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Pioneer’s Aspire Scholarship Exam

Solution

11th (Non – medical) 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 – A

MATHEMATICS (1 to 30), Section – B Physics (31 to 60) and Section – C 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.

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Section – A {Mathematics}

1. The set A {x :|2x 3| 7} is equal to the set

(a) D {x:0 x 5 7} (b) B {x: 3 x 7} (c) E {x: 7 x 7} (d) C {x: 13 2x 4}

Ans. (a)

Solution:

Given, set A = {x :|2x 3| 7}

Now, 2x 3 7 7 2x 3 7

7 3 2x 7 3 10 2x 4

5 x 2 0 x 5 7

2. If 2x

f(x) ,x 1,x 1

then the value of for which f(a) a, where a 0, is

(a) 1

1a

(b) 1

a (c)

11

a (d)

11

a

Ans. (c)

Solution:

2αa

f a a aa 1

2 2 1α. a a a α 1

α

3. If 1 2z 2 2(1 i) and z 1 i 3, then 2 31 2z z is equal to

(a) 128i (b) 64i (c) –64i (d) –128i

Ans. (d)

Solution:

Given, 1 2z 2 2 1 i and z 1 i 3

2 2 2 3 2 21 3z z [2 2 1 i ] [1 i 3] 8 1 i 2i

33[1 i 3 3.1.i 3 1 i 3 ]

= 8 1 1 2i [1 3 3 i 3 3i 1 i 3 ]

= 8 2i [1 9] 16i 8 128 i




4. If and are two different complex numbers with | | 1, then 1

is equal to

(a) 1/2 (b) 0 (c) –1 (d) 1

Ans. (d)

Solution:

Given, β 1

β α β α

1 aβ ββ aβ [ 1 ββ]

= 1 β α 1

1 1β 1β α

[ z z ]

5. The value of 2

1 i 3

11

i 1

is

(a) 20 (b) 9 (c) 5/4 (d) 4/5

Ans. (d)

Solution:

Let 2

1 i 3 2i 1 i 3 2i 3 4i1 i 3z

3 4i 25i 2

i 1

Now, |1 i 3||2i||3 4i| 2 2 5 4

z25 25 5

6. Argument of the complex number 1 3i

2 iis (In degrees is)

(a) 450 (b) 1350 (c) 2250 (d) 2400

Ans. (c)

Solution :

1 3i 1 3i 2 i

2 i 2 i 2 i

= 22 i 6i 3i

1 i4 1




Argument of 1 01 3i 1arg of 1 i tan 225

2 i 1

[since, the given complex number lies in IIIrd quadrant]

7. The value of 4 4(1 3 i) (1 3 i) is

(a) –16 (b) 16 (c) 14 (d) –14

Ans. (a)

Solution:

4 4 4 421 3 i 1 3 i 2ω 2ω

1 3i 1 3 iω and ω

2 2

= 2

8 4 3 2 316ω 16ω 16 ω ω 16 ω ω

= 216 ω ω 16 1 16 [ 21 ω ω 0]

8. If and are the roots of the quadratic equation 2x x 1 0, then the equation, whose roots are

19 and 7 , is

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

Ans. (d)

Solution:

Given equation is x2 + x + 1 = 0, 2α ωand β ω will satisfy the given equation .

Now, 19 19 7 14 2α ω ω, β ω ω

Required equation is

2 2 3x ω ω x ω 0

2x x 1 0

9. Let the two numbers have arithmetic mean 9 and geometric mean 4. Then, these numbers are the

roots of the quadratic equation

(a) 2x 18x 16 0 (b) 2x 18x 16 0 (c) 2x 18x 16 0 (d) 2x 18x 16 0

Ans. (b)

Solution:

Given, arithmetic mean = 9 and geometric = 4.

Let the two numbers be x1 and x2.




Given, 1 2x x

92

and 1 2x .x 16

1 2x x 18 and 1 2x . x 16

Hence, required equation is

x2 – (sum of roots) x + (product of roots) = 0

x2 – 18x + 16 = 0

10. If and are the roots of the quadratic equation 2x 4x 3 0, then the equation, whose roots are

2 and 2 , is

(a) 2x 12x 35 0 (b) 2x 12x 33 0 (c) 2x 12x 33 0 (d) 2x 12x 35 0

Ans. (d)

Solution:

Given, α and βand the roots of equation x2 + 4x + 3 = 0.

α β 4 and αβ 3

Now, 2α β α 2β 3 α β 12

and 2 22α β a 2β 2α 4αβ αβ 2β

= 2 2 2

α β αβ 2 4 3 35

Hence, required equation is

x2 – (sum of roots)x + (product of roots) = 0

x2 + 12x + 35 = 0

11. In PQR, R .2

If tan P

2and tan

Q

2are the roots of ax2 + bx + c = 0, a 0, then

(a) b = a + c (c) b = c (c) c = a + b (d) a = b + c

Ans. (c)

Solution:

Here, P Q b

tan tan2 2 a

and P Q c

tan tan2 2 a

..(i)

Also, P Q R π

2 2 2 2 P Q R π

P Q π

2 4

πR , given

2




P Qtan tan

P Q 2 2tan 1 1P Q2 2 1 tan tan2 2

bb ca 1 1

c a a1a

[from Eq. (i)]

c = a + b

12. A value of b for which the equations x2 + bx –1 = 0, x2 + x + b = 0 have one root in common, is

(a) – 2 (b) –i 3 (c) i 5 (d) 2

Ans. (b)

Solution :

We know that, if a1x2 + b1x + c1 = 0 and a2x2 + b2x + c2 = 0 have a common real root, then

(a1c2 – a2c1)2 = (b1c2 – b2c1) (a1b2 – a2b1)

Hence, x2 + bx – 1 = 0 and x2 + x + b have a common root.

2 21 b b 1 1 b

b2 + 2b + 1 = b2 – b3 + 1 – b

b3 + 3b = 0 b(b2 + 3) = 0

b 0, i 3

13. Let and be the roots of equation px2 + qx + r = 0, p 0. If p, q and r are in AP and 1 1

=4, then the

value of | |is

(a) 61

9 (b)

2 17

9 (c)

34

9 (d)

2 13

9

Ans. (d)

Solution:

Given, α and β are roots of px2 + qx + r = 0, p 0.

q r

α β , αβp p

..(i)

Since, p, q and r are in AP.

