (4)

12

1

lll



Sol. 1When key is at pointV1 = iR1 = xl1when key is at (3)V2 = i (R1 + R2) = xl2

21

1

RRR =

2

1

ll

2

1

RR

= 12

1

lll

3. A signal of frequency 20 kHz and peak voltage of 5 volt is used to modulate a carrier wave offrequency 1.2 MHz and peak voltage 25 volts. Choose the correct statement.(1) Modulation index = 5, side frequency bands are at 1400 kHz and 1000 kHz(2) Modulation index = 0.8, side frequency bands are at 1180 kHz and 1220 kHz(3) Modulation index = 0.2, side frequency bands are at 1200 kHz and 1180 kHz(4) Modulation index = 5, side frequency bands are at 21/2 kHz and 18.8 kHz

Sol. 3

Modulation idex = m = 0

m

VV

= 51

= 0.2

Frequency = 12 × 103 kHzF = 12.00 kHzF1 = 1200 – 20 = 1180 kHzF2 = 1200 + 20 = 1220 kHz

4. A single slit of width b is illuminated by a coherent monochromatic light of wavelength . If thesecond and fourth minima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6cm respectively from the central maximum, what is the width of the central maximum ? (i.e.,distance between first minimum on either side of the central maximum)(1) 4.5 cm (2) 1.5 cm (3) 6.0 cm (4) 3.0 cm

Sol. 4min.f sin = n

sin = 6n

n = 2

sin = 62

= tan 1 = Dx1

x = 4

sin 2 = 64

= Dx2

x2 – x1 = 64

– 62

= 62

= 3 cm

width of central max = 62

= 3 cm



5. A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontaltable. Two springs identical to the original spring are attached in parallel to an 8 kg block placedon the same table. So, the frequency of vibration of the 8 kg block is -

(1) 2 Hz (2) 41

Hz (3) 22

1 Hz (4)

21

Hz

Sol. 4

F = 2

1

mk

= 1

42 = mk

m = 1 m=1kg

k = 42

In parallel keq = 2k

F = 2

1

K 28

= 21

Hz 8kg

k

6. A magentic dipole in a constant magnetic field has -(1) maximum potential energy when the torque is maximum(2) zero potential energy when the torque is maximum(3) zero potential energy when the torque is minimum(4) minimum potential energy when the torque is maximum

Sol. 2PE = – PE cos = PE sin max when = 90°PE = 0

7. If the earth has no rotational motion, the weight of a person on the equation is W. Detrmine thespeed with which the earth would have to rotate about its axis so that the person at the equator

will weight 43

W. Radius of the earth is 6400 km and g = 10 m/s2.

(1) 0.63 × 10–3 rad/s (2) 0.28 × 10–3 rad/s(3) 1.1 × 10–3 rad/s (4) 0.83 × 10–3 rad/s

Sol. 1g' = g – 2R cos2

4g3

= g – 2R

w2R = 4g

w = R4g

= 31064410



= 10082

1

= 1600

1 =

161

× 10–2 = 0.6 × 10–3

8. An object is dropped from a height h from the ground. Every time it hits the ground it looses 50%of its kinetic energy. The total distance covered as t is -

(1) 2h (2) (3) 35

h (4) 38

h

Sol.21

mv'2 = 21

21

mvv2

v' = 2

v

v = eu

e = 2

1

H = 2

21 e1 e

= h

112112

= 3h

9. The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains 4Ccharge, its radius will be :

[Take : 04

1 = 9 × 109 N – m2/C2]

(1) 32 mm (2) 16 mm (3) 28 mm (4) 20 mmSol. 2

Energy of sphere = C2

Q2

4.5 = C21016 12

C = 91016 12

= 40R

R = 91016 12

× 04

1

= 9 × 109 × 9

16 × 10–12

= 16 × 10–3 = 16 mm



10. A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrappedover its rim and a body of mass m is tied to the free end of the string as shown in the figure. Thebody is released from rest. Then the acceleration of the body is -

