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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 1

JEE– Mains {12th Mega Test – II}

Examination Centre: Pioneer Education, Sector – 40-D General Instructions:-

The question paper contains 90 objective multiple choice questions.

There are three Sections in the question paper consisting of Section – I

MATHS (1 to 30), Section – II Physics (31 to 60) and Section – III

Chemistry (61 to 90).

Each right answer carries (4 marks) and wrong (–1marks)

The maximum marks are 360.

Maximum Time is 3Hrs.

Give your response in the Answer Sheet provided with the Question Paper.

Name: _____________________________________________________________________

Mobile:________________________ ROLL NUMBER:________________________



Section – I {Mathematics}

1. The range of f(x) = 15 x 20 3x2x 1 4x 5C C

is

(a) [1300, 4224] (b) {650, 1122} (c) {0, 4224} (d) None of these

Sol : (b)

f(x) to be defined

(i) 15 – x > 0 x < 15

(ii) 2x – 1 0 1

x2

(iii) 20 – 3x > 0 20 2

x 63 3

(iv) 5

4x 5 0 x4

(v) 16 1

15 x 2x 1 x 53 3

(vi) 25 4

20 3x 4x 5 x 37 7

(vii) 15 – x is an integer x must be an integer.

5 25

x and x N or I4 7

1 41 x 3

4 7

x = 2, 3

Now 15 x 20 3x2x 1 4x 5f x C C

13 143 3f 2 C C 650

12 115 7f 3 C C 1122

Rf x {f 2 , f 3 } = {650, 1122}

2. The Domain of 1 2sin [2x 5] , where [.] represent greatest integer function is

(a) 7

, 22

(b) 7

2,2

(c) 7 7

, 2 2,2 2

(d) None of these.



Sol : (d)

Fact. 1sin t2 2

21 t 1, t [2x 5]

21 [2x 5] 1

21 2x 5 2

2 72 x

2

72 x

2

x 2 and 7

x2

[ 2 1.414 (Approx.)]

7

2 = 1.8 (Approx.)

7 7

x 2 x 2 and x x2 2

3. The inverse of the function x x

x x

e 2ef x 1

e 2e

is

(a) 10

2xlog

2 x (b) 10

xlog

2 x

(c)

1/2

e

2xlog

2 x

(d)

1/22x

ln2 x

Sol : (c, d)

x x

x x

e 2ey 1

e 2e

2x

2x

y 1 e 2

1 e 2

Using componendo and dividendo rule

2xy 2e

y 2 4

2xe

2y 2ye 2x log

2 y 2 y

1 2y

x log2 2 y



1/2

1e

2yf y log

2 y

1/2 1/2

1e e

2x 2xf x log log

2 x 2 x

= 1/2

2xln

2 x

4. Let sin x

2

0

f x t dt then period of f’(x) is

(a) (b) 2 (c) 2

(d)

3

2

Sol : (b)

sin x

2

0

f x t dt

d

f ' xdx

(sin x) (sin x)2 –0

f’(x) = sin2x cos x = p(x) q(x)

Now period of sin2x is as

2 1 cos 2x 1 1sin x

2 2 2

cos 2x

period of sin2x is same as period of

1 1

2 2 cos 2x which is

2

2

=

period of p(x) is .

Again period of cos x i.e. period of q(x)

q(x) = cos (x)

q x cosx

q x cosx

q x 2 cosx

Period of cos x = 2

Now f’(x) = p(x) q(x)

Period of f’(x) will be the L. C. M. of the period of p(x) and q(x)

L. C. M. of , 2 is 2 .



5. If 5 3f x dx x ,then x f x dx is equal to [JEE – Main 2013]

(a) 3 3 2 31[x x x x dx] C

3 (b) 3 3 3 31

x x 3 x x dx C3

(c) 3 3 2 31x x x x dx C

3 (d) 3 3 3 31

[x x x x dx] C3

Sol : (c)

Given, f x dx x

Let 5 3I x f x dx

Put 3 2 dtx t x dx

3 ..(i)

1

I t f t dt3

= 1

[t t t dt]3

= 3 3 2 31[x x 3 x x dx] C

3 [ from Eq. (i)]

= 3 3 2 31x x x x dx C

3

6. The function f(x) = x

3 5t

2

t e 1 t 1 t 2 t 3 dt

has a local minima at x equals

(a) 0 (b) 2 (c) 3 (d) 1

Sol : (c, d)

x

3 5t

2

f x t e 1 t 1 t 2 t 3 dt

Differentiating with respect to x by using Leibnitiz rule.

tdf x x [t e 1 t 1

dx

3 5

x tt 2 t 3 ]

= 3 5xx e 1 x 1 x 2 x 3



f x 0 x 0 , 1, 2, 3 are set of point at which f(x) can attains max/minimum values.

Now f’(3+) > 0 & f’(3–) < 0 as well as f’(1) also changes its sign from –ve to +ve in the immediate nbd

(neighbourhood) of x = 1.

so x = 1, x = 3 are point at which f(x) attains minimum value.

7.

2y x2

a /2

sintcost dt dt

t

then dy

dx equals

(a) 2

2

2 sin x

x cos y (b)

2

2

2 sin x

x cos y (c)

2

2

2 cosy

yx 1 2 sin

2

(d) None of these

Sol : (a)

Given

2y x2

a /2

sintcost dt dt

t

(on differentiating w. r. t. x by using Leibnitz theorem)

2

2

2

d 2x sinxcos y y

dx x

2 2

2 2 2

dy 2x sinx dy 2 sin x

dx x cos y dx x cos y

8. 2x x x

x 0

(3 3) 9 3 1Lt

2 1 cos x

equals

(a) 8(log 3)2 (b) 2

4 2 log 3 (c) 2

8 2 log 3 (d) None of these

Sol : (d)

9. 1/m

7m 2m m 6m mx x x 2x 7x 14 dx equals

(a)

m 17m 2m mx x x

m 1

(b)

m 1

7m 2m m m1

2x 7x 14x14 m 1

(c) m 1

7m 2m m m2x 7x x

(d) None of these

Sol : (b)

