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CBSE 11thNon-Medical

The Aspire Scholarship

Test 2015

(Maths + Physics + Chemistry)

(Solved)

General Instructions:

The question paper contains 90 objective multiple choice questions.

There are three parts in the question paper consisting of

Section-A MATHEMATICS (1 to 30)

Section-B PHYSICS (31 to 60),

Section-C CHEMISTRY (61 to 90).

Each right answer carries 4 marks and Minus 1 for every wrong answer.

The paper consists of 90 questions. The maximum marks are 360.

Maximum Time 3Hrs.

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

http://www.pioneermathematics.com/




Section – I {Mathematics}

1. If X = n{4 3n 1 : n N} and Y = {9 n 1 : n N} , where N is the set of natural numbers, then X Y is

equal to [JEE – Main – 2014]

(a) N (b) Y – X (c) X (d) Y

Ans.(d)

Solution:

Set X contains elements of the form

4n – 3n – 1 = (1 + 3)n – 3n – 1

= n n n 1 n 2

n 1 23 C 3 ..... C 3

= n 2 n n 1 nn 1 29 3 C 3 ... C

Set X has natural numbers which are multiples of 9 (not all)

Set Y has all multiples of 9

X Y Y

2. A vertical tower CP subtends the equal angle at the point B on the horizontal plane through C, the foot

of the tower and at point A in the vertical plane. If the triangle ABC is equilateral with length on each side

equal to 4m, the height of the tower is

(a) 8 3 m (b) 4 3 m (c) 8

3m (d)

4 3

3m

Ans. (d)

Solution :

CBP = CAP = (angle in same segment) and PCBA is a cyclic quadrilateral

BAP + BCP = 1800

BAP = 900

0( BCP 90 )





= 900 – 600 = 300

If H be the height of the tower CP

then 0CPtan tan 30

CB

H 1

4 3

4 4 3H

33 m

3. If is cube root of unity then 200

200

1tan

4

equals

(a) 1 (b) 1

2 (c) 0 (d) None of these

Ans. (a)

Solution:

200

200

1tan

4

= 2tan tan4 4

( 21 0 )

= 3

tan 14

.

4. If (10)9 + 2(11)1 (10)8 + 3(11)2 (10)7 + …. + 10(11)9 = k(10)9, then k is equal to [JEE – Main 2014]

(a) 121

10 (b)

441

100 (c) 100 (d) 110

Ans. (c)

5. Number of irrational terms in the expansion of 60

5 102 3 are

(a) 54 (b) 61 (c) 30 (d) 31

Ans. (a)

Solution :

Given 601 160

5 10 5 102 3 2 3

Now L. C. M. of 5 and 10 is 10





Number of rational terms let us write

60 r r1 160 5 10

r 1 rT C 2 3

=

r r12

60 5 10rC 2 3

As 0 r 60

r = 0, 10, 20, 30, 40, 50, 60

Number of rational terms are 7

Number of irrational terms equals to

Total Number of terms – Number of rational terms

= 61 – 7 = 54

6. The digit in the unit position of the integer 1! + 2! + 3! + … + 98! Is

(a) 0 (b) 3 (c) 4 (d) 5

Ans. (b)

7. If is a root of 225cos 5 cos 12 0, ,2

then sin2 is equal to

(a) 24

25 (b) –

24

25 (c)

13

18 (d) –

13

18

Ans. (b)

Solution:

Since, is a root of 252cos 5cos 12 0.

225 cos 5 cos 12 0

5 cos 3 5 cos 4 0

4 3cos ,

5 5

But 2

ie, in second quadrant.

4

cos5

3sin

5





Now, sin 2 2 sin cos

= 3 4 24

25 5 25

8. In a triangle ABC, medians AD and BE are drawn. If AD = 4, DAB and ABE ,6 3

then the area of

the ABC is

(a) 8

3sq unit (b)

16

3sq unit (c)

32

3 3sq unit (d)

64

3 sq unit

Ans. (c)

Solution :

Given, AD = 4 and BD = DC

Since, the centroid G divides the line AD in the ratio 2 : 1.

8 4

AG and DG3 3

In AG

ABG, tan3 BG

BG AG cot3

8 1 8

BG3 3 3 3

Area of 1

ADB AD BG2

= 1 8 16

42 3 3 3 3

Since, median divides a triangle into two triangles of equal area. Therefore





Area of ABC 2 area of ADB

= 16 32

2 sq unit3 3 3 3

9. Let , be such that 3 . If sin 21 27

sin and cos cos ,65 65

then the value of cos

is2

(a) 3

130 (b)

3

130 (c)

6

65 (d) –

6

65

Ans. (a)

Solution:

Given that,

21

sin sin65

..(i)

and 27

cos cos65

..(ii)

On squaring and adding Eqs. (i) and (ii), we get

2 2 2 2sin sin 2 sin sin cos cos 2cos cos

= 2 2

21 27

65 65

2 2 cos cos sin sin

= 441 729

4225 4225

1170

2[1 cos ]4225

2 1170cos

2 4 4225

2 9cos

2 130

3

cos2 130

33

2 2 2





10. The possible values of 0, such that sin sin 4 sin 7 0 are

(a) 2 4 3 8

, , , , ,9 4 9 2 4 9

(b)

5 2 3 8, , , , ,

4 12 2 3 4 9

(c) 2 2 3 35

, , , , ,9 4 2 3 4 36

(d)

2 2 3 8, , , , ,

9 4 2 3 4 9

Ans. (a)

Solution :

A = sin2x + cos4x

2 4A 1 cos x cos x

= 4 2 1 3cos x cos x

4 4

= 2

2 1 3cos x

2 4

..(i)

where, 2

2 1 10 cos x

2 4

…(iii)

3

A 14

11. Let a1, a2, a3, … be terms of an AP. If 2

1 2 p 62

1 2 q 21

a a ... a ap, p q, then equals

a a .... a q a

(a) 7

2 (b)

2

7 (c)

11

41 (d)

41

11

Ans. (c)

Solution:

Given that, 2

1 2 p

21 2 q

a a ... a p

a a ... a q

21

2

1

p[2a p 1 d]

p2 ,q q[2a q 1 d]2

where d be a common difference of an AP.

