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 in its weight will:
A) Decreases by 0.5%.
B) Decreases by 2%.
C) Increases by 0.5%.
D) Increases by 1%.

Answer 1

The weight of the body is defined as the weight by which an object is pulled towards the centre of the earth. The weight of anybody changes with change in the height of the body. The weight of a body at height h will be different from the weight of the body at the surface of the earth.

Formula used: The formula of the acceleration due to gravity at height h is given by,
$\Rightarrow g' = g\left( {1 - \dfrac{{2h}}{R}} \right)$
Where apparent acceleration due to gravity is $g'$ the normal acceleration due to gravity is $g$ radius of earth is R and the distance above the surface of earth is h.
The formula of the acceleration due to gravity at depth h is given by,
$\Rightarrow g' = g\left( {1 - \dfrac{h}{R}} \right)$
Where apparent acceleration due to gravity is $g'$ the normal acceleration due to gravity is $g$ radius of earth is R and depth is equal to h.

Complete step by step solution:
It is given in the problem that 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 we need to tell the change in the weight of the body.
The weight of the body decreases by 1% when raised by height h.
The formula of the acceleration due to gravity at height h is given by,
$\Rightarrow g' = g\left( {1 - \dfrac{{2h}}{R}} \right)$.........eq. (1)
Where apparent acceleration due to gravity is $g'$ the normal acceleration due to gravity is $g$ radius of earth is R and the distance above the surface of earth is h.
The weight will be,
$\Rightarrow mg' = mg\left( {1 - \dfrac{{2h}}{R}} \right)$
Since the change in the weight is 1% therefore,
$\Rightarrow \dfrac{{mg - mg'}}{{mg}} = 1\%$
$\Rightarrow \dfrac{{g - g'}}{g} = 1\%$………eq. (2)
Replacing the value of $g'$ from the equation (1) in equation (2) we get.
$\Rightarrow \dfrac{{g - g'}}{g} = 1\%$
$\Rightarrow \dfrac{{g - g\left( {1 - \dfrac{{2h}}{R}} \right)}}{g} = 1\%$
$\Rightarrow 1 - \left( {1 - \dfrac{{2h}}{R}} \right) = 1\%$
$\Rightarrow 1 - 1 + \dfrac{{2h}}{R} = 1\%$
$\Rightarrow \dfrac{{2h}}{R} = 1\%$
$\Rightarrow \dfrac{h}{R} = 0 \cdot 5\%$………eq. (3)
Let us calculate the weight of the body at depth h.
The formula of the acceleration due to gravity at depth h is given by,
$\Rightarrow g' = g\left( {1 - \dfrac{h}{R}} \right)$
Where apparent acceleration due to gravity is $g'$ the normal acceleration due to gravity is $g$ radius of earth is R and depth is equal to h.
The weight of the body is equal to,
$\Rightarrow mg' = mg\left( {1 - \dfrac{h}{R}} \right)$
Replacing the value of $\dfrac{h}{R}$ from equation (3) in above relation.
$\Rightarrow mg' = mg\left( {1 - \dfrac{h}{R}} \right)$
$\Rightarrow mg' = mg\left( {1 - 0 \cdot 5\% } \right)$
$\Rightarrow mg' = mg\left( {1 - 0 \cdot 005} \right)$
$\Rightarrow mg' = \left( {0 \cdot 995} \right)mg$………eq. (4)
The percentage change of the weight is equal to,
$\Rightarrow \dfrac{{mg - mg'}}{{mg}} \times 100$
$\Rightarrow \dfrac{{mg - mg'}}{{mg}} \times 100$………eq. (5)
Replacing the value of $mg'$ from equation (4) into equation (5) we get.
$\Rightarrow \dfrac{{mg - mg'}}{{mg}} \times 100$
$\Rightarrow \dfrac{{mg - \left( {0 \cdot 995} \right)mg}}{{mg}} \times 100$
$\Rightarrow 1 - 0 \cdot 995 \times 100 = 0 \cdot 5\%$.

The weight of the body decreases by 0.5%. The correct answer for this problem is option A.

Note: It is advisable for students to understand and remember the formula of the apparent acceleration due to gravity for an object at some height and also for object at some depth. The weight of the body changes according to the acceleration due gravity but the mass of a body never changes and is constant everywhere.