2q = p + r ..(ii)

Also, 1 1 α β

4 4α β αβ




q 4r

α β 4αβp p

[from Eq. (i)]

q = – 4r

On putting the value of q in Eq. (ii), we get

2(–4r) = p +r p = – 9r

Now, q 4r 4r 4r

α βp p 9r 9

and r r 1

αβp 9r 9

2 2 16 4 16 36

α β α β 4αβ81 9 81

2 52 2

α β α β 1381 9

14. In a triangle, the lengths of two larger sides are 10 cm and 9 cm. If the angles of the triangle are in AP,

then the length of the third side is

(a) 5 6 (b) 5 6 (c) 5 6 (d)5 6

Ans. (d)

Let the figure, A > B > C

Since, angles are in AP.

C θ d, B θ and A θ d

As 0A B C 180

0θ d θ θ d 180

0 03θ 180 θ 60

Now, in ABC 2 2 2a c b

cos B2ac




2 2 20 10 x 9

cos 602 10 x

2 2

01 100 x 9 1cos 60

2 20x 2

2x 10x 19 0

10 100 76 10 24

x 5 62 2

15. If S1, S2 and S3 are the sum of n, 2n and 3n terms respectively of an arithmetic progression, then

(a) S3 = 2(S1 + S2) (b) S3 = S1 + S2 (c) S3 = 3(S2 –S1) (d) S3 = 3(S2 + S1)

Ans. (c)

Solution :

Let a and d be the first term and common difference respectively.

1

nS [2a n 1 d]

2

2

nS [2a 2n 1 d]

2

and 3

3nS [2a 3n 1 d]

2

Clearly, 2 1

3n3 S S [4a 4nd 2d 2a nd d]

2

= 3

3n 3n[2a 3nd d] [2a 3n 1 d] S

2 2

16. If a1,a2,….,an are in AP with common difference d 0, then (sin d)[sec a1 sec a2 + sec a2 sec a3 + ……+ sec

an–1 sec an] is equal to

(a) cot an – cot a1 (b) cot a1 – cot an (c) tan an – tan a1 (d) tan an – tan an–1

Ans. (c)

Solution:

Given, 2 1 3 2 4 3 n n 1d a a a a a a .... a a

1 2 2 3 n 1 nsin d [seca seca seca seca ... seca seca ]

= 1 2 2 3 n 1 n

sin d sin d sin d....

cos a cos a cos a cos a cos a cos a

= 2 1 3 2 n n 1

1 2 2 3 n 1 n

sin a a sin a a sin a a....

cos a cos a cos a cos a cos a cos a




= tan a2 – tan a1 + tan a3 – tan a2 + …. + tan an – tan an–1 s

= tan an – tan a1

17. The sum of first 20 terms of the sequence 0.7, 0.77, 0.777,…., is

(a) 207179 10

81 (b) 207

99 109

(c) 207179 10

81 (d) 207

99 109

Ans. (c)

Solution:

2 3

7 77 777S .... upto 20 terms

10 10 10

= 7 9 99 999

.... upto 20 terms9 10 100 1000

= 2 3

7 1 1 11 1 1 ... upto 20 terms

9 10 10 10

= 7

9(1[ + 1 + … + upto 20 terms]

– 2 3

1 1 1... upto 20 terms

10 10 10

=

201 1

110 107

2019 1

10

=

207 1 1

209 9 10

= 20

207 179 1 1 7179 10

9 9 9 10 81

18. The value of x which satisfies 21 cosx cos x ........8 64 in , is

(a) ,2 3

(b)3

(c) ,2 6

(d) ,6 3

Ans. (b)

Solution:

Given, 2

1

1 cos x cos x ..... 2 21 cos x8 8 8 8

[ cos x 1, as for cos x = 1 + 1 + 1 + ….. 2, and cos x 1]




1

2 2 cos x 11 cos x

1 π

cos x x2 3

19. How many 5-digit telephone numbers can be constructed using the digits 0 to 9, if each number starts

with 67 and no digit appears more than once?

(a) 335 (b) 336 (c) 337 (d) 338

Ans. (b)

Solution :

Since, telephone number start with 67, so two digits is already fixed. Now, we have to do arrangement

of three digits from remaining eight digits.

Possible number of ways = 8P3

= 8! 8!

8 7 6 366 days8 3 ! 5!

20. If Pm stands for mmP , then 1 + 1P1 + 2P2 + 3P3 + ……+ n . Pn is equal to

(a) n! (b) (n + 3)! (c) (n + 2)! (d) (n + 1)!

Ans. (d)

Solution:

1 + 1. P1 + 2. P2 + 3. P3 + … + nPn

= 1 + 1. (1!) + 2. (2!) + …. + n. (n!)

= 1 + (2 – 1)1! + (3 – 1)2! + … + [(n + 1) – 1]n!

= 1 + 2! – 1! + 3! – 2! + … + (n + 1)! = n! = (n + 1)!

21. In how many different ways can the letters of the word MATHEMATICS be arranged?

(a) 11! (b) 11!/2! (c) 11!/(2!)2 (d) 11!/(2!)3

Ans. (d)

Solution:

In the word ‘MATHEMATICS’ the letters are 2A, C, E, H, I, 2M, S, 2T.

Total number of different words = 3

11! 11!

2!2!2! 2!

22. The number of four-letter words that can be formed (the words need not be meaningful) using the

letters of the word ‘MEDITERRANEAN’ such that the first letter is E and the last letter is R, is

(a) 11!

2!2!2! (b) 59 (c) 56 (d)

11!

3!2!2!




Ans. (b)

Solution :

Word ‘MEDITERRANEAN’ has 2A, 2E, 1D, 1I, 1M, 2N, 2R, 1T.

Out of four letters E and R is fixed and rest of the two letters can be chosen in following ways :

Case I Both letters are of same kind i.e. 3C2 ways, therefore number of words = 32

2!C 3

2!