1

1

2a d pd p

2a d qd q

12a d p q 0

1

da

2





Now, 6 1

21 1

a a 5d

a a 20d

=

d5d

112d 4120d2

12. Let an be the nth term of an AP. If 100 100

2r 2r 1r 1 r 1

a and a ,

then the common difference of the AP is

(a) 200

(b) (c)

100

(d)

Ans. (c)

Solution:

Given, a2 + a4 + a6 + … + a200 = …(i)

and a1 + a3 + a5 + … + a199 = ..(ii)

Subtracting Eqs. (i) and (ii), we get

2 1 4 3 6 5 200 199a a a a a a ... a a

d + d + d + … + 100 times =

100d

d100

13. Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is

(a) a function (b) transitive (c) not symmetric (d) reflexive

Ans. (c)

Solution:

Given, R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}.

(a) Since, (2, 4) R and (2, 3) R R. So, R is not a function.

(b) Since, (2, 3) R but (3, 2) R . So, R is not symmetric.

(d) Since, (1, 1), (2, 2), (3, 3), (4, 4) R . So, R is not reflexive.

Hence, the option (c) is correct.

14. All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks

of 10 to each of the students. Which of the following statistical measures will not change even after the

grace marks were given ?

(a) Mean (b) Median (c) Mode (d) Variance





Ans. (d)

Solution:

If initially all marks were, xi, then

2

i21

x x

N

Now, each is increased by 10

2

i22

x 10 x 10 ]

N

= 21

So, variance will not change whereas mean, median and mode will increase by 10.

15. Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon.

If n 1 nT T 10, then the value of n is

(a) 7 (b) 5 (c) 10 (d) 8

Ans. (b)

Solution:

Given, n

n 3T C

n 1n 1 3T C

n 1 n

n 1 n 3 3T T C C 10

(given)

n n n

2 3 3C C C 10

n

2C 10

n = 5

16. The term independent of x in expansion of 10

2/3 1/3 1/2

x 1 x 1

x x 1 x x

is

(a) 4 (b) 120 (c) 210 (d) 310

Ans. (c)

Solution :

10

2/3 1/3 1/2

x 1x 1

x x 1 x x

=

1031/3 3 2

2/3 1/3

x 1 {( x ) 1}

x x 1 x( x 1)





= 10

1/3 x 1x 1

x

= 10

1/3 1/2x x

The general term is

10 r r

10 1/3 1/2r 1 rT C x x

= 10 r r

r10 3 2rC 1

For independent of x, put

10 r r0 20 2r 3r 0

3 2

20 5r r 4

10

5 4

10 9 8 7T C 210

4 3 2 1

17. If the equations x2 + 2x + 3 = 0 and ax2 + bx + c = 0, a, b, c R, have a common root, then a : b : c is

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

Ans. (a)

Solution:

Given equations ARE

x2 + 2x + 3 = 0 ..(i)

and ax2 + bx + c = 0 ..(ii)

Since, Eq. (i) has imaginary roots.

So, Eq. (ii) will also have both roots same as Eq. (i).

Thus, a b c

1 2 3

Hence, a : b : c is 1 : 2 : 3

18. If and are the roots of the equation x2 – x + 1 = 0, then 2009 2009 is equal to

(a) –2 (b) – 1 (c) 1 (d) 2

Ans. (c)

Solution :

Since, and are roots of the equation x2 – x + 1 = 0.

1, 1





1 3i

x2

1 3i 1 3i

x or2 2

2x or

Thus, 2 , then

or2, then 3where 1

Hence, 200920092009 2009 2

= 669 1337

3 2 3[ . . ]

= 2[ ] 1 1

19. If one root of the equation x2 + px + 12 = 0 is 4, while the equation x2 + px + q = 0 has equal roots, then

the value of q is

(a) 49

4 (b) 12 (c) 3 (d) 4

Ans. (a)

Solution:

Since, one of the roots of equation x2 + px + 12 = 0 is 4.

16 + 4p + 12 = 0

4p = – 28

p = – 7

So, the other equation is x2 – 7x + q = 0 whose roots are equal. Let the roots are and .

Sum of roots = 7

1

7

2

and product of roots = . q

2

7 49q q

2 4

20. In a geometrical progression consisting of positive terms, each term equals the sum of the next two

terms. Then, the common ratio of this progression equals

(a) 11 5

2 (b)

15

2 (c) 5 (d) 1

5 12





Ans. (d)

Solution:

Since, each term is equal to the sum of two proceeding terms.

n 1 n n 1ar ar ar

11 r

r

2r r 1 0

5 1 5 1

r r2 2

21. A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4

from the first five questions. The number of choices available to him is

(a) 140 (b) 196 (c) 280 (d) 346

Ans. (b)

Solution:

The number of choices available to him

= 5C4 × 8C6 + 5C5 × 8C5

= 5! 8! 5! 8!

4!1! 6!2! 5!0! 5!3!

= 8 7 8 7 6

5 12 3 2

= 5 × 4 × 7 + 8 × 7

= 140 + 56 = 196

22. How many ways are there to arrange the letters in the word GARDEN with the vowels in alphabetical

order ?

(a) 120 (b) 240 (c) 360 (d) 480

Ans. (c)

Solution :

Total number of ways in which all letters can be arranged in alphabetical order = 6!.

There are two vowels (A, E) in the word GARDEN. Total number of ways in which these two vowels

can be arranged = 2!.

Total number of required ways

= 6!

3602!





23. If z2 + z + 1 = 0, where z is complex number, then the value of

2 2 2 2

2 3 6

2 3 6

1 1 1 1z z z ... z is

z z z z

(a) 54 (b) 6 (c) 12 (d) 18

Ans. (c)

Solution :

Given equation is

z2 + z + 1 = 0

1 1 4 1 1

z2 1

21 3z z ,

2

Now, 2 2 2 2 2 2

2 3 4 5 6

2 3 4 5 6

1 1 1 1 1 1z z z z z z

z z z z z z

= (–1)2 + (–1)2 + (1 + 1)2 + (–1)2 + (-1)2 + (1 + 1)2

when, we put either 2z or z , we get the same result

= 1 + 1 + 4 +1 + 1 + 4 = 12

24. 5th term of a GP is 2, then the product of its 9 terms is

(a) 256 (b) 512 (c) 1024 (d) None of these

Ans. (b)

Solution:

Since, 5th term of a GP = 2

ar4 = 2

where a and r are the first term and common ratio of a GP.

