1.

The gravitational force between two objects is proportional to 1/R (and not as B
1/R2) where R is separation between them, then a particle in circular orbit under
such a force would have its orbital speed v proportional to

Ro
R1

Sol.

2. The distance between the centres of the Moon and the earth is D. The mass of D
the earth is 81 times the mass of the Moon. At what distance from the centre of
the Earth, the gravitational force will be zero?

Sol.

3. If g is same at a height h and at a depth d, then B

R = 2d
d = 2h
h=d
None
Sol.

4. If the radius of the earth decreases by 10%, the mass remaining unchanged, D
what will happen to the acceleration due to gravity?
Decreases by 19%
Increases by 19%
Decreases by more than19%
Increases by more than 19%
Sol.

5. In order to shift a body of mass m from a circular orbit of radius 3R to a higher C

orbit of radius 5R around the earth, the work done is

Sol.

6. Two satellites A and B of masses m1 and m2 (m1 = 2m2) are moving in circular D
orbits of radii r1 and r2 (r1 = 4r2), respectively, around the earth. If their periods
are TA and TB, then the ratio TA / TB is
4
16
2
8
Sol.

7. The breaking stress for a substance is 106 N/m2. What length of the wire of this C
substance should be suspended vertically so that the wire breaks under its own
weight? (Given: density of material of the wire = 4×103 kg/m3 and g =10 ms–2)
10 m
15 m
25 m
34 m
Sol.

8. When a certain weight is suspended from a long uniform wire, its length C
 increases by 1 cm. If the same weight is suspended from another wire of the
same material and length but having a diameter half of the first one, the increase
in length will be
0.5 cm
2 cm
4 cm
8 cm
Sol.

9. A smooth gate used like a wall containing liquid is kept in equilibrium by A

applying a horizontal force (shown in figure). What is the value of y so that no
horizontal reaction force acts at the pivot?

zero
Sol. The centre of liquid force passes through = .
10. An ideal fluid flows in the pipe as shown in the figure. The pressure in the fluid B
at the bottom P2 is the same as it is at the top P1. If the velocity of the v1 = 2
m/s. Then the ratio of area A1.A2 is

2:1
4:1
8:1
4:3
Sol.

11. A cylindrical vessel of 90 cm height is kept filled up to the brim. It has four B
 holes 1, 2, 3 and 4 which are respectively at heights of 20 cm, 30 cm, 45 cm and
50 cm from the horizontal floor PQ. The water falling at the maximum
horizontal distance from the vessel comes from

Hole number 4
Hole number 3
Hole number 2
Hole number 1
Sol. Horizontal range will be maximum when ℎ = = = 45 , . . , ℎ 3.
12. Two soap bubbles, one of radius 50 mm and the other of radius 80 mm, are B
brought in contact so that they have a common interface. The radius of the
curvature of the common interface is
0.003 m
0.133 m
1.2 m
8.9 m
Sol.

13. If T is surface tension of soap solution, the amount of work done in blowing a C
soap bubble from a diameter D to a diameter 2D is
2πD2T
4 πD2T
6 πD2T
8 πD2T
Sol.

14. Four identical rods are joined end to end to form a square. The mass of each rod D
is M. The moment of inertia of the square about the median line is
Sol.

15. In an experiment with a beam balance, an unknown mass m is balanced by two C

known masses of 16 kg and 4 kg as shown in the figure.

The value of the unknown mass m is

10 kg
6 kg
8 kg
12 kg
Sol.

16. A child is standing with folded hands at the centre of a platform rotating about B
its central axis. The kinetic energy of the system is K. The child now stretches
his arms so that the moment of inertia of the system doubles. The kinetic energy
of the system now is
2K

4K
Sol.
17. A block of mass m is attached to a pulley disc of equal mass m and radius r by D
means of a slack string as shown. The pulley is hinged about its centre on a
horizontal table and the block is projected with an initial velocity of 5 m/s. Its
velocity just after the string becomes taut will be

3 m/s
2.5 m/s
5/3 m/s
10/3 m/s
Sol.

18. A ring of radius R is first rotated with an angular velocity ω0 and then carefully D
 placed on a rough horizontal surface keeping it in vertical plane. The coefficient
of friction between the surface and the ring is µ. Time after which is angular
speed is reduced to half is
µ

µ
Sol.

