Weight of a body of mass m decreases by 1% wh | vedclass

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

Question

(A) The acceleration due to gravity at height $h$ is given by $g_h = g(1 - \frac{2h}{R})$.The fractional decrease in weight at height $h$ is $\frac{\D

Vedclass · Accepted Answer

decrease by $0.5\%$

Vedclass · Answer

(A) The acceleration due to gravity at height $h$ is given by $g_h = g(1 - \frac{2h}{R})$.The fractional decrease in weight at height $h$ is $\frac{\Delta g}{g} = \frac{2h}{R} = 1\% = 0.01$.Therefore,$\frac{h}{R} = 0.005$.The acceleration due to gravity at depth $h$ is given by $g_d = g(1 - \frac{h}{R})$.The fractional decrease in weight at depth $h$ is $\frac{\Delta g}{g} = \frac{h}{R} = 0.005 = 0.5\%$.Thus,the weight of the body decreases by $0.5\%$.

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

Similar Questions

At a height $R$ above the earth's surface,the gravitational acceleration is ($R$ = radius of earth,$g$ = acceleration due to gravity on earth's surface).

What effect occurs on the frequency of a pendulum if it is taken from the earth's surface to deep into a mine?

Assuming the Earth to be a sphere of uniform density, the acceleration due to gravity:

$A$ point mass of $10 \,kg$ is placed at the centre of the Earth. The weight of the point mass is

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 $..........\,N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module