A body weighs 72 N on the surface of the Earth. What is the gravitational force on it at a height equal to half the radius of the Earth? MCQ with Answer and Explanation
A body weighs 72 N on the surface of the Earth. What is the gravitational force on it at a height equal to half the radius of the Earth?
A. 32 N
B. 72 N
C. 28 N
D. 16 N
Answer: Option A
Solution (By JKSSB Mock Tests)
Weight on surface W = mg = 72 N. Acceleration due to gravity at height h is g' = g [R / (R+h)]². Here h = R/2. So, g' = g [R / (R + R/2)]² = g [1 / (3/2)]² = g (2/3)² = 4g/9. The new weight W' = mg' = m(4g/9) = (4/9)mg = (4/9) × 72 = 32 N.
Explanation:
Vector equilibrium: three equal forces at 120° form closed triangle. Third force equals magnitude of others, oriented 120° from each. Memory aid: 'Equilibrium: vector sum = 0; symmetric forces ⇒ symmetric angles'. Statics concept frequently tested in competitive exams.
Explanation:
Rainbow is formed by dispersion, refraction, and internal reflection of sunlight in water droplets. Two refractions and one internal reflection give primary bow. Secondary bow has two internal reflections. Not diffraction.
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