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 K \over \pi rev./min. The value of K is : (Assume the string is massless and unstretchable)

Circular Motion PYQ — JEE Main 2022 · 24 Jun · Shift 1 · Q50 with Solution

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Correct answer: C

Since the stone is moving in a horizontal circle, the tension provides the centripetal force:

$T = \frac{mv^2}{r}$

At maximum speed, tension is maximum:

$80 = \frac{0.1\,v^2}{2}$

$80 = 0.05v^2$

$v^2 = \frac{80}{0.05} = 1600$

$v = 40\,\text{m/s}$

Relation between linear speed and angular frequency:

$v = 2\pi r f$

where $f$ is frequency in revolutions per second.

So,

$f = \frac{v}{2\pi r} = \frac{40}{2\pi \cdot 2} = \frac{40}{4\pi} = \frac{10}{\pi}\,\text{rev/s}$

$f_{\text{rpm}} = \frac{10}{\pi} \times 60 = \frac{600}{\pi}\,\text{rev/min}$

Given maximum speed is $\dfrac{K}{\pi}$ rev/min, hence

$\frac{K}{\pi} = \frac{600}{\pi}$

So,

$K = 600$

Therefore, the correct option is C.