WallM

R

g

m

(1) Mm2

Mg2

(2) mM2

Mg2

(3) mM2

mg2

(4) Mm2

mg2

Sol. 4mg – T = maRT = I

RT = 2

MR2

.Ra

T = 2

Ma

mg – 2

Ma = ma

mg = a

m2M

mg = a

2

m2M

a = m2M

mg2

11. An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. Thethermal efficiency of the engine is - (Take Cv = 1.5 R, where R is gas constant)

B C

DA

P0

2P0

V0 2V0

V

P0

(1) 0.24 (2) 0.15 (3) 0.32 (4) 0.08



Sol. 2w = P0V0Heat given = QAB = QBC= nCVdTAB + nCPdTBC

= 23

(nRTTB – nRTA) + 25

(nRTTC – nRTB)

= 23

(2P0V0 – P0V0) + 25

(4P0V0 – 2P0V)

= 213

P0V0

n = w

Qgiven = 132

= 0.15

12. Time (T), velocity (3) and angular momentum (h) are chosen as fundamental quantities insteadof mass, length and time. In terms of these, the dimensions of mass would be -(1) [M] = [T–1 C–2h] (2) [M] = [T C–2h](3) [M] = [T–1 C–2h–1] (4) [M] = [T–1 C2h]

Sol. 1M Tx vy hz

M' L0T0 = (T')x (L1T–1)y (M1L2T–1)z

M1L0T0 = Mz Ly+2z + Tx-y-z

z = 1y + 2z = 0 x – y – z = 0y = – 2 x + 2 – 1 = 0

x = – 1M T–1 C–2 h1

13. In an experiment a sphere of aluminium of mass 0.20 kg is heated upto 150°C. Immediately, it is putinto water of volume 150 cc at 27°C kept in a calorimeter of water equivalent to 0.025 kg. Finaltemperature of the system is 40°C. The specific heat of aluminium is -(take 4.2 Joule = 1 calorie)(1) 434J/kg-°C (2) 378J/kg-°C (3) 315J/kg-°C (4) 476J/kg-°C

Sol. 1Q given = Q used0.2 × S× (150 – 40) = 150 × 1 × (40 – 27)

+ 25 × (40 – 27)0.2 × S × 110 = 150 × 13 + 25 × 13

S = 1102.0

72513

S = 434

14. There is a uniform electrostatic field in a region. The potential at various points on a small spherecentred at P, in the region, is found to vary between in limits 589.0V to 589.8 V. What is the potentialat a point on the sphere whose radius vector makes an angle of 60° with the direction of the field ?(1) 589.4 V (2) 589.5 V (3) 589.2 V (4) 589.6 V

Sol. 1V = E.d0.8 = Ed (max)V = Edcos = 0.8 × cos 60= 0.4589.4



15. Magnetic field in a plane electromagnetic wave is given by

Tj)tkxsin(BB 0

Expression for corresponding electric field will be

(1) m/Vk)tkxsin(cBE 0



(4) m/Vk)tkxsin(c

BE 0

Sol. 3

C = 0

0

EB

E = CB0= CB0

= CB0 sin (kx + t) i

16. According to Bohr's theory, the time averaged magnetic field at the centre (i.e. nucleus) of ahydrogen atom due to the motion of electrons in the nth orbit is proportional to :(n = principal quantum number)(1) n–3 (2) n–2 (3) n–4 (4) n–5

Sol. 2

B = r2I0

= r2qt0

r n2

B m–2

17. Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centreO and perpendicular to its plane is I0 as shown in the figure. A cavithy DEF is cut out from the lamina,where D,E,F are the mid points of the sides. Moment of inertia of the remaining part of lamina aboutthe same axis is -

C

F

A D B

EO

(1) 1615

I0 (2) 4I3 0 (3)

87

I0 (4) 32

I31 0



Sol. 1

I0 = kml2 BC = l

IDEF = K 4m

2

2l

= 16k

ml2

IDEF = 16

0

Iremain = I0 = 16I0

= 16

I15 0

18. The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n

falls on a metal surface. If the incident frequency is increased to 3n, the maximum velocity of the

ejected photoelectrons will be -

(1) more than 3 v (2) equal to 3 v

(3) v (4) less than 3 v

Sol. 2

E1 = hn –

E2 = 4hn –

E2 = 3 (E1 + ) –

E2 = 3 E1 + 2V

m0x 3 y

19. What is the conductivity of a semiconductor sample having electron concentration of

5 × 1018 m–3, hole concentration of 5 × 1019 m–3, electron mobility of 2.0 m2 V–1 s–1 and hole

mobility of 0.01 m2 V–1 s–1 ?

