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Q.

Calculate the percentage decrease in the weight of the body when it is taken 32 km below the surface of the earth.

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a

0.1%

b

0.01%

c

0.5%

d

0.05%

answer is C.

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Detailed Solution

Concept: Percentage decrease in weight :

\frac{mg - mg'}{mg'} \times 100

To obtain the value of the percentage drop in the value of weight, we will first determine the value of g' and then substitute this value in the previous equation.
Given that the earth's radius (R) is 6400 km and its height (h) is 32 km, the acceleration of gravity (g) is 9.8 m/s2.
We will start by calculating the acceleration caused by gravity at the specified height.
We can use the formula below to get the gravity at a specific height (g').

g^{'} = g \times (1 - \frac{h}{R})

=g’=9.8

\times (1 - \frac{32}{6400})

=

g^{'} = 9.8 \times (\frac{6400 - 32}{6400})

=

g^{'} = 9.8 \times 0.995 \frac{m}{s^{2}} = 9.751 m / s^{2}

We know that,

\frac{mg - mg'}{mg'} \times 100

\frac{g - g'}{g} \times 100

= \frac{9.8 - 9.751}{9.8} \times 100 = 0.5 %

Hence, the correct option is 3.

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