Question

If the Earth stops rotating, then the value of the
acceleration due to gravity at AIIMS campus, Delhi
will
(1) Increase
(2) Decrease
(3) Remains same (4) Becomes zero


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Answer 1

ANSWER

option (c)Remain same

Explanation

If the earth stop rotating now, then the value of g at the equator increases and become equal to the value at poles.

Cause, if earth stop rotating then the equatorial bulge will vanish and then earth will have a same radius all over the surface of Earth.

as,,,

$\sf\implies \: g \: \alpha \: \frac{1}{r {}^{2} }$

do the value of g will remain same through the all over the surface of Earth.

Answer 2

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

If the Earth stops rotating, then the value of the acceleration due to gravity at AIIMS campus, Delhi will?

$\huge{\bold{\underline{\underline{Answer}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Earth stops rotating.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

From Linear Velocity & Angular velocity ,

we know,

$\large{\boxed{\tt v = R \omega}}$

• v = Linear velocity
• ω = Angular velocity
• R = Radius of Earth.

$\large{\tt \leadsto v = R \omega}$

As the Earth Stops Rotating the Linear velocity will become Zero.

Substituting it,

$\large{\tt \leadsto 0 = R \omega}$

As we Know,

The radius (R) of Earth Cannot be Zero , As it is a Constant, Therefore Angular velocity will be Zero.

$\large{\tt \leadsto 0 = R \omega}$

Implies,

$\large{\leadsto{\underline{\underline{\tt \omega = 0 \: rad/sec}}} \: -----(1)}$

Therefore, Angular velocity will be zero.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From Variation of Acceleration due to gravity at Different Latitudes, we have,

$\large{\boxed{\tt g_{(\theta)} = g - R \omega^2 Cos^2 \theta}}$

• $\large{\tt g_{(\theta)}}$ = Acceleration due to gravity at a Latitude (AIIMS Campus)
• g = Acceleration due to gravity at surface of earth.
• ω = Angular velocity of Earth.
• R = Radius of earth.

$\large{\tt \leadsto g_{(\theta)} = g - R \omega^2 Cos^2 \theta}$

Substituting the value of Angular velocity from Equation (1).

$\large{\tt \leadsto g_{(\theta)} = g - R (0)^2 Cos^2 \theta}$

$\large{\tt \leadsto g_{(\theta)} = g - R \times 0 \times Cos^2 \theta}$

$\large{\tt \leadsto g_{(\theta)} = g - 0}$

$\huge{\boxed{\boxed{\tt g_{(\theta)} = g }}}$

Therefore, the Acceleration due to gravity above AIIMS Campus is Equal to the Acceleration due to gravity at surface of Earth when Earth has stop rotating.

So, the Acceleration due to gravity will remain same (Option- 3)

$\rule{300}{1.5}$