(Take charge of electron as 1.6 × 10–19 C)

(1) 1.83 (-m)–1 (2) 1.68 (-m)–1

(3) 1.20 (-m)–1 (4) 0.59 (W-m)-1

Sol. 2

s = e (nee + nnn)

= 1.6 × 10-19 (5 × 1018 × 2 + 5 × 1019 × 0.01)

= 1.6 × 10–19 (1019 + 0.05 × 1019)

= 1.6 × 1.05= 1.68



20. The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse biasresistance is :

1(mA)

20

15

10–10

.7 .8 V (volt)

(1) 100 (2) 106 (3) 10–6 (4) 10Sol. 3

f.B = 310101.0

l = 10

RB = 610

10 = 107

RBfB

= 10–6

21. Which graph corresponds to an object moving with a constant negative acceleration and a positivevelocity ?

(1) Vel

ocity

Distance

(2) Vel

ocity

time

(3) Vel

ocity

Distance

(4) Vel

ocity

time

Sol. 3a = –C

dxVdV

= – C

VdV = – CdX

2V2

= – Cx + k

x = – C2

V2

+ CK



22. A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop ofradius b. The two loops are in the same plane. The outer loop of radius b carries an alternatingcurrent I = I0 cos (t). The emf induced in the smaller inner loop is nearly ?

(1) )tsin(ba

I2

00 (2) )tcos(ba

.2I 200

(3) )tcos(a

bI 200

(4) )tsin(

ba

.2I 200

Sol. 4e = MdI

dT M = b2a2

0

= b2a2

0 = I0 lt

23.

A1

A2 A3

VB9V

A 9 V battery with internal resistance of 0.5 is connected across an infinite network as shownin the figure. All ammeters A1, A2, A3 and voltmeter V are ideal.Choose correct statement.(1) Reading of A1 is 18 A(2) Reading of V is 9 V(3) Reading of V is 7 V(4) Reading of A1 is 2 A

Sol. 4

A

x

B

1

1

4 x

x = x4

x4

+ 2

x = x4x68

4x + x2 = 8 + 6xx2 – 2x – 8 = 0

x = 2

)8)(1(442 =

2362



= 2

62 = 4

A' = 5.04

9

= 2

A' = 9

24. Let the refractive index of a denser medium with respect to a rarer medium be n12 and its criticalangle be C. At an angle of incidence A when light is travelling from denser medium to rarermedium, a part of the light is reflected and the rest is refracted and the angle between reflectedand refracted rays is 90°. Angle A given by -

(1) tan–1 (sin C) (2) )(sintan1

C1 (3) cos–1 (sin C) (4) )(sincos

1

C1

Sol. 1

A A90

r

= D

R

= 90sin

isin c

D

R

= sin ii

= D

R

= rsinAsin

= )A90sin(Asin =

AcosAsin

D

R

= tan A

tan A = sin CA = tan–1 (sin C)

25. The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s–1.

At, t = 0 the displacement is 5 m. What is the maximum acceleration ? The initial phase is 4

.

(1) 500 m/s2 (2) 750 2 m/s2 (3) 750 m/s2 (4) 500 2 m/s2

Sol. 4fmax = avmin = a



a

a = 10

w = 10x = a sin ( + /4)at f = 05 = a sin (/4)

a = 25max acc. = w2a

= 100 × 25

= 500 2

26. Two wires W1 and W2 have the same radius r and respective densities 1 and 2 such that 2 = 41.They are joined together at the point O, as shown in the figure. The combination is used as asonometer wire and kept udner tension T. The point O is midway between the two bridges. Whena stationary waves is set up in the composite wire, the joint is found to be a node. The ratio of thenumber of antinodes formed in W1 to W2 is -