1/m

7m 1 2m 1 m 1 6m mx x x x 2x 7x 14 dx

= 1/m

7m 1 2m 1 m 1 7m 2m mx x x 2x 7x 14x dx 1/m

mx x



= 1/m 7m 2m m1

f ' t f t dt f t 2x 7x 14x k14m

=

m 1

m1

f t k14 m 1

10. Let the function f(x) =

3 2sin xx 0

x

0 x 0

(a) is continuous but not derivable at x = 0

(b) neither continuous nor differentiable at x = 0

(c) continuous and differentiable at x= 0

(d) none of these

Sol : (c)

2

2 2

2x 0 x 0

sin xLt f x Lt x sin x

x

= 2

2 2

2x 0 x 0

sinxLt Lt x sin x 0 f 0

x

h 0

f(h x) f(x)f '(x) lim

h

and LHD at x = 0 =

3 2

n 0h 0

sin h0f 0 h 0 hLt lim

h h

RHD = 0, similarly L f’(0) = 0

f(x) is differentiable at x = 0

11. x

x aa

xLt f x

x a dx equals

(a) 2f(a) (b) f(a) (c) af(a) (d) 0

Sol : (c)

x

x aa

xLt f x

x a dx

=

x

a

x a

x f x dx

Lt 0 / 0x a

form [Apply L’ Hospital rule]



=

x x

a a

x a

d dx f x dx f x x

dx dxLt

1

=

x

a

x a

xf x f x

Lt1

= a f(a) + 0

12. The domain of the function

24 5 3f x log log log 18x x 77 is

(a) x 4,5 (b) x 0, 10 (c) x 8,10 (d) None of these

Sol : (c)

Method I:

Since log x is defined x > 0

24 5 3f x log (log (log (18x x 77)))

25 3log (log (18x x 77)) 0

2 03log 18x x 77 5

18x – x2 – 77 > 31

(x – 8) (x – 10) < 0

8 < x < 10 x (8, 10)

Method II:

218x x 77 0

2x 18x 77 0

2x 18x 77 0

2x 11x 7x 77 0

x(x 11) 7(x 11) 0

(x 7)(x 11) 0

13. Let y be an implicit function of x defined by

2x xx 2x cot y 1 0 . Then y’(1) equals :

(a) 1 (b) log 2 (c) – log 2 (d) – 1



Sol : (d)

2x xx 2x cot y 1 0 ..(i)

at x = 1 we have

1 – 2 cot y – 1 = 0

cot y = 0 y = /2

Differentiating (i) w. r. t. x, we have

2x x 2 dy2x 1 lnx 2[x cosec y

dx + cot y. xx (1 + ln x)] = 0

At P 1, /2 we have

p

dy2 1 ln1 2[1 1 0] 0

dx

p

dy2 2 0

dx

p

dy1

dx

14. The order and degree of the differential equation

233

3

dy d y1 3 4

dx dx

is

(a) 2

1,3

(b) 3, 1 (c) 3, 3 (d) 1, 2

Sol : (c)

233

3

dy d y1 3 4

dx dx

32 3

3

dy d y1 3 4

dx dx

highest order is 3 whose exponent is also 3.

15. If the function f(x) = 2x3 – 9ax2 + 12a2x + 1, a > 0 attains its maximum and minimum value at p and q

respectively such that p2 = q, then ‘a’ equals

(a) 1 (b) 2 (c) 1

2 (d) 3

Sol : (b)

For maximum and minima f’(x) = 0.

6x2 – 18ax + 12a2 = 0 and f’’(x) = 12x – 18a; f’(x) = 0.

x = a, 2a and f’’(a) < 0 and f’’(2a) > 0.



p = a (max)

q = 2a (minimum)

Now p = a and q = 2a. and p2 = q

a2 = 2a

a = 0, a = 2

16. A function y = f(x) has a second order derivative f’’(x) = 6(x – 1). If its graph passes through the point

(2, 1) and at that point the tangent to the curve is

y = 3x – 5, then the function is

(a) (x + 1)3 (b) (x – 1)3 (c) (x – 1)2 (d) (x + 1)2

Sol : (b)

Given f’’(x) = 6(x – 1)

26 x 1

f ' x c2

3 = 3 + c

f x y 3x 5

dym f ' x 3 x R

dx

c = 0

so f’(x) = 3(x – 1)2

f(x) = (x – 1)3 + c1 as curve passes through (2, 1)

1 = (2 – 1)3 + c1

c1 = 0

3

f x x 1

17. The normal to the curve x = a 1 cos ,y a sin at always passes through the fixed point

(a) (0, 0) (b) (0, a) (c) (a, 0) (d) (a, a)

Sol : (c)

The equation of normal at is

1 1

1y y x x

dy

dx

sin

y a sin x a 1 coscos

which passes through (a, 0)



18. A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a

rate of 50 cm3/min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice

decreases, is

(a) 1

18cm/min (b)

1

36cm/min (c)

5

6cm/min (d)

1cm / min.

54

Sol : (a)

34

v y 103

where y is thickness of ice

2dv dy

4 10 ydt dt

2

at y 5

dy 50

dt 4 15

3dv

as 50 cm / min.dt

= 1

18 cm/min.

19. If sin x

dx Ax Bsin x

log sin x + C then value of (A, B) is

(a) sin , cos (b) cos , sin (c) sin , cos (d) cos , sin

Sol : (b)

Ist – Method

sin(x )

sin(x )

sin(x )cos cos(x )sin

dxsin(x )

xcos sin log sin(x ) C

(A,B) (cos , sin )

II nd – Method

sin xdx Ax B log sin x C

sin x

Differentiating w. r. to x both sides



B cos xsin xA

sin x sin x

sin x = A sin x B cos x

sin x = A (sin x cos – cos x sin ) + B (cos x cos + sin x sin )

sin x = sin x (A sin + B sin ) + cos x (B cos – A sin )

Now solving A cos + B sin = 1 and B cos – A sin = 0

(A, B) = (cos , sin )

20.

2

2

log x 1

1 log x

dx is equal to

(a) 2

xC

x 1

(b)

2

logxC

log x 1

(c)

2

xC

logx 1

(d)

x

2

xeC

1 x

.

Sol : (c)

Method by cross check

Consider f(x) =

2

x

logx 1

2

22

2x log x1 log x .

xf ' x1 log x

22 2

2 22

1 log x 2 log x 1 log xf ' x

1 (log x)1 log x

2

2

1 log xdx f ' x dx f x

1 log x

2

2 2

1 log x xdx

1 log x 1 log x

Hence (c) is correct answer and we can check the other choices by the similar argument.

