MHT CET · Physics · Work Power Energy
A pendulum bob has a speed \(4 \mathrm{~m} / \mathrm{s}\) at its lowest position. The pendulum is 1 m long. When the length of the string makes an angle of \(60^{\circ}\) with the vertical, the speed of the bob at that position is (acceleration due to gravity, \(g=10 \mathrm{~m} / \mathrm{s}^2, \cos 60^{\circ}=0 \cdot 5\) )
- A \(6 \mathrm{~m} / \mathrm{s}\)
- B \(\sqrt{3} \mathrm{~m} / \mathrm{s}\)
- C \(\sqrt{6} \mathrm{~m} / \mathrm{s}\)
- D \(3 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{6} \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{2} m v_1^2 = \frac{1}{2} m v_2^2 + m g L (1 - \cos \theta)\) \(v_2 = \sqrt{v_1^2 - 2 g L (1 - \cos \theta)}\)
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