W2W1

(1) 4 : 1 (2) 1 : 2 (3) 1 : 1 (4) 1 : 3Sol. 2

n1 = n2T samer samel same

n = l2

p

drT2

n1 = n2

1

1

d

p =

2

2

d

p

2

1

pp

= 21

27. An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure(CP) and at constant volume (CV) is

(1) 57

(2) 6 (3) 27

(4) 25

Sol. 1

f = a

p

CC

= 1 + 12

= 7/5



28. Two deuterons udnergo nuclear fusion to form a Helium nucleus. Energy released in this process is :(given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV)(1) 23.6 MeV (2) 25.8 MeV (3) 30.2 MeV (4) 32.4 MeV

Sol. 1

1H2 + 1H

1 2Hc4

initiate 1.1 × 4 = 4.4final 4 × 7 = 28release 28 – 4.4 = 23.6

29. In a certain region static electric and magentic fields exist. The magnetic field is given by

k4j2iBB 0

. If a test charge moving with a velocity k2ji30 experience no force in

that region, then the electric field in the region, in SI units, is -

(1) k7jiBE 00

(2) k4j2i3BE 00

(3) 0 0ˆ ˆE B 14 j 7k

(4) 0 0

ˆ ˆE B 14j 7k

Sol. 4Fe = Fn = 0Fe = –Fm

= – q Bv

= – 0v0 k4j2ik2ji3

= – 0v0 k7i14

30. In a physical balance working on the principle of moments, when 5 mg weight is placed on the leftpan, the beam becomes horizontal. Both the empty pans of the balance are of equal mass. Whichof the following statements is correct ?(1) Every object that is weighted using this balance appears lighter than its actual weight(2) Left arm is shorter than the right arm(3) Both the arms are of same length(4) Left arm is longer than the right arm

Sol. 2



[CHEMISTRY]

1. If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelengthin Paschen series of He+ is :

(1) 5A9

(2) 36A

7(3)

36A5

(4) 9A5

Sol. 2Shortest wavelength is corresponding to best ine nL = 1 (Lyman series)

nH = (infinite)

1A

= r × (1)2 1 1

12 2

= R

Longest wavelength 1st Line = 3 nH = 4

1

= r × (2)2 2 21 13 4

= r 736

= 36A

7

2. Among the following, the essential amino acid is :(1) Valine (2) Alanine (3) Serine (4) Aspartic acid

Sol. 1

3. Identify the pollutant gases largely responsible for the discoloured and lustreless nature of marbleof the Taj Mahal.(1) SO2 and O3 (2) O3 and CO2 (3) SO2 and NO2 (4) CO2 and NO2

Sol. 3SO2 and NO2

4. Which of the following compounds will not undergo Friedel Craft's reaction with benzene ?

(1) Cl (2) Cl (3)

Cl

O

Cl

(4) COCl

Sol. 1Formation of carbocation is not possible in case of CH2 = CHCl

5. Which of the following is paramagnetic ?

(1) CO (2) NO+ (3) 22O (4) B2

Sol. 4No of e–

CO = 14, NO+ = 14O2

2– = 18 B2 = 10According to MOTB2 is paramagnetic



6. The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much,the temperature of reaction B should be increased from 300 K so that rate doubles if activationenergy of the reaction B is twice to that of reaction A.(1) 4.92 K (2) 9.84 K (3) 19.67 K (4) 2.45 K

Sol. 1

2 = EqR

1 1300 310

...(i)

2 = e2EaR

1 1300 T

...(ii)

2EaR

1 1300 T

= aER

1 1300 310

1300

+ 1

310 =

2T T =

300 310610

× 2

= 304.92

7. A solution containing a group-IV cation gives a precipitate on passing, H2S. A solution of thisprecipitate in dil. HCl produces a white precipitate with NaOH solution and bluish-white prcipitatewith basic potassium ferrocyanide. The cation is :(1) Mn2+ (2) Zn2+ (3) Ni2+ (4) Co2+

