JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2022
- Medium
- MCQ
A boy ties a stone of mass \(100 \,g\) to the end of a \(2\) \(m\) long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of \(80 \,N\). If the maximum speed with which the stone can revolve is \(\frac{ K }{\pi} rev\). / \(min\). The value of \(K\) is (Assume the string is massless and unstretchable)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(600\)
Step-by-step Solution
Detailed explanation
\(T = M \omega^{2} R\)
\(T =80 N \quad M =0.1 \quad \omega=? \quad R =2 m\)
\(80=0.1 \omega^{2}(2)\)
\(\omega^{2}=400\)
\(\omega=20\)
\(2 \pi f =20\)
\(f =\frac{10}{\pi} \frac{ rev }{ s }\)
\(=\frac{600}{\pi} \frac{ rev }{ min }\)
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