JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 24)
A solid disc of radius 20 cm and mass 10 kg is rotating with an angular velocity of 600 rpm, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in 10 s is ____________ $$\pi$$ $$\times$$ 10$$-$$1 Nm.
Answer
4
Explanation
$$\tau = {{\Delta L} \over {\Delta t}} = {{I({\omega _f} - {\omega _i})} \over {\Delta t}}$$
$$\tau = {{{{m{R^2}} \over 2} \times [0 - \omega ]} \over {\Delta t}}$$
$$ = {{10 \times {{(20 \times {{10}^{ - 2}})}^2}} \over 2} \times {{600 \times \pi } \over {30 \times 10}}$$
$$ = 0.4\pi = 4\pi \times {10^{ - 2}}$$
$$\tau = {{{{m{R^2}} \over 2} \times [0 - \omega ]} \over {\Delta t}}$$
$$ = {{10 \times {{(20 \times {{10}^{ - 2}})}^2}} \over 2} \times {{600 \times \pi } \over {30 \times 10}}$$
$$ = 0.4\pi = 4\pi \times {10^{ - 2}}$$
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