19. If a spherical ball rolls on a table without slipping, the fraction of its total B
energy associated with rotation is

Sol.

20. A solid cube ABCD of side a and mass M is placed on a rough horizontal A
surface. A bullet of mass m and speed v is shot at the top of the cube so that
cube rotates about point D. The bullet sticks to the cube , m <<M. Find the
angular velocity of the cube.

Sol.

21. A body weighs 64 N on the surface of the earth. What is the gravitational force 36
(in N) on it due to the earth at a height equal to one–third of the radius of the
earth?
Sol.

22. A cube of wood of mass 0.5 kg and density 800 kg m–3 is fastened to the free 5
end of a vertical spring of spring constant k = 50 Nm–1, fixed at the bottom.
Now, the entire system is completely submerged in water. The elongation or
compression of the spring in equilibrium is . The value of β is _____.
(Given, g = 10 ms–2)

Sol.

23. For the arrangement shown in the figure, Area of the cross section of the tank 1
= √5 and area of the orifice a = 4 cm2. If the time interval in seconds
after which the water jet ceases to cross the wall is N × 103 sec. The value of N
is ______. [Assume that the container remaining fixed].

Sol.

24. A disc of radius 0.5m has ω = 2 rad/s and speed vc = 2 m/s. The speed of the 3
plank for pure rolling (in m/s) is______.
Sol.

25. A solid cylinder of mass 3 kg is placed on a rough inclined plane of inclination 5

30o. If g = 10 ms-2, then the minimum frictional force (in N) required for it to
roll without slipping down the plane is

Sol.

26. The NH3 evolved due to complete conversion of N from 1.12 g sample of C
protein was absorbed in 45 mL of 0.4 N HNO3. The excess acid required 20 mL
of 0.1 N NaOH . The % N in the sample is:
8
16
20
25
Sol. W = weight of sample of organic compound.
W= 1.12 gm Meq = milli equivalent
Meq of HNO3 = N × V(mL)
= 0.4 N × 45 mL
= 18
Meq of Base = N(Base) × V(mL)
= 0.1 × 20
=2
.
∴ % N by kjeldhal method = (meq(acid) – meq(base)
.
%N= × (45 × 0.4 − 20 × 0.1)
.
 .
%N= ×(18 –2)
.
% N = 20
27. For the balanced chemical reaction A
xNHO3 + yH2S  aNO + bH2O + 3S
What are the value of x, y, a and b respectively ?
Write x + y + a + b = ?
11
16
38
42
Sol. 2NHO3 + 3H2S  2NO + 4H2O + 3S
28. Match List-I with List-II D
List-I List-II
(Name of oxo acid) (Oxidation state of ‘P’)
A. Hypophosphorous I. +5
B. Orthophosphoric acid II. +4
C. Hypophosphoric acid III. +3
D. Orthophosphorous acid IV. +1
Choose the correct answer from the options given below.
A-II, B-I, C-III, D-IV
A-II, B-III, C-I, D-IV
A-III, B-IV, C-I, D-II
A-IV, B-I, C-II, D-III
Sol.

29. Which of the following pair of molecules/ ions gave same value of bond order? C
O , NO, CN⊝
CN⊝ , N , N
N , NO, O
O , CN⊝ , N
Sol.

30. Match List-I with List-II B

Choose the most appropriate answer from the options given below.
A-IV, B-II, C-III, D-I
A-III, B-IV, C-II, D-I
A-II, B-I, C-IV, D-III
A-I, B-III, C-IV, D-II
31. Which of the following statement is correct regarding following process? C
 IE
(i) Cl EA
 Cl (ii) Cl  Cl
IE 2
IE
(iii) Cl   Cl (iv) Cl  Cl
|IE of process (iv) | = | IE of process (iii)|
|IE of process (iii) | = | IE of process (ii)|
|IE of process (ii) | = | EA of process (i)|
|IE of process (iv) | = | EA of process (i)|
Sol.

32. The forward rate constant for the reversible gaseous reaction, C2H6 ⇌ 2CH3 is B
3.14 × 102 s–1 at 200 K. If 10–5 moles of CH3 and 100 mol of C2H6 are present in
10 L vessel at equilibrium, then the rate constant for the backward reaction at
the temperature is .............. 1023 L mol–1 s–1.
298
314
367
430
Sol.

