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The height at which the weight of a body becomes 1/16 th, its weight on the surface of earth (radius R), is
Options
(a) 5 R
(b) 15 R
(c) 3 R
(d) 4 R
Correct Answer:
3 R
Explanation:
Let at h height, the weight of a body becomes 1/16th of its weight on the surface.
Wh = 16 Ws g’ = GM/(Re+h)²
Similarly, g = GM/Re²
Now, g’/g = Re²/(Re+h)²
⇒ g’=g(1+h/R)⁻²
⇒ Wh = 16 Ws
⇒ g’ = g/16 g’
= 16g(1+h/R)⁻² 1/16
= (1+h/R)⁻²
⇒ 4 = 1+h/R
⇒ h = 3 R
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Topics: Gravitation
(63)
Subject: Physics
(2479)
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