Friction — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Friction MCQs with step-by-step solutions (17 questions). Part of Laws of Motion. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Friction · medium
Calculate the acceleration of the block and trolly system shown in the figure. The coefficient of kinetic friction between the trolly and the surface is 0.05. ($g = 10$ m/s$^2$, mass of the string is negligible and no other friction exists).
A. $1.25$ m/s$^2$ ✓ Correct
B. $1.50$ m/s$^2$
C. $1.66$ m/s$^2$
D. $1.00$ m/s$^2$
Solution: For the 10 kg trolly, $T - \mu R = 10a \Rightarrow T - 0.05\times10\times10 = 10a \Rightarrow T - 5 = 10a$ ...(i). For the 2 kg block, $2g - T = 2a \Rightarrow 20 - T = 2a$ ...(ii). Adding, $20 - 5 = 12a \Rightarrow a = \dfrac{15}{12} = \dfrac{5}{4} = 1.25$ ms$^{-2}$.
Q2 — Friction · medium
Two particles $A$ and $B$ are moving in uniform circular motion in concentric circles of radii $r_A$ and $r_B$ with speed $v_A$ and $v_B$ respectively. Their time period of rotation is the same. The ratio of angular speed of $A$ to that of $B$ will be
A. $v_A : v_B$
B. $r_B : r_A$
C. $1 : 1$ ✓ Correct
D. $r_A : r_B$
Solution: The angular speed is $\omega = \dfrac{2\pi}{T}$. Since $T_A = T_B$, $\dfrac{\omega_A}{\omega_B} = \dfrac{2\pi/T_A}{2\pi/T_B} = \dfrac{T_B}{T_A} = \dfrac{1}{1}$, i.e. $1 : 1$.
Q3 — Friction · medium
A body of mass $m$ is kept on a rough horizontal surface (coefficient of friction $= \mu$). Horizontal force is applied on the body, but it does not move. The resultant of normal reaction and the frictional force acting on the object is given $F$, where $F$ is
A. $|F| = mg + \mu mg$
B. $|F| = \mu mg$
C. $|F| \le mg\sqrt{1+\mu^2}$ ✓ Correct
D. $|F| = mg$
Solution: The frictional force $f = \mu N = \mu mg$ (since $N = mg$). The resultant of $N$ and $f$ is $|F| = \sqrt{N^2 + f^2} = \sqrt{(mg)^2 + (\mu mg)^2} = mg\sqrt{1+\mu^2}$. This is the minimum force required to move the object; as the body is not moving, $|F| \le mg\sqrt{1+\mu^2}$.
Q4 — Friction · medium
Which one of the following statements is incorrect?
A. Frictional force opposes the relative motion
B. Limiting value of static friction is directly proportional to normal reaction
C. Rolling friction is smaller than sliding friction
D. Coefficient of sliding friction has dimensions of length ✓ Correct
Solution: The coefficient of sliding friction $\mu_s=\dfrac{N}{F_{sliding}}$ is a ratio of two forces, whose dimensions are the same, so $\mu_s$ is dimensionless (it is not a length). Hence statement (d) is incorrect.
Q5 — Friction · medium
A block $A$ of mass $m_1$ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block $B$ of mass $m_2$ is suspended. The coefficient of kinetic friction between the block and the table is $\mu_k$. When the block $A$ is sliding on the table, the tension in the string is
A. $\dfrac{(m_2+\mu_k m_1)g}{m_1+m_2}$
B. $\dfrac{(m_2-\mu_k m_1)g}{m_1+m_2}$
C. $\dfrac{m_1 m_2(1+\mu_k)g}{m_1+m_2}$ ✓ Correct
D. $\dfrac{m_1 m_2(1-\mu_k)g}{m_1+m_2}$
Solution: For block $A$: $T-m_1a=f_k=\mu_k m_1 g$ ...(i). For block $B$: $m_2g-T=m_2a$ ...(ii). Adding gives $a=\dfrac{(m_2-\mu_k m_1)g}{m_1+m_2}$. Then from (ii), $T=m_2(g-a)=\dfrac{m_1 m_2(1+\mu_k)g}{m_1+m_2}$.
