System of Particles and Rotational Motion MCQs for UPSC Prelims
100 practice questions covering System of Particles and Rotational Motion, organised into 0 topics.
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Sample questions
Q1
medium
The angular momentum of a wheel falls uniformly from $30\ \text{kg m}^2\text{/s}$ to $6\ \text{kg m}^2\text{/s}$ in $8\ \text{s}$. What is the magnitude of the average external torque acting on it?
A$0.75\ \text{N m}$
B$3\ \text{N m}$
C$4.5\ \text{N m}$
D$24\ \text{N m}$
Show answer and explanation
Correct answer: B - $3\ \text{N m}$
Since $\tau = \Delta L/\Delta t$, the torque is $(30 - 6)/8 = 3\ \text{N m}$. The value $24\ \text{N m}$ quotes the change in angular momentum itself without dividing by the time, $4.5\ \text{N m}$ adds the two angular momenta instead of subtracting them, and $0.75\ \text{N m}$ uses only the final value.
Q2
medium
A seesaw has a child weighing 300 N sitting 2 m from the pivot. If a second child sits on the opposite side 3 m from the pivot, what must be the weight of the second child to balance the seesaw?
A200 N
B250 N
C100 N
D150 N
Show answer and explanation
Correct answer: A - 200 N
To balance the seesaw, the moments must be equal: 300 N × 2 m = W × 3 m. Solving gives W = 200 N. Other values result from miscalculating the balance of moments.
Q3
medium
A uniform rod of length 10 m is freely pivoted at its midpoint. A force of 50 N acts downward at one end. What downward force must act at the other end to keep the rod in equilibrium?
A75 N
B100 N
C25 N
D50 N
Show answer and explanation
Correct answer: D - 50 N
The two ends lie on opposite sides of the pivot, so two downward forces turn the rod in opposite senses, exactly as two children balance a seesaw. Equal arms of 5 m mean equal forces, so 50 N is needed. An upward force at the far end would turn the rod the same way as the first force and make matters worse.
Q4
medium
A potter uses a wheel of radius 0.5 m spinning at 10 rad/s. If no torque is applied, how much time will it take for the wheel to come to rest due to a uniform angular deceleration of 2 rad/s²?
A20 s
B2.5 s
C5 s
D10 s
Show answer and explanation
Correct answer: C - 5 s
Using the equation ω = ω₀ + αt, with final ω = 0, initial ω₀ = 10 rad/s, and α = -2 rad/s² (since it's deceleration), we solve for t: 0 = 10 - 2t, giving t = 5 s. Alternatives arise from using wrong signs or incorrect formula application.
Q5
medium
A hollow metal ball, thin-walled enough to count as a spherical shell, has mass $3\ \text{kg}$ and radius $0.2\ \text{m}$. It is spun about a diameter at $10\ \text{rad/s}$. How much rotational kinetic energy does it carry?
A$8.0\ \text{J}$
B$0.4\ \text{J}$
C$2.4\ \text{J}$
D$4.0\ \text{J}$
Show answer and explanation
Correct answer: D - $4.0\ \text{J}$
For a thin spherical shell about a diameter, $I = \tfrac23MR^2 = \tfrac23(3)(0.04) = 0.08\ \text{kg m}^2$, so $K = \tfrac12I\omega^2 = \tfrac12(0.08)(100) = 4.0\ \text{J}$. The value $8.0\ \text{J}$ omits the factor $\tfrac12$, $2.4\ \text{J}$ uses $\tfrac25MR^2$, which belongs to a solid sphere, and $0.4\ \text{J}$ uses $\omega$ rather than $\omega^2$.
Q6
hard
Two particles of masses $2\ \text{kg}$ and $3\ \text{kg}$ lie $10\ \text{m}$ apart on a smooth horizontal surface and attract each other. Released from rest, they move towards each other. How far does the $2\ \text{kg}$ particle travel before the two meet?
A$5\ \text{m}$
B$6\ \text{m}$
C$10\ \text{m}$
D$4\ \text{m}$
Show answer and explanation
Correct answer: B - $6\ \text{m}$
Only internal forces act, so the centre of mass stays exactly where it was, and the particles can meet nowhere else. Its distance from the $2\ \text{kg}$ particle is $(3 \times 10)/(2+3) = 6\ \text{m}$, so that particle covers $6\ \text{m}$. $4\ \text{m}$ is the distance travelled by the $3\ \text{kg}$ particle, $5\ \text{m}$ assumes they meet at the midpoint, and $10\ \text{m}$ is the whole separation.
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