NEETPhysicsGravitation
Match List-I with List-II: List-I (Final position of mass m initially at Earth's surface) List-II (Change in gravitational potential energy, U ) (A) Height h = R (I) mgR (B) Height h = 3R (II) 0.5 mgR (C) Infinity (III) 0.75 mgR (D) Depth d = R 2 (IV) -0.375 mgR Choose the correct answer from the options given below:
Options
- A(A) (III), (B) (II), (C) (I), (D) (IV)
- B(A) (II), (B) (I), (C) (III), (D) (IV)
- C(A) (I), (B) (III), (C) (II), (D) (IV)
- D(A) (II), (B) (III), (C) (I), (D) (IV)
Correct answer
D. (A) (II), (B) (III), (C) (I), (D) (IV)
Step-by-step solution
The change in gravitational potential energy when a mass m is raised to a height h from the Earth's surface is U = mgh 1 + h R . For (A) h = R : U = mgR 1 + 1 = 0.5 mgR . So, (A) matches with (II). For (B) h = 3R : U = mg(3R) 1 + 3 = 3 4 mgR = 0.75 mgR . So, (B) matches with (III). For (C) Infinity ( h ): U = mgR . So, (C) matches with (I). For (D) Depth d = R 2 : The potential energy at depth d is U_d = - GMm 2R^3 (3R^2 - r^2) , where r = R - d = R 2 . U_d = - GMm 2R^3 (3R^2 - R^2 4 ) = - 11 8 GMm R = -1.375 mgR .