Sol. 2

ZnSHCl

ZnCl2NaOH

Zn(OH)2

k [Fe(CN) ]4 6

Zn [Fe(CN) ]2 6

8. Which of the following statements is not true about partition chromatography ?(1) Stationary phase is a finely divided solid adsorbent(2)Separation depends upon equilibration of solute between a mobile and a stationary phase(3) Paper chromatography is an example of partition chromatography(4) Mobile phase can be a gas

Sol. 4

9. Excess of NaOH (aq) was added to 100 mL of FeCl3 (aq) resulting into 2.14 g of Fe(OH)3. Themolarity of FeCl3(aq) is :(Given molar mass of Fe = 56 g mol–1 and molar mass of Cl = 35.5 g mol–1)(1) 0.3 M (2) 0.2 M (3) 0.6 M (4) 1.8 M

Sol. 23NaOH + FeCl3 Fe (CH)3 + NaCl

100 ml 2.14 gm m = ?

Moles of Fe(CH3) = 2.14107

= 2 × 10–2 mol

moles FeCl3 = 2 × 10–2 mol

M =22 10

100

× 1000 = 0.2 M



10. 5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.82°C.If Na2SO4 is 81.5% ionised, the value of x(Kf for water = 1.86°C kg mol–1) is approximately :(molar mass of S = 32 g mol–1 and that of Na = 23 g mol–1)(1) 25 g (2) 65 g (3) 15 g (4) 45 g

Sol. 4

Na2SO4 2Na+ + 24SO

x = 1 + (3 – 1) 0.815 = 2.63

3.82 = 1.86 × 2.63 ×

5 1000142 x

x =

1.86 2.63 5000

142 3.82= 45 gm

11. Consider the following standard electrode potentials (E° in volts) in aqueous solution :Element M3+/M M+/MAl –1.66 + 0.55TI + 1.26 –0.34Based on these data, which of the following statements is correct ?(1) T3+ is more stable than A3+ (2) A+ is more stable than A3+

(3) T3+ is more stable than A3+ (4) T+ is more stable than A+Sol. 4

G is –ve

12. The major product expected from the following reaction is :

OCH OH2

HO C2

OH

NH2 HCl(g)/CCl4

(1)

O

O

Cl

HO C2

(2)

O

O

OH

HO C2

(3)

O

O

OH

NH2

O

(4)

Cl

NH2

OCH OH2

HO C2



Sol. 3

13. Among the following, the incorrect statement is :(1) At low pressure, real gases show ideal behaviour.(2) At very low temperature, real gases show ideal behaviour.(3) At Boyle's temperature, real gases show idela behaviour.(4) At very large volume, real gases show ideal behaviour.

Sol. 2

14. The pair of compounds having metals in their highest oxidation state is :(1) MnO2 and CrO2Cl2 (2) [Fe(CN)6]

3– and [Cu(CN)4]2–

(3) [NiCl4]2– and [CoCl4]

2– (4) [FeCl4]– and Co2O3

Sol. 1MnO2 = + 4CrO2Cl2 = + 6

15. The IUPAC name of the following compound is

(1) 2-Ethyl-1, 1-dimethylcyclohexane(2) 1, 1-Dimethyl 1-2-ethylcyclohexane(3) 2, 2-Dimethyl-1-1-ethylcyclohexane(4) 1-Ethyl-2,2-dimethylcyclohexane

Sol. 1

12

3

45

6

16. A mixture containing the following four compounds is extracted with 1 M HCl. The compound thatgoes to aqueous layer is :

(1) IV (2) II (3) I (4) IIISol. 2



17. Consider the following ionization enthalpies of two elements ‘A’ and ‘B’.

Element

1st 2nd 3rd

A 899 1757 14847B 737 1450 7731

Ionization enthalpy (kJ/mol)

Which of the following statements is correct?(1) Both ‘A’ and ‘B’ belong to group-1 where ‘A’ comes below ‘B’.(2) Both ‘A’ and ‘B’ belong to group-2 where ‘A’ comes below ‘B’.(3) Both ‘A’ and ‘B’ belong to group-1 where ‘B’ comes below ‘A’.(4) Both ‘A’ and ‘B’ belong to group-2 where ‘B’ comes below ‘A’.