Case II Both letters are of different kinds i.e. 8C2 ways,

therefore number of words = 8C2 2! = 56

Hence, total number of words = 56 + 3 = 59

23. If a polygon of n sides has 275 diagonals, then n is equal to

(a) 25 (b) 35 (c) 20 (d) 15

Ans. (a)

Solution:

A polygon of n sides has number of diagonals

= n n 3

2752

[given]

n2 – 3n – 550 = 0

(n – 25) (n + 22) = 0

n = 25 [ n 22]

24. In the expansion ofn

3

2

1x ,n N,

xif the sum of the coefficients of x5 and x10 is 0, then n is equal to

(a) 25 (b) 20 (c) 15 (d) none of these

Ans. (c)

Solution:

rn r

n 3r 1 r 2

1T C x

x

= r rn 3n 3r 2r n 3n 5r

r rC x 1 x C x 1

For coefficient of x5 and x10, substitute 3n – 5r = 5 and 10 respectively, we get coefficient of

3n 55 n

53n 5

5

x C 1 and coefficient of x10 + 3n 10

.n5

3n 10

5

C 1 .

coefficient of x5 + Coefficient of x10 = 0

n 3n 5 3n 10

n3n 5 5 53n 10

5

C1 C 1

5




3n 5 3n 5

n n5 5

3n 5 3n 10

5 5

C 1 C 1

n n3n 5 3n 10

5 5

C C

3n 5 3n 10

n5 5

6n 15

n5

6n – 15 = 5n

n = 15

25. The equation of the base BC of an equilateral ABC is x + y = 2 and A is (2, –1). The length of the side of

the triangle is

(a) 2 (b) 1/2

3

2 (c)

/21

2 (d)

/22

3

Ans. (d)

Solution:

Length of perpendicular from A(2, – 1) to the line x + y – 2 = 0 is

2 1 2 1AD

1 1 2

In ABD, 0ADcos 30

AB

1/2

1 3 2 2AB

2 32 AB 3

26. If C is the reflection of A (2, 4) is X-axis and B is the reflection of C in Y-axis, then |AB| is equal to

(a) 20 (b) 2 5 (c) 4 5 (d) 4




Ans. (c)

Solution:

Here, coordinates are A(2, 4), B(–2, – 4) and C(2, – 4).

Now, 2 2

AB 2 2 4 4 16 64 80 4 5

27. The circumcentre of the triangle with vertices (8, 6), (8, –2) and (2, –2) is at the point

(a) (2, –1) (b) (1. –2) (c) (5, 2) (d) (2, 5)

Ans. (c)

Solution :

Let the vertices of a triangle are A(8, 6), B(8, - 2) and C(2 – 2).

Now, AB = 2 2 2(8 8) (16 2) 0 8 8

BC = 2 2

2 8 2 2 36 0 6

and 2 2 2 2CA 8 2 6 2 6 8

= 36 64 100 10

Now, AB2 + BC2 = (8)2 + (6)2 = 64 + 36 = 100 = AC2

So, ABC is a right angled triangle and right angled at B.




We know that, in a right angled, the circumcentre of the mid-point of hypotenuse.

Mid-point of AC = 8 2 6 2

, 5,22 2

Hence, the required circumcentre is (5, 2).

28. If orthocentre and circumcentre of a triangle are respectively (1, 1) and (3, 2), then the coordinates of

its centroid are

(a) 7 5

,3 3

(b) 5 7

,3 3

(c) (7, 5) (d) none of these

Ans. (a)

Solution:

We know that, centroid divides the line segment joining orthocentre and circumcentre in the ratio 2 :

1. Since, the coordinates of orthocentre are (1, 1) and (3, 2), respectively.

The coordinates of centroid are

2.3 1.1 2.2 1.1 7 5, ,

2 1 2 1 3 3

29. A straight line perpendicular to the line 2x + y = 3 is passing through (1, 1). Its y-intercept is

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

Ans. (d)

Solution:

A straight line perpendicular to 2x + y = 3 is

2y – x + c = 0 …(i)

Since, it passes through (1, 1).

2(1) – 1 + c = 0 c = – 1

On putting c = – 1 in Eq. (i), we get

2y x 1 0 2y x 1 y x

11/2 1

So, the y-intercept is 1

2.

30. If a ray of light along x + 3y 3 gets reflected upon reaching X-axis, then the equation of the

reflected ray is

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

Ans. (b)

Solution:

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




Equation of AB’, y – 0 = 1 0

x 30 3

3 y x 3

3 y x 3

Section – B {Physics}

31. The volume V of water passing any point of a uniform tube during t seconds is related to the

cross- sectional area A of the tube and velocity u of water by the relation V A u t which one of the

following will be tube?

(a) (b) (c) (d)

Ans. (b)

Solution:

The dimensions of the two sides of proportionality are

3 2α 1 β γ 2α β γ βL L (LT ) T L T

Equating the powers of dimensions on both sides, we have

2α β 3

γ β 0

which give 1

β γ and α2

3 β , i.e. α β γ .

Thus, the correct choice is (b).

32. When a wave traverses a medium, the displacement of a particle located at x at time is given by

y asin bt cx where a, b and c are constants of the wave. The dimensions of b are the same as those

by

(a) wave velocity (b) amplitude (c) wavelength (d) wave frequency

Ans. (d)

Solution:

Since the argument of a sine function (or any trigonometric function) must be dimensionless, bt and cx




are dimensionless. Since bt is dimensionless, the dimensions of b = dimensions of 1/t = T–1, which are

the dimensions of frequency. Hence, the correct choice is (d).

33. If velocity (V), force (F) and energy (E) are taken as fundamental units, then dimensional formula for

mass will be

(a) 2 0 1V F E (b) 0 2V FE (c) 2 0VF E (d) 2 0V F E

Ans. (d)

Solution:

Let α b cm V F E

1 1 a 2 b 2 2 c[M ] [LT ] [MLT ] [ML T ]

Equation the powers of M, L and T are solving, we get a = – 2, b = 0 an c = 1. Hence, the dimensional

formula of m = [V–2 F0 E] which is choice (d).

