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Answer key JEE Main - 2018 (Code-D) Date : 08-04-18MATHEMATICS

1. 12. 23. 44. 35. 16. 27. 38. 19. 310. 311. 412. 113. 414. 215. 416. 117. 218. 219. 320. 421. 322. 323. 224. 225. 426. 327. 428. 129. 430. 4

PHYSICS31. 332. 133. 234. 435. 436. 237. 338. 339. 440. 341. 342. 443. 444. 245. 246. 447. 448. 449. 250. 451. 252. 253. 354. 355. 456. 357. 158. 259. 460. 3

CHEMISTRY61. 162. 463. 364. 265. 466. 167. 468. 269. 470. 371. 472. 473. 274. 175. 176. 477. 1 and 378. 479. 180. 181. 282. 483. 484. 385. 186. 387. 488. 489. 390. 4


Vibrant A

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IIT JEE MAIN-2018 (CODE-D)SOLUTION

PART-A MATHEMATICS

1. If the curve y2 = 6x, 9x2 + by2 = 16 intersect each other at right angles, then the value of b is :

(1) 92

(2) 6 (3) 72

(4) 4

Ans. (1)Sol. 2yy' = 6

18x + 2byy' = 0m1m2 = –1

3 9x. 1y by

2

x b 127 6y

9b2

2. Let u

be a vector coplanar with the vectors ˆ ˆ ˆa 2i 3 j k

and ˆ ˆb j k

. If u is perpendicular to a

and

u . b 24 , tjhen 2u

is equal to :

(1) 84 (2) 336 (3) 315 (4) 256Ans. (2)

Sol. u a a b

= a.b a a.a b

= 2a 14b

= 2 a 7b

= ˆ ˆ ˆ2 2i 4 j 8k

u.b 24

–12 × 2 = 24= –1

ˆ ˆ ˆu 4i 8 j 16k

2u = 16 + 64 + 256

2u = 336


Vibrant A

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3. For each t R, let [t] be the greatest integer less than or equal to t. Then x 0

1 2 15lim x ........x x x

(1) does not exist (in R) (2) is equal to 0(3) is equal to 15 (4) is equal to 120

Ans. (4)

Sol.x 0

1 1 2 2 15 15lim x ....x x x x x x

x 0

finite

1 2 15120 lim x ....x x x

= 120 – 0 = 120.

4. If L1 is the line of intersection of the planes 2x – 2y + 3z – 2 = 0, x – y + z + 1 = 0 and L2 is the line ofintersection of the planes x + 2y – z – 3 = 0, 3x – y + 2z – 1 = 0, then the distance of the origin from the plane,containing the lines L1 and L2 is :

(1) 12

(2) 1

4 2(3)

13 2

(4) 1

2 2

Ans. (3)Sol. 2x – 2y + 3z – 2 +(x – y + z + 1) = 0

x + 2y – z – 3 + (3x – y + 2z – 1) = 0

2 2 3 21 3 2 1 2 3

1 5

1 2 3

5 + 5 = 2 + 3

32

2(x + 2y – z – 3) – 3(3x – y + 2z – 1) = 0–7x + 7y – 8z – 3 = 0

Distance = 3 3 1

162 9 2 3 2


Vibrant A

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5. The value of 22

x

2

sin x dx1 2

is :

(1) 4

(2) 8

(3) 2

(4) 4

Ans. (1)

Sol./2

2

/2

2I sin xdx

As sin2x is even function so,

2/

0

2 dxxsin22

/22

0

I sin x dx4

6. Let g(x) = cos x2, f(x) x , and , ( < ) be the roots of the quadratic equation 18x2 – 9x + 2 = 0. Then

the area (in sq. uinits) bounded by the curve y = (gof) (x) and the lines x = , x = and y = 0, is :

(1) 1 2 12

(2) 1 3 12

(3) 1 3 12

(4) 1 3 22

Ans. (2)

Sol. g(x) = cosx2 , f(x) = x

y = g(f(x)) = cosx18x2 – 9x + 2 = 018x2 – 6x – 3x + 2 = 0(6x – ) (3x – ) = 0

x = 6

, 3

= 6

, = 3

Req. Area = 21–3

21–

23]x[sindx.xcos 3/

6/

3/

6/


Vibrant A

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7. If sum of all the solutions of the equation 18cos x. cos x .cos x 16 6 2

in [0, ] is k, then k is

equal to :

(1) 209 (2)

23

(3) 139 (4)

89

Ans. (3)

Sol. 8cosx 121–x–

6cosx

6cos

8cosx

21–

6sin–xcos 22

8cosx

43–xcos2

= 1

8cos3x – 6cosx = 1

cos3x = 21

3x = 3

, 3–2

, 32

97,

95,

9x

sum = 913

8. Let 22

1f(x) xx

and 1g(x) xx

, x R – {–1, 0, 1}. If f(x)h(x)g(x)

, then the local minimum value of h(x)

is :

(1) 2 2 (2) 3

(3) –3 (4) 2 2

Ans. (1)


Vibrant A

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Sol. F(x) = x2 + 2x1

, g(x) = x – x1

h(x) = )x(g)x(F

h(x) =

x1–x

x1x 2

2

f() = 22

f() =

2

F() =

22–

+ 22

Minimum value will be 22

at – 2

= 0

= 2 which is possible as x – x1

R

9. The integral 2 2

5 3 2 3 2 5 2

sin xcos x dx(sin x cos x sin x sin xcos x cos x) is equal to :

(1) 3

1 C1 cot x

(2) 3

1 C3(1 tan x)

(3) 3

1 C3(1 tan x)

(4) 3

1 C1 cot x

(where C is a constant of integration)Ans. (3)

