The weight of a body on the surface of the earth is 100 N. The gravitational force on it when taken at a height from the surface of earth, equal to one-fourth the radius of the earth, is: [2023]

Question

The weight of a body on the surface of the earth is 100 N. The gravitational force on it when taken at a height from the surface of earth, equal to one-fourth the radius of the earth, is: [2023]

Smart Achievers · Accepted Answer

(2)

$\Rightarrow g' = \frac{g R^{2}}{r^{2}} = \frac{g R^{2}}{{(R + \frac{R}{4})}^{2}} = \frac{16 g}{25}$

$∴$ $Weight = \frac{16}{25} \times 100 = 64 N$

Solution

Q.
The weight of a body on the surface of the earth is 100 N. The gravitational force on it when taken at a height from the surface of earth, equal to one-fourth the radius of the earth, is: [2023]

1 100 N

2 64 N

3 50 N

4 25 N

Ans.
(2)

$\Rightarrow g' = \frac{g R^{2}}{r^{2}} = \frac{g R^{2}}{{(R + \frac{R}{4})}^{2}} = \frac{16 g}{25}$

$∴$ $Weight = \frac{16}{25} \times 100 = 64 N$

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