Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
A 0.5 kg mass is in contact against the inner wall of a cylindrical drum of radius 4 m rotating about its vertical axis. The minimum rotational speed of the drum to enable the mass to remain stuck to the wall (without falling) is 5 rad/s. The coefficient of friction between the drum's inner wall surface and mass is ________. (Take $g = 10$ m/s$^2$)
(1) 0.1
(2) 0.5
(3) 0.7
(4) 0.3
Show answer and solution
Answer: Option (1)
Key idea: Circular motion with friction - vertical cylinder rotation
Identify forces: Normal force $N$ (toward center), weight $mg$ (downward), friction $f$ (upward)
For circular motion, centripetal force: $N = m\omega^2 r = 0.5 \times 25 \times 4 = 50$ N
For mass not to fall, friction must balance weight: $f = mg = 0.5 \times 10 = 5$ N
At minimum speed, friction is maximum: $f_{max} = \mu N$
Solve for coefficient of friction: $\mu = \frac{mg}{N} = \frac{5}{50} = 0.1$
Two blocks of masses 2 kg and 1 kg respectively, are tied to the ends of a string which passes over a light frictionless pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in 2 s is _______ m. (Take $g = 10$ m/s$^2$) [Figure: A pulley fixed to ceiling with two masses of 2 kg and 1 kg hanging on either side]
(1) 3.33
(2) 3.12
(3) 2.22
(4) 1.42
Show answer and solution
Answer: Option (3)
Key idea: Atwood Machine and Center of Mass Motion
For the Atwood machine, find acceleration: $a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(2-1) \times 10}{3} = \frac{10}{3}$ m/s²
Calculate displacement of each mass in 2 s: $s = \frac{1}{2}at^2 = \frac{1}{2} \times \frac{10}{3} \times 4 = \frac{20}{3}$ m
The 2 kg mass moves down $\frac{20}{3}$ m, the 1 kg mass moves up $\frac{20}{3}$ m
Displacement of center of mass: $\Delta y_{cm} = \frac{m_1(+\frac{20}{3}) + m_2(-\frac{20}{3})}{m_1 + m_2}$
A spherical ball of mass 2 kg falls from a height of 10 m and is brought to rest after penetrating 10 cm into sand. The average force exerted by sand on the ball is ________ N. (Take $g = 10$ m/s$^2$)
(1) 1980
(2) 2020
(3) 2000
(4) 1000
Show answer and solution
Answer: Option (2)
Key idea: Work-Energy Theorem and Newton's Laws
Find velocity just before hitting sand: $v^2 = 2gh = 2(10)(10) = 200$ m²/s², so $v = 10\sqrt{2}$ m/s
Calculate deceleration in sand using $v^2 = u^2 + 2as$: $0 = 200 + 2a(0.1)$, giving $a = -1000$ m/s²
The time taken by a block of mass $m$ to slide down from the highest point to the lowest point on a rough inclined plane is 50 % more compared to the time taken by the same block on identical inclined smooth plane. Both inclined planes are at 45° with the horizontal. The coefficient of kinetic friction between the rough inclined surface and block is ______.
(1) $3/4$
(2) $2/3$
(3) $5/9$
(4) $4/9$
Show answer and solution
Answer: Option (3)
Key idea: Motion on inclined plane with friction; relating time of descent to coefficient of kinetic friction
For smooth plane: acceleration $a_s = g\sin 45° = \frac{g}{\sqrt{2}}$, and using $s = \frac{1}{2}a_s t_s^2$
At $t = 0$, a body of mass 100 g starts moving under the influence of a force $(5\hat{i} + 10\hat{j})$ N. After 2 s its position is $(2x\hat{i} + 5y\hat{j})$ m. The ratio $x : y$ is _______.
Show answer and solution
Answer: 1.25
Key idea: Newton's Second Law and Kinematics in 2D
Three masses $m_1 = 4$ kg, $m_2 = 4$ kg and $m_3 = 6$ kg are suspended from a fixed smooth frictionless pully as shown in the figure below. The value of $T_1/T_2$ is ______ (take $g = 10$ m/s²) [Figure: A pulley system with three masses suspended. Mass $m_1$ hangs on the left side with tension $T_1$, masses $m_2$ and $m_3$ hang on the right side with tensions $T_1$ and $T_2$, where $m_2$ is above $m_3$]
(1) 5/3
(2) 2/3
(3)
(4)
Show answer and solution
Answer: Option (1)
Key idea: Pulley systems with multiple masses, Newton's second law, constraint relations
Find acceleration: Net force = $(m_2 + m_3 - m_1)g = 60$ N, Total mass = 14 kg, so $a = \frac{60}{14} = \frac{30}{7}$ m/s²
For mass $m_1$ moving up: $T_1 - m_1g = m_1a$, giving $T_1 = m_1(g + a) = 4(10 + \frac{30}{7}) = \frac{400}{7}$ N
For mass $m_3$ moving down: $m_3g - T_2 = m_3a$, giving $T_2 = m_3(g - a) = 6(10 - \frac{30}{7}) = \frac{240}{7}$ N
A wedge $Y$ with mass of 10 kg and all frictionless surfaces and the inclined surface making 37° with horizontal. A block $X$ with mass 2 kg is placed at the highest point of the wedge as shown in figure is at rest. At $t = 0$ wedge $(Y)$ is pulled toward right with constant force $(f)$ of 24 N. Taking the block $X$ at rest at $t = 0$, the time taken by it to slide down 8.8 m on the slope, while $Y$ is on the move, is_____ s. (take tan $(37°) = 3/4$ and $g = 10$ m/s²) [Figure: A wedge Y with a block X on top. The wedge has an inclined surface at 37° with horizontal. A horizontal force f = 24 N is applied to the wedge toward the right.]