Sol. dx

)xcosxcos.xsinxsinxcosx(sinxcos.xsin

2523235

22

22233

22

))xcosx)(sinxcosx((sinxcos.xsin

dx

dx

)1x(tanxsec.xtan

23

22

Put tan3x + 1 = t

Ct3

1–tdt

31

2 C

)1x(tan31– 3


Vibrant A

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10. A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and thisball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at randomfrom the bag, then the probability that this drawn ball is red, is :

(1) 34

(2) 3

10 (3) 25 (4)

15

Ans. (3)Sol. Req. probability = RR + BR

= 124

106

126

104

4R6B

4R8B

6R6B

BR

R R

= 51

31 = 45

11. Let the orthocentre and centroid of a triangle be A(–3, 5) and B(3, 3) respectively. If C is the circumcentre ofthis triangle, then the radius of the circle having line segment AC as diameter is :

(1) 3 5

2(2) 10 (3) 2 10 (4)

532

Ans. (4)

Sol. 2 23 3AC AB 6 22 2

3 10

A B C

2 1

3 5r 10 32 2

12. If the tangent at (1, 7) to the curve x2 = y – 6 touches the circle x2 + y2 + 16x + 12y + c = 0 then the value ofc is :(1) 95 (2) 195 (3) 185 (4) 85

Ans. (1)Sol. x2 = y – 6

tangent at P(1, 7)

y 7x.1 62

2x – y + 5 = 0 (eq. of tangent)

r 64 36 c

r 100 c

Condition of tangency p = r

(–8, –6) 2 2

2( 8) ( 6) 5 100 c2 1

5 100 c

c = 95


Vibrant A

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13. If , C are the distinct roots, of the equation x2 – x + 1 = 0, then 101 + 107 is equal to :(1) 2 (2) –1 (3) 0 (4) 1

Ans. (4)Sol. Roots of x2 – x + 1 = 0 are (–, –2)

= –, = –2

101 + 107

(–)101 + (–2)107

–2 – = 1

14. PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles ofelevation of the top the tower at P, Q and R are respectively 45º, 30º and 30º then the height of the tower (inm) is :

(1) 50 2 (2) 100 (3) 50 (4) 100 3

Ans. (2)Sol. Let AB = h, QR = 2a

In ABQ, AB htan30 a h 3QB a

P

Q RB

A

In PBA, 2 2

AB htan45PB 200 a

2 2

h1200 3h

4h2 = 2002

h = 100

15. If 9 9

2i i

i 1 i 1(x 5) 9 and (x 5) 45

, then the standard deviation of the 9 items x1, x2, ..........., x9 is :

(1) 3 (2) 9 (3) 4 (4) 2Ans. (4)

Sol. Given 9)5–x(9

1ii

and 45)5–x(9

1i

2i

Variance 2

i2

i2 )5–x(

x1–)5–x(

x1

= 2

99–45

91

n = 9

2 = 5 – 1 = 4

Standard Devieration = 24


Vibrant A

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16. The sum of the co-efficients of all odd degree terms in the expansion of 5 53 3x x 1 x x 1 ,(x 1)

is :(1) 2 (2) –1 (3) 0 (4) 1

Ans. (1)

Sol. 25 5 5 3 3 5 1 30 2 4C x C x x 1 C x x 1

= 5 5 50 2 42 C C C

= 2 1 10 10

= 2

17. Tangents are drawn to the hyperbola 4x2 – y2 = 36 at the points P and Q. If these tangents intersect at thepoint T(0, 3) then the area (in sq. units) of PTQ is :

(1) 36 5 (2) 45 5 (3) 54 3 (4) 60 3

Ans. (2)

Sol.2 2x y 1

9 36

T(0,3)

QPy=-12

Equation of PQ(T=0) 0 – y 1

12

P & Q are (±3 5 , –12)

Area = 1 6 5 152

= 45 5

18. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arrangedin a row on a shelf so that the dictionary is always in the middle. The number of such arrangmenets is :(1) at least 750 but less than 1000 (2) at least 1000(3) less than 500 (4) at least 500 but less than 750

Ans. (2)Sol. 6 N1 3D

6C × 43C × 4! 1

Arrangements as DD N DD

Selecting 1 Dict.

Selecting 4 novels

= 15 . 3 . 24= 45 × 24= 1080


Vibrant A

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19. If the system of linear equationsx + ky + 3z = 03x + ky – 2z = 02x + 4y – 3z = 0

has a non-zero solution (x, y, z), then 2xzy

is equal to

(1) 30 (2) – 10 (3) 10 (4) – 30Ans. (3)Sol. For non-zero solution,

x + 11 y = – 3z .....(1)3x + 11y = 2z .....(2) –2x = – 5z ......(1) – (2)

x = 5 z2

11y = –3z –5 z2

= 11z2

y = z2

2xzy = 2

5 z.z2 10z4

20. If 2x 4 2x 2x2x x 4 2x (A Bx)(x A) ,2x 2x x 4

then the ordered pair (A, B) is equal to

(1) (4, 5) (2) (– 4, – 5) (3) (– 4, 3) (4) (– 4, 5)Ans. (4)

Sol.

x 4 2x 2x2x x 4 2x2x 2x x 4

R1 R1 + R2 + R3

5x 4 5x 4 5x 42x x 4 2x2x 2x x 4

1 1 1(5x 4) 2x x 4 2x

2x 2x x 4


Vibrant A

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C1 C1 – C2, C2 C2 – C3

0 0 0(5x 4) x 4 x 4 2x

0 x 4 x 4

= (x + 4)2(5x – 4) = (A + Bx) (x – A)2

A = –4, B = 5 (–4, 5) Ans.

