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An object is taken to a height above the surface of earth at a distance $\frac{5}{4} \mathrm{R}$ from the centre of the earth. Where radius of earth, $\mathrm{R=6400 \mathrm{~km}}$ . The percentage decrease in the weight of the object will be :
Option: 1
$36%$

Option: 2
$50%$

Option: 3
$64%$

Option: 4
$25%$

Answers (1)

best_answer

$\mathrm{R+h=\frac{5}{4}R}$ (from the centre of earth)

$\mathrm{h=\frac{R}{4}}$

$\mathrm{\frac{g_n}{g}=\frac{R^2}{(R+h)^2}=\frac{16 R^2}{25 R^2} }$

$\mathrm{g_n=\frac{16}{25} g}$

Weight on the eanth surface $\mathrm{= mg }$

Weight of the object at height $\mathrm{h=m g_n}$

$\mathrm{\frac{w}{w_n} =\frac{m g}{m g_n}=\frac{25}{16} }$

$\mathrm{ \frac{w-w_n}{W}=\frac{9}{25} }$

$\mathrm{\frac{\Delta W}{W} =36%}$
The percentage decrease in the weight of the object will be $36%$

Hence (1) is correct option.

Posted by

Ajit Kumar Dubey

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