Q6 — Friction · medium
A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches $30^\circ$, the box starts to slip and slides $4.0$ m down the plank in $4.0$ s. The coefficients of static and kinetic friction between the box and the plank will be, respectively
A. 0.6 and 0.6
B. 0.6 and 0.5 ✓ Correct
C. 0.5 and 0.6
D. 0.4 and 0.3
Solution: Static friction: $\mu_s=\tan30^\circ=\dfrac{1}{\sqrt3}=0.6$. For kinetic friction, using $s=ut+\dfrac12at^2$ with $u=0$ and $a=g(\sin\theta-\mu_k\cos\theta)$: $4=\dfrac12 g(\sin30^\circ-\mu_k\cos30^\circ)(4)^2\Rightarrow 5\sqrt3\,\mu_k=4.5\Rightarrow \mu_k\approx0.5$.
Q7 — Friction · medium
A system consists of three masses $m_1$, $m_2$ and $m_3$ connected by a string passing over a pulley $P$. The mass $m_1$ hangs freely and $m_2$ and $m_3$ are on a rough horizontal table (the coefficient of friction $=\mu$). The pulley is frictionless and of negligible mass. The downward acceleration of mass $m_1$ is (Assume, $m_1=m_2=m_3=m$)
A. $\dfrac{g(1-g\mu)}{9}$
B. $\dfrac{2g\mu}{3}$
C. $\dfrac{g(1-2\mu)}{3}$ ✓ Correct
D. $\dfrac{g(1-2\mu)}{2}$
Solution: For the hanging block $m_1$: $mg-T_1=ma$ ...(i). Treating $m_2$ and $m_3$ on the table as one system: $T_1-2\mu mg=2ma$ ...(ii). Adding (i) and (ii): $mg(1-2\mu)=3ma\Rightarrow a=\dfrac{g(1-2\mu)}{3}$.
Q8 — Friction · medium
Three blocks with masses $m$, $2m$ and $3m$ are connected by strings, as shown in the figure. After an upward force $F$ is applied on block $m$, the masses move upward at constant speed $v$. What is the net force on the block of mass $2m$? ($g$ is the acceleration due to gravity).
A. Zero ✓ Correct
B. $2mg$
C. $3mg$
D. $6mg$
Solution: The masses move at constant speed, so the acceleration is zero. Hence the net force on every block, including the $2m$ block, is zero.
Q9 — Friction · medium
The upper half of an inclined plane of inclination $\theta$ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by
A. $\mu=\dfrac{1}{\tan\theta}$
B. $\mu=\dfrac{2}{\tan\theta}$
C. $\mu=2\tan\theta$ ✓ Correct
D. $\mu=\tan\theta$
Solution: The block starts and ends at rest, so the net work done is zero: the loss in potential energy equals the work done against friction over the rough lower half. $mg\sin\theta\cdot L=\mu\,mg\cos\theta\cdot\dfrac{L}{2}\Rightarrow \mu=2\dfrac{\sin\theta}{\cos\theta}=2\tan\theta$.
Q10 — Friction · medium
A block of mass $m$ is in contact with the cart $C$ as shown in the figure. The coefficient of static friction between the block and the cart is $\mu$. The acceleration $\alpha$ of the cart that will prevent the block from falling satisfies
A. $\alpha>\dfrac{mg}{\mu}$
B. $\alpha>\dfrac{g}{\mu m}$
C. $\alpha\ge\dfrac{g}{\mu}$ ✓ Correct
D. $\alpha<\dfrac{g}{\mu}$
Solution: The cart's acceleration $\alpha$ produces a pseudo force $m\alpha$ that presses the block against the cart, giving normal reaction $R=m\alpha$. The block does not fall if friction $\mu R\ge mg\Rightarrow \mu m\alpha\ge mg\Rightarrow \alpha\ge\dfrac{g}{\mu}$.