21. Two sets A and B are as under :A = {(a, b) R × R} : |a – 5| < 1 and |b – 5| < 1 };B = {(a, b) R × R} : 4(a – 6)2 +9(b – 5)2 < 36 }. Then ;(1) neither A B nor B A (2) B A

(3) A B (4) A B = (an empty set)

Ans. (3)Sol. –1 < a – 5 < 1 a (4, 6), b (4, 6)

2 2(a 6) (b 5) 19 4

–3 a – 6 3 a (3, 9)–2 b –5 2 b (3, 7)

22. Tangent and normal are drawn at P(16, 16) on the parabola y2 = 16x, which intersect the axis of the parabolaat A and B, respectively. If C is the centre of the circle through the points P, A and B and CPB = , then avalue of tan is :

(1) 43 (2)

12

(3) 2 (4) 3

Ans. (3)Sol. A(–16, 0)

B(16 + 2(4), 0) = B(24, 0)AC = CB C(4, 0)C is the cimcumcentre

2 2BP 8 16 8 5 A C B

P(16,16)

2 2PC 12 16 20

4 5 1cos20 5

tan = 2


Vibrant A

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23. Let S = { t R : f(x) = |x – |. (e|x| – 1) sin|x| is not differentiable at t}. Then the set S is equal to(1) {0, } (2) (an empty set) (3) {0} (4) {}

Ans. (2)

Sol. |x|f(x) x e 1 sin | x |

f(0) = 0

h 0

f(h) f(0)f '(0) limh

h

h 0

( h) e 1 (sinh)lim 0

h

Differentiable at x = 0f() = 0

h 0

f( h) f(0)f '( ) limh

h

h 0

h e 1 sin( h)lim 0

h

Differentiable at x = S is an empty set.

24. The Boolean expression ~ (p q) (~ p q) is equivalent to

(1) ~ q (2) ~p (3) p (4) qAns. (2)Sol. ~(p q) (~p q)

(~p ~q) (~p q) Demorgan's Laws~p (~q q) Distribution Law~ p T~ p

25. A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is theorigin and the rectangle OPRQ is completed, then the locus of R is(1) 3x + 2y = 6xy (2) 3x + 2y = 6 (3) 2x + 3y = xy (4) 3x + 2y = xy

Ans. (4)

Sol.x y 1h k

P(h, 0)

(0, k) Q R (h, k)

(2,3)

O

2 3 1h k

2k + 3h = hk3x + 2y = xy


Vibrant A

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26. Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series12 + 2.22 + 32 + 2.42 + 52 + 2.62 + ........If B – 2A = 100, then is equal to(1) 496 (2) 232 (3) 248 (4) 464

Ans. (3)Sol. A = 12 + 22 + 32 + ..... + 202

22 + 42 + ..... + 202

A = 202 + 4·102

IIIg B = 402 + 4·202

B – 2A = 402 + 2·202 – 8·102

= 62111108–

64121202

6814140

= 640

(41 × 81 + 21 × 41 – 22 × 21) = 640

(4100 + 82 – 462)

= 640

(3720) = 24800

= 248

27. Let y = y(x) be the solution of the differential equation

dysin x ycos x 4x, x (0, )dx

. If y 02

, then y

6

is equal to

(1) 249

(2) 24

9 3 (3)

289 3

(4) 289

Ans. (4)

Sol. xsinx4xcoty

dxdy

Integrating factor = ecotx dx

= e+ ln |sin x|

= sin x ( x (0, ))

y·(sin x) = dx·)x(sinxsin

x4

y·(sinx) = 2x2 + c

2–c0

2y

2

6y

2–

62

6sin·y

22

2

98–y


Vibrant A

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28. The length of the projection of the line segement joining the points (5, – 1, 4) and (4, – 1, 3) on the plane,x + y + z = 7 is

(1) 23 (2)

23 (3)

23 (4)

13

Ans. (1)Sol. Foot of perpendicular of point A on plane

x 4 y ( 1) z 3 ( 1)1 1 1 3

A

B(4,–1,3)

(5, –1, 4)

A' B'

x + y + z = 7

A'(13/3, –2/3, 10/3)Similarly, B'(14/3, –4/3, 11/3)

A'B' = 2 2 2(14 / 3 13 / 3) ( 4 / 3 2 / 3) (11/ 3 10 / 3)

= 23

29. Let S = {x R : x 0 and 2 x 3 x x 6 6 0} . Then S

(1) contains exactly four elements(2) is an empty set(3) contains exactly one element(4) contains exactly two elements

Ans. (4)

Sol. 06x6–x3–x22

03–3–x3–x22

Let t3–x

t2 + 2t – 3 = 0t = 1 or –3 (Not possible)

4,2x

x = 4 or 16


Vibrant A

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30. Let a1, a2, a3, ........a49 be in A.P. such that 12

4k 1k 0

a 416

and a9 + a43 = 66. If 2 2 21 2 17a a .... a 140m ,

then m is equal to(1) 33 (2) 66 (3) 68 (4) 34

Ans. (4)Sol. a1 + a5 + ....... + a49 = 416

a1 + a1 + 4d + ....... + a1 + 48d = 416

13a1 + 4d

21312

= 416

13(a1 + 24d) = 416 a1 + 24d = 32 ......(1)

Also,a9 + a43 = 66 2a1 + 50d = 66a1 + 25d = 33 .......(2)

From (1) & (2), we getd = 1 & a1 = 8

a12 + a2

2 + ...... + a172 = 82 + 92 + ....... + 242

= 242 – 72

= 140 × 34

Vibrant A

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PART-B PHYSICS

31. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface chargedensities +, –and +respectively. The potential of shell B is :

(1) 0

a

cc–b 22

(2) 0

c

ab–a 22

(3) 0

c

bb–a 22

(4) 0

a

bc–b 22

Ans. (3)

Sol.

