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

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36%

25%

50%

64%

answer is A.

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Detailed Solution

$g_{eff} = \frac{g}{{(1 + \frac{h}{R})}^{2}}; g_{eff} = \frac{g}{{(1 + \frac{1}{4})}^{2}} = \frac{16 g}{25}$

$\begin{array}{l} change = \frac{g_{eff} - g}{g} \times 100 = \frac{\frac{16}{25} - 1}{1} \times 100 \\ = \frac{- 9}{25} \times 100 = - 36 % \end{array}$

Hence % decrease in the weight = 36%.

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