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Weight of a body decreases by $1.5 \%$, when it is raised to a height $h$ above the surface of earth. When the same body is taken to same depth $h$ in a mine, its weight will show ........
$0.75 \%$ increase
$3.0 \%$ decrease
$0.75 \%$ increase
$1.5 \%$ decrease
Solution
(c)
Given weight,
$1.5=\frac{\frac{ GM }{ R ^2}-\frac{ GM }{( R + h )^2}}{\frac{ GM }{ R ^2}}$
$=\frac{ R ^2+2 Rh + h ^2- R ^2}{( R + h )^2}$
$\Rightarrow \frac{2 Rh + h ^2}{( R + h )^2}=\frac{1.5}{100}$
At depth ' $h, g$ ' $=g\left(1-\frac{h}{R}\right)$
SO g' decreases, $\frac{g-g}{g}=\frac{h}{R}$
$\Rightarrow \frac{\frac{2 h }{ R }+\left(\frac{ h }{ R }\right)^2}{\left(1+\frac{ h }{ R }\right)^2}=\frac{1.5}{100}$
$\frac{2 h }{ R }=\frac{1.5}{100}$
$\frac{ h }{ R }=\frac{0.75}{100}$
But since $h \ll \ll R$
At depth ' $h$ there will be a decrese of $g$ by $0.75 \%$
