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A boy ties a stone of mass \mathrm{100\, g} to the end of a \mathrm{2\, m} long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of \mathrm{80\, N.} If the maximum speed with which the stone can revolve is \mathrm{\frac{K}{\pi }rev./min.} The value of \mathrm{K} is :

(Assume the string is massless and unstretchable)

Option: 1

400


Option: 2

300


Option: 3

600


Option: 4

800


Answers (1)

best_answer


Here tension (T) is providing centripetal force
\mathrm{\top=\frac{mV^{2}}{R}= MR\omega ^{2}}
\mathrm{\omega =\frac{K}{\pi}\left ( \frac{rev}{min} \right )= \frac{K\times 2\pi\, rad}{\pi\times 60\, s}}

\mathrm{\top _{max} =80= mR\omega _{max}^{2}}
                  \mathrm{80= \left ( 10^{-1} \right )\times \left ( 2 \right )\times \frac{K^{2}}{900}}
                \mathrm{K^{2}= 400\times 900}
                \mathrm{K= 20\times 30= 600}

The correct option is (3) 
 

Posted by

Rakesh

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