JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 21)
The value of the acceleration due to gravity is
g1 at a height h = $${R \over 2}$$ (R = radius of the earth) from the surface of the earth. It is again equal
to g1 at a depth d below the surface of the
earth. The ratio $$\left( {{d \over R}} \right)$$ equals :
$${5 \over 9}$$
$${1 \over 9}$$
$${7 \over 9}$$
$${4 \over 9}$$
Explanation
Given, $${g_{at\,high}} = {g_{at\,depth}}$$
We know, $${g_{depth}} = $$$$g\left( {1 - {d \over R}} \right)$$
$$ \therefore $$ $$g\left( {1 - {d \over R}} \right) = {{GM_e} \over {{{(R + h)}^2}}}$$
$$ \Rightarrow $$ $$g\left( {1 - {d \over R}} \right) =$$$${{G{M_e}} \over {{{\left( {R + {R \over 2}} \right)}^2}}}$$ = $${4 \over 9}{{G{M_e}} \over {{R^2}}}$$ = $${4 \over 9}g$$
$$ \Rightarrow $$ $${d \over R} = 1 - {4 \over 9} = {5 \over 9}$$
We know, $${g_{depth}} = $$$$g\left( {1 - {d \over R}} \right)$$
$$ \therefore $$ $$g\left( {1 - {d \over R}} \right) = {{GM_e} \over {{{(R + h)}^2}}}$$
$$ \Rightarrow $$ $$g\left( {1 - {d \over R}} \right) =$$$${{G{M_e}} \over {{{\left( {R + {R \over 2}} \right)}^2}}}$$ = $${4 \over 9}{{G{M_e}} \over {{R^2}}}$$ = $${4 \over 9}g$$
$$ \Rightarrow $$ $${d \over R} = 1 - {4 \over 9} = {5 \over 9}$$
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