–

a

b

c

2 2 2 2 2

B0 0 0 0

a b c a bV cb b c b

32. Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. Themoment of inertia of the arrangement about the axis normal to the plane and passing through the point P is:

P

(1) 2MR2

181(2) 2MR

219

(3) 2MR2

55(4) 2MR

273

Ans. (1)

Sol.2 2 2

2 2MR MR MR6 M(2R) 6 4.5MR2 2 2

I

2 25527.5MR MR2

I ; 2

2 2p

55 181MRMR (7M)(3R)2 2

I

Vibrant A

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33. From a uniform circular disc of radius R and mass 9 M, a small disc of radius 3R

is removed as shown in the

figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc andpassing through centre of disc is :

2R3

R

(1) 2MR9

37(2) 4 MR2 (3) 2MR

940

(4) 10 MR2

Ans. (2)

Sol.2

2

9M RM' M9R

2R3

R

R/3

2

22RM

(9M)R 2R3 M2 2 3

I

I = 4MR2

34. The reading of the ammeter for a silicon diode in the given circuit is :

20

3V(1) 13.5 mA (2) 0 (3) 15 mA (4) 11.5 mA

Ans. (4)

Sol.3 0.7 2.3 11.5mA

200 200

I

Vibrant A

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35. Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed

behind A. The intensity of light beyond B is found to be 2I

. Now another identical polarizer C is placed

between A and B. The intensity beyond B is now found to be 8I

. The angle between polarizer A and C is :

(1) 60º (2) 0º (3) 30º (4) 45ºAns. (4)Sol. Polarizing axis of A & B polarizer are parallel to each other

21 cos

2

II

21 cos

8

II

4cos8 2

I I

1cos2

45º

36. For an RLC circuit driven with voltage of amplitude m and frequencye 01LC

the current exhibits resonance.

The quality factor, Q is given by :

(1) 0

CR (2) 0L

R

(3) 0RL

(4)

0

R( C)

Ans. (2)

Sol. Q-factor of series R-L-C circuit 0L LxR R

where xL is inductive reactance

Vibrant A

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37. Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string over a frictionless pulley, aremoving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight mthat should be put on top of m2 to stop the motion is :

m2

m1

m g1

T

Tm

(1) 10.3 kg (2) 18.3 kg (3) 27.3 kg (4) 43.3 kgAns. (3)Sol. m1g = µ (m2 + m)g

m = 23.3 kgvalue greater than this and miminum in the given options is 27.3 kg

38. In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If thefinal total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocitybetween the two particles, after collision is :

(1) 0v2

(2) 0v4

(3) 02v (4) 0v2

Ans. (3)Sol. From C.O. L.M. v0

m

v1 v2

mv0 = mv1 + mv2 ...(1)

and f i3K.E K.E.2

2 2 21 2 0

1 1 3 1mv mv mv2 2 2 2

2 2 21 2 0

3v v v2

...(2)

Solving equation (1) and (2) we get

vrel = v2 – v1 = 02v

Vibrant A

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39. A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportionalto the nth power of R. If the period of rotation of the particle is T, then :

(1) T Rn/2 (2) T R3/2 for any n (3) n 12T R

(4) T R(n+1)/2

Ans. (4)Sol. For circular motion

2

n

mv cR R

n 1

cvmR

Time period 2 RTv

n 12T R

40. Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10. The internalresistances of the two batteries are 1and 2 respectively. The voltage across the load lies between :(1) 11.7 V and 11.8 V (2) 11.6 V and 11.7 V (3) 11.5 V and 11.6 V (4) 11.4 V and 11.5 V

Ans. (3)

Sol. eq

12 13371 2E

1 1 31 2

R

12V

10

2

1

13V

Ieqeq

1 1 1 3 2rr 1 2 2 3

37373

2 32103

I

voltage across the load = 37 370(10) 11.56 volt32 32

Vibrant A

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41. In an a.c. circuit, the instantaneous e.m.f. and current are given bye = 100 sin30t

i = 20 sin 30t –4

.

In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively :

(1) 50, 0 (2) 50, 10 (3) 1000

2,10 (4)

502

,0

Ans. (3)Sol. e = 100 sin (30t)

i 20sin 30t4

rms rms100 20e volt, i amp

2 2

Power factor 1cos cos

4 2

< P > = erms irms cos

100 20 12 2 2

1000 watt2

wattless current = irms sin

20 1 10amp2 2

Vibrant A

cademy


42. An EM wave from air enters a medium. The electric fields are

1 01zÊ E xcos 2 – tc

in air and 2 02 Ê E xcos

[k(2 z – ct)] in medium, where the wave number k and

frequency refer to their values in air. The medium is non-magnetic. If 1 2r rand refer to relative permittivities

of air and medium respectively, which of the following options is correct?

(1) 1

2

r

r

12

(2) 1

2

r

r

= 4 (3) 1

2

r

r

= 2 (4) 1

2

r

r

14

Ans. (4)

Sol. 01z Ê E cos 2 t x in airc

2kc

speed = c

2 02 Ê E cos k(2z ct) x

2 022 Ê E cos (2z ct) xc

speed c2

10 r

1c

20 r

c 12

2 2

1 1

r r

r r

2 4

1

2

r

r

14

Vibrant A

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43. A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized fortransmission. How many telephonic channels can be transmitted simultaneously if each channel requires abandwidth of 5 kHz?(1) 2 × 106 (2) 2 × 103 (3) 2 × 104 (4) 2 × 105

Ans. (4)Sol. Carrier frequency = 10 × 109 Hz

10% is utilised 910 10 10100

= 109 Hzband width = 5 k Hz

no. of channel required 9

3

105 10

= 2 × 105

44. A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The densityof granite is 2.7 × 103 kg/m3 and its Young’s modulus is 9.27 × 1010 Pa. What will be the fundamentalfrequency of the longitudinal vibrations?(1) 7.5 kHz (2) 5 kHz (3) 2.5 kHz (4) 10 kHz

Ans. (2)Sol. = 60 cm

= 2.7 × 103 kg/m3

y = 9.27 × 1010 Pa

For fundemental frequency 2 = 2

v 1 Yf f2

10

3

1 9.27 10f2 0.6 2.7 10

f 5 kHz

Vibrant A

cademy


45. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of itsenergy is pd ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc.The values of pd and pc are respectively :(1) (0, 1) (2) (89, 28) (3) (28, 89) (4) (0, 0)

