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Question: Weight of a body decreases by \(1.5%\), when it is raised to a height h above the surface of the Ear...

Weight of a body decreases by 1.51.5%, when it is raised to a height h above the surface of the Earth. When the same body is taken to same depth h in a mine, its weight will show
a) 0.75% increase
b) 3.0% decrease
c) 0.75% decrease
d) 1.5% decrease

Explanation

Solution

When a body moves upwards the mass of the body remains the same but the weight decreases as the gravitational pull of the Earth towards the Center decreases.

Complete step by step answer:
Given, the weight of a body decreases by 1.51.5% when it is raised to a height h above the surface of the Earth.
We know that the weight of a particle at surface level is,
W=GMR2W = \dfrac{{GM}}{{{R^2}}}
where G is the gravitational constant , M = mass of the particle and R = radius of Earth
Therefore,
1.5% = decrease percentage in weight of the particle when it is raised to height h
1.5%=GMR2GM(R+h)2GMR2\Rightarrow 1.5\% = \dfrac{{\dfrac{{GM}}{{{R^2}}} - \dfrac{{GM}}{{{{(R + h)}^2}}}}}{{\dfrac{{GM}}{{{R^2}}}}}
On taking GM common we get,
1.5%=1R21(R+h)21R2\Rightarrow 1.5\% = \dfrac{{\dfrac{1}{{{R^2}}} - \dfrac{1}{{{{(R + h)}^2}}}}}{{\dfrac{1}{{{R^2}}}}}
After adding using LCM we get,
1.5%=(R+h)R2(R+h)2R2R2\Rightarrow 1.5\% = \dfrac{{\dfrac{{{{(R + h)}^{}} - {R^2}}}{{{{(R + h)}^2}{R^2}}}}}{{{R^2}}}
On simplifying we get,
1.5%=R2+2Rh+h2R2(R+h)2\Rightarrow 1.5\% = \dfrac{{{R^2} + 2Rh + {h^2} - {R^2}}}{{{{(R + h)}^2}}}
From gravitation, we know that,
At depth h, g=g(1hR)g' = g(1 - \dfrac{h}{R})
As g’ decreases, ggg=hR\dfrac{{g - g'}}{g} = \dfrac{h}{R}
2hR+(hR)2(1+hR)2=1.5100\Rightarrow \dfrac{{\dfrac{{2h}}{R} + {{(\dfrac{h}{R})}^2}}}{{{{(1 + \dfrac{h}{R})}^2}}} = \dfrac{{1.5}}{{100}}
After simplification we get,
2hR=1.5100\Rightarrow \dfrac{{2h}}{R} = \dfrac{{1.5}}{{100}}
On cross multiplying 2 we get,
hR=0.75100\Rightarrow \dfrac{h}{R} = \dfrac{{0.75}}{{100}}
Since h < < r at depth h, there will be decrease in the weight of the particle by 0.75%
Hence, the required option for the given question is c) 0.75% increase.
Additional information:
Climbing stairs and lifting objects is work in both the scientific and everyday sense i.e. it is work done against the gravitational force. When there is work, there is transformation of energy.

Note: Students need to know the concepts used in the sum, they also need to know when to simplify the given equation and when to equate it with other equations derived from different formulas. They must also know the decreasing and increasing effect of the Gravitational pull of the Earth when the body moves above or below the Earth’s surface.