34. The acceleration a of a particle moving with an initial velocity u varies with distance x as a = k x

where k is a constant. The distance covered by the particle when its velocity becomes 3u is given by

(a)

2/323u

k (b)

4/323ku (c)

2/326u

k (d)

2/32u

3k

Ans. (c)

Solution:

dν dν dx dν

a νdt dx dt dx

ν dν adx k x dx

Integrating

3u

u

νdν k x dx

3/23u2 x

u

ν k

2 3/ 2

2 2 3/21 2k9u u x

2 3

2 3/22k4u x

3

2/326ux

k




35. The velocity v of a particle moving in a straight line varies with distance x as v = k x where k is a

positive constant. The distance x varies with time t as

(a) x t (b) x t (c) 3/2x t (d) 2x t

Ans. (d)

Solution:

ν k x

Acceleration dν dν dx

adt dx dt

= dν

νdx

= d

k x k xdx

= 2

1/2 1/2k kk x x

2 2

2k

dν dt2

Integrating

2k

d ν dt2

2k

ν t2

2dx k

tdt 2

2k

dx t dt2

Intergating

2k

dx t dt2

2 2k t

x2

or 2x t

36. A police jeep moving at a constant speed v on a straight road is chasing a thief riding on motor-cycle.

The thief starts from the rest when the police jeep is at a distance x away and accelerates at a constant

rate a. The police will be able to catch the thief if




(a) v ax (b) v 2ax (c) v 2 ax (d) ax

v2

Ans. (c)

Solution:

Suppose the thief is caught at a time t after the starting of the motorcycle. The distance travelled by the

motorcycle in this time t is

21S at

2 (i)

During this time, the jeep must travel a distance

S + x = νt (ii)

Using (i) and (ii)

21at x νt

2

2at 2νt 2x 0

The roots of this quadratic equation are

2v ν 2ax

ta

Since t must be real and positive, we must have

2ν 2ax ν 2ax

37. The driver of a train moving at a speed v1 sights another train at a distance d ahead of him, moving in

the same direction with a slower speed v2. He immediately applies brakes to achieve a constant

retardation a. There will be no collision if d is greater than

(a)

2

1 2v v

a (b)

2 21 2v v

a (c)

2

1 2v v

2a (d)

2 21 2v v

2a

Ans. (c)

Solution :

The two trains will not collide if the initial relative velocity 1 2u ν ν reduces to zero relative

velocity ν 0 in a minimum distance S = dmin under a retardation (–a). Using 2 2ν u 2a S,we have

2

1 2 min0 ν ν 2 a d

2

1 2min

ν νd

2a

So, the correct choice is (c).




38. Two bodies are projected horizontally in opposite directions with velocities u1 and u2 from the top of a

building. If their velocities become perpendicular to each other after falling through a distance h, then

(a) 1 2u uh 2

g (b) 1 22u u

hg

(c) 1 22u uh

2g (d) 1 2u u

h2g

Ans. (d)

Solution:

Let u1 be along the positive x-axis and u2 along the negative x-axis. Their velocities at any time t are

1 1ν u i gt j

2 2ν u i gt j

Let t be the time at which 1ν becomes perpendicular to 2ν , then

1 2ν . ν 0

1 2u i gt j . u i gt j 0

2

1 2u u gt 0

1/2

1 2

1t u u

g

2 1 2 1 22

u u u u1 1h gt g

2 2 2gg

39. A body is projected with a speed u at an angle with the horizontal. The speed of the body when its

velocity vector makes an angle with vertical is

(a)ucos

cos (b)

ucos

sin (c)

usin

cos (d)

usin

sin

Ans. (b)

Solution:

Let the speed of the body be ν when its velocity vector makes an angle βwith the vertical fig.

Horizontal component of u is u cos α and the horizontal component of ν is ν sin β . Since the horizontal




component of velocity remains constant,

u cos α ν sinβ

ucos α

ν ,sin β

which is choice (b)

40. A body is projected with kinetic energy K at an angle of 600 with the vertical. The kinetic energy of the

body when it is at the highest point on the trajectory is

(a) zero (b) 3K

4 (c)

K

2 (d)

K

4

Ans. (b)

Solution:

Angle with the horizontal is 0 0θ 90 60 = 300. At the highest point, the horizontal component of

velocity is u cos θ and the vertical component is zero.

Given 21mu K.

2At the highest point,

2 2 21 1K' m u cos θ mu cos θ

2 2

= 2 0 3KK cos 30

4

41. A force F is applied horizontal to block A of mass m1 which is in contact with a block B of mass m2, as

shown in fig. If the surface are frictionless, the force exerted by A on B is given by

(a) 1

2

mF

m (b) 2

1

mF

m (c) 1

2 2

m F

(m m ) (d) 2

2 1

m F

m m

Ans. (d)

Solution:

The contact force exerted by A on B is F2 = 2

1 2

m F

m m, which is choice (d).

42. A block of mass 10 kg is placed at a distance of 5 m from the rear end of a long trolley as shown in fig..

The coefficient of friction between the block and the surface below is 0.2. Starting from rest, the trolley

is given a uniform acceleration of 3 ms–2. At what distance from the starting point will the block fall off

the trolley ? Take g = 10 ms–2.




(a) 15 m (b) 20 m (c) 25 m (d) 30 m

Ans. (a)

Solution:

Since the block is placed on the trolley, the acceleration of the block = acceleration of the trolley a = 3

ms–2. Therefore, the force acting on the block is

F = ma = 10 3 = 30 N

The weight mg of the block is balanced by the normal reaction R. As the trolley accelerates in the

forward direction, it exerts a reaction force F = 30 N on the block in the backward direction, as shown

in the figure. The force of friction will oppose this force and will act in a direction opposite to that of F.

The force of limiting friction f is given by

f f

μR mg

or f μmg 0.2 10 10 20N

Thus, the block is acted upon by two forces-force F = 30N towards the right and frictional force f = 20 N

towards the left see fig. The net force on the block towards the right, i.e. towards the rear end of the

trolley is

F’ = F – f = 30 – 20 = 10 N.

Due to this force, the block experiences an acceleration towards the rear end which is given by

2F' 10a' 1 ms

m 10

Let t be the time taken for the block to fall from the rear end of the trolley. Clearly, the block has to

travel a distance S’ = 5 m to fall off the trolley. Since the trolley starts from rest, initial velocity u = 0.

Now t can be obtained from the relation

21s ut at

2

Putting s = 5m, u = 0 and a = a’ = 1 ms–2, we get t = 10 s.

The distance covered by the trolley in time t = 10 s is ( u 0)




21 1s' ut at 0 3 10 15m

2 2

Hence, the correct choice is (a).

