JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Laws of Motion

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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 the maximum tension of 80 N . If the maximum speed with which the stone can revolve is

\frac{K}{π} rev ./ min

. The value of K is : (Assume the string is massless and unstretchable)

(A)400
(B)300
(C)600✓ Correct
(D)800

Answer: (C)

Solution

The centripetal force is provided by tension:

T = M ω^{2} R

Given

T = 80 N

M = 0.1 kg

R = 2 m

80 = 0.1 \times ω^{2} \times 2

ω^{2} = 400

ω = 20 rad/s

Frequency

f = \frac{ω}{2 π} = \frac{20}{2 π} = \frac{10}{π} rev/s = \frac{600}{π} rev/min

Therefore

K = 600

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Single correct option

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