A ball rolls down an inclined plane,as shown | vedclass

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(D) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{K^2}{R^2}}$. Since the angle of inclinat

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$iii$ only

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(D) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{K^2}{R^2}}$. Since the angle of inclination $	heta$ and the radius of gyration $K$ are constant for the same ball,the acceleration $a$ is constant throughout the motion. Thus,statement $(ii)$ is incorrect.Using the equation of motion $S = ut + \frac{1}{2}at^2$,with $u = 0$,we get $S = \frac{1}{2}at^2$,which implies $t = \sqrt{\frac{2S}{a}}$.Let $S_P$ be the distance $PO$ and $S_Q$ be the distance $QO$. Since $S_P = \frac{h}{\sin 	heta}$ and $S_Q = \frac{2h}{\sin 	heta}$,we have $S_Q = 2S_P$.Therefore,the time taken $t_Q = \sqrt{\frac{2(2S_P)}{a}} = \sqrt{2} t_P$. Thus,statement $(i)$ is incorrect.By the law of conservation of energy,the potential energy at the starting point is converted into kinetic energy at the bottom $O$. $K.E._P = mgh$ and $K.E._Q = mg(2h) = 2mgh$. Therefore,$K.E._Q = 2 K.E._P$. Thus,statement $(iii)$ is correct.

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