Solution
PHYSICS
Class 11 - Physics
1. (a) 4
Explanation: There are three rules on determining how many significant figures are in a number:
Non-zero digits are always significant.
Any zeros between two significant digits are significant.
A final zero or trailing zeros in the decimal portion ONLY are significant.
So keeping these rules in mind, there are 4 significant digits.
2. (a) Unit of volume
Explanation: We know that the units of physical quantities which can be expressed in terms of fundamental units are called
derived units. Mass, length and time are fundamental units but volume is a derived unit (as V = L3)
3.
(c) [ML-1T-1]
Explanation: [η] = [ FΔx
]
AΔv
2
[MLT ][L]
=
2 −1
[L ][LT ]
= [ML-1T-1]
4. (a) moment of force
Explanation: [Moment of force] = [Torque] = [ML2T-2]
5.
(d)
Explanation: As slope of s − t graph decreases with t, so v − t will decrease. At the top of the graph slope is zero, so velocity is
zero. In the downward journey slope of s − t graph increases negatively. So velocity represented by it will be negative.
6. (a) 6.61
Explanation: We know that, for a particle moves in a straight line with a constant acceleration ax, its motion is described by
the kinematic equation given by:
1
Δx = (vxi + vxf ) Δt
2
Rearranging it, we have
2Δx
vi = − vf
Δt
Substituting the values of Δx, Δt and vf into this equation, we get
2(40.0m)
vi = − (2.80m/s) = 6.61m/s
8.50s
7. (a) 6.61
Explanation: Let initial velocity is given by = u
Final velocity is given by v = 2.80 m/s
Distance covered is ,s = 40.0 m
Time taken is, t = 8.50 s
We know,
v = u + at
⇒ v-u = at .....(1)
1/8
Also
1 2
s = ut + at
2
From (1) put value of at, we get
⇒ s = ut +
1
t(v − u)
2
⇒ s = ut + 1
2
tv −
1
2
ut
1 1
⇒ s= 2
ut +
2
tv
Put all the given values, we get
1 1
⇒ 40 = 2
× u × 8.5 +
2
× 2.8 × 8.5
⇒ 80 - 23.8 = 8.5u
⇒ 8.5u = 56.2
⇒ u = 6.61 m/s
8.
(d) its velocity is constant
Explanation: its velocity is constant
9.
–
(b) 1 : √3
Explanation:
Slope of displacement-time graph gives velocity.
v1 tan θ1 ∘
∴
v2
=
tan θ2
= tan 30
∘
tan 45
=
1
√3
10.
–
(b) √3v 0
Explanation: 0 − v = 2(-g)h 2 2
0
or v = 2gh i.e., v ∝ h
2
0
2
0
′2
∴ v ∝ 3h
0
′2
v
0 3h
∴ = = 3
2 h
v
0
–
or v ′
0
= √3v0
11. (a)
Explanation: Except for graph, other graphs show more than one velocity of the particle at single instant of time which is not
possible for realistic situation.
12. (a) c1 + 2c2
Explanation: v(t) = c2t2 + c1t + c0
a= dv
dt
= 2c2t + c1
At t = 1s, a = c1 + 2c2
13.
(d) 27°
2/8
Explanation: Horizontal range,
2
u sin 2θ
R=
g
82×82 sin 2θ
∴ 560 =
10
5600
sin 2θ = = 0.82
6724
or 2θ = 53.8 ∘
⇒ θ = 27
∘
14.
(c) Impulse
Explanation: Since force is a vector quantity, the impulse is also a vector in the same direction. Impulse applied to an object
produces an equivalent vector change in its linear momentum, also in the same direction.
15.
(b) tan −1 1
Explanation: At θ = 45°,
2 2 ∘ 2
u sin 45 u
y = =
2g 4g
2 ∘ 2
1 u sin 90 u
x = ⋅ =
2 g 2g
y 1
∴ tan β = =
x 2
−1 1
⇒ β = tan ( )
2
16.
(b) 5 m/s2
2 2
Explanation: a = rw 2
= r(
2π
T
) =
4π
2
r
= 5 m/s2
2 −2
4π ×5× 10
=
2
(0.2π)
−−
4
−−−−
17. (a) √ v
2
+ a
2
Explanation: Resultant acceleration
−−−−−−−−−−−−−−
2 2
= √a + a
tangential radial
−−−−−−−−−
2
2 −−
4
−−−−
v v
2 2
= √a + ( ) = √ + a
r 2
r
18. (a) 60 ∘
Explanation: Suppose the particle is projected with velocity u at an angle theta with the horizontal. Horizontal component of
its velocity at all height will be u cosθ.