33. 5.75 g of a dibasic acid was dissolved in water and diluted to 250 mL. 35 mL of D
this solution was neutralised by 15 mL of 1 N NaOH solution. The molecular
weight of the acid will be ............ g/mol.
53.6
98
102
107
Sol.

34. A solution contains Pb+2 ion. In order to precipitate Pb+2 ions, sodium sulphate C
solution is required to be added. What is the concentration of sulphate ion
required to reduce the concentration of Pb+2 to 2 × 10–6 mole per litre? (Ksp for
PbSO4 = 1.8 × 10–8)
5 × 10–3 M
4 × 10–3 M
9 × 10–3 M
6 × 10–3 M
Sol. The cocentration of sulphate ion required to determined as follows
Ksp = [Pb+2] [SO4–2]
1.8 × 10–8 = 2 × 10–6 × [SO4–2]
[ SO4–2] = 9 × 10–3 M
35. Consider the following balanced chemical reaction C
3A + 4B + 5C  2D
If 7 moles of A, 9 moles of B and 8 moles of C are given initially, the mole
fraction of D present after completion of reaction is
0.24
0.65
0.40
0.08
Sol. 3A + 4B + 5C  2D
Mole 7 9 8
After reaction : 2.2 2.6 – 3.2
 C is limiting reagent
Mole fraction of D = 0.4.
36. The Go at 300 K is 2494.2 J for the reaction 2A ⇌ B + C. At a particular time, C
the composition of the reaction mixture is [A] = , [B] = 2 and C = . The
reaction proceeds in the :
[R = 8.314 J/K – mol, e = 2.718] {Given antilog (–0.44) = 0.36}
Reverse direction because Q < KC
Forward direction because Q > KC
Reverse direction because Q > KC
 Forward direction because Q < KC
Sol.

37. Which of the following is the wrong statement? D

ONCl and ONO– are not isoelectronic
O3 molecule is bent
Ozone is violet-black in solid state
Ozone is paramagnetic gas
Sol.

38. The equilibrium constants at 298 K for a reaction A + B ⇌ C + D is 100. If the D

initial concentration of all the four species were 1 M each, the equilibrium
concentration of D (in mol L–1) will be
1.182
0.182
0.818
1.818
Sol.

39. A(g) + 2B(s) ⇌ 2C(g). A

Initially 2 mol A(g), 4 mole of B(s) and 1 mole of an inert gas are present in a
closed container. After equilibrium has established total pressure of container
becomes 9 atm.
If A(g) is consumed 50% at equilibrium then calculate Kp for above reaction
9 atm
36/5 atm
12 atm
6 atm
40. Which of the following is the correct prediction about observed B – F bond C
length in BF3 molecule?
B – F bond length in BF3 is found to be less than theoretical value because the
electronegativity values of B(2.04) and F (4.0) suggest the bond length to ionic
and hence, the attraction between oppositely charged ions must decrease the
bond length.
BF3 and [BF4]– have equal B – F bond length.
The decrease in the B – F bond length in BF3 is due to delocalized p – p
bonding between vacant 2p orbital of B and filled 2p orbital of F
The correct B – X bond length order is B – F > B – Cl > B – Br > B – I
Sol.

It has partial double character in B – F bond due to p – p back bonding.

41. The dissociation equilibrium of a gas AB2 can be represented as, D
2AB2(g) ⇌ 2AB(g) + B2(g)
The degree of dissociation is x and is small as compared to 1. The expression
relating the degree of dissociation (x) with equilibrium constant KP and total
 pressure P is :
1
2
(2K p / P)
Kp/P
2Kp/P
1
3
(2K p / P)
Sol.

42. 0.25 mol of formic acid (HCO2H) is dissolved in enough water to make one litre C
of solution. The pH of that solution is 2.19. The Ka of formic acid is
6.5 × 10–3
4.3 × 10–4
1.7 × 10–4
5.3 × 10–2
Sol.
43. What weight of H2C2O4.2H2O (Molecular weight = 126 g/mol) should be B
dissolved to prepare 250 ml of centinormal solution to be used as reducing
agent?
0.635 g
0.1575 g
0.1263 g
0.835 g
Sol.

