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MHT CET 2025 (23 Apr, Shift 2) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from MHT CET 2025 (23 Apr, Shift 2)

Q5MHT CET 2025 · 23 APR · SHIFT 2Medium

Centripetal Acceleration & Force

Two stones of masses m and 3 m are whirled in horizontal circles, the heavier one in a radius (r3)\left(\frac{r}{3}\right) and lighter one in a radius r. The tangential speed of lighter stone is ' nn ' times the value of heavier stone. When the magnitude of centripetal force becomes equal the value of nn is

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Answer: (b)

  1. Equal forces: mv12r=3mv22r/3=9mv22r\frac{m v_1^2}{r} = \frac{3m v_2^2}{r/3} = \frac{9 m v_2^2}{r}.
  2. v12=9v22⇒v1=3v2v_1^2 = 9 v_2^2 \Rightarrow v_1 = 3 v_2
  3. So n=3n = 3. The answer is (b) 3.
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