Alternative Method:

2

22

log x 1dx.

1 log x

Let log x = z ez dz = dx



2

z

2 22 2

log x 1 1 2zdx e dz

1 z1 log x 1 z

ze f z f ' z dz where 2

1f z

1 z

= z z

2

1e f z e C

1 z

=

log x

2 2

e xC

1 log x 1 log x

.

21. Suppose cubic equation x3 – px + q = 0 has three distinct & real roots (p, q > 0) then which of the

following is true for f(x) = x3 – px + q

(a) maximum at p/3 & minimum at p/3

(b) minimum at both p/3 and p/3

(c) maximum at both p/3 and p/3

(d) minimum at p/3 and maximum at p/3

Sol : (a, d)

f(x) = x3 – px + q = 0 (p, q > 0)

y’ = f’(x) = 3x2 – p, f’’(x) = 3(2x)

for maximum & minimum values f’(x) = 0

p p

x , x3 3

Now 2

2p

x3

d y pf '' x y'' 6 0 as p 0

dx 3

p

f3

is minimum at x = p

3

Similarly p p

f " 63 3

< 0 (as p > 0)

p

at x3

, f(x) is maximum.



22. How many real solutions does the equation x7 + 14x5 + 16x3 + 30x – 560 = 0 have?

(a) 5 (b) 7 (c) 1 (d) 3

Sol: (c)

Let f(x) = x7 + 14x5 + 16x3 + 30x – 560

f’(x) = 7x6 + 70x4 + 48x2 + 30 f ' x 0 x R

i.e. f(x) is an strictly increasing function.

So it can have at the most one solution. It can be shown that it has exactly one solution.

23. The two curves x3 – 3xy2+ 2 = 0 and 3x2y – y3 – 2 = 0 [AIEEE – 2002]

(a) cut at right angle (b) touch each other

(c) cut at an angle 3

(d) cut at an angle

4

Sol :

The equations of two curves are

x3 – 3xy2 + 2 = 0 ..(i)

and 3x2y – y3 – 2 = 0 ..(ii)

on differentiating Eqs. (i) and (ii) w. r. t. x, we get

1

2 2

c

dy x y

dx 2xy

and 2

2 2c

dy 2xy

dx x y

Now,

1 2

2 2

2 2c c

dy dy x y 2xy

dx dx 2xy x y

= – 1

Hence, the two curves cut at right angle.

24. If

1 3

P 1 3 3

2 4 4

is the adjoint of a 3 × 3 matrix A and A 4 , then is equal to [JEE – Main 2013]

(a) 4 (b) 11 (c) 5 (d) 0

Sol : (b)

Given, P =

1 3

1 3 3

2 4 4



P 1 12 12 4 6 3 4 6

= 2 6

P adj A (given)

2

P adj A A 16

2 6 16 2 22

11

25. A ray of light along x 3y 3 gets reflected upon reaching x-axis, the equation of the reflected ray

is [JEE – Main 2013]

(a) y x 3 (b) 3 y x 3 (c) y 3 x 3 (d) 3y x 1

Sol : (b)

Take any point B(0, 1) on given line.

Equation of AB’

1 0y 0 x 3

0 3

3y x 3

x 3 y 3

3y x 3

26. If x, y and z are in AP and tan–1x, tan–1 y and tan–1 z are also in AP then [JEE – Main 2013]

(a) x = y = z (b) 2x = 3y = 6z (c) 6x = 3y = 2z (d) 6x = 4y = 3z

Sol : (a)

Since, x, y and z are in AP.

2y = x + z

Also, tan–1x, tan–1y and tan–1z are in AP.



2 tan–1y = tan–1 x + tan–1(z)

1 1

2

2y x ztan tan

1 y 1 xz

2

2

x z x zy xz

1 y 1 xz

Since x, y and z are in AP as well as in GP.

x = y = z

27. The system of equations

x y z 1

x y z 1

x y z 1

has no solution, if is [AIEEE – 2004,5]

(a) 1 (b) not – 2 (c) either –2 or 1 (d) – 2

Sol : (d)

The system of given equations has no solution, if

1 1

1 1 0

1 1

Apply 1 1 2 3C C C C

2 1 1

2 1

2 1

= 0

Applying 2 2 1 3 3 1R R R , R R R

1 1 1

2 0 1 0 0

0 0 1

2

2 1 0

1, 2

But 1 makes given three equations same. So, the equation have infinite solution. So, answer is

2 for which the system of equations has no solution.



28. The value of sin xdx

2

sin x4

is [AIEEE – 2008]

(a) x log cos x c4

(b) x log cos x c

4

(c) x log sin x c4

(d) x log sin x c

4

Sol : (d)

Let sin x

I 2 dx

sin x4

Put x t dx dt4

sin t dt4

I 2sin t

= 1 1

2 cot t dt2 2

= log sin t t c = x + log sin x c4

29. Consider the differential equation 2 1y dx x dy 0

y

. If y(1) = 1, then x is given by

[AIEEE – 2011]

(a) y

11 e1y e

(b) y

12 e4y e

(c) y

11 e3y e

(d) 1

y1 11 e

y e

Sol : (d)

Here, 2 3

dx 1 1. x

dy y y (linear differential equation in x)

2

1 1dy

y yIF e e

Complete Solution is ,

1 1

y y

3

1x.e . e dy

y

1 1

y y

2

1 1x.e . .e dy,

y y



Put 1

ty

2

1dy dt

y

1

tyxe t.e dt

1

t tyxe {t.e 1.e dt} c

1

t tyxe te e c

1 1 1

y y y1xe .e e c,

y

as y (1) = 1

1 1 1 1e e e c c

e

1 1 1

y y y1 1xe . e e

y e

1

y1 1x 1 .e

y e

30. The solution of the differential equation dy x y

dx x

satisfying the condition y(1) = 1 is

(a) y = x log x + x (b) y = log x + x [AIEEE – 2008]

(c) y = x log x + x2 (d) y = x 1xe

Sol : (a)

Given equation can be rewritten as

dy 1. y 1

dx x

Now, 1

dxlog xx

1IF e e

x

Required solution is

1 1y dx log x c

x x

Since, y(1) = 1 1 = log 1 + c

c = 1

y = x log x + x



Section – II {Physics}

31. An oil drop of n excess electrons is held stationary under a constant electric field E in Millikan’s oil

drop experiment. The density of oil is . The radius of the drop is

(a)

1

23neE

2 g

(b)

1

23n g

2 eE

(c)

1

33neE

4 g

(d)

1

33n .g

4 eE

Sol : (c)

The weight of the oil drop is balanced by the electrostatic force.

