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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 :

Option: 1

64 N


Option: 2

25 N


Option: 3

100 N


Option: 4

50 N


Answers (1)

best_answer

using newton’s formula F  =\frac{\mathrm{GMm}}{\mathrm{r}^2}

at surface of earth, 100 =\frac{\mathrm{GM}_e \mathrm{~m}}{\mathrm{Re}^2}               ......(1)

at        r=R_e+\frac{R_e}{4}=\frac{5}{4} R_e

\begin{aligned} & F^{\prime}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\left(\frac{5}{4} \mathrm{R}_{\mathrm{c}}\right)^2}=\frac{16}{25} \times \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_c{ }^2} \\ & F^{\prime}=\frac{16}{25} \times 100=64 \mathrm{~N} \end{aligned}

 

 

Posted by

Ritika Harsh

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