In the given figure,a ring of mass m is kept | vedclass

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(A) Let $a$ be the acceleration of the centre of mass of the ring and $\alpha$ be the angular acceleration. Since the ring rolls without slipping,$a =

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Statement $(i)$ and $(ii)$ only

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(A) Let $a$ be the acceleration of the centre of mass of the ring and $\alpha$ be the angular acceleration. Since the ring rolls without slipping,$a = R\alpha$.For the hanging mass $m$,the equation of motion is: $mg - T = ma_h$,where $a_h$ is the acceleration of the hanging mass. Since the string is attached to the top of the ring,$a_h = a + R\alpha = a + a = 2a$.So,$mg - T = m(2a) \implies T = m(g - 2a)$.For the ring,the forces are tension $T$ (forward) and friction $f$ (assume forward). The equations are:$T + f = ma$$(T - f)R = I\alpha = (mR^2)\alpha = mRa \implies T - f = ma$.Adding the two equations: $2T = 2ma \implies T = ma$.Substituting $T = ma$ into $mg - T = 2ma$,we get $mg - ma = 2ma \implies 3ma = mg \implies a = g/3$.Then $a_h = 2a = 2g/3$.From $T + f = ma$,$ma + f = ma \implies f = 0$.Since the calculated acceleration $a = g/3$ and $a_h = 2g/3$ match statements $(i)$ and $(ii)$,the correct option is $A$.

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