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MHT CET 2026 (19 Apr, Shift 2) · Vertical Circle: Speed & Tension

Vertical Circle: Speed & Tension question from MHT CET 2026 (19 Apr, Shift 2)

Q2MHT CET 2026 · 19 APR · SHIFT 2Easy

Vertical Circle: Speed & Tension

A stone of mass 4040 g is tied to a light inextensible string of length 5050 cm and whirled in a vertical circle at the rate of 3030 revolutions per minute. The tension in the string when it is at the lowest point of the circle is

(Take gravitational acceleration, g=10g = 10 m/s2^{2} and π2=10\pi^2 = 10)

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Answer: (c)

  1. ω=2π(30)60=π\omega = \frac{2\pi(30)}{60} = \pi rad/s, ω2=10\omega^2 = 10
  2. T=m(g+ω2r)=0.04(10+10×0.5)T = m(g + \omega^2 r) = 0.04(10 + 10 \times 0.5)
  3. T=0.04×15=0.6T = 0.04 \times 15 = 0.6 N
  4. So the answer is (c) 0.60.6 N.
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