1

334 3neE

r g ne E, So, r3 4 .g

32. In the basic CsCl crystal structure, Cs+ and Cl– ions are arranged in a bcc configuration as shown in fig.

The net electrostatic force exerted by the eight Cs+ ions on the Cl– ion is :

[AIIMS]

(a) 2

20

1 4e

4 3a (b)

2

20

1 16e

4 3a (c)

2

20

1 32e

4 3a (d) zero

Sol : (d)

Symmetry indicates the net electric field of Cs+ ions at the centre is 0.

33. Three charges are placed at the vertices of an equilateral triangle of side ‘a’ as shown in fig. The force

experienced by the charge placed at the vertex A in a direction normal to BC is [AIIMS]

(a) 2 20Q / 4 a (b) 2 2

0Q C / 4 a ) (c) zero (d) 2 20Q / 2 a



Sol : (c)

AB ACF F is such that the components normal to BC cancel off. Hence the force at A normal to BC is 0.

34. The electric field components in the figure are 1/2x y zE . x , E E 0. Calculate the charge within the

cube of side a.

(a) 5/20a (b) 5/2

0a 2 1 (c) 5/20a 3 1 (d) 5/2

0a 2

Sol : (b)

As Ey and Ez are zero, the flux for all faces except the ones normal to the x-axis is 0. Now,

net left right

= 1 1

2 0 2 02 2a a cos 180 2a a cos 0

= 5 1 5

2 2 2a 2 . a

= 5

2a 2 1



Now, INnet IN net 0

0

qor q .

5

2IN 0q a ( 2 1)

35. In the circuit shown in figure, the equivalent capacitance between the points A and B is

(a) 10

F3 (b)

15F

14 (c)

2F

5 (d)

25F

9

Sol : (a)

The circuit can be redrawn as

As 2 F 4 F

3 F 6 F

, the circuit represents a balanced Wheat stone bridge. The 15 F can be open-circuited.

The circuit becomes

eq

2 2 3 6C F F

2 4 3 6

= 8 10

F 2 F F6 3 .

36. By the superposition principle, the electric field due to a system of n charges is

(a) n

iip3

i 10 ip

q1r

4 r (b)

ni

ip2i 10 ip

q1r

4 r (c)

ni

2i 10 ip

q1

4 r (d)

ni

3i 10 ip

q1

4 r



Sol : (a)

p 1 2 nE E E ... E

n1 2 i

1 2 i3 3 3i i0 1 2 0 i

q q q1 1r p r p .. rp

4 r p r p 4 r p

37. Three charges-q1, +q2 and –q3 are placed as shown in the figure. The x-component of the force on –q1 is

proportional to [AIEEE – 2003]

(a) 322 2

qqcos

b a (b) 32

2 2

qqsin

b a (c) 32

2 2

qqcos

b a (d) 32

2 2

qqsin

b a

Sol : (b)

Force on (–q1) due to 1 22 2

0

q qq

4 b

1 21 2

0

q qF

4 b

along (q1q2)

Force on (–q1) due to (–q3) = 1 3

20

q q

4 a

1 32 2

0

q qF

4 a

as shown

F2 makes an angle of 090 with (q1/q2)

Resolved part of E2 along a1q2

= 0 1 32 1 22

0

q q sinF cos 90 along q q

4 a

Total force on 1q



= 1 31 22 2

0 0

q q sinq q

4 b 4 a

along x-axis.

x-component of force 322 2

qqsin

b a

.

38. Let 4

QP r r

R

be the charge density distribution for a solid sphere of radius R and total charge Q.

For a point ‘p’ inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric

field is [AIEEE – 2009]

(a) 0 (b) 2

0 1

Q

4 r (c)

21

40

Qr

4 R (d)

21

40

Qr

3 R

Sol : (c)

If the charge density, 4

Qr

R

,

The electric field at the point p distance r1 from the centre, according to Gauss’s theorem is

21

0

1E.4 r dV

1r

2 21 4

0 0

1 QrE.4 r . 4 r dr

R

2

14

0

QrE

4 R

.

39. An early model for an atom considered to have a positively charged point nucleus of charge Ze,

surrounded by uniform density of negative charge up to a radius R. The atom as a whole is neutral. The

negative charge density is

(a) 3

Ze

4 R

(b)

3

3Ze

4 R

(c)

3

Ze

2R

(d)

3

4Ze

3 R

Sol : (b)



Since the atom is neutral, 34p R Ze 0

3

3

3Zep

4 R

.

40. A regular hexagon of side 10 cm has charge 5 C at each of its vertices. The potential at the centre of

the hexagon is

(a) 1.8 × 106 V (b) 2.2 × 106 V (c) 2.7 × 106 V (d) 3.6 × 106 V

Sol :

9 6

0

6 9 10 5 101 qV 6

4 r 0.1

= 2.7 × 106 V

41. A parallel place capacitor with air between the plates has a capacitance of 9 pF. The separation

between its plates is d. The space between the plates is now filled with two dielectrics. One of the

dielectrics has dielectric constant k1 = 3 and thickness d/3 while the other one has dielectric constant

k2 = 6 and thickness 2d/3. Capacitance of the capacitor is now [AIEEE – 2008]

(a) 20.25 pF (b) 1.8 pF (c) 45 pF (d) 40.5 pF.

Sol : (d)

120AC 9 10 F

d

With dielectric, C = 0kA

d

01

A.3C 9C

d /3

;

02

A.6C 9C

2d /3

1 2total

1 2

C CC

C C

as they are in series.



= 9C 9C 9

C18C 2

or 129

9 10 F2

totalC 40.5 pF .

