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Weight of a body of mass ‘m’ decreases by $1%$ when it is raised to a height ‘h’ above the earth’s surface. If the body is taken to a depth ‘h’ in a mine, the change in its weight is

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increases by

1%

increases by

0 .5%

decreases by

2%

decreases by

0 .5%

answer is B.

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Detailed Solution

x = \frac{Δ W_{h}}{w} \times 100 = \frac{2h}{R} \times 100 \to (1)

$y = \frac{Δ W_{d}}{w} \times 100 = \frac{h}{R} \times 100 \to (2)$

$from (1) & (2)$

$x = 2 y$

$y = \frac{x}{2} = \frac{1}{2} = 0.5 %$

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