43. A given object takes n times as much time to side down a 450rough incline as it takes to side down a

perfectly smooth 450 incline. The coefficient of kinetic friction between the object and the incline given

by

(a) 2k 1/ 1 n (b) 2

k 1 1/ n (c) 2k 1/ 1 n (d) 21 1/ n

Ans. (b)

Solution :

The square of the time of slide is inversely proportional to the acceleration. The acceleration in the two

cases are

01 2

ga g sin 45 and a

2

= 0 0kg sin 45 μ gcos 45

= k

g1 μ

2

2

22 12

2 k1

t a 1n

a 1 μt

or k 2

1μ 1

n.

Hence, the correct choice is (b).

44. Two blocks A and B are connected to each other by a string and a spring of a force constant k, as shown

in Fig., The string passes over a frictionless pulley as shown. The block B slides over the horizontal top

surface of a stationary block C and the block A slides along the vertical side of C both the same uniform

speed. The coefficient of friction between the surfaces of the block B and C is . If the mass of block A

is m, what is the mass of block B?




(a) m

(b) m

(c) m (d) m

Ans. (b)

Solution:

Since the blocks slide at the same uniform speed, no net force acts on them. If M is the mass of block B,

then the tension in the string is T = μ M g. Also T = mg. Equating the two, we get μM m or m

Mμ

,

which is choice (b).

45. A smooth inclined plane of angle of inclination 300 is placed on the floor of a compartment of a train

moving with a constant acceleration a. When a block is placed on the inclined plane, it does not slide

down or up the plane. The acceleration a must be

(a) g (b) g

2 (c)

g

2 (d)

g

3

Ans. (d)

Solution:

Refer to fig. The component of acceleration vector a along the plane is a cos θ . The component of

acceleration due to gravity g along the plane is g sin θ . The block will stay at rest if a cos θ = g sin θ

or a = g tan θ

Now 0θ 30 .Therefore, a = 0 gg tan 30

3. Hence the correct choice is (d).

46. A body is sliding down a rough inclined plane of angle of inclination for which the coefficient of

friction varies with distance x as x kx , where k is a constant. Here x is the distance moved by the

body down the plane. The net force on the body will be zero at a distance x0 given by

(a) tan

k (b) k tan (c)

cot

k (d) k cot

Ans. (a)

Solution:




The net downward force on the body at a distance x is

f x mg sin θ μmg cos θ

= mg sin θ μ cos θ

= mg sin θ kx cos θ

f x 0 at a value of x = x0 given by

0sin θ kx cos 0

which gives 0

tanθx

k

Thus, the correct choice is (a).

47. The displacement x of a particle of mass m moving in a straight line varies with time t as x = kt3/2 under

the action of a force F where k is a constant. The work done by the force is proportional to

(a) t (b)t (c) 3/2t (d)t2

Ans. (b)

3/2x kt

Velocity 1/2dx 3ν kt

dt 2 (i)

Acceleration 1/2 1/2dν 3 1 3ka k t t

dt 2 2 4

Force 1/23kmF ma t

4 (ii)

Work done W Fdx (iii)

From Eq (i) dx = 1

23

k t dt2

(iv)

Using (ii) and (iv) in (iii) we have

1

1/2 23km 3

W t kt dt4 2




= 2 29k m 9k m

dt t8 8

Hence W t .

48. A uniform chain of mass M and Length L has a part l of its length hanging over the edge of the table. If

the friction between the chain and the edge is neglected, the word done to pull the length l on the table

is

(a) 2Mgl

L (b)

2Mgl

2L (c)

2MgL

l (d)

2MgL

2l

Ans. (b)

Solution:

Mass per unit length = M

L. Mass of part of length l is m =

Ml

L. Force (weight) of this part is

Mlg

F mgL

Work done l l 2

0 0

Mg MglW Fdl ldl

L 2L

49. One end of an elastic spring of natural length L and spring constant k is fixed to a wall and the other

end is attached to a block of mass m lying on a horizontal frictionless table (fig.,). The block is moved to

a position A so that the spring is compressed to half its natural length and then the released. What is

the velocity of the block when it reaches position B which is at a distance 3L

4from the wall

(a)L k

4 m (b)

L 2k

2 m (c)

L 3k

4 m (d)

L k

2 m

Ans. (c)

Solution:

At position A, the compression of the spring is L

.2

At position B, the compression is 3L L

L4 4

.

Therefore, loss potential energy as the block moves from A to B is given by




2 2 21 L 1 L 3kL

k k2 2 2 4 32

From conservation of energy, loss in P.E. = gain in K.E., i.e.

2

23kL 1mν

32 2

L 3k

ν4 m

50. A car of mass m starts from rest at time t =0 and is driven on straight horizontal road by the engine

which exerts a constant force F. If friction is negligible, the car acquires kinetic energy E at time t and

develops a power P. Which of the following is correct?

(a) E t (b) 2E t (c) 3/2P t (d) 2P t

Ans. (b, c)

Solution:

Since F is constant, acceleration a = F

mis constant. At time t, the velocity of the car is

Ft

ν u at 0 at atm

.

Kinetic energy E at time

2 2

2 21 1 Ft Ftt mν m t .

2 2 m m

Since F is constant, 2E t .

Power P at time t = 2Ft F

Fν F tm m

Thus P t . So the correct choices are (b) and (c).

51. Two identical balls marked 2 and 3, in contact with each other and at rest on a horizontal frictionless

table, are hit head-on by another identical ball marked 1 moving initially with speed v as shown in fig..

What is observed, if collision is elastic?

(a) Ball 1 comes to rest and balls 2 and 3 roll out with speedv

2each.

(b) Balls 1 and 2 come to rest and ball 3 rolls out with speed v each.




(c) Balls 1, 2 and 3 roll out with speed v

3each.

(d) Balls 1, 2 and 3 come to rest.

Ans. (b)

Solution:

The system consists of three identical balls marked 1, 2 and 3. Let m be the mass of each ball. Before

the collision,

KE of the system = KE of 1 + KE of 2 + KE of 3

= 2 21 1mν 0 0 mν

2 2

Case (a) After the collision,


= 2 2

1 ν 1 ν0 m m

2 2 2 2

= 21mν

4

Case (b)


= 2 21 10 0 mν mν

2 2

Case (c)


= 2 2 2

1 ν 1 ν 1 νm m m

2 3 2 3 2 3

= 21mν

6

Case (d)

KE of the system = 0

Now, in an elastic collision, the kinetic energy of the system remains unchanged. Hence choice (b) is

the only possible result of the collision.

