JEE Mains · Physics · STD 11 - 3.2 motion in plane
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)
- A \(400\)
- B \(300\)
- C \(600\)
- D \(800\)
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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