Ans. (2)Sol. conservation of momentum

mv0 + 0 = mv1 + 2mv2

v0 = (v1 + 2v2) ... (i) 1H

2n

m m2m 2m

v0

rest v1 v2

2 1

0

v ve

v

(e = 1)

v2 – v1 = v0 ... (ii)v0 – 2v1 = 2v0 + v1

3v1 = –v0 01

vv

3

fractional loss of its K.E. =

22 0

0

20

v1 1mv m2 2 3

1mv2

8 0.889

0.89

Neutron colliding with carbonconservation of momentum

m m12m 12m

v0

rest v1 v2

Cmv0 = mv1 + 12mv2v1 + 12v2 = v0 ... (i)

e = 1 = 2 1

0

v vv

v2 – v1 = v0 ... (ii)v1 + 12v1 = v0 – 12v0

011v13

fractional loss in K.E. =

22

0 0

20

1 1 11mv m v2 2 13

1mv2

= 121 481169 169

0.28

Vibrant A

cademy


46. The density of a material in the shape of a cube is determined by measuring three sides of the cube and itsmass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximumerror in determining the density is :(1) 6% (2) 2.5% (3) 3.5% (4) 4.5%

Ans. (4)

Sol. 3

m

d dm d3m

d dm d100 100 3 100m

= 1.5 + 3 × 1 = 4.5 %

47. Two moles of an ideal monoatomic gas occupies a volume V at 27ºC. The gas expands adiabatically to avolume 2 V. Calculate (a) the final temperature of the gas and (b) chane in its internal energy.(1) (a) 195 K (b) 2.7 kJ (2) (a) 189 K (b) 2.7 kJ(3) (a) 195 K (b) –2.7 kJ (4) (a) 189 K (b) –2.7 kJ

Ans. (4)Sol. n = 2, T1 = 27°C = 300 K, Vi= V, Vf = 2V

In adiabatic conditionT1V1

– 1 = T2V2–1

300 × V(5/3–1) = T2(2V)5/3–1

T2 189U = nCvTas temp decreases so U is –ve

RU 2 T1

= –2.7 kJ

Vibrant A

cademy


48. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindricalcontainer. A massless piston of area a floats on the surface of the liquid, covering entire cross section ofcylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the

fractional decrement in the radius of the sphere, drr

, is :

(1) mgKa

(2) Kamg (3)

Ka3mg (4)

mg3Ka

Ans. (4)

Sol.mg / aK

VV

and 34V r

3

V dr3

V r

So, mg / aKdr3r

dr mgr 3Ka

49. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material

of dielectric constant 35K is inserted between the plates, the magnitude of the induced charge will be

(1) 0.9 nC (2) 1.2 nC (3) 0.3 nC (4) 2.4 nCAns. (2)Sol. q = CV = 3 × 10–9 ; here C = KC0

qind = 9 13 10 1

K

qind = 9 33 10 1

5

= 9 23 105

qind = 1.2 × 10–9

Vibrant A

cademy


50. The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loopis B1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre

of the loop is B2. The ratio 2

1

BB

is

(1) 21

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

Ans. (4)Sol. I(r2) = M, 2M = I (r')2

r' = 2 r

o1B

2r

I

o2B

2 2 r

I

1

2

B 2B 1

51. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let n,g be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let n be thewavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B areconstants)

(1) 2n (2) 2

nn

BA

(3) nn BA (4) 2n

2n BA

Ans. (2)Sol. De Broglie wavelength n

2

0

h h (const)nmv me

2nh

For wavelength of emitted photon

2n

hc 113.6 1 eVn

1

n 2

hc 11 units13.6 n

2

hc 11 units13.6 n

2n

BA

as n n

Vibrant A

cademy


52. The mass of a hydrogen molecule is 3.32 × 10–27 kg. If 1023 hydrogen molecules strike, per second, a fixedwall of area 2 cm2 at an angle of 45º to the normal, and rebound elastically with a speed of 103 m/s, then thepressure on the wall is nearly(1) 4.70 × 102 N/m2 (2) 2.35 × 103 N/m2 (3) 4.70 × 103 N/m2 (4) 2.35 × 102 N/m2

Ans. (2)Sol. Force = rate of change of momentum

(perpendicular to area)= n (2 mu cos )

pressure

Force n(2mucos )Area A

13 3 2

4

3.32 10 3.32 10 2.35 10 N / m1.41102

53. All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

(1) time

velocity

(2) position

velocity

(3) time

distance

(4) time

position

Ans. (3)Sol. If retardation is uniform, graphs 1, 2, 4 are consistent. Speed first decreases to zero and then increases. But

in (3) speed first increases and then decreases and speed is initially zero.

54. An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits ofradii re, rp, r respectively in a uniform magnetic field B. The relation between re, rp, r is(1) re < r < rp (2) re > rp = r (3) re < rp = r (4) re < rp < r

Ans. (3)

Sol.mv 2mkrBq Bq

pp

e e

mrr m mp > me

rp > re

pp p

p p

mr 2q1

r q 4m

rp = r

Vibrant A

cademy


55. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. Theresistance of their series combination is 1 k. How much was the resistance on the left slot before interchangingthe resistances?(1) 910 (2) 990 (3) 505 (4) 550

Ans. (4)Sol. x + y = 1000

xy 100

G

100–

x y

G

110––10

xy

y 10x 110

( 10)(100 )(110 )

= 1

2 – 10 = 11000 – 100 – 110 + 2

200= 11000 = 55

x 55y 45

x = 550

56. In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminalsof the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of5 , a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance ofthe cell.(1) 2.5 (2) 1 (3) 1.5 (4) 2

Ans. (3)

Sol.

