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(D) For a body rolling down an inclined plane,the force of friction $f$ acts in the upward direction along the incline.$1$. Since the sphere is moving

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All of these

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(D) For a body rolling down an inclined plane,the force of friction $f$ acts in the upward direction along the incline.$1$. Since the sphere is moving downward,the friction $f$ acts upward,which opposes the translational motion.$2$. The torque produced by the friction $f$ about the center of the sphere is $	au = f 	imes R$. This torque causes an angular acceleration $\alpha$ in the direction of rolling,meaning friction supports the rotational motion.$3$. The equation of motion for translation is $mg \sin 	heta - f = ma$,and for rotation is $fR = I\alpha$. For a hollow sphere,$I = \frac{2}{3}mR^2$. Solving these,we find $f = \frac{mg \sin 	heta}{1 + \frac{mR^2}{I}} = \frac{mg \sin 	heta}{1 + \frac{3}{2}} = \frac{2}{5} mg \sin 	heta$.$4$. Since $f = \frac{2}{5} mg \sin 	heta$,it is clear that $f \propto \sin 	heta$. Therefore,if $	heta$ decreases,$\sin 	heta$ decreases,and consequently,the frictional force $f$ decreases.Thus,all the given statements are correct.

$A$ hollow sphere of mass $m$ and radius $R$ is rolling downward on a rough inclined plane of inclination $\theta$. If the coefficient of friction between the hollow sphere and incline is $\mu$,then ........

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