Question

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 weigh will be:
(A). Decrease by $5\%$
(B). Decrease by $2\%$
(C). Decrease by $0.5\%$
(D). Increase by $1\%$

Answer 1

Hint: This problem can be solved by using the formulas for the variation of acceleration due to gravity $g$ with height above the surface of the earth to find out a relation of $h$. Then this relation can be used to find out the variation in the weight of the body due to the variation in acceleration due to gravity at the depth $h$ below the surface of the earth.

Formula used:
The gravitational force applied by the earth on an object of mass $m$ at a height $h$ above its surface is given by
${{F}_{g}}=\dfrac{G{{M}_{E}}m}{{{\left( {{R}_{E}}+h \right)}^{2}}}=mg=\text{ weight of the object}$
where ${{M}_{E}}$ is the mass of the earth, ${{R}_{E}}$ is the radius of the earth and G is the universal gravitational constant equal to $6.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}$.
The acceleration due to gravity $g'$ a depth $d$ below the surface of the earth is given by
$g'=g\left( 1-\dfrac{d}{{{R}_{E}}} \right)$
where${{R}_{E}}$ is the radius of the earth and $g$is the acceleration due to gravity on the surface of the earth.

Complete step by step answer:
The gravitational force applied by the earth on an object of mass $m$at a height $h$ above its surface is given by
${{F}_{g}}=\dfrac{G{{M}_{E}}m}{{{\left( {{R}_{E}}+h \right)}^{2}}}=mg=\text{ weight of the object}$ --(1)
where ${{M}_{E}}$ is the mass of the earth, ${{R}_{E}}$ is the radius of the earth and G is the universal gravitational constant equal to $6.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}$.
The acceleration due to gravity $g'$ a depth $d$ below the surface of the earth is given by
$g'=g\left( 1-\dfrac{d}{{{R}_{E}}} \right)$ --(2)
where${{R}_{E}}$ is the radius of the earth and $g$is the acceleration due to gravity on the surface of the earth.
Therefore, let us analyze the question.
Since, by the condition the weight of the body decreases by $1\%$, therefore the weight at height $h$ will be $\dfrac{99}{100}mg$ where $m$ is the mass of the body and $mg$ its weight at the surface.
Hence, using (1), we get,
$\dfrac{\dfrac{99}{100}mg}{mg}=\dfrac{\dfrac{G{{M}_{E}}m}{{{\left( {{R}_{E}}+h \right)}^{2}}}}{\dfrac{G{{M}_{E}}m}{{{\left( {{R}_{E}} \right)}^{2}}}}$
\[\therefore \dfrac{99}{100}={{\left( \dfrac{{{R}_{E}}}{{{R}_{E}}+h} \right)}^{2}}\]
$\therefore \dfrac{100}{99}={{\left( \dfrac{{{R}_{E}}+h}{{{R}_{E}}} \right)}^{2}}$
Square rooting both sides, we get,
$\sqrt{\dfrac{100}{99}}=\sqrt{{{\left( \dfrac{{{R}_{E}}+h}{{{R}_{E}}} \right)}^{2}}}$
$\therefore 1.005=\dfrac{{{R}_{E}}+h}{{{R}_{E}}}$
$\therefore 1.005{{R}_{E}}={{R}_{E}}+h$
$\therefore \left( 1.005-1 \right){{R}_{E}}=h=0.005{{R}_{E}}$ --(3)
Now, the body is taken to a mine of depth $h$.
We can find out the variation in weight with the formula of the variation in acceleration due to gravity with depth below the earth’s surface.
Hence, using (2), we get,
$mg'=mg\left( 1-\dfrac{h}{{{R}_{E}}} \right)$
$\therefore \dfrac{mg'}{mg}=\left( 1-\dfrac{0.005{{R}_{E}}}{{{R}_{E}}} \right)=1-0005=0.995$
Therefore, percentage of weight variation
$\left( \dfrac{mg'}{mg} \right)\times 100=99.5\%$
Hence, there is a decrease of $\left( 100-99.5 \right)=0.5\%$ in the weight of the body.
Hence, the correct option is C) decrease by $0.5\%$.

Note: This problem could also be solved by directly using the formula for variation in acceleration due to gravity at a height $h$ above the surface.
$g'=g\left( 1-\dfrac{2h}{{{R}_{E}}} \right)$ .
From this formula it can be clearly seen that for the same height and depth since, this formula has a factor of 2 in it, the weight will always decrease twice the amount of the decrease in weight when the body is taken to the same depth under the surface as above the surface.
Therefore, this shortcut could also have been used in this question. Since the decrease in weight above the surface was $1%$, for the same depth the decrease would be half of that, that is $0.5%$ which is the correct answer.