(C) The centripetal force required for circular motion is provided by the tension in the string: $T = M \omega^{2} R$.Given: $T = 80 \,N$,$M = 100 \,g

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(C) The centripetal force required for circular motion is provided by the tension in the string: $T = M \omega^{2} R$.Given: $T = 80 \,N$,$M = 100 \,g = 0.1 \,kg$,$R = 2 \,m$.Substituting the values: $80 = 0.1 \times \omega^{2} \times 2$.$80 = 0.2 \omega^{2} \implies \omega^{2} = 400 \implies \omega = 20 \,rad/s$.Since $\omega = 2 \pi f$,where $f$ is the frequency in $rev/s$: $2 \pi f = 20 \implies f = \frac{10}{\pi} \,rev/s$.To convert frequency to $rev/min$,multiply by $60$: $f = \frac{10}{\pi} \times 60 = \frac{600}{\pi} \,rev/min$.Comparing this with $\frac{K}{\pi} \,rev/min$,we get $K = 600$.

Class 11 Physics 3-2.Motion in Plane Mix Examples-Motion in Plane

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$A$ 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 a maximum tension of $80 \,N$. If the maximum speed with which the stone can revolve is $\frac{K}{\pi} \,rev/min$,find the value of $K$. (Assume the string is massless and unstretchable)

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A
$400$
B
$300$
C
$600$
D
$800$

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