Sol. 4

18. sp3d2 hybridization is not displayed by :(1) SF6 (2) BrF5 (3) PF5 (4) [CrF6]

3–

Sol. 3SF6 = Sp3d2 BrF5 = SP3d2

[CrF6]3– = sp3d2 PFs = sp3d

19. The number of S = O and S–OH bonds present in peroxodisulphuric acid and pyrosulphuric acidrespectively are :(1) (2 and 4) and (2 and 4) (2) (4 and 2) and (4 and 2)(3) (4 and 2) and (2 and 4) (4) (2 and 2) and (2 and 2)

Sol. 2Peroxodisulphuric acid

H2S2O8HO—S—O—O—S—OH

O

O O

O

Pyrosulphuric acid

H2S2O7 : HO—S—O—S—OH

O

O O

O

20. Among the following, correct statement is :

(1) Sols of metal sulphides are lyophilic.

(2) Brownian movement is more pronounced for smaller particles than for bigger-particles

(3) One would expect charcoal to adsorb chlorine more than hydrogen sulphide.

(4) Hardy Schulze law states that bigger the size of the ion is, the greater is its coagulatingpower.

Sol. 3



21. The major product of the following reaction is :

(1) (2)

(3) (4)

Sol. 1

22. What is the standard reduction potential (E°) for Fe3+ Fe ?Given that :

Fe2+ + 2e– Fe; 2Fe /FeE 0.47V

Fe3+ + e– Fe2+; 3 2Fe /FeE 0.77 V

(1) –0.057 V (2) + 0.30 V (3) – 0.30 V (4) + 0.057 VSol. 1

23. The reason for “drug induced poisoning” is :(1) Binding irreversibly to the active site of the enzyme(2) Binding at the allosteric sites of the enzyme(3) Binding reversibly at the active site of the enzyme(4) Bringing conformational change in the binding site of enzyme

Sol. 2


(1) (2)

(3) (4)



Sol. 1


(1) CH3CH = CH – CH = CHCH3 (2) CH2 = CHCH = CHCH2CH3

(3) CH3CH = C = CHCH2CH3 (4) CH2= CHCH2CH = CHCH3

Sol. 1

26. In which of the following reactions, hydrogen peroxide acts as an oxidizing agent ?

(1) HOCl + H2O2 H3O+ + Cl– + O2

(2) I2 + H2O2 + 2OH– 2I– + 2H2O + O2

(3) PbS + 4H2O2 PbSO4 + 4H2O

(4) 2MnO4– + 3H2O2 2MnO2 + 3O2 + 2H2O + 2OH–

Sol. 3PbS + 4H2O2 PbSO4 + 4H2O+2 +4

27. For a reaction, A(g) A(); H = –3RT.

The correct statement for the reaction is :

(1) H = U = 0 (2) |H| < |U|

(3) |H| > |U| (4) H = U 0

Sol. 3



28. The enthalpy change on freezing of 1 mol of water at 5°C to ice at –5°C is :

(Given fus H = 6 kJ mol–1 at 0°C,

Cp(H2O, ) = 75.3 J mol–1 K–1,

Cp (H2O, s) = 36.8 J mol–1 K–1)

(1) 6.00 kJ mol–1 (2) 5.81 kJ mol–1

(3) 5.44 kJ mol–1 (4) 6.56 kJ mol–1

Sol. D

29. Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of pH 6. If ionisationconstant of HA is 10–5, the ratio of salt to acid concentration in the buffer solution will be :

(1) 10 : 1 (2) 4 : 5 (3) 5 : 4 (4) 1 : 10

Sol. 1

30. A metal ‘M’ reacts with nitrogen gas to afford ‘M3N’. ‘M3N’ on heating at high temperature givesback ‘M’ and on reaction with water produces a gas ‘B’. Gas ‘B’ reacts with aqueous solution ofCuSO4 to form a deep blue compound. ‘M’ and ‘B’ respectively are :

(1) Li and NH3 (2) Na and NH3 (3) Ba and N2 (4) Al and N2 Sol. 1

Li + N2 Li3N

H2O

LiOH + NH3

COSO4

[Cu(NH3)4] SO4

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