A solid sphere of mass 500,g and radius 5,cm | vedclass

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(B) The moment of inertia of a solid sphere about its diameter is $I_d = \frac{2}{5}mR^2$.The angular momentum of the sphere about its diameter is $L

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(B) The moment of inertia of a solid sphere about its diameter is $I_d = \frac{2}{5}mR^2$.The angular momentum of the sphere about its diameter is $L = I_d \omega = \frac{2}{5}mR^2 \omega$.The moment of inertia of the sphere about its tangent is given by the parallel axis theorem: $I_t = I_{cm} + mR^2 = \frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2$.According to the problem,$I_t = (x 	imes 10^{-2}) 	imes L$.Substituting the expressions: $\frac{7}{5}mR^2 = (x 	imes 10^{-2}) 	imes (\frac{2}{5}mR^2 \omega)$.Canceling $\frac{1}{5}mR^2$ from both sides: $7 = (x 	imes 10^{-2}) 	imes 2 \omega$.Given $\omega = 10\,rad\,s^{-1}$,we have: $7 = (x 	imes 10^{-2}) 	imes 2 	imes 10$.$7 = x 	imes 10^{-2} 	imes 20$.$7 = x 	imes 0.2$.$x = \frac{7}{0.2} = 35$.

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