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MHT CET 2026 (15 May, Shift 1) · Vertical Circle: Speed & Tension

Vertical Circle: Speed & Tension question from MHT CET 2026 (15 May, Shift 1)

Q3MHT CET 2026 · 15 MAY · SHIFT 1Medium

Vertical Circle: Speed & Tension

The maximum and minimum tensions in the string whirling in a circle of radius 2.52.5 m are in the ratio 5:35 : 3 then its velocity is (g=10g = 10 m/s2^{2})

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Answer: (a)

  1. Uniform speed in vertical circle: Tmax=mv2r+mgT_{max} = \frac{mv^2}{r} + mg, Tmin=mv2r−mgT_{min} = \frac{mv^2}{r} - mg
  2. v2/r+gv2/r−g=53⇒v2r=4g\frac{v^2/r + g}{v^2/r - g} = \frac{5}{3} \Rightarrow \frac{v^2}{r} = 4g
  3. v2=4(10)(2.5)=100⇒v=10v^2 = 4(10)(2.5) = 100 \Rightarrow v = 10 m/s
  4. So the answer is (a) 1010 m/s.
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