A body weighs 300 N on the surface of the ea | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula: $g_d = g(1 - \frac{d}{R})$.Given that the d

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(D) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula: $g_d = g(1 - \frac{d}{R})$.Given that the depth $d = \frac{R}{2}$,we substitute this value into the equation:$g_d = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.The weight of the body at the surface is $W = mg = 300 \ N$.The weight of the body at depth $d$ is $W_d = mg_d$.Substituting $g_d = \frac{g}{2}$,we get $W_d = m(\frac{g}{2}) = \frac{1}{2} 	imes mg = \frac{1}{2} 	imes 300 \ N = 150 \ N$.

$A$ body weighs $300 \ N$ on the surface of the earth. How much will it weigh at a distance $\frac{R}{2}$ below the surface of the earth (in $N$)? ($R$ is the radius of the earth.)

Similar Questions

$A$ body weighs $72 \, N$ on the surface of the earth. When it is taken to a height of $h = 2R$,where $R$ is the radius of the earth,it would weigh ........ $N$.

If two planets have their radii in the ratio $x: y$ and densities in the ratio $m: n$,then the ratio of the acceleration due to gravity on them is

How does the acceleration due to gravity change with altitude?

$Assertion$ : In a free fall,the weight of a body becomes effectively zero.
$Reason$ : The acceleration due to gravity acting on a body in free fall is zero.

$A$ body weighs $500 \, N$ on the surface of the earth. How much would it weigh halfway below the surface of the earth (in $, N$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module