In the given figure,two wheels P and Q are co | vedclass

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(C) Let the radius of wheel $Q$ be $R$ and the radius of wheel $P$ be $3R$. Since they are connected by a belt,their tangential speeds at the rim are

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Let the radius of wheel $Q$ be $R$ and the radius of wheel $P$ be $3R$. Since they are connected by a belt,their tangential speeds at the rim are equal,so $v = \omega_{P} (3R) = \omega_{Q} R$. This implies $\omega_{P} = \frac{\omega_{Q}}{3}$. The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^{2}$. Given that the rotational kinetic energies are equal,we have: $\frac{1}{2} I_{P} \omega_{P}^{2} = \frac{1}{2} I_{Q} \omega_{Q}^{2}$ $I_{P} \left(\frac{\omega_{Q}}{3}\right)^{2} = I_{Q} \omega_{Q}^{2}$ $I_{P} \left(\frac{1}{9}\right) = I_{Q}$ $\frac{I_{P}}{I_{Q}} = 9$. Thus,the ratio is $9: 1$,and the value of $x$ is $9$.

In the given figure,two wheels $P$ and $Q$ are connected by a belt $B$. The radius of $P$ is three times as that of $Q$. In case of same rotational kinetic energy,the ratio of rotational inertias $\left(\frac{I_{P}}{I_{Q}}\right)$ will be $x: 1$. The value of $x$ will be $.....$ .

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