JEE Main 2022 · 24 Jun · Shift 1 · Physics Q37

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PhysicsSingle correct+4 / -1Question 37 of 90

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 $\dfrac K\pi$ rev./min. The value of $K$ is :

(Assume the string is massless and unstretchable)

400

300

600

800

Answer and worked solution

Official answer: C

Tension supplies the centripetal force (string nearly horizontal): $T=m\omega^2L$ .

Maximum angular speed. $80=0.1\,\omega^2(2)$ , so $\omega=20$ rad s $^{-1}$ .

In rev/min. $\dfrac{20\times60}{2\pi}=\dfrac{600}\pi$ , so $K=600$ .

(C)

Topics:Circular motion dynamics

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JEE Main 2022 · 24 Jun · Shift 1, Physics question 37