52. A body of mass m = 1 kg is dropped from a height h = 40 cm on a horizontal platform fixed to one end

of an elastic spring, the other being fixed to a base, as shown in fig.. As a result the spring is

compressed by an amount x = 10 cm. What is the force constant of the spring. Taken g = 10 ms–2.




(a) 600 Nm–1 (b) 800 Nm–1 (c) 1000Nm–1 (d) 1200 Nm–1

Ans. (c)

Solution:

Since the platform is depressed by an amount x, the total work done on the spring is mg (h + x). This

work is stored in the spring in the form of potential energy 21kx

2. Equating the two, we have

21kx mg h x

2

or 2

2mg h xk

x

Given, h = 0.4 m, x = 0.1 m, m = 1 kg and g = 10 ms–2. Substituting these values, we get k = 1000 Nm–1.

Hence the correct choice is (c).

53. A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction

which varies with the distance x of the particle from the origin as F(x) = –kx + ax3. Here k and a are

positive constants. For x 0 , the functional form of the potential energy U(x) of the particle is (see fig.)




Ans. (d)

Solution:

The potential energy of the particle is given by

3U Fdx kx ax dx

or 2 4 2

2x x xU k a 2k ax

2 4 4 (i)

From Eq. (i) it follows that U = 0 at two values of x which are x = 0 and x = 2k / a . Hence graphs (b)

and (c) are not possible. Also U is maximum or minimum at a value of x given by dU

dx = 0, i.e.

2 4d kx ax

0dx 2 4

= kx – ax3 = x(k – ax2)

or x k / a .

At this value of x, U is maximum if 2

2

d U0

dx,

Now 2

3 2

2

d U dkx ax k 3ax

dxdx.

At x = k / a ,

2

2

d U kk 3a k 3k 2k

adx,

which is negative.

Hence U is maximum at x = k / a .

Hence graph (a) is also not possible. Also U is negative for x 2k / a . Therefore, the correct graph is

(d).

54. Two particles of masses m and 4m have linear momenta in the ratio 2 : 1. What is the ratio of their

kinetic energies?

(a) 2 (b) 2 (c) 4 (d) 16

Ans. (d)

Solution:

Given 1 1 1p m ν 2p and 2 2 2p m ν p , so that

1 1

1 2

m ν2

m ν




The ratio of their kinetic energies is

22 21 1

1 1 1 22 2

2 12 222 2

1m νKE m ν m2 .

1KE mm νm ν2

But 1 12 1

2 2

m νm 4 m and 2.

m ν Therefore,

21

2

KE2 4 16

KE

Hence the correct choice is (d).

55. Three particles of the same mass lie in the x – y plane. The (x, y) coordinates of their positions are

(1, 1), (2, 2) and (3, 3) respectively. The (x, y) coordinates of the centre of mass are

(a) (1, 2) (b) (2, 2) (c) (4, 2) (d) (6, 6)

Ans. (b)

Solution:

The x and y co-ordinates of the centre of mass are

1 1 2 2 3 3

1 2 3

m x m x m xx

m m m

= 1 2 3

1x x x

2 1 2 3m m m

= 1

1 2 3 23

and 1 2 3

1 1y y y y 1 2 3 2.

3 3

Hence the correct choice is (b).

56. A particle is moving in a circle of radius r under the action of a force F = 2αr which is directed towards

centre of the circle. Total mechanical energy (Kinetic energy + potential energy) of the particle is

(take potential energy = 0 for r = 0) :

(a) 31αr

2 (b) 35

αr6

(c) 34αr

3 (d) 3αr

Ans. (b)

Solution:

dU F.dr

r 23

0

αrU αr dr

3




22mv

αrr

2 2 3m v mαr

312m KE αr

2

Total 3 2

3αr αr 5E αr

3 2 3

57. A vector A is rotated by a small angle θ radians θ 1 to get a new vector B . In that case B A

is :

(a) A θ (b) B θ A (c) 2θ

A 12

(d) 0

Ans. (a)

Solution :

Arc length = Radius × Angle

B A A θ

58. A uniform thin rod AB of length L has linear mass density bx

μ x aL

, where x is measure from A. If

the CM of the rod lies at a distance of 7

12L from A, then a and b are related as :

(a) a = 2b (b) 2a = b (c) a = b (d) 3a = 2b

Ans. (b)

Solution:

L 2

0cm L

0

bx(ax )dx

Lx

bx(a )dx

L




a b7L 2 3

b12a

2

b =2a

59. Diameter of a steel ball is measured using a vernier calipers which has divisions of 0.1 cm on its main

side (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main scale. Three such

measurements for a ball are given as :

S. No. MS (cm) VS divisions

1 0.5 8

2 0.5 4

3 0.5 6

If the zero error is – 0.03 cm, then mean corrected diameter is :

(a) 0.53 cm (b) 0.56 cm (c) 0.59 cm (d) 0.52 cm

Ans. (c)

Solution:

Least count = 0.01 cm

d1 = 0.5 + 8 × 0.01 + 0.03 = 0.61 cm

d2 = 0.5 + 4 × 0.01 + 0.03 = 0.57 cm

d3 = 0.5 + 6 × 0.01 + 0.03 = 0.59 cm

Mean diameter = 0.61 0.57 0.59

0.59 cm2

60. A block of mass 0.1 kg is connected to a spring of unknown spring constant k. It is compressed

to a distance x from rest. After approaching half the distance x

2from equilibrium position, it hits

another block and comes to rest momentarily, while the other block moves with a velocity 3ms–1, the

total initial energy of the spring is

(a) 0.3 J (b) 0.6 J (c) 1.5 J (d) 0.8 J

Ans. (b)

Solution :

Apply momentum conservation

0.1u + m(0) = 0.1(0) + m(3)

221 10.1u m 3

2 2

Solving u = 3




2

2 21 1 x 1kx K 0.1 3

2 2 2 2

23Kx 0.9

4

21Kx 0.6 J

2

Section – C {Chemistry}

61. A 5 M solution of H2SO4 is diluted from 1 litre to 10 litres. What is the normality of the solution?

(a) 0.25 N (b) 1 N (c) 2N (d) 7 N

Ans. (b)

Solution:

M1V1= M2V2

2 25 1 M 10 M 0.5

Normality = 0.5 2 = 1N.

