Consider a particle of mass $m$ suspended vertically by a string at the equator. Let $R$ and $M$ denote the radius and the mass of the earth, respectively. If $ω$ is the angular velocity of earth rotation about its own axis, the tension in the string is equal to

\frac{G m M}{2 R^{2}}

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\frac{G m M}{R^{2}}

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\frac{G m M}{R^{2}} - m ω^{2} R

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\frac{G m M}{R^{2}} + m ω^{2} R

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Solution

The correct option is C $\frac{G m M}{R^{2}} - m ω^{2} R$
As there is rotational motion so centrifugal force will act outward from the earth surface. Thus tension on the string is $T =$ gravitational force on $m -$ centrifugal force
$T = \frac{G M m}{R^{2}} - m ω^{2} R$

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Consider a particle of mass m suspended vertically by a string at the equator. Let R and M denote the radius and the mass of the earth, respectively. If ω is the angular velocity of earth rotation about its own axis, the tension in the string is equal to

The correct option is C GmMR2−mω2RAs there is rotational motion so centrifugal force will act outward from the earth surface. Thus tension on the string is T= gravitational force on m− centrifugal forceT=GMmR2−mω2R

Consider a particle of mass $m$ suspended vertically by a string at the equator. Let $R$ and $M$ denote the radius and the mass of the earth, respectively. If $ω$ is the angular velocity of earth rotation about its own axis, the tension in the string is equal to

The correct option is C $\frac{G m M}{R^{2}} - m ω^{2} R$
As there is rotational motion so centrifugal force will act outward from the earth surface. Thus tension on the string is $T =$ gravitational force on $m -$ centrifugal force
$T = \frac{G M m}{R^{2}} - m ω^{2} R$