MHT CET · Physics · Rotational Motion
A solid sphere at rests rolls down an inclined plane of vertical height \(h\) without sliding. Its speed on reaching the bottom of plane is (\(g=\) acceleration due to gravity)
- A \(\left(\frac{9 g h}{11}\right)^{\frac{1}{2}}\)
- B \(\left(\frac{10 g h}{7}\right)^{\frac{1}{2}}\)
- C \(\left(\frac{8 g h}{7}\right)^{\frac{1}{2}}\)
- D \(\left(\frac{6 g h}{7}\right)^{\frac{1}{2}}\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{10 g h}{7}\right)^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
\(mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\) \(mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2\)
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