44. Which of the following represents the correct order of increasing first ionization A
enthalpy for Ca, Ba, S, Se and Ar?
Ba < Ca < Sr < S < Ar
Ca < Ba < S < Se < Ar
Ca < S < Ba < Se < Ar
S < Se < Ca < Ba < Ar
Sol. Ba < Ca < Sr < S < Ar
Ba, Ca belong to gr IIA (Alkaline earth metals)
Se and S belong to gr VIA (Chalcogens) and as we go down the group
ionaization enthalpy deceases with increasing atomic size. So,
Ba < Ca and Se < S
Ar has stable completely filled electronic configuration and, therefore, has
highest ionization enthalpy. Hence the correct order is
Ba < Ca < Se < S < Ar
–1
in KJmol 503 590 941 999 1520
45. The electronic configuration of few elements is given below. Mark the B
statement which is not correct about these elements.
(i) 1s2 2s2 2p6 3s1
(ii) 1s2 2s2 2p5
(iii) 1s2 2s2 2p6
(iv) 1s2 2s2 2p3
(i) is an alkali metal
(iii) is a noble metal
(i) and (ii) form ionic compounds
(iv) has high ionization enthalpy than accepted
Sol. As (iii) has 1s2 2s2 2p6 type of noble gas configuration so it is an inert gas Neon
not a metal.
46. Among the triatomic molecules/ions, 4
BeCl2, N3–, N2O, NO2+, O3, SCl2, ICl2–, I3– and XeF2,
the total number of linear molecules(s)/ion(s) where the hybridization of the
 central atom does not have contribution from the d-orbital(s) is
[Atomic number : S = 16, Cl = 17, I = 53 and Xe = 54]
Sol.

47. At 200°C, PCl5 dissociates as PCl5(g) ⇌ PCl3(g) + Cl2(g). It was found that the 68
equilibrium vapours are 62 times as heavy as hydrogen. The percentage
dissociation of PCl5 at 200°C is.
[Atomic mass : P = 31, Cl = 35.5]
(Write Nearest Integer)
Sol. M  M 208.5  124
 0   0.681
(n  1).M (2  1).124
48. A student was given 0.01 mole of a weak monoprotic organic acid and told to 5
determine the Ka of the acid. He prepared 100 ml of an aqueous solution
containing the sample. The volume of 50 ml of this solution was then titrated
with NaOH to the equivalence point. The titrated solution was then mixed with
the other 50 ml of the solution of same acid and the pH was determined. A
value of 4.80 was obtained for the mixed solution. What is the value of pKa for
the acid?
(Write Nearest Integer)
Sol. In final solution [HA] = [A–]
49. 0.41 gm of silver salt of a dibasic organic acid (Ag2A) left a residue 0.216 gm of 196
Ag on complete ignition. Then gram molecular weight of a dibasic acid (H2A ;
where A is anionic group of dibasic organic acid) is
(Atomic mass of Ag = 108g/mol):
Sol.

50. Total number of species in which atleast one atom have same hybridization as 5
in central atom of azide ion.
N2O, C2H2, CO2, C3O2, BeF2, NO2, PF3
Sol. N2O, C2H2, CO2, C3O2, BeF2
51. If the locus of the midpoint of the line segment from the point (−2, 3) to a point C
on the circle + − 2 − 2 − 2 = 0 is a circle of radius r, then r =
2
 4
1
3
Sol.

52. The area of the triangle formed by the tangents drawn from a point (2, 3) to the C
circle + + 4 + 6 + 4 = 0 and its chord of contact is (in square units) :
( )

( )

( )

9
Sol.

53. If the circles + + 6 + 8 + 16 = 0 and + + 2 3 − √3 + A

2 4 − √6 = + 6√3 + 8√6, > 0 touch internally at the point ( , ),
then + √3 + + √6 is :
25
16
49
100
Sol.

54. Consider the circle + = 25 and a point A(1, 2) lying inside it. Consider D
secants of the circle passing through the point A. It turns out that the midpoint
of the secants, lie on another circle of centre (a, b) and radius r, then the value of
+ +2 =
 1
2
3
4
Sol.

55. If a chord of the circle + − 4 − 2 − = 0 is trisected at the points A

, and , , then the radius of the circle will be
5
7
9
None of these
Sol.

56. If + = 16 and + = 36 are two circles and P and Q move B

respectively on these circles such that PQ = 4, then the locus of the midpoint of
PQ is a circle of radius is r, then =
19
22
24
21
Sol.

57. Let z and w be two complex numbers such that ( ) = | − 2|, ( )= C

| − 2| and arg( − ) = , then lm( + ) is
√3
√

√
Sol.

