JEE Advanced Physics Friction PYQ | Previous Year Questions with Solutions

Q 1 :

A block of mass $m$ is on an inclined plane of angle $θ$ . The coefficient of friction between the block and the plane is $μ$ and $\tan θ > μ$ . The block is held stationary by applying a force $P$ parallel to the plane. The direction of force pointing up the plane is taken to be positive. As $P$ is varied from $P_{1} = m g (\sin θ - μ \cos θ)$ to $P_{2} = m g (\sin θ + μ \cos θ)$ , the frictional force $f$ versus $P$ graph will look like [2010]

[IMAGE 46]
[IMAGE 47]
[IMAGE 48]
[IMAGE 49]

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(1)

According to question, $\tan θ > μ$ , so block has a tendency to move down the incline. Force $P$ is applied upwards along the incline to keep the block stationary.

Here, at equilibrium, $P + f = m g \sin θ \Rightarrow f = m g \sin θ - P$

Now as $P$ increases, $f$ decreases linearly with respect to $P$ .

[IMAGE 50]

When $P = m g \sin θ$ , $f = 0$ .

When force $P$ is increased further, the block has a tendency to move upwards along the incline and hence frictional force acts downwards along the incline.

Here, at equilibrium, $P = f + m g \sin θ$

$\Rightarrow f = P - m g \sin θ$

Now as $P$ increases, $f$ increases linearly w.r.t $P$ .

Hence graph (1) correctly depicts the situation.

Q 2 :

A block of base $10 cm \times 10 cm$ and height $15 cm$ is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$ . The inclination $θ$ of this inclined plane from the horizontal plane is gradually increased from $0 °$ . Then [2009]

at $θ = 30 °$ , the block will start sliding down the plane
the block will remain at rest on the plane up to certain $θ$ and then it will topple
at $θ = 60 °$ , the block will start sliding down the plane and continue to do so at higher angles
at $θ = 60 °$ , the block will start sliding down the plane and on further increasing $θ$ , it will topple at certain $θ$

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(2)

[IMAGE 51]

Maximum angle not to slide the block, angle of inclination = angle of repose,

i.e., $\tan^{- 1} μ = \tan^{- 1} \sqrt{3} = 60 °$

For the block to topple, the condition of the block has been shown in the figure.

In $∆ P O M$ , $\tan θ = \frac{P M}{O M} = \frac{10 / 2}{15 / 2} = \frac{5 cm}{7.5 cm} = \frac{2}{3}$

So, $θ < 60 °$ .

From this we can conclude that the block will topple at a lesser angle of inclination. Clearly the block will remain at rest on the plane up to a certain angle $θ$ and then it will topple.

Q 3 :

What is the maximum value of the force $F$ such that the block shown in the arrangement does not move? [2003]

[IMAGE 52]

20 N
10 N
12 N
15 N

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(1)

Since the block is not moving forward for the maximum force $F$ applied, therefore

[IMAGE 53]

$F \cos 60 ° = f = μ N$

(Horizontal direction)

For vertical equilibrium of the block,

$N = m g + F \sin 60 °$

$∴$ $F \cos 60 ° = μ N = μ (F \sin 60 ° + m g)$

$\Rightarrow F = \frac{μ m g}{\cos 60 ° - μ \sin 60 °}$ $= \frac{\frac{1}{2 \sqrt{3}} \times \sqrt{3} \times 10}{\frac{1}{2} - \frac{1}{2 \sqrt{3}} \times \frac{\sqrt{3}}{2}} = \frac{5}{\frac{1}{4}} = 20 N$

Q 4 :

An insect crawls up a hemispherical surface very slowly (see fig.). The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle $α$ with the vertical, the maximum possible value of $α$ is given by [2001]

[IMAGE 54]

$\cot α = 3$
$\tan α = 3$
$\sec α = 3$
$cosec α = 3$

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(1)

[IMAGE 55]

The two forces acting on the insect are $m g$ and $N$ .

Two components of $m g$ are

$m g \cos α balances N$

$m g \sin α is balanced by the frictional force.$

$∴$ $N = m g \cos α$

$f = m g \sin α$

But, $f = μ N = μ m g \cos α$

$∴$ $μ m g \cos α = m g \sin α \Rightarrow \cot α = \frac{1}{μ} \Rightarrow \cot α = 3$

Q 5 :

A block is moving on an inclined plane making an angle $45 °$ with the horizontal and the coefficient of friction is $μ$ . The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define $N = 10 μ$ , then $N$ is [2011]

(5)

[IMAGE 56]

For upward moving of block, pushing force $F_{1} = m g \sin θ + f$

$∴$ $F_{1} = m g \sin θ + μ m g \cos θ = m g (\sin θ + μ \cos θ)$

The force required to just prevent it from sliding down or block just remains stationary,

$F_{2} = m g \sin θ - μ N = m g (\sin θ - μ \cos θ)$

Given, $F_{1} = 3 F_{2}$

$∴$ $\sin θ + μ \cos θ = 3 (\sin θ - μ \cos θ)$

$∴$ $1 + μ = 3 (1 - μ)$ $[∵ \sin θ = \cos θ]$

$∴$ $4 μ = 2 \Rightarrow μ = 0.5$

$∴$ $N = 10 μ = 10 \times 0.5 = 5 N$

Q 6 :

A small block of mass of $0.1 kg$ lies on a fixed inclined plane $P Q$ which makes an angle $θ$ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure.

