JEE Advanced 2018 · Paper 1, Physics question 9

Accelerations. For rolling without slipping, $a=\dfrac{g\sin\theta}{1+I/mR^2}$:

• ring ($I=mR^2$): $a_r=\dfrac{g\sin\theta}2$;
• disc ($I=\tfrac12mR^2$): $a_d=\dfrac{2g\sin\theta}3$.

Times. The path length is $s=h/\sin\theta$, and $t=\sqrt{2s/a}$:
$$t_r=\frac2{\sin\theta}\sqrt{\frac hg},\qquad t_d=\frac{\sqrt3}{\sin\theta}\sqrt{\frac hg}.$$

Difference. With $\sin60^\circ=\tfrac{\sqrt3}2$:
$$t_r-t_d=\frac{2-\sqrt3}{\sqrt3/2}\sqrt{\frac h{10}}=\frac{2-\sqrt3}{\sqrt{10}}\implies\sqrt{\frac{4h}{30}}=\frac1{\sqrt{10}}\implies h=0.75\text{ m}.$$

Answer: 0.75