WBJEE · Physics · Rotational Motion
A uniform solid spherical ball is rolling down a smooth inclined plane from a height \(h\). The velocity attained by the ball when it reaches the bottom of the inclined plane is v. If the ball is now thrown vertically upwards with the same velocity \(v\), the maximum height to which the ball will rise is
- A \(\frac{5 h}{8}\)
- B \(\frac{3 h}{5}\)
- C \(\frac{5 h}{7}\)
- D \(\frac{7 h}{9}\)
Answer & Solution
Correct Answer
(C) \(\frac{5 h}{7}\)
Step-by-step Solution
Detailed explanation
We know that total kinetic energy of a body rolling without slipping \(K_{\text {total }}=K_{\text {rot }}+K_{\text {trans }}\) For solid spherical ball, \(I=\frac{2}{5} m R^{2}\) (along to diameter) and \(v=R \omega\) where \(R\) is radius of spherical ball So.…
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