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A solid sphere of radius R rolls down from the top of an inclined plane from a height 'h'. If it is kept on the frictionless incline and released from the same height, its speed is 'k' times greater at the bottom as compared to the first case. So, the value of 'k' will be:
Option: 1
$\frac{3}{4}$

Option: 2
$\frac{4}{3}$

Option: 3
$\frac{5}{7}$

Option: 4
$\frac{7}{5}$

Answers (1)

best_answer

For Case 1:

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2$

$mgh = \frac{1}{2}mv^2 + \frac{1}{2} * \frac{2}{5}mr^2 \omega^2$

$gh = \frac{1}{2}v^2 + \frac{1}{5}v^2$

$gh = \frac{7}{10}v^2$

$v= \sqrt{\frac{10gh}{7}}$

$mgh = \frac{1}{2} mv'^{2}$

$v' = \sqrt{2gh}$

$v' = kv$

$\sqrt{2gh} = k \sqrt{\frac{10gh}{7}}$

$k = \frac{2gh}{10gh} * 7 = \frac{7}{5}$

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Riya

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