Question

If a uniform solid sphere and a disc of same mass and same radius rolls down on an inclined smooth plane from rest to the same distance, then the ratio of the time taken by them will be :

• A. $15:14$
• B. $\sqrt{14}:\sqrt{15}$
• C. $14:15$
• D. ${15}^2:{14}^2$

Answer 1

Answer: (B)

$\sqrt{14}:\sqrt{15}$

By the relation,
$t=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}(1+\frac{K^2}{R^2})}$
We have
$\frac{t_S}{t_D}=\sqrt{\frac{(1+\frac{K^2}{R^2}{)}_S}{(1+\frac{K^2}{R^2}{)}_D}}$
$=\sqrt{\frac{1+\frac{2}{5}}{1+\frac{1}{2}}}=\sqrt{\frac{\frac{7}{5}}{\frac{3}{2}}}=\sqrt{\frac{14}{15}}$
Hence $t_S:t_D=\sqrt{14}:\sqrt{15}$