Verified Solution • 2025–26 Session moderate

QJEEM22Ph8de7: JEE Main 2022 Physics - Class 11 JEEM22Ph8de7 Solution: A boy ties a stone of mass 100 g to the end of a 2 m long…

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 Kπ{K \over \pi } rev./min. The value of K is :

(Assume the string is massless and unstretchable)

Multiple Choice Options

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

Step-by-Step Verified Solution

T=mω2rT = m{\omega ^2}r

⇒80=0.1×(2π×Kπ×160)2×2\Rightarrow 80 = 0.1 \times {\left( {2\pi \times {K \over \pi } \times {1 \over {60}}} \right)^2} \times 2

⇒8002=K2900\Rightarrow {{800} \over 2} = {{{K^2}} \over {900}}

⇒K=30×20=600\Rightarrow K = 30 \times 20 = 600

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