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

Suppose that the angular velocity of rotation of the earth is increased. Then,as a consequence:

If the radius of the Earth were to shrink by half,its mass remaining the same,what would be the weight of an object on its surface?

At what height above the earth's surface does the acceleration due to gravity fall to $1\%$ of its value at the earth's surface (in $,R$)?

Weight of a body of mass $m$ decreases by $1\%$ when it is raised to height $h$ above the Earth's surface. If the body is taken to a depth $h$ in a mine,then its weight will

At what depth $d$ below the surface of the Earth does the acceleration due to gravity become $\frac{1}{n}$ times its value at the surface? ($R$ = radius of the Earth)

Vedclass Test Series

Exam Paper Generator

Online Exam Module