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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)

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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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