52 cm

Gr

E = k(52)E

40 cm

GrE

R = 5

E 5 K 40r 5

2r + 10 = 13 r = 1.5

Vibrant A

cademy


57. If the series limit frequency of the Lyman series is L , then the series limit frequency of the Pfund series is:

(1) L/25 (2) 25 L (3) 16 L (4) L/16Ans. (1)

Sol. 2L

1 1h 13.6z1

2 1 1h ' 13.6z25

L'25

58. The angular width of the central maximum in a single slit diffraction pattern is 60º. The width of the slit is 1m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it,Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringewidth is 1 cm, what is slit separation distance ?(i.e. distance between the centres of each slit.)(1) 100 m (2) 25 m (3) 50 m (4) 75 m

Ans. (2)

Sol.

30°30°

a sin =

sin30° = a

(a = 1 µm)

= a2

= 12

µm

d

D = 50cm

= 2D 1 10d

= 6 2

2

1 10 50 102 1 10

d

d = 25 µm

Vibrant A

cademy


59. A particle is moving in a circular path of radius a under the action of an attractive potential U = – 2

k2r

. Its total

energy is :

(1) – 2

3 k2 a

(2) – 2

k4a

(3) 2

k2a

(4) Zero

Ans. (4)

Sol. U = 2

k2r

F = 3du kdr r

This force is providing centripetal acceleration.

2

3mv k

a a ( r = a)

mv2 = k/a2

22

1 kmv2 2a

T.E. = 2 2

k k 02a 2a

60. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec.What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 andAvagadro number = 6.02 × 1023 gm mole–1)(1) 5.5 N/m (2) 6.4 N/m (3) 7.1 N/m (4) 2.2 N/m

Ans. (3)Sol. kx = m2x

k = m2

= 23

1086.02 10

× (2× 1012)2 × 10–3

k 7.1N / m


Vibrant A

cademy

PART-C CHEMISTRY61. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point ?

(1) [Co(H2O)3Cl3].3H2O (2) [Co(H2O)6]Cl3 (3) [Co(H2O)5Cl]Cl2.H2O (4) [Co(H2O)4Cl2]Cl.2H2OAns. (1)Sol. Tf = ikfm

Tfsolvent – Tfsolution

= ikfmfor highest freezing point Van't hoff factor (i) should be lowest.

62. Hydrogen peroxide oxidises [Fe(CN)6]4– to [Fe(CN)6]

3– in acidic medium but reduces [Fe(CN)6]3– to [Fe(CN)6]

4–

in alkaline medium. The other products formed are, respectively :(1) H2O and (H2O + OH–) (2) (H2O + O2) and H2O(3) (H2O + O2) and (H2O + OH–

) (4) H2O and (H2O + O2)Ans. (4)Sol. Reaction in acidic medium :

H2O2 + 2H+ + 2e– 2H2OH2O2 oxidises [Fe(CN)6]4– to [Fe(CN)6]3– and get reduced to H2O.Reaction in basic medium :H2O2 + 2OH– O2 + 2H2O + 2e–

H2O2 reduces [Fe(CN)6]3– to [Fe(CN)6]4– in alkaline medium and oxidised itself to O2

63. Which of the following compounds will be suitable for Kjeldahl's method for nitrogen estimation ?

(1)

N2Cl–+

(2) N

(3)

NH2

(4)

NO2

Ans. (3)Sol. Diazonium, nitro and ring containing nitrogen do not show Kjeldahl method because it is not interconvert into

ammonium ion so the answer will be (3)

Ph — NH2 2 4H SO (NH4)2SO4 Kjeldahl method

64. Glucose on prolonged heating with HI gives :(1) 6-iodohexanal (2) n-Hexane (3) 1-Hexene (4) Hexanoic acid

Ans. (2)

Sol.

CHO

CH OH2

OH

OHOH

HHO

HH

HI

n-Hexane

H (Reduction H H )


Vibrant A

cademy

65. An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correctcombination ? Base Acid End point(1) Strong Strong Pink to colourless(2) Weak Strong Colourless to pink(3) Strong Strong Pinkish red to yellow(4) Weak Strong Yellow to pinkish red

Ans. (4)Sol. Acid added in alkali so color of methyl orange changes from yellow to pinkish red.

Working pH range of MeOH 3.1 to 4.3

66. The predominant form of histamine present in human blood is (pKa, Histidine = 6.0)

(1) N

NH

NH3 (2) N

NH

NH2

(3) N

NH

H

NH3

(4) N

NH

HNH2

Ans. (1)

Sol. N

NH

NH3

5.74

9.8H

OH N

NH

NH3

+H O2

Isoelectronic point = 77.72

8.974.5

Histamine has two basic center which are Aliphatic amino group and imidazole ring. In blood, the aliphaticamino group (pKa = 9.4) will be exist whereas the second nitrogen of imidazole ring (pKa = 5.8) will not beprotonated in blood which pH is 7.4.

pH = [Salt]pKa log[Acid]

7.4 =

It should be high for balancing

[Salt]7.77 log[Acid]


Vibrant A

cademy

67. The increasing order of basicity of the following compounds is :

(a) NH2 (b) NH (c) NH2

NH(d) NHCH3

(1) (d) < (b) < (a) < (c) (2) (a) < (b) < (c) < (d) (3) (b) < (a) < (c) < (d) (4) (b) < (a) < (d) < (c)Ans. (4)

Sol.NH2

NH > NHCH3

> NH2 > NH

Amidine 2° Amine 1° Amine Imine

68. Which of the following lines correctly show the temperature dependence of equilibrium constant, K, for anexothermic reaction?

ln K A

(0,0)1

T(K)...........................

B

DC

(1) A and D (2) A and B (3) B and C (4) C and DAns. (2)

Sol. lnK = lnA – H

RT

,

For exothermic Reaction

H = –ve slope = H

R = +ve.