58. The mirror image of the curve arg = in the real axis is A

arg =
arg =
arg =
arg =
Sol.

59. If a,b,c are integers, no two of them are equal then minimum value of | + B
+ | is (where w is complex cube root of unity)
1
√3
2
3
Sol.

60. z is a complex number such that | | = 1 and = then ( )= D

| |

| |
√
| |
0
Sol

61. Consider four complex numbers = 2 + 2 , = 2 − 2 , = −2 − 2 , = C

−2 + 2
Statement – 1 : , , and constitute the vertices of a square of the
complex plane.
Statement – 2 : The non-zero complex numbers , ̅, − , − ̅ always form a
square.
Both Statement – 1and Statement – 2 are true and Statement – 2 is the correct
explanation of Statement – 1.
Both Statement – 1and Statement – 2 are true and Statement – 2 is not the
correct explanation of Statement – 1
 Statement – 1 is true but Statement – 2 is false
Statement – 1 is false but Statement – 2 is true
Sol

62. Suppose + + = 0, where , , are in A.P. be a normal to a family of A

circles. The equation of the circle of the family which intersects the circle +
− 4 − 4 − 1 = 0 orthogonally is ‘S’ then radius of ‘S’ is
√8
√7
√6
8
Sol.

63. The vertex A of ∆ is (3, -1). Equation of median BE and internal angular B
bisector CF are 6 + 10 − 59 = 0 and − 4 + 10 = 0 respectively. Then
the −coordinate of B is
-8
8
0
4
Sol.

64. The set of all values of for which the line + = 0 bisects two distinct A
 chords drawn from a point , on the circle 2 +2 − (1 + ) −
(1 − ) = 0 is equal to
(8, ∞)
(0, 4)
(4, ∞)
(2, 7)
Sol

65. Let sides of an isosceles triangle are 7 − − 4 = 0 and + + 1 = 0 if (1,2) D

is on the base then the perpendicular distance from origin to the base is
2
0
3/√10
5/√10
Sol.

66. The ortho centre of a triangle lies on the variable line (1 + 2 ) − (2 + ) = A

4 + 5 and circum centre lies on (1 + 2 ) − (2 + ) = 4 + 5 ∀ , ∈ ,
The centroid of this triangle is ( , ) then + is

0
1
Sol.

67. In a ∆ , + + 2 = 0 is the perpendicular bisector of side AB and it C

meets AB at (-1, -1). If − − 1 = 0 is ⊥ bisector of side AC and it meets
AC at (2, 1) and P is mid point of BC then distance of P from ortho centre of
 ∆ is √ then k is
5
√8
13
7
Sol.

68. = max{( + 2) + ( − 3) } and = min{( + 2) + ( − 3) } where (x, B

y) lies on the circle + + 8 − 10 − 59 = 0. Then + =
212
216
215
None of these
Sol.

69. Let B be the centre of the circle + − 2 + 4 + 1 = 0. Let the tangent at B

Two points P and Q on the circle intersect at the point A(3, 1). Then
∆
8 = then is
∆
1
2
3
4
Sol.

70. The circle passing through the intersection of the circles + −6 = D

0, + − 4 = 0, having its centre on the line − + 1 = 0, then the
radius of the circle is
 √

3
9
√

Sol.

71. Three coins of unit radius are placed in an equilateral triangle as shown in the 13
following figure. Then, the area of the equilateral triangle is + √ , then +
+ =

Sol.

72. If tangent at (1, 2) to the circle + = 5 intersects the circle + = 9 at 27

P and Q and the tangents at P and Q to the second circle meet at the point R (a,
b), then 5(a + b) =
Sol.

73. Let ≠ 1 be a complex cube root of unity. If (4 + 5 + 6 ) + 33

(6 + 5 + 4 ) + (5 + 6 + 4 ) = 0 and ∈ . And ∈
[1,100] then number of values of n is.
Sol.

74. The complex number z satisfies the condition + | | = 2 + 8 . The value of | | 17

is
Sol.

75. A straight line cuts off the intercepts OA = a and OB = b on the positive 130
directionof x-axis and y-axis respectively if the perpendicular from origin O to
this line makes an angle of with positive direction of y-axis and the area of
∆ is √3 . Then [ − ] is equal to (where [.] is G.I.F.)
Sol.