JEE Advanced Previous Year Questions with Solutions on Friction

When an object is placed on a rough surface and is given a push, it does not move. This shows there is another horizontal force opposing the applied force. This opposing force is called the frictional force exerted by the rough surface on the object. When the force applied is gradually increased, the objects start moving. This means there is a limit to the frictional force. If the applied force exceeds the maximum frictional force, the object will start accelerating according to Newton’s law.

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Question 1) When a disc slides on a smooth inclined surface from rest, the time taken to move from A to B is t₁. When the disc performs pure rolling from rest then the time taken to move from A to B is t₂. If

\(\begin{array}{l}\frac{t_{1}}{t_{2}}=\sqrt{\frac{3}{x}}\end{array} \)

find x.

JEE Advanced Previous Year Questions with Solutions on Friction

(a) 2

(b) 1

(c) 5

(d) 7

Answer:(a) 2

Solution:

When disc slides a₁ = gsinθ

So,

\(\begin{array}{l}S = ut_{1}+\frac{1}{2}a_{1}t_{1}^{2}=\frac{1}{2}gsin\theta t_{1}^{2}…………(1)\end{array} \)

When disc do pure rolling

\(\begin{array}{l}a_{2}=\frac{gsin\theta }{1+\frac{k^{2}}{R^{2}}}=\frac{gsin\theta }{1+1/2}=\frac{2}{3}gsin\theta\end{array} \)

So,

\(\begin{array}{l}S= ut_{2}+\frac{1}{2}a_{2}t_{2}^{2}=\frac{1}{2}\times \frac{2}{3}gsin\theta t_{2}^{2}——-(2)\end{array} \)

From (1) and (2)

\(\begin{array}{l}\frac{t_{2}}{t_{1}}=\sqrt{\frac{3}{2}}\end{array} \)

So, x = 2

Question 2) For a body in pure rolling, its rotational kinetic energy is 1/2 times of its translational kinetic energy. The body should be?

(a)Solid cylinder

(b) Ring

(c) Solid sphere

(d) Hollow sphere

Answer: (a)

Solution:

Rotational Kinetic energy = (½) Translational energy

(½) Iω²= (½) x (½ mv²)

In pure rolling, V = rω

(½) Iω²= (½) x (½ mr²ω²)

I = ½ mr²

Hence, it is a solid cylinder.

Question 3) A horizontal force F is applied at the center of mass of a cylindrical object of mass m and radius R, perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is μ. The center of mass of the object has an acceleration a. The acceleration due to gravity is g. Given that the object rolls without slipping, which of the following statement(s) is(are) correct?

(a) For the same F, the value of a does not depend on whether the cylinder is solid or hollow

(b) For a solid cylinder, the maximum possible value of a is 2μg

(c) The magnitude of the frictional force on the object due to the ground is always μmg

(d) For a thin-walled hollow cylinder, a = F/2m

Answer (B, D)

Solution:

For solid cylinder,

\(\begin{array}{l}F\times R= \frac{3}{2}mR^{2}\times \frac{a}{R}\end{array} \)

⇒ a = 2F/3m

For hollow cylinder,

\(\begin{array}{l}F\times R= 2mR^{2}\times \frac{a}{R}\end{array} \)

⇒ a = F/3m

For solid cylinder

f = F – m x (2F/3m) = (F/3) ≤ μmg

⇒ F≤ 3μmg

Therefore,

a ≤ (2/3m) x (3μmg)

⇒ a_max= 2μg

Question 4) A given object takes n times more time to slide down a 45⁰ rough inclined plane as it takes to slide down a perfectly smooth 45⁰ incline. The coefficient of kinetic friction between the object and the incline is

\(\begin{array}{l}(a)\ \sqrt{1-\frac{1}{n^{2}}}\end{array} \)

\(\begin{array}{l}(b)\ 1-\frac{1}{n^{2}}\end{array} \)

(c) 1/(2 – n²)

\(\begin{array}{l}(d)\ \sqrt{\frac{1}{1-n^{2}}}\end{array} \)

Answer: (b)

\(\begin{array}{l}1-\frac{1}{n^{2}}\end{array} \)

Solution:

The coefficient of kinetic friction between the object and the incline.

\(\begin{array}{l}\Rightarrow \mu =\left ( 1-\frac{1}{n^{2}} \right )(\because \theta =45^{0})\end{array} \)

Question 5) Consider a cylinder of mass M resting on a rough horizontal rug that is pulled out from under it with acceleration ‘a’ perpendicular to the axis of the cylinder. What is F_friction at point P? It is assumed that the cylinder does not slip.

(a) Mg

(b) Ma

(c) Ma/2

(d) Ma/3

Answer: (d) Ma/3

Solution:

Force of friction at point P,

F_Friction = (⅓)Ma Sin θ

= (⅓)Ma Sin 90⁰

= Ma/3

Question 6) A body starts from rest on a long inclined plane of slope 45⁰. The coefficient of friction between the body and the plane varies as μ = 0.3x, where x is the distance travelled down the plane. The body will have maximum speed (for g = 10 m/s²) where x =

(a) 9.8 m

(b) 27 m

(c) 12 m

(d) 3.33 m

Answer: (d) 3.33 m

Solution:

When the body has maximum speed then

μ = 0.3x = tan 45°

Therefore, x = 3.33 m

Question 7) A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is 5m/s², the frictional force acting on the block is ………newtons.

Answer: 5 N

Solution:

The frictional force is responsible to move the block of mass 1 kg with an acceleration of 5m/s².

Therefore, frictional force,

f = m x a = 1 x 5 = 5N

Question 8) When a person walks on a rough surface, the frictional force exerted by the surface on the person is opposite to the direction of his motion. Say true or false

Answer: False

Solution:

Friction force opposes the relative motion of the surface of contact.

As the feet push the surface in the backward direction, so frictional force exerted by the surface on the person is in the direction of his motion.

Question 9) A block of mass 2 kg rests on a rough inclined plane making an angle of 30⁰ with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is

(a) 9.8 N

(b) 0.7 x 9.8 x √3 N

(b) 9.8 x √3 N

(d) 0.7 x 9.8 N

Answer: (a) 9.8 N

Solution:

The block is at rest

For equilibrium, frictional force,

f = mg sin θ = mg sin 30⁰

= 2 x 9.8 x (½) = 9.8 N

Also Read:

Friction JEE Main Previous Year Questions With Solutions

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