Now, required product

= a × ar × ar2 × ar3 × ar4 × ar5 × ar6 × ar7 × ar8

= a9r36 = (ar4)9

= 29 = 512 [from Eq. (i)]

25. Three straight lines 2x + 11y – 5 = 0, 24x + 7y – 20 = 0 and 4x – 3y – 2 = 0

(a) form a triangle

(b) are only concurrent

(c) are concurrent with one line bisecting the angle between the other two

(d) None of the above





Ans. (c)

Solution :

The angle bisector for the given two lines 24x + 7y –20 = 0 and 4x – 3y – 2 = 0, are

24x 7y 20 4x 3y 2

25 5

Taking positive sign, we get

2x + 11y – 5 = 0

This equation of line is already given.

Therefore the given three lines are concurrent with one line bisecting the angle between the other two.

26. The equation of the directrix of the parabola y2 + 4y + 4x + 2 = 0 is

(a) x = – 1 (b) x = 1 (c) x = – 3/2 (d) x = 3/2

Ans. (d)

Solution:

Given equation of parabola can be rewritten as

2 1

y 2 4 x2

Let y + 2 = Y and 1

x X2

2Y 4X

Here, a = 1

Equation of directrix is X = a

1 3x 1 x

2 2

27. Let A(h, k), B(1, 1) and C(2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the

area of the triangle is 1, then the set of values which ‘k’ can take is given by

(a) {1, 3} (b) {0, 2} (c) {–1, 3} (d) {–3, –2}

Ans. (c)

Solution :

Since, A(h, k), B(1, 1) and C(2, 1) are the vertices of a right angled triangle ABC.





Now, AB = 2 2

AB 1 h 1 k

2 2

BC 2 1 1 1 1

and 2 2

CA h 2 k 1

From Pythagoras theorem

AC2 = AB2 + BC2

4 + h2 – 4h + k2 + 1 – 2k

= h2 + 1 – 2h + k2 + 1 – 2k + 1

5 – 4h = 3 – 2h

h = 1 ..(i)

Now, given that area of triangle is 1.

Then, area 1

ABC AB BC2

2 21

1 1 h 1 k 12

2 2

2 1 h 1 k ..(ii)

2

2 k 1 [From Eq. (i)]

4 = k2 + 1 – 2k

k2 – 2k – 3 = 0

(k – 3) (k + 1) = 0

k = –1, 3

Thus, the set of values of k is {–1, 3}.





28. The point diametrically opposite to the point P(1, 0) on the circle x2 + y2 + 2x + 4y – 3 = 0 is

(a) (3, 4) (b) (3, – 4) (c) (–3, 4) (d) (–3, – 4)

Ans.(d)

Solution:

Given equation can be rewritten as

(x + 1)2 + (y + 2)2 = 2

2 2

Let required point be Q , .

Then, mid point of P(1, 0) and Q , is the centre of the circle.

ie, 1 0

1 and 22 2

3 and 4

Required point is (–3, – 4).

29. The perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y-intercept –4. Then, a

possible value of k is

(a) – 4 (b) 1 (c) 2 (d) –2

Ans.(a)

Solution :

Since, slope of

4 3 1PQ

1 k 1 k

Slope of AM = (k – 1)

Equation of AM is

7 k 1

y k 1 x2 2





For y-intercept, x = 0 , y = – 4

7 k 1

4 k 12 2

215 k 1

2 2

2k 1 15

k2 = 16

k = 4

30. Equation of the ellipse whose axes are the axes of coordinates and which passes through the point

(–3, 1) and has eccentricity2

5 is

(a) 5x2 + 3y2 – 48 = 0 (b) 3x2 + 5y2 – 15 = 0

(c) 5x2 + 3y2 – 32 = 0 (d) 3x2 + 5y2 – 32 = 0

Ans. (d)

Solution:

Given,

2 2

2 2

x y1

a b

Passes through P(–3,1) and e= 2

.5

22

2 2 2

9 1 b1 and e 1

a b a

2 2

9 51

a 3a

and

2

2

2 b1

5 a

2

2 2

27 5 b 31 and

3a a 5





2 232 32a and b

3 5

Equation on ellipse

2 23x 5y

132 32

2 23x 5y 32

Section – II {Physics}

31. The amplitude of a damped oscillator of mass m varies with time t as at /m

0A A e

The dimensions of a are

(a) ML0T–1 (b) M0LT–1 (c) MLT–1 (d) ML–1T

Ans. (a)

Solution :

The exponent is a dimensionless number. Hence at/m is dimensionless. Therefore,

Dimension of a = dimension of m M

dimension of t T

= 0 1M L T

32. The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured

with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale.

The reading on the main scale is 2.5 mm in which 20 divisions of circular scale is coinciding. If the

measured mass of the ball has a relative error of 2%, the relative percentage error in the density is

(a) 0.9% (b) 2.4% (c) 3.1% (d) 4.2%

Ans. (c)

Solution :

Least count of screw gauge

= pitch

number of divisions on circular scale

= 0.5 mm

0.01 mm50

diameter of ball = 2.5 mm + 20 × 0.01 mm

= 2.7 mm

Density 3

M

4r

3

Maximum relative error in is

M 3 r

M r





= 2% +3 ×0.01

1002.7

= 2% + 1.1% = 3.1%

33. A body starts from rest at time t = 0 and undergoes an acceleration as shown in fig.

Which of the graph shown in fig, represents the velocity-time (v – t) graph of the motion of the body

from t = 0 s to t = 4 s?

Ans. (d)

Solution :

Since the body is at rest at t = 0, u = 0. Now from t = 0 to t = 2 s, acceleration a = + 3 ms–2. Therefore,

during this time interval (i.e. from t = 0 to t = 2 s), the velocity of the body is 0 at or 3t.

Thus at t = 1s, = 3 ms–1 and at t = 2s, = 6 ms–1. Hence from t = 0 to t = 2 s, the velocity-time graph is

linear with a positive slope = + 3. Portion AB of graph (d) represents this.

For the next time interval from t = 2 s to t = 4 s, the acceleration a = – 3 ms–2. For this time interval, the





initial velocity u = velocity at the end of the first time interval = 6 ms–2. Therefore, velocity in the time

interval from t = 2 s to t = 4 s is given by

u at 6 3t

Therefore, velocity at t = 3 s (i.e. 1 s after t = 2 s)

= 6 – 3 × 1 = 3 ms–1 and velocity at t = 4 s (i.e. 2s after t = 2s) = 6 – 3 × 2 = 0. The slope of the velocity

time graph from t = 2 to t = 4 s is negative = – 3.

The graph is again linear but its slope is negative = – 3. This is represented by the portion BC in fig (d).

Hence the correct choice is (d).