42. In the given network, The currents i1, i2, i3 are respectively:

(a) 5 5 15

A, A, A2 8 8

(b) 5 5 12

A, A, A3 8 8

(c) 5 15 5

A, A, A2 8 3

(d) 5 5 15

A, A, A3 2 3

Sol : (a)

Using the junction rule the currents in the branches are :

Loop ADCA,

1 2 2 3 1 14 i i 2 i i i i 10 0

1 2 37i 6i 2i 10 ..(1)

Loop ABCA,

2 2 3 14i 2 i i i 10 0

1 2 3i 6i 2i 10 ..(2)

Loop BCD B,

2 3 2 3 15 2 i i 2 i i i 0

1 2 32i 4i 4i 5 ..(3)



Solving (1), (2), (3), we get

1 2 3

5 5 15i A, i A, i A.

2 8 8

43. Two moving coil meters, M1 and M2 have the following particulars [DPMT]

31 1 1R 10 , N 30, A 3.6 10 m2, B1 = 0.25 T R2 = 14 , N2 = 42,

A2 = 1.8 × 10–3 m2, B2 = 0.50 T

(The spring constants are identical for the two meters) The ratio of voltage sensitivity of M2 to M1 is

(a) 1 (b) 1.2 (c) 1.4 (d) 1.6

Sol : (a)

NAB

I K

Voltage sensitivity = I NAB

V I V IR K.R

2 1 1 1 2

1 2 2 21

Voltage sensitivity N A B R

Voltage sensitivity R N A B

=

3

3

30 3.6 10 0.25 141.

10 42 1.8 10 0.50

44. Two identical rings, each of radius r, are co-axially placed. The distance between their centres is r.

Same charge Q is placed on each ring. The work done in moving a test charge from the centre of one

ring to that of the other is

(a) 0

0

2q Q1.

4 r (b) zero

(c) 0

0

( 2 1)q Q1.

4 2r

0

0

2q Q1.

4 ( 2 1)r

Sol : (b)

The centre of the rings in the system are at symmetrical positions. Thus C1 C2V V

Hence 0 C2 C1W q V V 0.

45. In a metre bridge experiment null point is obtained at 20 cm from one end of the wire when resistance

X is balanced against another resistance Y. If X < Y, then where will be the new position of the null

point from the same end, if one decides to balance a resistance of 4X against Y?

(a) 50 cm (b) 80 cm (c) 40 cm (d) 70 cm.



Sol : (a)

For meter bridge experiment,

1 1 1

2 2 1

R l l

R l 100 l

In the first case, X 20 20 1

Y 100 20 80 4

In the second case,

4X l 4 l

l 50 cmY 100 l 4 100 l

46. A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b

respectively. When a current I passes through the coil, the magnetic field at the centre is [IIT – JEE]

(a) 0NI

b

(b) 02 NI

a

(c)

0NI b

ln2 b a a

(d)

0IN a

ln2 b a b

Sol : (c)

Number of turns in elementary ring of thickness dr, dN =

N

b a dr

Magnetic field due to this elementary ring

0I N

dB . dr2r b a

b

0

a

NI drB

2 b a r

0NI b

B . ln2 b a a

.

47. Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding.

The charges on the two rings are +Q and –Q. The potential difference between the centers of the two

rings is



(a) zero (b) 2 2

0

Q 1 1

4 R R d

(c)

20

QR

4 d (d)

2 20

Q 1 1

2 R R d

Sol : (d)

A 2 20 0

1 Q 1 QV

4 R 4 R d

B 2 2

0 0

Q1 1 QV

4 R 4 R d

A B 2 20

1 Q 2 2V V

4 R R d

= 2 2

0

Q 1 1

2 R R d

.

48. A galvanometer coil has a resistance of 15 and the meter shows full scale deflection for a current of

4 mA. To convert the meter into an ammeter of range 0 to 6 A, the shunt required is

(a) 2 m (b) 4 m (c) 6 m (d) 10 m

Sol : (d)

Shunt is put in parallel to the galvanometer resistance (Rg).

g gI . G I I . S

(4 mA) (15 ) = (6 A – 4 mA). S 4

S 15 10m5996

.

49. A wire loop PQRSP formed by joining two semicircular wires of radii R1 and R2 carries a current I as

shown in fig. The magnitude of magnetic induction at centre C is [IIT Karnataka]



(a) 0

2 1

1 1/ 4 I

R R

(b) 0

1 2

1 1I

4 R R

(c) 0

1 2

1 1I

R R

(d) 0 1I 1/R

Sol : (b)

0arc

IB

4 R

0 0 0centre

1 1 2

I I 1 1B I

4 R 4 R 4 R R

50. An electric charge 310 C is placed at the origin (0, 0) of X – Y co-ordinate system. Two points A and B

are situated at 2, 2 and (2, 0) respectively. The potential difference between the points A and B

will be [AIEEE – 2007]

(a) 4.5 volt (b) 9 volt (c) zero (d) 2 volt

Sol : (c)

1r 2i 2j

2 2

1 1r r 2 2 2

2r 2i 0j

or 2 2r r 2

Potential at point A is

A

0 1

1qV

4 r



= 3 6

0

1 10 10

4 2

Potential at point B is

3 6

B

0 2 0

1 q 1 10 10V

4 r 4 2

A BV V 0.

51. The current in a copper wire is increased by increasing the potential difference between its ends.

Which one of the following statements regarding n, the number of charge carriers per unit volume in

the wire and vd, the drift velocity of the charge carriers is correct ?

(a) n is unaltered but vd is decreased

(b) n is unaltered but vd is increased

(c) n is increased but vd is decreased

(d) n is increased but vd is decreased

(e) Both n and vd are increased

Sol : (b)

n the number density, depends on the atomic structure of the material.

d

VI neA.v

R

As V increases, I increases, vd increases.

52. A particle of mass M and charge Q moving with velocity v describes a circular path of radius R when

subjected to a uniform transverse magnetic field of induction B. The work done by the field when the

particle completes one full circle is [AIEEE – 2003]

(a) 2Mv

2 RR

(b) zero (c) BQ 2 R (d) BQv 2 R .

Sol : (b)

Workdone by the field = zero.

53. A horizontal overhead powerline is at a height of 4 m from the ground and carries a current of 100 A

from east to west. The magnetic field directly below it on the ground is 7 10 4 10 Tm A

(a) 2.5 × 10–7 T northward (b) 2.5 × 10–7 T southward [AIEEE – 2008]

(c) 5 × 10–6 T northward (d) 5 × 10–6 T southward.