62. If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3(PO4) the maximum number of Ba3(PO4)2 that can be

formed is

(a) 0.7 (b) 0.5 (c) 0.3 (d) 0.1

Ans. (d)

Solution:

3BaCl2 + 2Na3PO4 Ba3(PO4)2 + 6NaCl

3 mole 2 mole 1 mole

BaCl2 and Na3PO4 react in the molar ratio 3 : 2 respectively. The given amount is in the ratio 5 : 2, hence

Na3PO4 is limiting reagent.

Ba3(PO4)2 obtained = (1/2) 0.2 = 0.1 mole

63. At a given temperature, density of gas x is twice as that of has y and molar mass of x is one third of gas

y. The ratio of their pressure will be

(a) x

y

P 1

P 4 (b)

x

y

P4

P (c)

x

y

P6

P (d)

x

y

P 1

P 6

Ans. (c)

Solution:

We know, PM = dRT

So, PxMx = dyRT




Dividing one by other

x x x

y y y

P M d RT

P M d RT

x y

x y

given d 2d and

1M M

3

x

y

P6.

P

64. Which of the following plots is incorrect?

(a) (b) (c) (d)

Ans. (a)

Solution:

1V

P at constant temperature.

65. Which of the following statements is not true?

(a) The pressure of a gas is due to collision of the gas molecular with the wall of the container (b) The

molecular velocity of any gas is proportional to the square root of the absolute temperature.

(c) The rate of diffusion of a gas is directly proportional to the density of the gas at constant pressure.

(d) Kinetic energy of an ideal has is directly proportional to the absolute temperature.

Ans. (c)

Solution:

Rate of diffusion of a gas is inversely proportional to the density of the gas at constant pressure.

66. The van der Waals equation reduces itself to the ideal gas equation at

(a) High pressure and low temperature (b) Low pressure and low temperature

(c) Low pressure and high temperature (d) High pressure alone.

Ans. (c)

Solution:

At low pressure and high temperature, the real gases tend to behave ideally.

67. The radius of which of the following orbits is same as that of the first Bohr’s orbit of hydrogen atom?

hydrogen atom is

(a) He+ (n=2) (b) Li2+ (n=2) (c) Li2+ (n=3) (d) Be3+ (n=2)

Ans. (d)

T= constant

P

V P= constant

T

V

V= constant

T

P PV

T




Solution :

Applying, 2

0n 0

r nr where r

Z is the radius of 1st Bohr orbit or ground state of H-atom

As given,2

00

r nr

Z

i.e. n2 = Z which is possible only for Be3+

For Be3+,

n = 2, Z = 4

n2 = Z = 4.

68. The energy of hydrogen atom in its ground state is -13.6 eV. The energy of the level corresponding to

the quantum number n=5 is

(a) –0.54 eV (b) –5.40 eV (c) –0.58 eV (d) –2.72 eV

Ans. (a)

Solution :

0n 2

EE

n for H-atom

s 2

13.6E 0.54eV.

5

69. The wavelength associated with a golf ball weighting 200 g and moving at a speed of 5 m/h is of the

order

(a) 10–10 m (b) 10–20 m (c) 10–30 m (d) 10–40 m

Ans. (c)

Solution :

hλ

mv

m = 200 g = 0.2 kg, v = 5 m/h = 5

3600 m/s

34306.6 10 3600

λ 2.37 10 m.0.2 5

70. The electrons identified by quantum number n and l,

(i) n = 4, l = 1 (ii) n = 4, l = 0

(iii) n = 3, l = 2 (iv) n = 3, l = 1

can be placed in order of increasing energy, from the lowest to highest, as

(a) (iv) < (ii) < (iii) < (i) (b) (ii) < (iv) < (i) < (iii)

(c) (i) < (iii) < (ii) < (iv) (d) (iii) < (i) < (iv) < (ii)




Ans. (a)

Solution :

By applying (n + l) rule, energies of different electrons are in the order (iv) < (ii) < (iii) < (i).

71. Uncertainty in position and momentum are equal, Uncertainty in velocity is

(a)h

(b)h

2 (c)

1 h

2m (d) none of these

Ans. (c)

Solution:

hx. m v but x m v

4π (given).

2 2

2

h h 1 hm v v , v .

4π 2m π4πm

72. Which of the following represents the given mode of hybridization sp2 – sp2 –sp – sp from left to right?

(a) H2C=CH–C N (b) HC C – C H

(c) H2C=C=C=CH2 (d)

Ans. (a)

Solution:

2 2sp sp sp sp

2H C CH C N.

73. Which of the following are isoelectronic and isostructural?

23 3 3 3NO ,CO ,ClO ,SO

(a) 23 3NO ,CO (b) 3 3SO ,NO (c) 2

3 3ClO ,CO (d) 23 3CO ,SO

Ans. (a)

Solution:

Number of electrons in 3NO = 7 + 3 + 8 1 = 32

Number of electrons in 23CO = 6 + 3 8 + 2 = 32

Number of electrons in 3ClO = 17 + 3 8 + 1 = 42

Number of electrons in SO3 = 16 + 3 8 = 40

Thus, 23 3NO and CO are isoelectronic.

CH2

CH2




O O

3NO 23CO

O N

O-

O

C

O

Thus, 23 3NO and CO are isostructural also.

74. According to molecular orbital theory which of the following statements about the magnetic character

and bond order is correct regarding 2O .

(a) Paramagnetic and bond order < O2 (b) Paramagnetic and bond order > O2

(c) Diamagnetic and bond order < O2 (d) Diamagnetic and bond order > O2

Ans. (b)

Solution :

2O 15 electrons :

2 2 2 2 2 1 0z x y x yKK σ2s σ*2s σ2p π2p π2p π*2p π*2p

one unpaired electron, hence paramagnetic.

B.O. = b a

1 8 3N N 2.5

2 2

but B.O. of O2 = 2, Thus 22

OOB.O. B.O.

75. Which of the following reaction corresponds to the definition of enthalpy of formation?

(a) 22 g 2 g l

1H O H O

2 (b) 22 g 2 s l

1H O H O

2

(c) 22 g 2 l l

1H O H O

2 (d) 22 g 2 g l

H O 2H O

Ans. (a)

Solution:

Heat of formation is the heat change when 1 mole of a substance is formed from its constituent

elements in gaseous state.