34. The angle between two vectors A and B is . Vector R is the resultant of the two vectors. If R makes an

angle

2with A, then

(a) A = 2B (b) A =B

2 (c) A = B (d) AB = 1

Ans. (c)

Solution :

The angle which the resultant R makes with A is given by

B sintan

A B cos

given . Hence2

B sintan

2 A B cos

or

sin 2B sin cos2 2 2

A B coscos

2

which gives A = B.

35. Rain is falling vertically with a speed of 4 ms–1. After some time, wind starts blowing with a speed of 3

ms–1 in the west to east direction. In order to protect himself from rain, a man standing on the ground

should hold his umbrella at an angle given by

(a) 1 3

tan4

with the vertical towards south

(b) 1 3

tan4

with the vertical towards north

(c) 1 3

cot4

with the vertical towards south

(d) 1 3

cot4

with the vertical towards north

Ans. (b)

Solution:





Velocity of rain 1r 4 ms vertically downwards. Velocity of wind 1

w 3 ms from north to south

direction. A rain drop is acted upon by two velocities r wand as shown in fig. From the triangle law,

the resultant velocity of the rain drop is OW. In order to protect himself from rain, he must hold his

umbrella at an angle with the vertical (towards north) given by

w

r

RW 3tan

OR 4

Thus the correct choice is (b).

36. A body is projected with a velocity u at an angle with the horizontal. The velocity of the body will

become perpendicular to the velocity of projection after a time t given by

(a) 2u sin

g

(b)

u sin

g

(c)

2u

g sin (d)

u

g sin Ans. (d)

Solution :

Velocity of projection is 0v u cos i u sin j. At time t, the velocity of the body is

v u cos i u sin gt j

The dot product of v0 and v is

2 20v .v u cos u sin u sin -gt

or 20v .v u u sin gt (i)

Since v is perpendicular to v0, v0. v= 0. Using this in (i), we have

2 u0 u u sin gt or t

g sin

Hence the correct choice is (d).

37. A block P of mass m = 1 kg is placed over a plank Q of mass M = 6 kg, placed over a smooth horizontal

surface as shown in fig. Block P is given a velocity v =2 ms–1 to the right. If the coefficient of friction

between P and Q is 0.3, find the acceleration of Q relative to P.





(a) 1.5 (b) 2.5 (c) 3.5 (d) 4.5

Ans. (c)

Solution:

Frictional force between P and Q is f = mg which will retard P and accelerate Q.

Retardation of P is

p

f mga g

m m

Acceleration of Q is

Q

f mga

M M

Acceleration of Q relative to P is

QP Q P

mga a a g

M

=

mg 1

M

=

10.3 10 1

6

= 3.5 ms–2

38. 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)

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

force and will act in a direction opposite to that of F. The force and will act in a direction opposite to

that of F. The force of limiting friction f is given by

f f

R mg

or f mg 0.2 10 10

= 20 N

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

20N 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 = 5 m, u = 0 and a = a’ = 1 ms–2, we get t 10 s.

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

21s' ut at

2

= 1

0 3 10 15m2

39. A body of mass M kg is on the top point of a smooth hemisphere of radius 5 m. It is released to slide

down the surface of the hemisphere. It leaves the surface when its velocity is 5 m/s. At this instant the

angle made by the radius vector of the body with the vertical is :

(Accelerating due to gravity = 10 ms–2)

(a) 300 (b) 450 (c) 600 (d) 900

Ans. (c)

Solution:

Let the body leave the surface at point B as shown in fig. When the body is between points A and B, we

have

2MMg cos N

r

When the body leaves the surface at point B, the normal relation N becomes zero. Thus

2MMg cos

r

or

22 5cos

rg 5 10





= 01or 60

2

Hence the correct choice is (c).

40. A constant power P is supplied to a car of mass m = 3000 kg. The velocity of the car increases from

u = 2 ms–1 to v = 5 ms–1 when the car travels a distance x = 117 m. Find the value of P. Neglect friction.

(a) 1 kw (b) 2k w (c) 3kw (d) 4kw

Ans. (a)

Solution:

PP F ma a

m

Now

d d dx d

a .dt dx dt dx

d P

dx m

2 Pd dx

m

x

2

u 0

Pd dx

m

3 31 Pxu

3 m

3 3m uP

3x

=

3 33000 5 2

3 117

= 1000 W = 1 kW

41. A particle at the origin is under the influence of a force F = kx, where k is a positive constant. If the

potential energy U is zero at x = 0, the variation of potential energy with the coordinate x is

represented by





Ans. (a)

Solution :

Potential energy function is

x x

2

0 0

1U x F dx k x dx k x

2

The value of U(x) is always negative for both positive and negative values of x. Thus the variation of

potential energy with x is an inverted parabola as shown in choice (a).

42. A stone is tied to a string of length L and whirled in a vertical circle with the other end of the string at

the centre. At a certain instant of time, the stone is at the lowest position and has a speed u. The

magnitude of the change in its velocity as it reaches a position where the string is horizontal is

(a) 2u 2gL (b) 2gL (c) 2u gL (d) 22 u gL

Ans. (d)

Solution:





From conservation of energy,

2 21 1mu m mgL

2 2

2u 2gL

Change in velocity u u . Thus is the resultant of and u . It shows from fig. (b)

that

22 u

= 2 2 2u 2gL u 2 u gL

43. The bob of a pendulum is released from a horizontal position A as shown in fig. The length of the

pendulum is 2 m. If 10% of the initial energy of the bob is dissipated as heat due to the friction of air,

what would be the speed of the bob when it reaches the lowermost point B? Take g = 10 ms–2

(a) 3 ms–1 (b) 4 ms–1 (c) 5 ms–1 (d) 6 ms–1

Ans. (d)

Solution :

PE at A = mgh. Since 10% of this energy is lost, KE at point B = mgh × 90

0.9 mgh.100

Therefore,

21m 0.9 mgh

2

or 2 = 1.8 gh = 1.8 × 10 × 2 = 36

which gives = 6 ms–1. Hence the correct choice is (d).

44. A stationary horizontal uniform disc of mass M and radius R is free to rotate about an axis passing

through its centre and perpendicular to its plane. A torque a b is applied to it, where is the

angular displacement and a and b are positive constants. Obtain the expression for the angular

velocity of the disc as a function of .

(a) 2

2

a 2b

MR

(b)

2

2

2 a 2b

MR

(c)

2

2

4 a 2b

MR

(d)

2

2

4 a b

MR

Ans.

