If the acceleration due to gravity at the surface of theearth is g, th

1.	If the acceleration due to gravity at the surface of theearth is g, the work done in slowly lifting a body of mass m from the earth's surface to a height R equal to the radius of the earth is.(a) 1/2 mgR(b) 2mgR(c) mgR (d) 1/4 mgR.
Answer» The correct answer is (a) 1/2 mgR EXPLANATION: The distance of the body from the center of the earth at a height R from the earth's surface = 2R. The acceleration due to gravity at distance x from the center of the earth ( for x>R) = GM/x² Force on the body by the earth =mGM/x² Since the body is lifted slowly, hence the moving force is ≈ mGM/x² Hence the work done by the force in taking the body from R to 2R = ∫m(GM/x²)dx (limit of integration from R to 2R) =mGM[-1/x] =mGM[-1/2R+1/R] =mGM/2R =½m(GM/R²)R = ½mgR

If the acceleration due to gravity at the surface of theearth is g, the work done in slowly lifting a body of mass m from the earth's surface to a height R equal to the radius of the earth is.(a) 1/2 mgR(b) 2mgR(c) mgR (d) 1/4 mgR.

Answer»

The correct answer is (a) 1/2 mgR

EXPLANATION:

The distance of the body from the center of the earth at a height R from the earth's surface = 2R. The acceleration due to gravity at distance x from the center of the earth ( for x>R) = GM/x²

Force on the body by the earth =mGM/x²

Since the body is lifted slowly, hence the moving force is ≈ mGM/x²

Hence the work done by the force in taking the body from R to 2R

= ∫m(GM/x²)dx

(limit of integration from R to 2R)

=mGM[-1/x]

=mGM[-1/2R+1/R]

=mGM/2R

=½m(GM/R²)R

= ½mgR

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