The time period of a simple pendulum on the surface of the moon compared to its time period on Earth is: MCQ with Answer and Explanation
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Question #1
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A body is thrown vertically upward. At the highest point, its acceleration is
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A body falls freely from rest. The ratio of distances covered in 1st, 2nd, and 3rd seconds is
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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:
Correct Answer: Option C
Explanation:
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.
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.
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