Rotational Motion Question 181

Question: A wheel having angular momentum $2\pi kg-m^{2}/s$ about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel’s rotation in 30 sec would be

Options:

A) $\frac{\pi }{18}Nm$

B) $\frac{2\pi }{15}Nm$

C) $\frac{\pi }{12}Nm$

D) $\frac{\pi }{15}Nm$

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Answer:

Correct Answer: D

Solution:

[d]

$\tau \times \Delta t=L _{0}$

${\because \sin ceL _{f}=0}$

or $\tau \times 30=2\pi $

$\tau =\frac{\pi }{15}N-m$



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