Kinematics

JEE Main level, single correct, medium

296 of 308

Question 296 of 308

A boy ties a stone of mass 100g100g to the end of a 2m2m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80N80N. If the maximum speed with which the stone can revolve is Kπrevmin⁡−1\dfrac{K}{\pi }rev{\min }^{-1}. The value of KK is :
(Assume the string is massless and un-stretchable)

Show answer and solution

Answer: (C) 600600

The maximum angular velocity of the stone is ω=rpm×2π60=K30rads−1\omega =rpm\times \dfrac{2\pi }{60}=\dfrac{K}{30}rad{s}^{-1}
The tension in the string will provide the required centripetal force. Therefore,
T=mω2l⇒80=0.1×(K30)2×2⇒K2=360000⇒K=600T=m{\omega }^{2}l \Rightarrow 80=0.1\times {\left(\dfrac{K}{30}\right)}^{2}\times 2 \Rightarrow {K}^{2}=360000\Rightarrow K=600

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Projectile motion simulationDrag the sliders and watch the numbers in this question change.Revise Kinematics & Projectile Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Kinematics (Uniform Circular Motion), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.