Question

The weight of a body of mass $\mathrm{m}$ decreases by $1\%$ when it is raised to height $\mathrm{h}$ above the earth's surface. If the body is taken to a depth $\mathrm{h}$ in a mine. Change in its weight is

• A. $2\%$ decreases
• B. $0.5\%$ decreases
• C. $1\%$ increases
• D. $0.5\%$ increases

Answer 1

The correct option is B

$0.5\%$ decreases

Step 1. Given data

The weight of a body mass $\mathrm{m}$ decreases by $1\%$ when it is raised to height $\mathrm{h}$ above the earth's surface.

The weight of the body decreases by $1\%$ when it increased by height $\mathrm{h}$.

Step 2. Formula to be used

The formula of the acceleration due to gravity at height $\mathrm{h}$ is,

$\mathrm{g}\text{'}=\mathrm{g}\left(1-\frac{2\mathrm{h}}{\mathrm{R}}\right)$ ------ $\left(1\right)$

Here, $\mathrm{g}\text{'}$ is apparent acceleration due to gravity, $\mathrm{g}$ is acceleration due to gravity, $\mathrm{R}$ is the radius of the earth and $\mathrm{h}$ is the distance above the surface of earth.
The formula of the acceleration due to gravity at depth $\mathrm{h}$ is,

$\mathrm{g}\text{'}=\mathrm{g}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)$
Step 3. Calculate change in weight

As we know that,

$\mathrm{g}\text{'}=\mathrm{g}\left(1-\frac{2\mathrm{h}}{\mathrm{R}}\right)$

So, the weight is,

$\mathrm{mg}\text{'}=\mathrm{mg}\left(1-\frac{2\mathrm{h}}{\mathrm{R}}\right)$
Thus, the change in the weight is 1%

Therefore,

$\frac{\mathrm{mg}-\mathrm{mg}\text{'}}{\mathrm{mg}}=1\%$
$\frac{\mathrm{g}-\mathrm{g}\text{'}}{\mathrm{g}}=1\%$ ------ $\left(2\right)$
Replace the value of $\mathrm{g}\text{'}$ from the equation $\left(1\right)$ in $\left(2\right)$ we get,

$\frac{\mathrm{g}-\mathrm{g}\text{'}\left(1-{\displaystyle \frac{2\mathrm{h}}{\mathrm{R}}}\right)}{\mathrm{g}}=1\%$
$1-1+\frac{2\mathrm{h}}{\mathrm{R}}=1\%$

$\frac{2\mathrm{h}}{\mathrm{R}}=1\%$

$\frac{\mathrm{h}}{\mathrm{R}}=0.5\%$ ------- $\left(3\right)$
Step 4. Calculate the weight of the body at depth $\mathrm{h}$
Now, we will calculate the weight of the body at depth $\mathrm{h}$.
As we know,
$\mathrm{g}\text{'}=\mathrm{g}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)$

So,
The weight of the body is

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

Replace the value of $\frac{\mathrm{h}}{\mathrm{R}}$ from equation $\left(3\right)$ in above relation, we get,

$\mathrm{mg}\text{'}=\mathrm{mg}\left(1-0.5\%\right)$

$=\mathrm{mg}\left(1-0.005\right)$

$=\left(0.995\right)\mathrm{mg}$ ------- $\left(4\right)$

Step 5. Calculate the percentage change of the weight

The percentage change of the weight is,

$\frac{\mathrm{mg}-\mathrm{mg}\text{'}}{\mathrm{mg}}\times 100$ ------- $\left(5\right)$
Replace the value of $\mathrm{mg}\text{'}$from equation $\left(4\right)$ into $\left(5\right)$ ,we get.

$\frac{\mathrm{mg}-\left(0.995\right)\mathrm{mg}}{\mathrm{mg}}\times 100$

$1-0.995\times 100=0.5\%$

The weight decreases by $0.5\%$.

Hence, option ($\mathrm{B}$) is the correct answer.