At the greatest height, the vertical component of velocity is zero, so the resultant velocity is
v1 = u cos θ
At half the greatest height during upward motion,
h
y= 2
, ay = -g, uy = u sin θ
Using v − u = 2a y 2
y
2
y y
We get, v − u sin θ = 2(−g)
2
y
2 2 h
2
2 2 2 2 2 2
or v 2
y
= u
2
sin
2
θ− g ×
u sin
2g
θ
=
u sin
2
θ
[∵ h =
u sin
2g
θ
]
u sin θ
or v y =
√2
Hence, resultant velocity at half of the greatest height is
−−−−−−−
2 2
v2 = (√v + (vy ))
1
−−−−−−−−−−−−−−−−
2 2
2 2 u sin θ
= (√u cos θ+ ( ))
2
−
−
v1
Given, v2
= (√
2
5
)
2
v 2 2
1 u cos θ 2
∴ = =
2 2 2 5
v 2 2 u sin θ
2 u cos θ+( )
2
or 1
1
=
2
5
2
1+ tan θ
2
or 2 + tan 2
θ= 5 or tan 2
θ= 3
3/8
–
or tan θ = (√3)
∘
∴ θ = 60
19.
(c) bob will go down in a parabolic path
Explanation: At the mean position, the bob has a horizontal velocity. So when the string is cut, it will fall along a parabolic
path under the effect of gravity.
20. (a) 1929 N
Explanation: For car : f1 = μm1g = 0.001 × 400 × 10 = 4 N
4500 - T - 4 = 400 × a
For coach : f2 = 0.001 × 300 × 10 = 3 N
T - 3 = 300 × a
∴ 4500 - 7 = 700a
or a = 4493
ms
700
−2
Hence T = 300a + 3
4493
= 300 × + 3
700
= 1929 N
21. (a) 1000 N
Explanation: v = 36 km h-1 = 36 × 5
18
ms
−1
= 10 ms-1
2
F =
mv
r
=
500×10×10
50
= 1000 N
22.
(b) 15.65 m/s
Explanation: As the centripetal force is equal to weight (mg) and normal reaction (R), we have
2
mv
r
= R + mg
At a minimum speed, R= 0, Therefore, we have
2
mv
r
= mg
v = √−
−
rg = √25 × 9.8 = 15.65ms
− −1
−−−−−−
23.
(b) radius
2
Explanation: a = v
r
i.e. a ∝ 1
24. (a) 1 m/s2
Explanation: When the body starts moving with acceleration,
P - fk = ma
μs mg − μk mg = ma
a = (μs − μk ) g
= (0.5 - 0.4) × 10 = 1 m/s2
25.
(c) 3.2 ms-2
Explanation: 3.2 ms-2
26.
(c) car will cover less distance before stopping
Explanation: Being lighter than a truck, the car has less kinetic energy. On applying brakes with the same force, the car will
cover less distance before coming to rest.
27.
(b) zero
Explanation: mA = 10 kg and mB = 40 kg
In this case, fL = 0.5 × 40 × 10 = 200N
But F = 40 N
Hence, F < fL
4/8
Hence, mB cannot move with respect to mA.
28.
2ma
(d) g+a
Explanation:
When the balloon descends down with acceleration a,
mg - U = ma
When the balloon moves up with acceleration a,
U - (m - m0)g = (m - m0)a
On adding the two equations,
⇒ mg - mg + m0g = ma + ma - m0a
2ma
⇒ m0 =
g+a
29. (a) 50 J
Explanation: Mass of the body, m = 0.5 kg
3
The velocity of the body v = ax 2
−1
a=5m 2 s-1
Initial velocity at x = 0 is u = 0
–
Final velocity at x = 2 m is v = 10√2 m/s
work done = Change in kinetic energy
W = Kf - Ki
W = mv2 - 0
1
0.5 × (10√2)2 =
–
W= 1
2
×
1
2
× 0.5 × 200 = 50J
30.
(c) 6 J
2 2
Explanation: W = ∫ F dx = ∫ (x + x3) dx = [
2 4
x x
+ ]
2 4
0
0
=6J
31.
(d) 100%
Explanation: Let m be the mass of the body and v1 and v2 be the initial and final velocities of the body respectively
∴ Initial kinetic energy = 1
2
mv
2
1
Final kinetic energy = 1
2
mv
2
2
Initial kinetic energy is increased 300% to get the final kinetic energy
1 2 1 300 2
∴ mv = (1 + ) mv
2 2 2 100 1
v2
⇒ v2 = 2v1 or v1
= 2 ...(i)
Initial momentum = P1 = mv1
Final momentum = p2 = mv2
p2 mv2 v2
∴
p1
=
mv1
=
v1
=2
∴ p2 = 2p1 = (1 + 100
100
) p1
So momentum has increased 100%.
5/8
32. (a) 6
Explanation: W = F ⃗ ⋅ s ⃗
6 = (3^i + c^j + 2k
^
) ⋅ (-4^i + 2^j + 3k
^
)
6 = -12 + 2c + 6
c=6
33.
(c) 3.61 m/s
Explanation: ΔK = W
F = 36N
f = μR = μmg = 0.3 × 4.3 × 9.8 = 12.642N
Fnet = F - f = 36 - 12.642 = 23.358N
s = 1.2 m
mv - 0 = Fnet × s
1 2
2
× 4.3 × v2 = 23.358 × 1.2
v
2
=
23.358×1.2×2
4.3
= 13.03
−−−−
v= √13.03 = 3.61m/s
34.
