Consider a particle of mass m suspended by a string at the equator. Let R and M denote radius and mass of the earth. If ω is the angular velocity of rotation of the earth about its own axis, then the tension on the string will be (cos 0° = 1) (a) (GMm)/(R²) (b) (GMm)/(2R²) (c) (GMm)/(2R²) + mω² R (d) (GMm)/(R²) - mω² R

Answer: (d). At the equator the particle moves in a circle of radius R with the earth. Net inward force: (GMm)/(R²) - T = mω² R T = (GMm)/(R²) - mω² R So the answer is (d).

Consider a particle of mass m suspended by a string at the equator… | MHT CET 2019 (3 May, Shift 1)

Q113MHT CET 2019 · 3 MAY · SHIFT 1Medium

Acceleration Due to Gravity

Consider a particle of mass $m$ suspended by a string at the equator. Let $R$ and $M$ denote radius and mass of the earth. If $\omega$ is the angular velocity of rotation of the earth about its own axis, then the tension on the string will be $(\cos 0^\circ$ $= 1)$

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Answer: (d)

At the equator the particle moves in a circle of radius $R$ with the earth.
Net inward force: $\frac{GMm}{R^2} - T = m\omega^2 R$
$T = \frac{GMm}{R^2} - m\omega^2 R$
So the answer is (d).

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Acceleration Due to Gravity question from MHT CET 2019 (3 May, Shift 1)

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