TS EAMCET · Physics · Rotational Motion
A solid sphere of mass \(2 \mathrm{~kg}\) rolls on a smooth horizontal surface at \(10 \mathrm{~m} / \mathrm{s}\). It then rolls up a smooth inclined plane of inclination \(30^{\circ}\) with the horizontal. The height attained by the sphere before it stops is [take \(g=10 \mathrm{~m} / \mathrm{s}^2\) ]
- A \(70 \mathrm{~cm}\)
- B \(701 \mathrm{~cm}\)
- C \(7.0 \mathrm{~m}\)
- D \(70 \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(7.0 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
The given situation is shown in the following figure If \(h\) be height attained by the sphere before it stopped, then according conservation of energy, \(K_{\text {rot }}+K_{\text {Trans }}=m g h\)…
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