Question

A solid cylinder rolls down an inclined plane of height 3 m and reaches the bottom of plane with angular velocity, of 2 $ \sqrt {2} rads ^{-1} $ . The radius of cylinder must be (Take g = 10 $ ms ^{-2}$ )

• A. 5 cm
• B. 0.5 cm
• C. $ \sqrt{10}$ cm
• D. $ \sqrt 5 $ m

Answer 1

The Correct Option is C

$\begin{array}{l}
v=\sqrt{\frac{2 g h}{1+k^{2} / R^{2}}}=\sqrt{\frac{2 g h}{1+(1 / 2)}}=\sqrt{\frac{4 g h}{3}}=\sqrt{\frac{4 \times 10 \times 3}{3}} \\
2 \sqrt{10} \\
\omega=\frac{v}{R} \Rightarrow R=\frac{v}{\omega}=\frac{2 \sqrt{10}}{2 \sqrt{2}}=\sqrt{5} m
\end{array}$