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Question

The height from earth's surface at which acceleration due to gravity becomes $$\dfrac {g}{4}$$ is (where $$g$$ is acceleration due to gravity on the surface of earth and $$R$$ is radius of earth).


A
2R
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B
R
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C
R2
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D
2R
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Solution

The correct option is B $$R$$
We know that  $$g=\dfrac{GM_e}{R^2}$$

And at height h from the earth surface $$ g_1=\dfrac{GM_e}{(R+h)^2}$$

And here $$g_1=\dfrac{g}{4}$$

So, $$ \dfrac{g}{4}=\dfrac{GM_e}{(R+h)^2}$$

so, $$\dfrac{GM_e}{4R^2}=\dfrac{GM_e}{(R+h)^2}$$

Hence, $$\dfrac{1}{4R^2}=\dfrac{1}{(R+h)^2}$$

$$ \implies\dfrac{(R+h)^2}{R^2}=4$$

$$\implies\dfrac{R+h}{R}=2$$

$$\implies R+h=2R$$

$$\implies R=h$$

Hence, at height equal to R from earth surface gravity will become $$\dfrac{g}{4}$$.
 Answer-(B)




Physics

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