76. Calculate H of the reaction,

H

C ClH

Cl

(g)C(g) + 2H(g) + 2Cl(g)

The average bond energies of C–H bond and C – Cl bond are 416 kJ and 325 kJ mol–1 respectively.




(a) 1482 kJ (b) –1482 kJ (c) 1282 kJ (d) None of these

Ans. (a)

Solution:

We know that

Reactant ProductH B.E. B.E.

= C H C Cl[2 B.E. 2 B.E. ]

= 2 416 2 325 1482 kJ.

77. Consider the reaction, N2(g) + 3H2(g) 2NH3(g) carried out at constant temperature and pressure. If H

and U are enthalpy and internal energy change for the reaction, which of the following expressions is

true?

(a) H 0 (b) H U (c) H U (d) H U

Ans. (c)

Solution:

2 g 2 g 3 gN 3H 2NH

gH U n RT

gn 2 1 3 2

H U 2RT. Hence H U.

78. Kp/Kc for the reaction g 2 g 2 g

1CO O CO

2

(a) RT (b) (RT)1/2 (c) Kp (d) 1

RT

Ans. (d)

Solution:

gn

p cK K RT

For g 2 g 2 g

1CO O CO

2

g

1 1n 1 1

2 2

1/2

p cK K RT

p

c

Κ 1

K RT.




79. When pressure is applied to the equilibrium system

Ice Water

which of the following phenomenon will happen?

(a) More ice will be formed (b) Water will evaporate

(c) More water will be formed (d) Equilibrium will not be formed.

Ans. (c)

Solution:

Increase of pressure favours the melting of ice which is associated with decrease of volume.

80. The conjugate base of OH– is

(a) O2 (b) H2O (c) O– (d) O2–

Ans. (d)

Solution:

2

acid conjugatebase

OH O H .

81. pH of solution produced when an aqueous solution of pH = 6 is mixed with an equal volume of an

aqueous solution of pH = 3 is about

(a) 3.3 (b) 4.3 (c) 4.0 (d) 4.5

Ans. (a)

Solution:

[H+] in solution having pH of 6 = 10–6 M

[H+] in solution having pH of 3 = 10–3 M

After mixing total [H+] = 6 3

410 105.005 10

2

4pH log [H ] log (5.005 10 )

= 4 log 5.005 3.3

82. The solubility of A2X3 is ‘s’ mol dm–3. Its solubility product is

(a) 6 s4 (b) 64 s4 (c) 36 s5 (d) 108 s5

Ans. (d)




Solution:

Let 3 22 3A X 2A 3X

3 2 2 3spK [A ] [X ]

Since solubility = y mol dm–3

2 2 5spK 2y 3y 108 y .

83. In the standardization of Na2S2O3 using K2Cr2O7 by iodomentry, the equivalent weight of K2Cr2O7 is

(a) (Molecular weight)/2 (b) (Molecular weight)/6

(c) (Molecular weight)/3 (d) Same as molecular weight

Ans. (b)

Solution:

K2Cr2O7 acts as an oxidising agent in presence of dil. H2SO4.

2 32 7 2Cr O 14H 6e 2Cr 7H O

Equivalent weight of K2Cr2O7 = Molecular weight M

Number of electrons gained 6.

84. The reaction, 3aq aq aq3ClO ClO 2Cl is an example of

(a) Oxidation reaction (b) Reduction reaction

(c) Disproportion reaction (d) Decomposition reaction

Ans. (c)

Solution :

1 5

33ClO C lO 2Cl

In the given reaction chlorine is getting oxidised 1 5

3ClO ClO as well as reduced 1

ClO Cl

simultaneously. Therefore it is a disproportionation reaction.

85. The maximum number of hydrogen bonds in which water molecule can participate is

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

Ans. (d)

Solution :

A water molecule can participate in 4 hydrogen bond formation. One each by two H-atoms and one




each by two long pairs present on O-atom.

O

H H

H H

H-bond

O

H

O

H

86. Which of the following statements is incorrect?

(a) H2O2 is a pale blue viscous liquid

(b) H2O2 can act as an oxidizing as well as a reducing agent.

(c) In H2O2 the two hydroxyl groups lie on the same plane.

(d) H2O2 has an ‘open-book’ structure.

Ans. (c)

Solution:

H2O2 has non-planar structure. Two hydroxyl groups lie in different planes.

87. The ionic mobility of alkali metal ions in aqueous solution in maximum for

(a) K (b) Rb (c) Li (d) Na

Ans. (b)

Solution:

The alkali metal ion undergoes hydration in the aqueous solution. The degree of hydration, decreases

with ionic size as we go down the group. Since the mobility of ions is inversely proportional to the size

of the their hydrated ions, hence the increasing order of ionic mobility is

Li+ < Na+ < K+ < Rb+.

88. Thermal stability of alkaline earth metal carbonates decreases in the order

(a) BaCO3 > SrCO3 > CaCO3 > MgCO3 (b) BaCO3 > SrCO3 > MgCO3 > CaCO3

(c) CaCO3 > SrCO3 > MgCO3 >BaCO3 (d) MgCO3 > CaCO3 > SrCO3 > BaCO3

Ans. (a)

Solution:

Thermal stability of alkaline earth metal carbonates increases down the group. Thus order of stability

should be :

BaCO3 > SrCO3 > CaCO3 > MgCO3

89. Amongst LiCl, RbCl, BeCl2 and MgCl2 the compounds with greater and the least ionic character,

respectively are

(a) LiCl and RbCl (b) RbCl and BeCl2 (c) RbCl and MgCl2 (d) MgCl2 and BeCl2




Ans. (b)

Solution:

Smaller cation with large magnitude of charge (Be2+) leads to more covalent character i.e., less ionic

character and large cation with small magnitude of charge (Rb+) leads to less covalent character i.e.,

more ionic character according to Fajan rules. Thus RbCl is most ionic and BeCl2 is least ionic.

90. IUPAC name of

CH3

OH

(a) 3-Methylhexan-2-enol (b) 4- Methylhexan-2-enol

(c) 4-Methyhpentan-3-enol (d) 4-Methylpentan-2-enol

Ans. (d)

Solution :

CH3

OH4

3

2

15

4-Methylpent-2-en-1-ol

–Good Luck

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