Friction HC Verma Exercise Questions Solutions Ch-6 Concept of Physics Vol-1 for ISC Class-11. Step by Step Solution of Exercise Questions of Ch-6 Friction HC Verma Concept of Physics . Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

## FrictionExercise Questions Solutions HC Verma Ch-6 Concept of Physics Vol-1 for ISC Class-11

 Board ISC and other board Publications Bharti Bhawan Publishers Ch-6 Friction Class 11 Vol 1st writer H C Verma Book Name Concept of Physics Topics Solution of Exercise Questions Page-Number 97, 98, 99

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### FrictionExercise Questions Solutions

(HC Verma Ch-6 Concept of Physics Vol-1 for ISC Class-11)

Page-97

#### Question-1 :-

A body slipping on a rough horizontal plane moves with a deceleration of 4.0 m/s2. What is the coefficient of kinetic friction between the block and the plane?

Let m be the mass of the body.

From the free body diagram,
R − mg = 0
(where R is the normal reaction force and g is the acceleration due to gravity)
⇒ R = mg                 (1)
Again ma − μkR = 0
(where μk is the coefficient of kinetic friction and a is deceleration)
or ma = μkR
From Equation (1),
ma =  μkmg
⇒ a = μkg
⇒ 4 = μkg

Hence, the coefficient of the kinetic friction between the block and the plane is 0.4 .

From the above diagram:
R − mg cos θ = 0
⇒ R = mg cos θ                     (1)
For the block, u = 0 m/s, s = 8 m and t = 2 s.
According to the equation of motion

Again,
μkR + ma − mg sin θ = 0
(where μis the coefficient of kinetic friction)
From Equation (1):
μkmg cos θ + ma − mg sin θ = 0
⇒ m (μkg cos θ + a − g sin θ) = 0
⇒ μk × 10 × cos 30° = g sin 30° − a

Therefore, the coefficient of kinetic friction between the block and the surface is 0.11.

From the above diagram:
F − ma − μkR + mg sin 30° = 0
4 − 4a − μkR + 4g sin 30° = 0               (1)
R − 4g cos 30° = 0                               (2)
⇒ R = 4g cos 30° = 0
Substituting the values of R in Equation (1) we get
4 − 4a − 0.11 × 4g cos 30° + 4g sin 30° = 0

= 6.41 N

Because F = 6.41 N, the body will move down the incline with acceleration, hence the force required is zero.

g = 10 m/s2, m = 2 kg, θ = 30 and μ = 0.2
R − mg cos θ − F sin θ = 0
⇒ R = mg cos θ + F sin θ              (1)
and
mg sin θ + μR − F cos θ = 0
⇒ mg sin θ + μ(mg cos θ + F sin θ) − F cos θ = 0
⇒ mg sin θ + μmg cos θ + μF sin θ − F cos θ = 0

Therefore, while pushing the block to move up on the incline, the required force is 17.5 N.

From the above diagram:
R − mg cos θ = 0
⇒ R = mg cos θ                      (1)
and
ma + mg sin θ − μR = 0

For the first half metre, u = 0, s = 0.5 m and t = 0.5 s.
According to the equation of motion,
v = u + at
= 0 + (0.5)4 = 2 m/s

⇒ a = 4 m/s2
For the next half metre, u = 2 m/s, a = 4 m/s2 and s = 0.5.

Therefore, the time taken to cover the next half metre is 0.21 s

(where λ is the angle of friction)
When F = μR, F is the limiting friction (maximum friction). When applied force increases and the body still remains still static then the force of friction increases up to its maximum value equal to limiting friction (μR).

#### Question-11 :-            (Friction HC Verma Exercise)

Consider the situation shown in the following figure. Calculate (a) the acceleration of the 1.0 kg blocks, (b) the tension in the string connecting the 1.0 kg blocks and (c) the tension in the string attached to 0.50 kg.

From the above diagrams:
T + ma − mg = 0
T + 0.5a − 0.5 g = 0                       (1)
μR + ma + T1 − T = 0
μR + 1a + T1 − T = 0                      (2)
μR + 1a − T1 = 0
μR + a = T1                                    (3)
From Equations (2) and (3) we have
μR + a = T − T1
⇒ T − T1 = T1
⇒ T = 2T1
So, Equation (2) becomes
μR + a + T1 − 2T1 = 0
⇒ μR + a − T1 = 0
⇒ T1 = μR + a
= 0.2g + a                       (4)

and Equation (1) becomes
2T1 + 0.5a − 0.5g = 0

= 0.25g – 0.25a                    (5)
From Equations (4) and (5)
0.2g + a = 0.25g − 0.25a

= 0.4 x 10 m/s2                     [g = 10 m/s2] Therefore,
(a) the acceleration of each 1 kg block is 0.4 m/s2,
(b) the tension in the string connecting the 1 kg blocks is
T1 = 0.2g + a + 0.4 = 2.4 N
​    and
(c) the tension in the string attached to the 0.5 kg block is
T = 0.5g − 0.5a
= 0.5 × 10 − 0.5 × 0.4
= 4.8 N.

#### Question-12 :-

If the tension in the string in the following figure is 16 N and the acceleration of each block is 0.5 m/s2, find the friction coefficients at the two contact with the blocks.