Sol : (d)



By Amper’s theorem, 0B. 2 d i

770i 4 10 100A

B 50 10 T2 d 2 4m

B = 5 × 10–6 T southwards.

*It is assumed that this is a direct current. If it is a.c, the current at the given instant is in the given

instant is in the given direction.

54. Curie temperature is the temperature of transition from

(a) paramagnetism to diamagnetism

(b) ferromagnetism to paramagnetism

(c) diamagnetism to ferromagnetism

(d) ferromagnetism to diamagnetism

Sol : (b)

At high enough temperature a ferromagnet becomes a paramagnet. This temperature of transition is

called Curie temperature.

55. Two short bar magnets of magnetic moments M each are arranged at the opposite corners of a square

of side d such that their centres coincide with the corners and their axes are parallel. If the like poles

are in the same direction, the magnetic induction at any of the other corners of the square is

(a) 03

M

4 d

(b) 0

3

2M

4 d

(c) 0

3

M

4 2d

(d)

20

3

M

4 2d

Sol : (a)

P is axial point for 2M and equatorial point for 1M .



2 10 0

p3 3

M MB 2

4 d 4 d

2 10p 3

2M MB

4 d

As 2 1M M M

0p 3

MB

4 d

or B = 0

3

M

4 d

56. A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a

permanent magnet such that B is in plane of the coil. If due to a current i in that triangles a torque

acts on it, the side l of the triangle is :

(a)

1

22

Bi3

(b)

1

2

23Bi

(c) 2

Bi3

(d) 1

Bi3

Sol : (b)

Area of equilateral triangle = 23l

4

23M i l

4

Here, 0MB sin 90

1

22 2i 3 4

l B or l l 24 3iB 3iB

57. A current carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III and IV,

arrange them in the decreasing order of potential energy

(a) I > III > II > IV (b) I > II > III > IV (c) I > IV > II > III (d) III > IV > I > II



Sol : (c)

U MB MB cos

Hence, M = magnetic moment of the loop

θ = angle between M and B

U is maximum when θ = 1800 and minimum when 00 . So, as θ decreases from 1800to 00 its PE also

decreses.

58. A straight conductor of length 4 metre moves at a speed of 10 ms–1 when the conductor makes an

angle of 300 with the direction of magnetic field of induction

0.1 T. Then the induced emf is

(a) 1 volt (b) 2 volt (c) 4 volt (d) 8 volt.

Sol : (b)

0mBlv sin 0.1T 4 M 10 sin 30

s

= 1

0.1 4 10 V 2V2

.

59. Two coaxial solenoids are made by winding thin insulated wire over a pipe of

cross-sectional area A = 10 cm2 and length = 20 cm. If one of the solenoids has 300 turns and the other

400 turns, their mutual inductance is 7 10 4 10 T mA [AIEEE – 2008]

(a) 42.4 10 H (b) 52.4 10 H (c) 44.8 10 H (d) 54.8 10 H .

Sol : (a)

20 1 2 1M n n r l.

From 22 1 0 2r ni n I.

2 21 1 2A r 10 cm , l 20cm, N 300, N 400 .

7 40 1 2N N A 4 10 300 400 10 10

Ml 0.20

= 42.4 10 H

60. In an a. c. circuit the voltage applied is 0E E sin t . The resulting current in the circuit is

0I I sin t2

. The power consumption in the circuit is given by [AIEEE – 2007]

(a) 0 0P 2E I (b) 0 0E IP

2 (c) P = zero (d) 0 0E I

P2



Sol : (c)

Given : 0E E sin t

0I I sin t2

Since the phase difference between voltage and current is 2

.

Power factor cos cos 02

Power consumption = rms rmsE I cos 0 .



Section – III {Chemistry}

61. Which of the following fcc structure contains cation in alternate tetrahedral voids? [IIT – 2004]

(a) NaCl (b) ZnS (c) Na2O (d) CaF2.

Sol : (b)

In ZnS, Zn2+ occupies alternate tetrahedral voids.

62. Aluminium oxide may be electrolysed at 1000C to furnish aluminium metal

(At. mass = 27 amu, 1 faraday = 96500 coulombs). The cathode reaction is 3Al 3e Al

To prepar 5.12 kg of aluminium metal by this method would require [AIEEE – 2005]

(a) 5.49 × 104 C of electricity (b) 5.49 × 101C of electricity

(c) 5.49 × 107 C of electricity (d) 1.83 × 107C of electricity

Sol : (c)

3Al 3e Al

3F 27 g

27 g of Al get deposited by = 3 × 96500 C

5120 g of Al get deposited by = 3 96500

512027

= 5.49 × 107C

63. Schottky defect generally appears in [DEC – 2004]

(a) CaCl (b) KCl (c) NaCl (d) All.

Sol : (d)

Schottky defect is common in highly ionic compounds with high coordination number and size of two

ions (cations and anions) are almost same. e.g. NaCl, CsCl, KCl and KBr.

64. The equivalent conductance of NaCl, HCl and C2H5COONa at infinite dilution are 126.45, 426.16 and 91

ohm–1 cm2 respectively. The equivalent conductance of C2H5COOH is [MP CET – 1997]

(a) 201.28 ohm–1 cm2 eq–1 (b) 390.71 ohm–1 cm2 eq–1.

(c) 698.28 ohm–1 cm2eq–1 (d) 540.48 ohm–1 cm2 eq–1.

Sol : (b)

0 00 0

2 5 2 5C H COOH C H COONa HCl NaCl

= 91 + 426.16 – 126.45

= 390.71 ohm–1 cm2 eq–1



65. Consider the following E0 values

3 2 2

o 0

Fe /Fe Sn /SnE 0.77 V, E 0.14 V

Under standard conditions, the potential for the reaction

3 2 2

s aq aq aqSn 2Fe 2Fe Sn is [AIEEE – 2004]

(a) 1.68 V (b) 0.63 V (c) 0.91 V (d) 1.40 V.

Sol : (c)

3 2 2

sSn 2Fe 2Fe Sn

o o ocell cathode anodeE E E 0.77 0.14 0.91 V.

66. In a solution of CuSO4 how much time will be required to precipitate 2 g copper by 0.5 ampere

current? [CBIT Ranchi 1996]

(a) 12157.48 sec (b) 102 sec (c) 510 sec (d) 642 sec.

