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# The height at which the acceleration due to gravity becomes g / 9 (where g = theacceleration due to gravity on the surface of the earth) in terms of R, the radius of(A) $R / 2$ (B) $\sqrt{2} R$ (C) 2 R(D) $\dfrac{R}{\sqrt{2}}$

Last updated date: 13th Jun 2024
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Hint: We know that in mechanics, acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities (in that they have magnitude and direction). The orientation of an object's acceleration is given by the orientation of the net force acting on that object. Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity. Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity. As velocity is an example of a vector, it has direction and magnitude, a change in both speed and direction.

${\prime}^{\prime} \mathrm{g}^{\prime}=\mathrm{g}\left(\dfrac{\mathrm{R}}{\mathrm{R}+\mathrm{H}}\right)^{2}$
where $\mathrm{g} \rightarrow$ on the surface of the earth.
$\mathrm{R} \rightarrow$ Radius of earth.
As $\mathrm{g}^{\prime}=\mathrm{g} / 9,$ we get:
$\Rightarrow \dfrac{\mathrm{g}}{9}=\mathrm{g}\left(\dfrac{\mathrm{R}}{\mathrm{R}+\mathrm{H}}\right)^{2}$
$\Rightarrow \dfrac{\mathrm{R}}{\mathrm{R}+\mathrm{H}}=\dfrac{1}{3} \Rightarrow \mathrm{h}=2 \mathrm{R}$