69. How long (approximate) should water be electrolysed by passing through 100 amperes current so that theoxygen released can completely burn 27.66 g of diborane? (Atomic weight of B = 10.8 u)(1) 1.6 hours (2) 6.4 hours (3) 0.8 hours (4) 3.2 hours

Ans. (4)Sol. B2H6 + 3O2(g) B2O3 + 3H2O ()

moles = 27.66

127.6


Vibrant A

cademy

1 mole of B2H6 required 3 moles of O2

2H2O () O2(g) + 4H+(aq) + 4e–

It96500

= mole of O2 × nf

100 t 360096500

= 3 × 4 t = 3.2 hrs.

70. Consider the following reaction and statements :[Co(NH3)4Br2]

+ + Br– [Co(NH3)3Br3] + NH3(I) Two isomers are produced if the reactant complex ion is a cis-isomer(II) Two isomers are produced if the reactant complex ion is a trans-isomer.(III) Only one isomer is produced if the reactant complex ion is a trans-isomer.(IV) Only one isomer is produced if the reactant complex ion is a cis-isomer.(1) (II) and (IV) (2) (I) and (II) (3) (I) and (III) (4) (III) and (IV)

Ans. (3)

Sol.Co

Br NH (b)3

NH3(d)Br

NH (a)3

NH (c)3

Cis isomer

+ Br–

+

If we replace (a) & (c)

Cis or Fac isomer

CoBr NH3

NH3Br

Br

NH3

or

Cis or Fac isomer

CoBr NH3

NH3Br

NH3

Br

CoBr NH (b)3

NH3(d)Br

NH (a)3

NH (c)3

Cis isomer

+ Br–

+

If we replace (b) & (d)Co

Br Br

NH3Br

NH3

NH3

Trans or Mer-isomer

CoBr NH3

BrBr

NH3

NH3

Trans or Mer-isomer

or

Trans-isomer

CoH N3 Br

NH3Br

NH3

NH3

+ Br–

CoH N3 Br

NH3Br

Br

NH3

or CoH N3 Br

BrBr

NH3

NH3

Co CoH N3 BrBr Br

NH3 NH3Br Br

NH3 NH3

Br NH3

or or

All are same (Trans isomers)


Vibrant A

cademy

71. Phenol reacts methyl chloroformate in the presence of NaOH to form product A. A reacts with BNr52 to formproduct B. A and are respectively :

(1)

O

OH

OCH3 and

O

OH

OCH3

Br

(2)

O

OH

OCH3 and

O

OH

OCH3

Br

(3)

O O

O and

O O

O

Br

(4)

O O

O and

O O

OBr

Ans. (4)

Sol.

OH

NaOH

OCl–C–OMe

O O–C–OMe

O

Br–Br

O–C–OMe

O

Br

Electrophilic aromatic substitution


Vibrant A

cademy

72. An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of Na2SO4

is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is1 × 10–10. What is the original concentration of Ba2+ ?(1) 1.0 × 10–10 M (2) 5 × 10–9 M (3) 2 × 10–9 M (4) 1.1 × 10–9 M

Ans. (4)

Sol. BaSO4(s) Ba2+(aq) + SO42–(aq)

Na2SO4(aq) 2Na2+(aq) + SO42–

(aq)

1 50500

= 0.1M

Ksp = [Ba2+]final [SO42–]final

1 × 10–10 = 2initial

450[Ba ]500

× [0.1]

[Ba2+] = 1.1 × 10–9 M

73. At 518°C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr,was 1.00 Torr s–1 when 5% had reacted and 0.5 Torr s–1 when 33% had reacted. The order of the reaction is:(1) 0 (2) 2 (3) 3 (4) 1

Ans. (2)

Sol. rate = – dPdt

= 3

x

CH CHOk P

x

1x

2

363 95r 1 100r 0.5 67363

100

2 = (1.41)x 2 = x( 2) x 2

74. The combustion of benzene (l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constantvolume is –3263.9 kJ mol–1 at 25°C; heat of combustion (in kJ mol–1) of benzene at constant pressure will be:(R = 8.314 JK–1 mol–1)(1) –3267.6 (2) 4152.6 (3) –452.46 (4) 3260

Ans. (1)

Sol. C6H6() + 152

O2(g) 6CO2(g) + 3H2O()

ng = 6 – 7.5 = –1.5H = U + ngRT

H = – 3263.9 – 1.5 8.314 298

1000

= –3267.6 kJ mol–1


Vibrant A

cademy

75. The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6 : 1. If one molecule of the abovecompound (CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHYcompletely to CO2 and H2O. The empirical formula of compound CXHYOZ is :(1) C2H4O3 (2) C3H6O3 (3) C2H4O (4) C3H4O2

Ans. (1)

Sol. Mole ratio of C : H 6 1:

12 11 : 2

CxHy + yx4

O2 xCO2 + y2

H2O

according to question 2z = 2yx4

.........(1)

x 1y 2 ......(2)

from (1) & (2)

3xz2

CxHyOz CxH2x 3x2

O C2H4O3

76. The trans-alkenes are formed by the reduction of alkynes with :(1) Sn -HCl (2) H2-Pd/C, BaSO4 (3) NaBH4 (4) Na/Liq. NH3

Ans. (4)

Sol.

Na/NH3Birch reduction

H

H

.

. .

e

Na Na + 1e + –

H

.

NH3Na

H

..

NH3


Vibrant A

cademy

77. Which of the following are Lewis acids ?(1) BCl3 and AlCl3 (2) PH3 and BCl3 (3) AlCl3 and SiCl4 (4) PH3 and SiCl4

Ans. (1) and (3)Sol. BCl3 and AlCl3 both are hypovalent compounds due to incomplete octet on central atom. Thus both can act

as lewis acid.SiCl4 can also act as lewis acid due to presence of vacant 3d-orbitals on silicon atom in its vacant d-orbital.Option (1) and (3) both are correct.