Solution : I





or I

d a b

dt I

d d a b

d dt I

a b a b

d d d dI I I

Integrating

0 0 0

a bd d d

I I

2 2a b

2 I 2 I

21a 2b

I

2

2

2a 2b

MR

21I MR

2

45. One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. It is made to

rotate about a line perpendicular to its plane and passing through the center of the original disc.

It s moment of inertia about the axis of rotation is fig.

(a) 21MR

2 (b) 21

MR4

(c) 21MR

8 (d) 22MR

Ans. (a)

Solution :

Area of complete disc = 2R . Area of one quarter sector OAB = 21R .

4 Mass of this sector = M.

Mass per unit area of the sector is

0 2 2

M 4Mm

R /4 R

(1)

Divide the sector into a large number of cylindrical shells. Consider an element of mass dm at a

distance r from the centre O and having thickness dr (see fig.) Then 0

2 rdrdm m

4





The moment of inertia of the sector about the axis of rotation is

R R 42 30 0

0 0

2 m m RI r dm r dr

4 8

(2)

Using (1) in (2), we get 21I MR ,

2 which is choice (a).

46. A uniform disc of radius R is rolling (without slipping) on a horizontal surface with an angular speed

as shown in fig. O is the centre of the disc, points A and C are located on its rim and point B is at a

distance R

2from O. During rolling, the points A , B and C lie on the vertical diameter at a certain instant

of time. If A B Cv , v and v are the linear speeds of points A, B and C respectively at that instant, then

(a) A B C (b) A B C

(c) A C B

40,

3 (d) A C B0, 2

Ans. (c)

Solution :

The disc is rolling about the point O. Thus the axis of rotation passes through the point A and is

perpendicular to the plane of the disc. From the relation r where r is the distance of the point on

the rim about the axis of rotation, we have





A B

3R0, AB

2

and C AC 2R

Hence B

C

3R 1 3.

2 2R 4

Thus the correct choice is (c).

47. A uniform this rod of mass M and length L is hinged by a frictionless pivot at its end O, as shown fig. A

bullet of mass m moving horizontally with a velocity strikes the free end of the rod and gets

embedded in it. The angular velocity of the system about O just after the collision is

(a)

m

L M m

(b)

2m

L M 2m

(c)

3m

L M 3m

(d)

m

LM

Ans. (c)

Solution :

Let be the angular velocity acquired by the system (rod + bullet) immediately after the collision.

Since no external torque acts, the angular momentum of the system is conserved. Thus

m L I (1)

where I is the moment of inertia of the system about an axis through O and perpendicular to the rod.

Thus

I = M.I. of rod about O + M.I. of bullet stuck at its lower end about O

= 2 2 21 1ML ml M 3m L

3 3 (2)

Using Eq. (1) in Eq. (2), we have

21m L M 3m L

3

or

3m

L M 3m


48. A binary star systems consists of two stars of masses M1 and M2 revolving in circular orbits of radii R1

and R2 respectively. If their respective time periods are T1 and T2, then

(a) T1> T2 if R1> R2 (b) T1 > T2 if M1> M2





(c) T1 = T2 (d)

3/2

1 1

2 2

T R

T R

Ans. (c)

Solution:

In a binary star system, the two stars move under their mutual gravitational force. Therefore, their

angular velocities and hence their time periods are equal. Thus the correct choice is (c).

49. A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched

into an orbit of radius 1.01 R. The period of the second satellite is longer than that of the first by

approximately

(a) 0.5% (b) 1.0% (c) 1.5% (d) 3.0%

Ans. (c)

Solution:

According to Kepler’s law of period, T2 = kR3 where k is a constant. Taking logarithm of both sides, we

have

2 log T = log k + 3 log R

Differentiating, we get

T R2 0 3

T R

or T 3 R 3 1.01R R

100T 2 R 2 R

= 1.5%


50. A uniform sphere of mass M and radius R exerts a force F on a small mass m situated at a distance of

2R from the centre O of the sphere. A spherical portion of diameter R is cut from the sphere as shown

in fig. The force of attraction between the remaining part of the sphere and the mass m will be

(a) 7F

9 (b)

2F

3 (c)

4F

9 (d)

F

3 Ans. (a)

Solution:

The force of attraction between the complete sphere and mass m is

2 2

GmM GmMF

4R2R (i)





Mass of complete sphere is M = 34R .

3

Mass of the cut out portion is

3

0

4 Rm .

3 2

Thus,

0

Mm .

8

The distance between the centre of the cut out portion and mass m = 2R – R 3R

.2 2

Hence the force of attraction between the cut out portion and mass m is

02 2 2

G M/8 mGm m GmM 2f

9R /4 4R 93R /2

Using (i), we get 2F

f .9

Therefore, the force of attraction between the remaining part of the sphere and

mass m = F – f = F –2F 7F

9 9 which is choice (a).

51. Figure shows the P-V diagram for a fixed mass of an ideal gas undergoing cyclic process.

AB represents isothermal process and CA represents adiabatic process. Which of the graphs shown in

fig. represents the P-T diagram of the cyclic procee?

Ans. (a)

Solution:

Since AB is an isothermal process, the temperature of the gas remains constant between AB. Hence the

P-T diagram must be perpendicular to the T-axis between A and B. Hence the correct choice is (a).

52. A spherical body of emissivity , placed inside a perfectly black body (emissivity = 1), is maintained at

absolute temperature T. The energy radiated by a unit area of the body per second will be

( is Stefan’s constant)





(a) 4T (b) 4T (c) 41 T (d) 41 T

Ans. (b)

Solution:

The correct choice is (b). According to Stefan’s law, the energy radiated per second by a unit area of a

body of emissivity is 4T , irrespective of the surroundings.

53. An insulated box containing a diatomic gas of molar mass M is moving with a velocity . The box is

suddenly stopped. The resulting change in temperature is (R is the gas constant)

(a)

2M

2R

(b)

2M

3R

(c)

2M

5R

(d)

22M

5R

Ans. (c)

Solution:

Let n be the number of moles of the gas in the box. The kinetic energy of the gas = 21

n M2

.

When the box is suddenly stopped, this energy is used up in changing the internal energy, as a result of

which the temperature of the gas rises. The change in internal energy is given by

5U nC T n R T

2

For a diatomic gas 5

C R.2

Hence 25R 1n T n M

2 2

or

2MT ,

5R

which is choice (c)

54. An ideal gas is expanded isothermally from a volume V1 to volume V2 and then compressed

adiabatically to original volume V1. The initial pressure is P1 and the final pressure is P3. If the net work

done is W, then

(a) P2> P1, W > 0 (b) P3< P1, W < 0 (c) P3> P1, W < 0 (d) P3 = P1, W = 0

Ans. (c)

Solution:

For an isothermal process: PV = constant and for an adiabatic process: PV = constant, where

is the ratio of the two specific heats pC /C of the gas. When a gas is compressed from a volume V to a

volume V V , the increase in pressure is

adia

VPP

V

for an adiabatic compression

and iso

VPP

V

for an isothermal compression.