Sol : (a)

Applying W = Z × I × t

eq.wt. 63.5/ 2 31.75Z

96500 96500 96500

2 96500 1t 12157.48 sec.

31.75 0.5

67. How many unit cells are present in a cubeshaped ideal crystal of NaCl of mass 1.00 g?

(a) 2.57 × 1021 unit cells (b) 5.14 × 1021 unit cells [AIEEE – 2006]

(c) 1.28 × 1021 unit cells (d) 1.71 × 1021 unit cells.

Sol : (a)

Each unit cell consists of 4 NaCl units

Mass of one unit cell = Mass of one NaCl unit × Number of NaCl units present

=23

58.54

6.023 10

Number of unit cells in a mass of 1.0 g

= 23

216.023 102.57 10

58.5 4

unit cells.

68. Which of the following plots does not represent the behaviour of an ideal binary liquid solution of A

and B ?

(a) Plot of pB versus XB is linear

(b) Plot of pA versus XA is linear



(c) Plot of pTotal versus XA or XB is linear

(d) Plot of pTotal versus XA is non-linear.

Sol : (d)

Plot of pTotal versus XA or XB is linear.

69. For the electrochemical cell, [IIT – 2009]

oM M X X, E M /M 0.44 V and oE X / X = 0.33 V. From this data, one can deduce that

(a) M X M X is the spontaneous reaction

(b) M X M X is the spontaneous reaction

(c) Ecell = 0.77 V

(d) Ecell = – 0.77 V

Sol : (b)

M M X X , Cell reaction should be M + X M X

o o oCell cathode anodeE E E 0.33 – 0.44 = – 0.11 V

Reverse cell reaction should be spontaneous i.e.

M X M X .

70. The correct statement(s) regarding defects in solids is (are)

(a) Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion

(b) Frenkel defect is a dislocation defect

(c) Trapping of an electron in the lattice leads to the formation of F-center

(d) Schottky defects have no effect on the physical properties of solids.

Sol : (b, c)

When an ion is missing from its normal position and occupies an interstitial site between the lattice



points, Frenkel defect arises, hence it is a dislocation defect.

The electrons trapped in anion vacancies are referred to as F-centers.

Schottky defects arise when some atoms or ions are missing from their normal lattice points. Due to

the presence of large number of vacancies in crystals, its density (i.e., physical property) is lowered.

71. 1 dm3 of 2 M CH3COOH is mixed with 1 dm3 of 3 M ethanol to form ester. The decreases in the initial

rate if each solution is diluted with an equal volume of water would be [Karnataka CET – 2001]

(a) 2 items (b) 4 times (c) 0.25 times (d) 0.5 times

Sol : (c)

Rate 1r k [CH3COOH] [CH3CH2OH]

Concentration becomes half on diluting with equal volume of water.

rate 2 3 3 2

1 1r k CH COOH CH CH OH

2 2

= 1

k4

[CH3COOH] [CH3CH2OH]

2 1

1r r 0.25

4 1r .

72. An aqueous solution freezes at – 0.1860C (Kf = 1.860Cm–1 ; Kb = 0.5120Cm–1).

What is the elevation in boiling point ? [AIEEE – 2002]

(a) 0.186 (b) 0.512 (c) 0.80 (d) 0.0512.

Sol : (d)

0 0f 0T 0.186 C, T 0 C

f 0 fT T T 0.186

f fT K m

f

f

T 0.186m 0.1

K 1.86

Now, b bT K m 0.512 0.1 0.0512.

73. A standard hydrogen electrode has zero electrode potential because

(a) Hydrogen is easier to oxidise

(b) This electrode potential is assumed to be zero

(c) Hydrogen atom has only electron

(d) Hydrogen is the highest element.

Sol : (b)



Electrode potential of standard hydrogen electrode is assumed to be zero to measure the absolute

value of other electrodes.

74. Monoclinic crystal has dimension [DEC – 2000]

(a) 0a b c; 90

(b) 0a b c; 90

(c) 0a b c; 90

(d) 0 0a b c; 90 ; 90 .

Sol : (d)

For monoclinic crystals axial ratios are not equal i.e. a b c and axial angles 090 .

75. When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of benzene (Kf = 1.72 K kg mol–1), a freezing

point depression of 2 K is observed. The van’t Hoff factor (i) is [IIT – 2007]

(a) 0.5 (b) 1 (c) 2 (d) 3

Sol : (a)

For electrolytic solution, f fT iK m

i.e. 2 = i × 1.72 × 20 1000

172 50 (Mol. wt. of C11H8O2 = 172)

i = 0.5

76. Which has minimum osmotic pressure ? [DEC – 2005]

(a) 200 ml of 2 M NaCl solution

(b) 200 ml of 1 M glucose solution

(c) 200 ml of 2 M urea solution

(d) All have same.

Sol : (b)

Osmotic pressure is a colligative property which depends upon the number of particles. Concentration

of particles is least in 1 M glucose solution. So 1 M glucose solution has minimum osmotic pressure.

77. 25 mL of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid

gave a titre value of 35 ml. The molarity of barium hydroxide solution was [AIEEE – 2003]

(a) 0.07 (b) 0.14 (c) 0.28 (d) 0.35.

Sol : (a)

2 22Ba OH 2HCl BaCl 2H O



1 1 2 2

1 2

M V M V

n n

11

M 25 0.1 35 0.1 35M 0.07 M.

1 2 2 25

78. Benzene and toluene form nearly ideal solutions. At 200C, the vapour pressure of benzene is 75 torr

and that of toluene is 22 torr. The partial vapour pressure of benzene at 200C for a solution containing

78 g of benzene and 46 g of toluene in torr is [AIEEE – 2005]

(a) 50 (b) 25 (c) 37.5 (d) 53.5

Sol : (a)

Molecular mass of benzene (C6H6) = 78

No. of moles of benzene 78

78

Molecular mass of toluene (C6H5 – CH3) = 92

No. of moles of toluene 46

92 0.5

Benzene

1 2X

1 0.5 3

0Benzene Benzene BenzeneX

275 50

3 torr

79. Under the same reaction conditions, initial concentration of 1.386 mol dm–3 of a substance becomes

half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio

1

0

k

k

of the rate constant for first order (k1) and zero order (k0) of the reactions is [IIT –

2008]

(a) 0.5 mol–1 dm3 (b) 1.0 mol dm–3 (c) 1.5 mol dm–3 (d) 2.0 mol–1 dm3.

