Back to Directory
NEET PHYSICSMedium

A solid cylinder of mass 2 kg and radius 4 cm rotating about its axis at the rate of 3 rpm. The torque required to stop after 2π2\pi revolutions is

1

2×10−6 N m2 \times 10^{-6} \text{ N m}

2

2×10−3 N m2 \times 10^{-3} \text{ N m}

3

12×10−4 N m12 \times 10^{-4} \text{ N m}

4

2×106 N m2 \times 10^{6} \text{ N m}

Step-by-Step Solution

Using the work-energy theorem: W=12I(ωf2−ωi2)W = \frac{1}{2}I(\omega_f^2 - \omega_i^2). Here θ=2π\theta = 2\pi revolutions =2π×2π=4π2 rad= 2\pi \times 2\pi = 4\pi^2 \text{ rad}. Initial angular velocity ωi=3×2π60 rad/s\omega_i = 3 \times \frac{2\pi}{60} \text{ rad/s}. The work done by torque is W=−τθW = -\tau\theta. Substituting values: −τ(4π2)=12×(12mr2)×(02−ωi2)-\tau(4\pi^2) = \frac{1}{2} \times (\frac{1}{2}mr^2) \times (0^2 - \omega_i^2). Solving this yields τ=2×10−6 N m\tau = 2 \times 10^{-6} \text{ N m}.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started