Hence P3 will be greater than P1. Therefore, the P-V diagrams of isothermal expansion from V1, P1 to V2,

P2 and adiabatic compression for V2, P2 to V1, P3 are as shown in fig. Let W1 and W2 be the work done in

isothermal expansion and adiabatic compression respectively. Therefore, net work done is

W = W1 + (–W2) = W1 – W2





Now, the area under the adiabatic curve is more than that under the isothermal curve.

Hence W2> W1. Therefore, W < 0. Hence the correct choice is (c).

55. An enclosure of volume V contains a mixture of 8 g of oxygen, 14g of nitrogen and 22 g of carbon

dioxide at absolute temperature T. The pressure of the mixture of gases is (R is universal gas constant

(a) RT

V (b)

3RT

2V (c)

5RT

4V (d)

7RT

5V Ans. (c)

Solution:

The pressure exerted by a gas is given by

nRTP

V

= mass RT

molecular weight V

Pressure exerted by oxygen 1

8 RT 1 RTP

32 V 4 V

Pressure exerted by oxygen 2

14 RT 1 RTP

28 V 2 V

Pressure exerted by carbon dioxide

3

22 RT 1 RTP

44 V 2 V

From Daltons’ law of partial pressures, the total pressure exerted by the mixture is given by

P = P1 + P2 + P3

= 1 RT 1 RT 1 RT

4 V 2 V 2 V

= 5RT

,4V

which is choice (c).

56. In the first experiment, two identical conducting rods are joined one after the other and this

combination is connected to two vessels, one containing water at 1000C and the other containing ice at

00C see fig. In the second experiment, the two rods are placed one on top of the other and connected to

the same vessels. If q1 and q2 (in gram per second) are the respectively rates of melting of ice in the





two cases, then the ratio 1

2

q

qis

(a)1

2 (b)

2

1 (c)

1

4 (d)

1

8 Ans. (d)

Solution:

If a steady temperature difference 1 2 is maintained between the ends of a conducting rod of

length L and cross-sectional area A, the rate of flow of heat through the rod is given by

1 2kAq

L

where k is the coefficient of thermal conductivity of the material of the rod. In the first experiment, the

two rods are connected in series. If two rods of equal cross-sectional areas and of lengths L1 and L2 and

conductivities k1 and k2 are joined in series, the equivalent conductivity ks is given by

1 2 1 2

s 1 2

L L L L

k k k

(1)

For two identical rods, L1 = L2 = L and k1 = k2 = k, in which case, Eq. (1) gives ks = k. Further, when two

identical rods are joined in series, the length of the composite rod is (2L) but its cross-sectional area if

A, the same as that of each rod. Hence the rate of flow of heat in this case is given by

1 2

1

kAq

2L

(2)

In the second case, the two rods are connected in parallel. If two rods of equal lengths and equal cross-

section areas having conductivities k1 and k2 are joined in parallel, the equivalent conductivity of the

composite rod is given by

p 1 2k k k

For two identical rods, k1 = k2 = k. Hence kp = (2k). Furthermore, the cross-sectional area of the

composite rod is (2A). Therefore, in the second case, the rate of flow of heat is given by





1 2

2

2k 2Aq

L

(3)

Dividing (2) by (3), we get 1

2

q 1.

q 8 Now, the rate of melting of ice is proportional to the rate of flow of

heat. Hence the correct choice is (d).

57. Calculate the work done (or energy needed) to split a spherical drop of mercury of diameter 1 cm into

8 identical drops. The surface tension of mercury = 0.035 N m–1.

(a) 1.3 × 10–5 (b) 1.2 × 10–5 (c) 1.2 × 10–5 (d) 1.0 × 10–5

Ans.

Solution:

As, r = R

2

Surface area of the big drop = 24 R

Surface area of 8 small drops = 28 4 r

2

2R8 4 8 R

2

Increase in surface area 2 2 2A 8 R 4 R 4 R

Work done = A

= 0.035 × 4 × 3.14 × (0.5 × 10–2)2

= 1.1 × 10–5 J

58. A container of a large uniform cross-sectional area A resting on a horizontal surface holds two

immiscible, non-viscous and incompressible liquids of densities d and 2d, each of height H

2as shown in

fig. The lower density liquid is open to atmosphere. A homogeneous solid cylinder of length

HL L ,

2

cross-sectional area

A

5 is immersed such that it floats with its axis vertical to the liquid

interface with length L

4in the denser liquid. The density of the solid is

(a) 3d

2 (b)

4d

3 (c)

5d

4 (d) 3d

Ans. (c)





Solution :

Mass of solid cylinder = cross-sectional area × length × density

= A 1

L D ALD5 5

Weight of solid cylinder (W) = 1

ALDg5

Buoyant force acting on the cylinder is

FB = weight of denser liquid displaced + weight of lighter liquid displaced

= A L A 3L

2d g d g5 4 5 4

= 1 3 1

ALdg ALdg10 20 4

For the principle of floatation,

BW F

or 1 1 5d

ALDG ALdg or D5 4 4

, Which is choice (c).

59. A wooden block with a coin placed on its top, floats in water as shown in fig. The distances l and h are

shown. After some time the coin falls into the water. Then

(a) l decrease and h increase (b) l increases and h decreases

(c) both l and h increase (d) both l and h decrease

Ans. (d)

Solution:

When the coin falls into water, the downward force reduces causing the block to rise thus decreasing l.

Since the density of the coin is greater then that of water, the volume of coin will be less than the

volume of the liquid displaced, causing a decrease in h. Hence the correct choice is (d).

60. A syringe containing water is held horizontally with its nozzle at a height h above the ground as shown in fig. The cross-sectional areas of the piston and the nozzle are A and a respectively. The piston is pushed with a constant speed V. The horizontal range R of the stream of water on the ground is.