Sol : (a)

For first order reaction

1/2

1

0.693t

k

0.693k

40



For zero order reaction 01/2

At

2k ,

0

1/2

A 1.386k

2 t 2 20

Now, 1 31

0

k 0.693 400.5 mol dm .

k 40 1.386

80. The half cell reactions for the corrosion are 02 22H 1/ 2 O 2e H O; E 1.23 V [IIT – 2005]

2 o

sFe 2e Fe ; E 0.44 V

Find the oG (in kJ) for the over all reaction

(a) – 76 (b) – 322 (c) – 151 (d) – 152.

Sol : (b)

The reduction potential of two half reactions suggest that reduction of H+ and oxidation of Fe take

place, so

o o ocell cathode anodeE E E 1.23 0.44 1.67 V

Applying 0 0G nFE n = 2

0G = – 2 × 96500 × 1.67 = – 322310 J = – 322 . 31 kJ.

81. For a first order reaction A P, the temperature (T) dependent rate constant (k) was found to follow

the equation 1

log k 2000 6.0T

. The pre-exponential factor A and the activation energy aE ,

respectively are [IIT – 2009]

(a) 1.0 × 106 s–1 and 9.2 kJ mol–1

(b) 6.0 s–1 and 16.6 kJ mol–1

(c) 1.0 × 106 s–1 and 16.6 kJ mol–1

(d) 1.0 × 106 s–1 and 38.3 kJ mol–1.

Sol : (d)

According to Arrhenius equation,

aElog k log A

2.303 RT ..(i)

Given, log k = – (2000) 1

6.0T or log k = +6.0–(2000)

1

T …………(ii)

Comparing equations (i) and (ii).

log A = 6 A = 1 × 106 s–1



aa

E 2000E

2.303 RT T

= 2000 × 2.303 × 8.314

aE = 38294 or 1aE 38.3 kJ mol .

82. Match List I with List II and select the correct answer using the codes given below the lists:

List List II [DEC – 2007]

A. Coagulation 1. Scattering

B. Lyophilization 2. Washing of precipitates

C. Peptization 3. Purification of colloids

D. Tyndall effect 4. Electrolyte

(a) A-4; B-3; C-2; D-1 (b) A-2; B-1; C-3; D-4

(c) A-3; B-1; C-2; D-4 (d) A-4; B-3; C-1; D-2.

Sol : (a)

Coagulation : Electrolyte

Lyophilization : Electrolyte

Lyophilization : Purification of colloids

Peptization : Washing of precipitates

Tyndall effect : Scattering

83. Among the electrolytes Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective agent for Sb2S3 sol is

[IIT – 2009]

(a) Na2SO4 (b) CaCl2 (c) Al2(SO4)3 (d) NH4Cl.

Sol : (c)

Sb2S3 is a negative sol and according to Hardy-Schulze rule

(i) Ions carrying charge opposite to that of sol particles are effective in causing coagulation.

(ii) Coagulating power of an electrolyte is directly proportional to the valency of the active ions.

Out of the given options, the most effective coagulating agent is Al2(SO4)3 or Al3+ ion.

84. The order of reactivities of the following alkyl halides for SN2 reaction is [IIT– 2000]

(a) RF > RCl > RBr > RI (b) RF > RBr > RCl > RI

(c) RCl > RBr > RF > RI (d) RI > RBr > RCl > RF.

Sol : (d)

R – I > R – Br > R – Cl > R – F. Greater the bond length, laser the bond strength hence higher is the

reactivity.



85. Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solution containing 1 mol

of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this

solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) of X and

Y in their pure states will be respectively [AIEEE – 2009]

(a) 200 and 300 (b) 300 and 400 (c) 400 and 600 (d) 500 and 600

Sol : (c)

0 0T X X Y YP p x p x

where PT = Total pressure

0p x = Vapour pressure of X in pure state

0p y = Vapour pressure of Y in pure state

xx = Mole fraction of X = 1/4

xy = Mole fraction of Y = 3/4.

(i) When T = 300 K, PT = 550 mm Hg

0 0X Y

1 3550 p p

4 4

0 0x yp 3p 2200 ..(i)

(ii) When at T = 300 K, 1 mole of Y is added, PT = (550 + 10) mm Hg

X Yx 1/5 and x 4 /5

0 0x Y

1 4560 p p

5 5

or 0 0x xp 4p 2800 ..(2)

On solving equations (1) and (2), we get

0Yp = 600 mm Hg and 0

xp = 400 mm Hg.

86. Identify the set of reagents/reaction conditions ‘X’ and ‘Y’ in the following set of transformations.

X3 2 2CH CH CH Br Product Y

3 2|

Br

CH C H CH [IIT– 2002]

(a) X = Dilute aqueous NaOH, 200C; Y = HBr/acetic acid 200C

(b) X = Concentrated alcoholic NaOH, 800C; Y = HBr/acetic acid 200C

(c) X = Dilute aqueous NaOH, 200C; Y = Br2/CHCl3, 00C

(d) X = Concentrated alcoholic NaOH, 800C; Y = Br2/CHCl3, 00C.



Sol : (b)

87. Which halide among the following is used as methylating agent ? [Karnataka CET – 2000]

(a) CH3I (b) C2H5Cl (c) C2H5Br (d) C6H5Cl.

Sol : (a)

CH3I is used as methylating agent since C – I bond is a weak bond.

88. Out of the following compounds, which one will have a zero dipole moment ? [IIT– 1987]

(a) 1, 1-Dichloroethylene

(b) cis-1, 2-Dichloroethylene

(c) trans-1, 2-dichloroethylene

(d) None of these compounds.

Sol : (c)

Trans symmetrical alkenes are non-polar in nature i.e. 0 .

89. During debromination of meso-dibromobutane, the major compound formed is [IIT– 1997]

(a) n-Butane (b) 1-Butene (c) cis-2-Butene (d) trans-2-Butene.

Sol : (d)

Dehalogenation is trans-elimination.



90. The reaction between alkyl halides and sodium metal is called

(a) Wurtz reaction (b) Kolbe’s reaction

(c) Clemmensen’s reaction (d) Finkelstein reaction.

Sol : (a)

R – X + 2Na + X – R Dry ether R – R + 2NaX

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