Question

A solid cylinder and a solid sphere, having same mass M and radius R, roll down the same inclined plane from top without slipping They start from rest The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be :

• A. \(\sqrt{\frac{5}{3}}\)
• B. \(\sqrt{\frac{4}{5}}\)
• C. \(\sqrt{\frac{3}{5}}\)
• D. \(\sqrt{\frac{14}{15}}\)

Answer 1

The Correct Option is D

In pure rolling, the point of contact stays stationary relative to the surface, meaning friction causes no displacement. 
This eliminates the work done by friction, making conservation of mechanical energy a valid approach.

\(V = \sqrt{\frac{2gH}{1+K^2/R}}\)

\(\frac{V_{cylinder}}{V_{sphere}}=\sqrt{\frac{1+k^2/R^2_{sphere}}{1+k^2/R^2_{cylinder}}}\)

= \(\sqrt{\frac{\frac{1+2}{5}}{\frac{1+1}{2}}}\)

= \(\sqrt{\frac{7}{5}\times\frac{2}{3}}\)

= \(\sqrt{\frac{14}{15}}\)

Therefore the correct option is \(\sqrt{\frac{14}{15}}\)