(a) 2h

R Vg

(b) g

R V2h

(c) aV 2h

RA g

(d) AV 2h

Ra g

Ans. (d) Solution : Let be the horizontal speed of water when it emerges from the nozzle. From the equation of continuity, we have

AV a or

AV (1)

Let t be the time taken by the stream of water to strike the ground. The horizontal and vertical distances covered in time t are R t (2)

21h gt

2 (3)

From Eq. (3) we have t =2h

.g

Using this value in Eq. (2), we get

2h

Rg

(4)

Using Eqs. (1) and (4), we have

AV 2h

R ,a g

which is choice (d).

Section – III {Chemistry}

61. The compound 1, 2-butadiene has

(a) only sp hybridized carbon atoms

(b) only sp2 hybridized carbon atoms

(c) both sp and sp2 hybridized carbon atoms

(d) sp, sp2 and sp3 hybridized carbon atoms

Ans. (d)

Solution:





The 1, 2-butadiene is 1 2 3 4

2 3CH C CH CH

The states of hybridization are C1–sp2, C2–sp, C3–sp2 and C4–sp3.

62. Which of the following sets of quantum numbers is possible for an electron in an atom?

(a) n = 3, l = 2, m = + 1, ms = + 1/2 (b) n = 2, l = 3, m = 0, ms = 0

(c) n = 3, l = 4, m = 4, ms = – 1/2 (d) n = 4, l = 4, m = 5, ms = + 1

Ans. (a)

Solution:

The possible quantum numbers satisfy the following requirements

n = 1, 2, 3, …….. m = 0, 1, 2,...... l = 0, 1, 2, …., (n – 1) ms = +1/2 or –1/2

63. Which of the following spectral series of hydrogen atom lies in the visible region of electromagnetic

radiation?

(a) Lyman (b) Balmer (c) Paschen (d) Brackett

Ans. (b)

Solution:

Balmer series lies in the visible region of electromagnetic radiation.

64. The oxidation number of Fe in Fe2(CO)9 is

(a) – 1 (b) 0 (c) +2 (d) +3

Ans. (b)

Solution:

CO is a neutral ligand, hence, the oxidation number of Fe is zero.

65. Pyrophosphoric acid (H4P2O7) is

(a) tribasic reducing acid (b) tetrabasic reducing acid

(c) tetrabasic nonreducing acid (d) tribasic nonreducing acid

Ans. (c)

Solution:

The structure of H4P2O7 is

66. Under the conditions of low pressure and high temperature, the inversion temperature of a gas is

given by

(a) Ti = 2b/Ra (b) Ti = 2R/ab (c) Ti = 2a/Rb (d) Ti = 2ab/R

Ans. (c)





Solution :

Inversion temperature, iT 2a/Rb

67. Which of the following statements correctly represent the functioning of a fuell cell?

(a) Hydrogen is oxidized at the anode and oxygen is reduced at the cathode

(b) Hydrogen is reduced at the cathode and oxygen is oxidized at the anode

(c) Hydrogen is oxidized at the cathode and oxygen is reduced at the anode

(d) Hydrogen is reduced at the anode and oxygen is oxidised at the cathode.

Ans. (a)

Solution:

In a fuel cell, the reaction occurring is 2 2 2

1H g O g H O.

2 . There occurs oxidation of H2(g)

to H+ and O2(g) is changed to OH–.

68. The pH of water at 700C

(a) will be equal to 7.0 (b) will be more than 7.0

(c) will be less than 7.0 (d) will be more than 7.5 but less than 8.0

Ans. (c)

Solution:

There will be more ionization of water, hence more will be H+ as compared to 10–7 M at 250C.

Thus pH will be less than 7.0

69. As ideal gas undergoes an adiabatic reversible expansion. Which of the following conditions does not

hold good?

(a) pV constant (b) 1TV constant

(c) / 1

pT constant

(d) 1TV constant

Ans. (d)

Solution:

The relation 1TV = constant does not hold good.

70. The number of nodes in the radial part of the wave function of 4s orbital is equal to

(a) one (b) two (c) three (d) four

Ans. (c)

Solution:

The nodes are n – l – 1 = 4 – 0 – 1 = 3





71. If o 10 o 12sp sp 2 3K AgCl 1.7 10 ;K Ag CO 8.4 10 , and o 18

sp 3 4K Ag PO 2.7 10 , which of the following

order regarding the solubility of the given salts is correct ?

(a) AgCl > Ag2CO3> Ag3PO4 (b) AgCl > Ag3PO4> Ag2CO3

(c) Ag2CO3> Ag3PO4> AgCl (d) Ag2CO3> AgCl > Ag3PO4

Ans. (c)

Solution:

2 10 5

ss

AgCl Ag Cl ; s 1.7 10 M; s 1.3 10 M

22 12 2 4

2 3 32s s

Ag CO 2Ag CO ; 2s s 8.4 10 M ; s 1.28 10 M

33 18 4 5

3 4 43s s

Ag PO 3Ag PO ; 3s s 2.7 10 M ; s 1.78 10 M

72. The structure of XeF4 is

Ans. (a)

Solution :

There six paired of electrons ground Xe. Thus, Xe involves d2sp3 hybridization. The structure of XeF4 is

73. The combustion reaction occurring in an automobile is 2C8H18(s) + 25O2(g) 16CO2(g) + 18H2O(g).

This reaction is accompanied with

(a) H ve; S ve; G ve (b) H ve; S ve; G ve

(c) H ve; S ve; G ve (d) H ve; S ve; G ve

Ans. (c)

Solution:

The combustion reaction is an exothermic process. Hence, H ve.

The reaction occurs with gv positive. Hence, S ve

Since G H T S ve T ve ve





74. The radius of first orbit in hydrogen atom is 52.9 pm. The radius of third orbit of Li2+ ion will be

(a) 26.45 pm (b) 52. 9 pm (c) 105.8 pm (d) 158.7 pm

Ans. (d)

Solution :

The dependence of radius of an orbit on the nuclear charge Z is r 2n /Z. The value of n2/Z for the

third orbit of Li2+ will be three times that of the first orbit of hydrogen atom. Hence,

r = 3 × 52.9 pm = 158.7 pm

75. Given that E0 values of 2 3Ag | Ag,K |K,Mg |Mg and Cr |Cr

are 0.80 V, –2.93V, –2.37V and –0.74V,

respectively. Which of the following orders regarding the reducing ability of the metal is correct ?

(a) Ag > Cr > Mg > K (b) Ag < Cr < Mg < K (c) Ag > Cr > K > Mg (d) Cr > Ag > Mg > K

Ans. (b)

Solution:

The larger the reduction potential , more easily the ion is reduced and thus the reduction tendency of

the correspond metal is decreased.

