Weight of a body decreases by $1.5$%, 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

Given weight,

$1.5=\dfrac { \dfrac { GM }{ { R }^{ 2 } } -\dfrac { GM }{ { (R+h) }^{ 2 } } }{ \dfrac { GM }{ { R }^{ 2 } } } \\ =\dfrac { { R }^{ 2 }+2Rh+{ h }^{ 2 }-{ R }^{ 2 } }{ { \left( R+h \right) }^{ 2 } } \\ \Rightarrow \dfrac { 2Rh+{ h }^{ 2 } }{ { \left( R+h \right) }^{ 2 } } =\dfrac { 1.5 }{ 100 }$

At depth $'h', g'=g(1-\dfrac{h}{R})$

SO g' decreases, $\dfrac{g-g'}{g}=\dfrac{h}{R}$

$\Rightarrow \dfrac { \dfrac { 2h }{ R } +{ \left( \dfrac { h }{ R } \right) }^{ 2 } }{ { \left( 1+\dfrac { h }{ R } \right) }^{ 2 } } =\dfrac { 1.5 }{ 100 } \\ \dfrac { 2h }{ R } =\dfrac { 1.5 }{ 100 } \\ \dfrac { h }{ R } =\dfrac { 0.75 }{ 100 }$

But since $h<<<R$

At depth $'h'$ there will be a decrese of $g$ by $0.75$%