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MHT CET 2026 (15 Apr, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from MHT CET 2026 (15 Apr, Shift 1)

Q13MHT CET 2026 · 15 APR · SHIFT 1Medium

Acceleration Due to Gravity

What should be the angular velocity of earth due to rotation about its own axis so that the weight at thth

equator becomes (35)\left(\frac{3}{5}\right) of initial value? (g=g = acceleration due to gravity, R = radius of earth)

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Answer: (b)

  1. At the equator: g′=g−ω2Rg' = g - \omega^2 R
  2. g−ω2R=35g⇒ω2=2g5Rg - \omega^2 R = \frac{3}{5}g \Rightarrow \omega^2 = \frac{2g}{5R}
  3. ω=(2g5R)1/2\omega = \left(\frac{2g}{5R}\right)^{1/2}
  4. So the answer is (b).
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