78. When metal 'M' is treated with NaOH, a white gelatinous precipitate 'X' is obtained, which is soluble inexcess of NaOH. Compound 'X' when heated strongly gives an oxide which is used in chromatography is anadsorbent. The metal 'M' is :(1) Fe (2) Zn (3) Ca (4) Al

Ans. (4)Sol. According to given question

Al(s) + NaOH(aq) Al(OH)3(X)

Whitegelatinous ppt

Excess NaOH Na[Al(OH) ]4

Soluble

Compound (X) Al(OH)3 on heating gives Al2O3 which is used in chromatography as an adsorbent. Thusmetal M is aluminium (Al).

79. According to molecular orbital theory, which of the following will not be a viable molecule ?(1) H2

2– (2) He22+ (3) He2

+ (4) H2–

Ans. (1)

Sol. Since bond order of 2–2H is zero, thus it does not exist (by MOT).

80. The major product formed in the following reaction is :

O

O HIHeat

(1) OH

I(2)

OH

OH(3)

I

I(4)

OH

I

Ans. (1)


Vibrant A

cademy

Sol.

O

O HIHeat

O

O

H

IS 2N

OH

O

H

OH

O

H

OH

I

OHS 1N

I

C H2 5I +

81. Phenol on treatment with CO2 in the presence of NaOH followed by acidification produces compound X asthe major product. X on treatment with (CH3CO)2O in the presence of catalytic amount of H2SO4 produces

(1)

O

O

CO H2

CO H2

CH3

(2)

O

O

CO H2

CH3

(3)

O

O CH3

CO H2

(4)

O CH3CO

OOH

Ans. (2)

Sol.NaOHCO2

OH OC

ONa

O

H

OHCOOH

(CH CO) O/H3 2

O–C–CH3

COOH

O

Asprine

SNAE


Vibrant A

cademy

82. Which of the following compunds contain(s) no covalent bond(s) ?KCl, PH3, O2, B2H6, H2SO4(1) KCl, B2H6 (2) KCl, B2H6, PH3 (3) KCl, H2SO4 (4) KCl

Ans. (4)Sol. KCl is an ionic compound while PH3, O2, B2H6, H2SO4 contains covalent bonds in them.

83. Which type of 'defect' has the presence of cations in the interstitial sites ?(1) Metal deficiency defect (2) Schottky defect(3) Vacancy defect (4) Frenkel defect

Ans. (4)Sol. In Frenkel defect cations are present in interstitial sites.

84. The major product of the following reaction is :

BrNaOMe

MeOH

(1)

OMe

(2)

OMe

(3) (4)

Ans. (3)

Sol.

Br

OCH3H

E2

85. The compound that does not produce nitrogen gas by the thermal decomposition is(1) (NH4)2SO4 (2) Ba(N3)2 (3) (NH4)2Cr2O7 (4) NH4NO2

Ans. (1)

Sol. (NH4)2SO4 NH3 + H2SO4

Ba(N3)2 Ba + 3N2

(NH4)2Cr2O7 N2 + Cr2O3 + 4H2O

(Green)

NH4NO2 N2 + 2H2O


Vibrant A

cademy

86. An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constants for the formation ofHS– from H2S is 1.0×10–7 and that of S2– from HS– ions is 1.2 × 10–13 then the concentration of S2– ions inaqueous solution is(1) 5 × 10–19 (2) 5 × 10–8 (3) 3 × 10–20 (4) 6 × 10–21

Ans. (3)Sol. HCl(aq) H+(aq) + Cl–(aq)

0.2 M

H2S(aq) H+(aq) + HS– (aq), k1 = 1 × 10–7

HS– (aq) H+(aq) + S–2 (aq), k2 = 1.2 × 10–13

H2S(aq) 2H+(aq) + S–2(aq) Keq = k1 × k2

0.1M

Keq. = 1.2 × 10–20 = 2 2[0.2] [S ]

0.1

[S–2] = 3 × 10–20 M

87. The oxidation states of Cr in [Cr(H2O)6]Cl3, [Cr(C6H6)2], and K2[Cr(CN)2(O)2(O2)(NH3)] respectively are :(1) +3, 0, and +4 (2) +3, +4, and +6 (3) +3, +2, and +4 (4) +3, 0, and +6

Ans. (4)Sol. [Cr(H2O)6]Cl3 Let x be oxidation state of chromium

x + 6(0) = +3x = +3

[Cr(C6H6)2]x + 2(0) = 0x = 0

k2[Cr(CN)2(O)2(O2)(NH3)]x – 2 + 2(–2) + 1(–2) + 0 = –2x – 2 – 4 – 2 = –2x – 6 = 0x = +6

88. The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is required tomake teeth enamel harder by converting [3Ca3(PO4)2

.Ca(OH)2] to :(1) [3{Ca(OH)2}.CaF2] (2) [CaF2] (3) [3(CaF2).Ca(OH)2] (4) [3Ca3(PO4)2

.CaF2]Ans. (4)Sol. The F– ion makes the enamel on teeth much harder by converting hydroxyapatite [3Ca3(PO4)2.Ca(OH)2] into

much harder fluorapatite [3Ca3(PO4)2.CaF2].


Vibrant A

cademy

89. Which of the following salts is the most basic in aqueous solution?(1) Pb(CH3COO)2 (2) Al(CN)3 (3) CH3COOK (4) FeCl3

Ans. (3)Sol. CH3COOK + H2O CH3COOH + KOH (salt of weak acid and strong base)

Option (3) is most basic.

90. Total number of line pair of electrons in 3– ion is

(1) 12 (2) 3 (3) 6 (4) 9Ans. (4)

Sol.

..

..I ..

: :I..

: :I..

–

Total 9 lone pair.

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