(1) 2
(2) 4
(3) $\sqrt{2}$
(4) $2\sqrt{2}$
Show answer and solution
Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Two blocks ($P$ and $Q$) with respectively masses 2 kg and 1.5 kg are joined by a massless thread. These blocks are mounted on a frictionless pully which is fixed on the edge of a cube ($S$), as shown in the figure below. Block $P$ is positioned on the top surface which has no friction and block $Q$ is in contact with side-surface, having coefficient friction $\mu$. The cube ($S$) moves towards the right with acceleration of $\frac{g}{2}$, where $g$ is gravitational acceleration. During this movement the block $P$ and $Q$ remain stationary. The value of $\mu$ is ________. (take $g = 10$ m/s$^2$) [Figure: A cube with block P (2 kg) on top surface connected by thread over pulley to block Q (1.5 kg) on side surface]
(1) 0.33
(2) 0.67
(3) 1
(4) 0.5
Show answer and solution
Answer: Option (2)
Key idea: Non-inertial reference frames, pseudo forces, friction in accelerating systems
Work in the reference frame of the accelerating cube (non-inertial frame with pseudo forces)
For block $P$ on top: Tension $T = m_P \times a = 2 \times \frac{g}{2} = 10$ N
For block $Q$ on side: Pseudo force (horizontal) $= m_Q \times a = 1.5 \times 5 = 7.5$ N pushes $Q$ into wall
Normal force from wall: $N_Q = 7.5$ N (balances pseudo force)
Vertical equilibrium for $Q$: $T + f = m_Q g$, so friction $f = 15 - 10 = 5$ N (upward)
A lift of mass 1600 kg is supported by thick iron wire. If the maximum stress which the wire can withstand is $4 \times 10^8$ N/m$^2$ and its radius is 4 mm, then maximum acceleration the lift can take is ________ m/s$^2$. (take $g = 10$ m/s$^2$ and $\pi = 3.14$)
(1) 2.56
(2) 3.89
(3) 4.32
(4) 5.16
Show answer and solution
Answer: Option (1)
Key idea: Stress-Strain and Newton's Laws in accelerating systems
Calculate the cross-sectional area of the wire: $A = \pi r^2 = 3.14 \times (4 \times 10^{-3})^2 = 5.024 \times 10^{-5}$ m²
Find maximum tension the wire can withstand: $T_{max} = \sigma_{max} \times A = 4 \times 10^8 \times 5.024 \times 10^{-5} = 20096$ N
Apply Newton's second law for upward acceleration: $T = m(g + a)$
At maximum acceleration: $T_{max} = m(g + a_{max})$
A block takes $t$ time to slide down a plane inclined at $45°$ to the horizontal. If the surface is made smooth (frictionless), the block takes time $\frac{t}{2}$ to slide down the plane. The coefficient of friction between the block and the inclined plane is $\left(\frac{\alpha}{100}\right)$. The value of $\alpha$ is _______.
Show answer and solution
Answer: 75
Key idea: Motion on inclined plane with and without friction, kinematic equations
For motion with friction: acceleration $a_1 = g\sin 45° - \mu g\cos 45° = \frac{g}{\sqrt{2}}(1-\mu)$
Two masses of 3.4 kg and 2.5 kg are accelerated from an initial speed of 5 m/s and 12 m/s, respectively. The distances traversed by the masses in the 5$^{th}$ second are 104 m and 129 m, respectively. The ratio of their momenta after 10 s is $\frac{x}{8}$. The value of $x$ is __________.
Show answer and solution
Answer: 9
Key idea: Kinematics and Momentum
Use the formula for distance in $n^{th}$ second: $s_n = u + a(n - \frac{1}{2})$ to find accelerations
For mass 1: $104 = 5 + a_1(4.5) \Rightarrow a_1 = 22$ m/s²
For mass 2: $129 = 12 + a_2(4.5) \Rightarrow a_2 = 26$ m/s²
Find velocities after 10 s: $v_1 = 5 + 22(10) = 225$ m/s and $v_2 = 12 + 26(10) = 272$ m/s
