When a body is taken from equator to pole, its weight
A. Decreases
B. Becomes zero
C. Remains same
D. Increases
Answer: Option D
Solution (By JKSSB Mock Tests)
g is slightly higher at poles (9.83 m/s²) than equator (9.78 m/s²) due to Earth's rotation (centrifugal effect) and equatorial bulge. Weight = mg, so weight increases slightly. Mass constant.
Explanation:
Newton's law of gravitation: F = G·m₁m₂/r². Thus force is directly proportional to the product of the masses (m₁m₂) and inversely proportional to square of distance (r²). Option D incorrectly states 'directly proportional to square of distance' – it's inverse square. This fundamental law governs celestial mechanics. Memory tip: 'F ∝ m₁m₂ and F ∝ 1/r²'. Competitive exams often test precise wording of physical laws; distractors may reverse proportionality or misstate dependencies. Always recall the exact mathematical form.
Explanation:
Coulomb's law: E = kq/r² ⇒ E ∝ 1/r². Inverse square law fundamental to electrostatics and gravitation. Memory tip: 'Point charge field: inverse square; double distance ⇒ field becomes 1/4'. Electrostatics concept frequently tested in competitive exams.
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