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MHT CET 2024 (15 May, Shift 2) · Acceleration Due to Gravity

Acceleration Due to Gravity question from MHT CET 2024 (15 May, Shift 2)

Q23MHT CET 2024 · 15 MAY · SHIFT 2Easy

Acceleration Due to Gravity

The density of a new planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If RR is the radius of earth, then radius of the planet would be

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Answer: (b)

  1. g=43πGρRg = \frac{4}{3}\pi G\rho R, so equal gg needs equal ρR\rho R.
  2. 2ρ⋅Rp=ρR⇒Rp=R22\rho \cdot R_p = \rho R \Rightarrow R_p = \frac{R}{2}
  3. So the answer is (b) R/2R/2.
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