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CGP EDU Academic Team
Published on: September 12, 2026
A 2 kg stone at the end of a string 1 m long is whirled in a vertical circle at a constant speed. The speed of the stone is 4 m/sec. The tension in the string will be 52 N, when the stone is
Text Solution
Verified by ExpertsThe correct answer is:
B
\(mg = 20\,N\) and \(\frac{mv^2}{r} = \frac{2 \times (4)^2}{1} = 32N\)
It is clear that 52 N tension will be at the bottom of the circle. Because we know that \(\Gamma_{\text{bottom}} = mg + \frac{mv^2}{r}\)
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