The acceleration due to gravity on Earth's surface is g. At a height equal to Earth's radius above the surface, g' will be: MCQ with Answer and Explanation

The acceleration due to gravity on Earth's surface is g. At a height equal to Earth's radius above the surface, g' will be:
A. g/4
B. 2g
C. g/9
D. g/2
Answer: Option A
Solution (By JKExamLibrary)
Acceleration due to gravity at height h: g' = gR²/(R+h)², where R is Earth's radius. Given h = R, so g' = gR²/(2R)² = gR²/4R² = g/4. This inverse square law dependence arises from Newton's law of gravitation. At depth, variation differs, but for height, this formula applies. Memory aid: Double the distance from center ⇒ gravity becomes one-fourth. This standard result is frequently tested in gravitation sections of competitive exams. Always verify whether height or depth is specified, as formulas differ significantly.

This question belongs to: Science Physics

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Question #1 Report Error
A ball is dropped from a height of 20 m. Time to reach ground (g=10) is
A. 3 s
B. 1 s
C. 4 s
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Correct Answer: Option D


Explanation:
h = ½gt² => t = √(2h/g) = √(40/10) = √4 = 2 s. Free fall formula. Final velocity v = gt = 20 m/s.

This question belongs to: Science Physics
Question #2 Report Error
A car starts from rest and accelerates uniformly at 2 m/s² for 10 s. Distance covered in 10th second is
A. 19 m
B. 10 m
C. 20 m
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Correct Answer: Option A


Explanation:
s_nth = u + (a/2)(2n-1) = 0 + (2/2)(20-1) = 19 m.

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Question #3 Report Error
The dimensional formula of angular momentum is:
A. [M⁰L⁰T⁰]
B. [ML²T⁻²]
C. [MLT⁻¹]
D. [ML²T⁻¹]

Correct Answer: Option D


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This question belongs to: Science Physics