Earth is assumed to be a sphere of radius R. If ' gφ ' is value of effective acceleration due to gravity at latitude 30° and ' g ' is the value at equator, then the value of |g - gφ| is ( ω is angular velocity of rotation of earth, cos 30° = √(3)/2) (a) 1/4ω² R (b) 3/4ω² R (c) ω² R (d) 1/2ω² R

Answer: (a). Effective gravity at latitude φ: gφ = g₀ - ω² Rcos ²φ Equator ( φ = 0): g = g₀ - ω² R; at 30°: gφ = g₀ - 3/4ω² R |g - gφ| = ω² R(1 - 3/4) = 1/4ω² R So the answer is (a).

Earth is assumed to be a sphere of radius R. If ' gφ ' is value of… | MHT CET 2023 (14 May, Shift 2)

Q70MHT CET 2023 · 14 MAY · SHIFT 2Medium

Acceleration Due to Gravity

Earth is assumed to be a sphere of radius R. If ' $g_\phi$ ' is value of effective acceleration due to gravity at latitude $30^\circ$ and ' $g$ ' is the value at equator, then the value of $|g - g_\phi|$ is ( $\omega$ is angular velocity of rotation of earth, $\cos 30^\circ$ $= \frac{\sqrt{3}}{2}$ )

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Answer: (a)

Effective gravity at latitude $\phi$ : $g_\phi = g_0 - \omega^2 R\cos^2\phi$
Equator ( $\phi = 0$ ): $g = g_0 - \omega^2 R$ ; at $30^\circ$ : $g_\phi = g_0 - \frac{3}{4}\omega^2 R$
$|g - g_\phi| = \omega^2 R\left(1 - \frac{3}{4}\right) = \frac{1}{4}\omega^2 R$
So the answer is (a).

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Acceleration Due to Gravity question from MHT CET 2023 (14 May, Shift 2)

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