(b) 100 J
Explanation: Total work done = Work done against friction + Increase in P.E.
300 J = W + 2 × 10 × 10
W = 300 - 200 = 100 J
35.
(b) 1.30 × 105J
Explanation: Work is done in giving kinetic energy is equal to change in kinetic energy
W = ΔK
1 2
W = mv − 0
2
W= 1
2
× 800 × 18 × 18 = 129600J = 1.3 × 105J
36. (a) 2.5 sec
m
Explanation: By conservation of momentum,
5 × 10 + 20 × 0 = (10 + 20) × v
= 2.5 ms-1
50
∴ v= 20
37.
(d) 3.5 J
2 2
2
x
Explanation: W = ∫ dW = ∫ (2 + x)dx = [2x + 2
]
1
1
= 2(2 - 1) + 1
2
(22 - 12)
3
=2+ 2
= 3.5 J
38.
(b) t3/2
Explanation: We know that
Power, P = Fv = k (constant)
Considering the dimensions,
[P] = [F][v]
⇒ [P] = [MLT-2][LT-1]
⇒ [P] = [ML2T-3]
So, [ML2T-3] = k
⇒ [L2T-3] = k ...[M is constant]
⇒ [L2/T3] = k
6/8
⇒ [L2] = k[T3]
⇒ [L2] ∝ [T3]
⇒ [L] ∝ [T3/2]
Hence, L ∝ t3/2
39.
(b) 1.70
hρgV
Explanation: Power = W
t
=
PV
t
=
t
−3 3 −3
150× 10 ×13.6× 10 ×10×5× 10
=
60
W
−3
=
150×13.6×5×10
6
W
= 1700 × 10-3W = 1.70 W
40.
(b) 5 : 1
Explanation: 5 : 1
41.
∑ m1 ⋅ ri
(d) R = ∑ m1
Explanation: Let us consider a system consisting of N – particles of masses m1, m2,….. mN having position vectors
r1 ,...... r ⃗ respectively.
⃗ , r⃗
2 N
The total mass M of the system is given by
M = m1 +m2 +…….+ mN
We can generalize the definition of the position of centre of mass consisting of N particles, hence the position vector of centre
of mass is given below:-
N N
∑ m r⃗ ∑ m r⃗
i i i i
m r ⃗ +m r ⃗ +………+m r⃗ i=1 i=1
⃗ 1 1 2 2 N N
R = = =
m1 +m2 +……+m N M
N
∑ m
i
i=1
42.
(c) zero
Explanation: As there is no external force and the two bodies move due to mutual force of attraction, so vCM = 0.
43. (a) zero
Explanation:
As there is no external force, the centre of mass of the system does not shift.
44. (a) ( 14
17
,
24
17
)
Explanation:
2×0+3×2+4×2+8×0 14
xCM = =
2+3+4+8 17
2×0+3×0+4×2+8×2 24
yCM = =
2+3+4+8 17
45.
(b) v > Rω
cm
Explanation: Here v CM T > 2πR
7/8
2πR
⇒ vCM >
T
⇒ vCM > ωR
46.
(c) 3
5
^
i +
4
5
^
j
⃗ +m r ⃗ +m r ⃗
m1 r 1
Explanation: r ⃗ CM =
2 2 3 3
m1 + m2 + m3
^ ^ ^ ^
2( i )+4( i + j )+4( j )
=
2+4+4
^ ^
6 i +8 j 3 4
^ ^
= = i + j
10 5 5
47.
(b) Due to explosion CM traces its path back to origin
Explanation: Due to explosion CM traces its path back to origin
48.
(b) 0
Explanation: As there is no external force acting on the system, the centre of mass continues to remain at rest.
49.
(d) zero
Explanation: No external force is acting on the centre of mass of system. It remains at rest. The speed of the CM is zero.
50. (a) the product of the total mass of the system and the velocity of its centre of mass
Explanation: Let us consider a system of n particles of masses m1, m2, ….mN. If M is the total mass of the system.
M = m1 + m2 +……..+ mN
If R⃗ is the position vector of the centre of mass and ⃗ , r2
r1 ⃗ , r3
⃗ . . . . . . rn
⃗ those of constituent particles then
⃗ + m r ⃗ +....+ m
m1 r 1 ⃗ ⃗ +m r ⃗ +m ⃗
N rN m1 r 1 N rN
⃗
=
2 2 2 2
R =
m1 + m2 +.....+ mN M
Differentiating both sides w.r.t. time t, we get
⃗ ⃗
dr 1 ⃗
dr 2 ⃗
dr N
dR
dt
= M
1
[ m1
dt
+ m2
dt
+. . . . . . . . . . + mN
dt
]
⃗ →
dR
Let the velocity of centre of mass is dt
= VCM
→ → →
dr1 dr2 drn →
= v,⃗ ⃗ ....
= v. = vn
dt dt dt
1 2,
→ N
M VCM = m1 v ⃗ + m2 v+.
⃗ . . . + mN v ⃗ = ∑ mi vi⃗
1 2 N
i=1
Hence the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its
centre of mass.
8/8