Q11 — Friction · medium
A block $B$ is pushed momentarily along a horizontal surface with an initial velocity $v$. If $\mu$ is the coefficient of sliding friction between $B$ and the surface, block $B$ will come to rest after a time
A. $\dfrac{v}{g\mu}$ ✓ Correct
B. $\dfrac{g\mu}{v}$
C. $\dfrac{g}{v}$
D. $\dfrac{v}{g}$
Solution: Friction provides the retardation: $\mu mg=ma\Rightarrow a=\mu g$. The block stops after time $t=\dfrac{v}{a}=\dfrac{v}{\mu g}$.
Q12 — Friction · medium
The coefficient of static friction, $\mu_s$, between block $A$ of mass 2 kg and the table as shown in the figure, is 0.2. What would be the maximum mass value of block $B$, so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless ($g=10$ m/s$^2$)
A. 2.0 kg
B. 4.0 kg
C. 0.2 kg
D. 0.4 kg ✓ Correct
Solution: Let B have mass $M$. For B in equilibrium: $T=Mg$. For A (2 kg) at the point of slipping: $T=f_s=\mu_s R=\mu_s mg$. Hence $Mg=\mu_s mg\Rightarrow M=\mu_s m=0.2\times2=0.4$ kg.
Q13 — Friction · medium
A block of mass 10 kg is placed on a rough horizontal surface having coefficient of friction $\mu=0.5$. If a horizontal force of 100 N is applied on it, then the acceleration of the block will be (Take $g=10$ m/s$^2$)
A. $15$ m/s$^2$
B. $10$ m/s$^2$
C. $5$ m/s$^2$ ✓ Correct
D. $0.5$ m/s$^2$
Solution: By Newton's second law, $F-\mu mg=ma\Rightarrow a=\dfrac{F-\mu mg}{m}=\dfrac{100-(0.5)(10)(10)}{10}=\dfrac{50}{10}=5\ \text{m/s}^2$.
Q14 — Friction · medium
A block has been placed on an inclined plane with the slope angle $\theta$, block slides down the plane at constant speed. The coefficient of kinetic friction is equal to
A. $\sin\theta$
B. $\cos\theta$
C. $g$
D. $\tan\theta$ ✓ Correct
Solution: Sliding down at constant speed means the incline is at the angle of repose. Then $f=mg\sin\theta$ and $R=mg\cos\theta$, so $\mu=\dfrac{f}{R}=\dfrac{mg\sin\theta}{mg\cos\theta}=\tan\theta$.
Q15 — Friction · medium
Consider, a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (Take $g=10$ m/s$^2$)
A. 30 m
B. 40 m ✓ Correct
C. 72 m
D. 20 m
Solution: Deceleration $a=\mu g=0.5\times10=5\ \text{m/s}^2$. With $u=72\ \text{km/h}=72\times\dfrac{5}{18}=20\ \text{m/s}$ and $v=0$, $s=\dfrac{u^2}{2a}=\dfrac{20^2}{2\times5}=40\ \text{m}$.
Q16 — Friction · medium
A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is
A. 20% ✓ Correct
B. 25%
C. 35%
D. 15%
Solution: Friction on the part still on the table must balance the weight of the hanging part. If $M$ is the mass of the whole chain of length $L$ and $x$ hangs over: $\mu\dfrac{M}{L}(L-x)g=\dfrac{M}{L}xg\Rightarrow \mu(L-x)=x\Rightarrow \dfrac{x}{L}=\dfrac{\mu}{1+\mu}=\dfrac{0.25}{1.25}=\dfrac15=20\%$.
Q17 — Friction · medium
Starting from rest, a body slides down a $45^\circ$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is
A. 0.80
B. 0.75 ✓ Correct
C. 0.25
D. 0.33
Solution: While sliding down with friction $a=g(\sin\theta-\mu\cos\theta)$, so $t_1=\sqrt{\dfrac{2s}{g(\sin\theta-\mu\cos\theta)}}$. Without friction $t_2=\sqrt{\dfrac{2s}{g\sin\theta}}$. Given $t_1=2t_2\Rightarrow t_1^2=4t_2^2$, so $\sin\theta=4\sin\theta-4\mu\cos\theta$, giving $\mu=\dfrac{3}{4}\tan\theta=\dfrac{3}{4}\tan45^\circ=\dfrac{3}{4}=0.75$.