[IMAGE 57]

The block remains stationary if (take $g = 10 {m/s}^{2}$ ) [2012]

$θ = 45 °$
$θ > 45 °$ and a frictional force acts on the block towards P
$θ > 45 °$ and a frictional force acts on the block towards Q
$θ < 45 °$ and a frictional force acts on the block towards Q

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Select one or more options

(1, 3)

[IMAGE 58]

The various forces acting on the block are as shown in the figure.

When $θ = 45 °$ , $\sin θ = \cos θ$

The block will remain stationary and the frictional force is zero.

When $θ > 45 °$ , $\sin θ > \cos θ$

Therefore a frictional force acts towards Q.

When $θ < 45 °$ , $\cos θ > \sin θ$ .

Therefore a frictional force acts towards P.

Q 7 :

A block of mass $m_{1} = 1 kg$ and another mass $m_{2} = 2 kg$ , are placed together (see figure) on an inclined plane with angle of inclination $θ$ . Various values of $θ$ are given in List-I. The coefficient of friction between the block $m_{1}$ and plane is always zero. The coefficient of static and dynamic friction between the block $m_{2}$ and the plane are equal to $μ = 0.3$ . In List-II expressions for the friction on block $m_{2}$ are given. Match the correct expression of the friction in List-II with the angles given in List-I, and choose the correct option. The acceleration due to gravity is denoted by $g$ .

Useful information: $\tan (5.5 °) \approx 0.1; \tan (11.5 °) \approx 0.2; \tan (16.5 °) \approx 0.3$ [2014]

[IMAGE 59]

List - I List - II

P. $θ = 5 °$ 1. $m_{2} g \sin θ$

Q. $θ = 10 °$ 2. $(m_{1} + m_{2}) g \sin θ$

R. $θ = 15 °$ 3. $μ m_{2} g \cos θ$

S. $θ = 20 °$ 4. $μ (m_{1} + m_{2}) g \cos θ$

Code:

P-1, Q-1, R-1, S-3
P-2, Q-2, R-2, S-3
P-2, Q-2, R-2, S-4
P-2, Q-2, R-3, S-3

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(4)

Block will not slip or will be at rest if $(m_{1} + m_{2}) g \sin θ \leq μ m_{2} g \cos θ$

[IMAGE 60]

$\tan θ \leq \frac{μ m_{2} g}{(m_{1} + m_{2}) g}$

$\Rightarrow \tan θ \leq \frac{μ m_{2}}{m_{1} + m_{2}}$

$\Rightarrow \tan θ \leq \frac{0.3 \times 2}{1 + 2} \leq \frac{1}{5}$

$\Rightarrow \tan θ \leq 0.2 i.e., θ \leq 11.5 °$

i.e., If the angle $θ < 11.5 °$ , the frictional force is less than $μ N_{2} = μ m_{2} g = 0.3 \times 2 \times g = 0.6 g$

and is equal to $(m_{1} + m_{2}) g \sin θ$ .

Blocks will not slip on the inclined plane and friction is static.

At $θ > 11.5 °$ , the bodies start moving on the inclined plane and friction is kinetic and equal to $μ m_{2} g \cos θ$ .

Topic Question Set

Q 2 :

A block of base $10 cm \times 10 cm$ and height $15 cm$ is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$ . The inclination $θ$ of this inclined plane from the horizontal plane is gradually increased from $0 °$ . Then [2009]

Q 3 :

What is the maximum value of the force $F$ such that the block shown in the arrangement does not move? [2003]

[IMAGE 52]

Q 6 :

A small block of mass of $0.1 kg$ lies on a fixed inclined plane $P Q$ which makes an angle $θ$ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure.

[IMAGE 57]

The block remains stationary if (take $g = 10 {m/s}^{2}$ ) [2012]

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	List - I		List - II
P.	$θ = 5 °$	1.	$m_{2} g \sin θ$
Q.	$θ = 10 °$	2.	$(m_{1} + m_{2}) g \sin θ$
R.	$θ = 15 °$	3.	$μ m_{2} g \cos θ$
S.	$θ = 20 °$	4.	$μ (m_{1} + m_{2}) g \cos θ$

Topic Question Set

Q 2 : A block of base 10 cm×10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is 3. The inclination θ of this inclined plane from the horizontal plane is gradually increased from 0°. Then [2009]

Q 3 : What is the maximum value of the force F such that the block shown in the arrangement does not move? [2003] [IMAGE 52]

Q 5 : A block is moving on an inclined plane making an angle 45° with the horizontal and the coefficient of friction is μ. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N=10μ, then N is [2011]

Q 6 : A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle θ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure. [IMAGE 57] The block remains stationary if (take g=10 m/s2) [2012]

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Q 2 :

A block of base $10 cm \times 10 cm$ and height $15 cm$ is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$ . The inclination $θ$ of this inclined plane from the horizontal plane is gradually increased from $0 °$ . Then [2009]

Q 3 :

What is the maximum value of the force $F$ such that the block shown in the arrangement does not move? [2003]

[IMAGE 52]

Q 6 :

A small block of mass of $0.1 kg$ lies on a fixed inclined plane $P Q$ which makes an angle $θ$ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure.

[IMAGE 57]

The block remains stationary if (take $g = 10 {m/s}^{2}$ ) [2012]