76. For the chemical equations

2 5 2 2

1N O 2NO O

2 rate constant = k

2 5 2 22N O 4NO O rate constant = k’

which of the following expressions is correct ?

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

Ans. (b)

Solution:

We will have 2 5 2 52 5 2 5

d[N O ] d[N O ]1k[N O ] and k'[N O ]

dt 2 dt

To have the same value of – d [N2O5]/dt, we will have k = 2k’.

77. The number of lone pair of electrons around Xe in XeOF4 is

(a) one (b) two (c) three (d) zero

Ans. (a)

Solution:

Total number of valence electrons = 8 + 6 + 4 × 7 = 42. Out of these 40 electrons are involved around O

and 4F atoms. Two electrons left will appear as lone pair of electrons around Xe. Hence, there will be

one lone pair of electrons.





78. The IUPAC name of the compound

is

(a) 1-amino-1-phenyl-2-methylpropane (b) 1-amino-2-methyl-1-phenylpropane

(c) 2-methyl-1-amino-1-phenylpropane (d) 1-isopropyl-1-phenylmethyl amine

Ans. (b)

79. The epoxide is

(a) three membered cyclic ethers (b) four membered cyclic ethers

(c) five membered cyclic ethers (d) noncyclic ethers.

Ans. (a)

Solution :

The epoxide are three membered cyclic ethers.

80. The stoichiometric number of species from left to right in the chemical reaction

2 22 4 2Mn PbO H MnO Pb H O when balanced respectively are

(a) 1, 4, 8, 1, 4, 4 (b) 1, 3, 4, 1, 3, 2 (c) 2, 5, 4, 2, 5, 2 (d) 2, 7, 12, 2, 7, 6

Ans. (c)

Solution:

The balanced chemical equation is 2 2

2 4 22Mn 5PbO 4H 2MnO 5Pb 2H O

81. The correct order of bond angles (smallest first) in H2S, NH3, BF3 and SiH4 is

(a) H2S < SiH4< NH3< BF3 (b) NH3< H2S < SiH4< BF3

(c) H2S < NH3< SiH4< BF3 (d) H2S < NH3< BF3< SiH4

Ans. (c)

Solution:

Bond angle is H2S is near to 900

Bond angle in NH3 is less than 1090 28’ (a perfect tetrahedral angle due to the electronic repulsion

between lone pair and bonded pair

Bond angle in SiH4 is 1090 28’

Bond angle in BF3 is 1200

Therefore, the choice is correct.





82. The IUPAC name of the compound is

(a) 3, 3-dimethyl-1-hydroxycyclohexane

(b) 1, 1-dimethyl-3-hydroxycyclohexane

(c) 3,3-dimmethyl-1-cyclohexanol

(d) 1,1-dimethyl-3-cyclohexanol.

Ans. (c)

Solution:

The IUPAC name of the compound is 3, 3-dimethyl-1-cyclohexanol.

Therefore, the choice c is correct.

83. Which of the following species is diamagnetic in nature?

(a) 2H (b) 2H

(c) 2He (d) 2H

Ans. (d)

Solution :

The molecular configurations are

1

2H : 1s 2 1

2He : 1s *1s

2 1

2H : 1s *1s 2

2H : 1s

A diamagnetic substance does not possess unpaired electron. Dihydrogen does not possess unpaired

electron’ hence, it is diamagnetic.

84. The correct order of the thermal stability of hydrogen halides (H – X) is

(a) HI > HBr > HCl > HF (b) HF > HCl > HBr > HI

(c) HCl < HF < HBr < HI (d) HI > HCl < HF < HBr

Ans. (b)

Solution:

The thermal stability of H–X (hydrogen halides) decreases with increase in the atomic number of X.





85. In which of the following arrangements the order is not according to the property indicated against it?

(a) Increasing ionic size Al3+< Mg2+< Na+< F–

(b) Increasing first ionization energy B < C < N < O

(c) Increasing electron gain enthalpy I < Br < F < Cl

(d) Increasing metallic radius Li < Na < K < Rb

Ans. (b)

Solution:

The first ionization energy of N (1s22s22p3) is more than that of O(1s22s32p4) due to the stable half-

filled configuration of N atom.

86. In the dichromate ion

(a) 3Cr–O bonds are equivalent (b) 4Cr–O bonds are equivalent

(c) 6Cr–O bonds are equivalent (d) 7Cr–O bonds are equivalent

Ans. (c)

Solution:

The structure of 2

2 7Cr O ion is

87. The hybridization of atomic orbitals of nitrogen in 2 3 4NO , NO and NH respective are

(a) sp, sp3 and sp2 (b) sp, sp2 and sp3

(c) sp2, sp and sp3 (d) sp2, sp3 and sp

Ans. (b)

Solution:

2NOis linear, 3NO

is planar and 4NHis tetrahedral.

88. Which of the following orders regarding the size of hybrid orbital of carbon is correct?

(a) sp > sp2> sp3 (b) sp < sp2< sp3

(c) sp > sp2< sp3 (d) sp < sp2> sp3

Ans. (b)

Solution:

The larger percentage contribution from s orbital makes the orbital size smaller.





89. The simplest formula of the compound containing 32.5% K, 0.839% H, 26.7% S and 39.9% O by mass

is

(a) KHSO2 (b) KHSO3 (c) KHSO4 (d) K2H2S2O7

Ans. (b)

Solution:

Ratio of atoms K : H: S : O = 32.5 0.039 26.7 39.9

: : : 0.833 : 0.839 : 0.834 : 2.49439 1 32 16

= 1 : 1 : 1 : 3

Empirical formula = KHSO3

90. The expression to compute pH of NH4Cl solution is

(a) 0 0w b 4

1 1 1pH pK pK NH OH log([salt]/M)

2 2 2

(b) 0 0w b 4


2 2 2

(c) 0 0w b 4


2 2 2

(d) 0 0b 4 w

1 1 1pH pK NH OH pK log ([salt]/M)

2 2 2

Ans. (a)

Solution:

4 2 4NH H O NH OH H

w 4b

b 4

K [NH OH][H ]K

K [NH ]

or 0

2 w 40

b

K [NH ][H ]

K [NH4OH]

Also we know that

0 0w w0 0b b

02 w

0b

K K[H ] h C C C

K K

C K[H ]

K

Taking logarithm on both sides, 0 0w b2log[H ] logK logK logC

0 0w b 4

1 1 1log[H ] logK logK (NH OH) log [salt / M]

2 2 2

0 0w b

1 1 1pH pK pK log ([salt]/M)

2 2 2


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