Question

The weight of a body decreases by $1.5$ when it is raised to a height $h$ above the surface of the Earth. When the same body is taken to the same depth $h$ in a mine, its weight will show:

• A. 3.0% Decrease
• B. 1.5% Decrease
• C. 0.75% Decrease
• D. 0.75% Increase

Answer 1

The correct option is C

0.75% Decrease

The explanation for the correct option:

Step 1: Given data

The weight of a body decreases by $1.5$ .

Height $h$.

Step 2: To find

We have to find the weight of the body.

Step 3: Calculation

The weight of a particle at surface level is given by,

$W=\frac{GM}{R^2}$

Where $G$ is the gravitational constant, $M$ is the mass of the particle, and $R$ is the radius of earth.

The weight of a particle at height $h$ above the earth surface is given by,

$W\text{'}=\frac{G{M}^2}{{\left(R+h\right)}^2}$

Given that the weight of particle decreases $1.5\%$ when it is raised to height $h$.

$\begin{array}{rcl}\frac{W-W\text{'}}{W}& =& 1.5\%\\ & \Rightarrow & \frac{{\displaystyle \frac{GM}{R^2}}-{\displaystyle \frac{GM}{{\left(R+h\right)}^2}}}{{\displaystyle \frac{GM}{R^2}}}=1.5\%\\ & \Rightarrow & \frac{{\displaystyle \frac{1}{R^2}-\frac{1}{{\left(R+h\right)}^2}}}{{\displaystyle \frac{1}{R^2}}}=1.5\%\\ & \Rightarrow & \frac{{\displaystyle \frac{{\left(R+h\right)}^2-R^2}{{\left(R+h\right)}^2{R}^2}}}{{\displaystyle \frac{1}{R^2}}}=1.5\%\\ & \Rightarrow & \frac{R^2+2Rh+h^2-R^2}{{\left(R+h\right)}^2}=1.5\%\end{array}$

At depth $h$, the expression for gravity is,

$g\text{'}=g\left(1-\frac{h}{R}\right)$

As $g\text{'}$ decreases, $\frac{g-g\text{'}}{g}=\frac{h}{R}$

$\Rightarrow \frac{{\displaystyle \frac{2h}{R}}+{\left({\displaystyle \frac{h}{R}}\right)}^2}{{\left(1+{\displaystyle \frac{h}{R}}\right)}^2}=\frac{1.5}{100}$

On further simplification,

$$\begin{gathered} \Rightarrow \frac{2h}{R}=\frac{1.5}{100} \\ \Rightarrow \frac{h}{R}=\frac{0.75}{100} \end{gathered}$$

Since, $h<<R$ at depth $h$, there will be a decrease in the weight of the particle by $0.75\%$.

Hence, the correct option is (C).