NEET 2019 Odisha · PhysicsWork donePrevious Year Question
Taking gravitational potential energy at infinity to be zero, the change in PE (final − initial) of a mass m raised to height h above the earth's surface (radius R) is:
Answer: (B) GMmh/(R(R+h)). Correct Answer (B): GMmh/(R(R+h)) Solution: ΔU = −GMm/(R+h) − (−GMm/R) = GMm[1/R − 1/(R+h)] = GMmh/(R(R+h)).
- A.GMm/(R+h)
- B.GMmh/(R(R+h))✓
- C.mgh
- D.−GMm/(R+h)
Correct Answer
(B) GMmh/(R(R+h))
Solution & Explanation
Correct Answer (B): GMmh/(R(R+h)) Solution: ΔU = −GMm/(R+h) − (−GMm/R) = GMm[1/R − 1/(R+h)] = GMmh/(R(R+h)).
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