JEE MAIN - Physics (2020 - 5th September Evening Slot - No. 7)
The acceleration due to gravity on the earth’s
surface at the poles is g and angular velocity of
the earth about the axis passing through the
pole is $$\omega $$. An object is weighed at the equator
and at a height h above the poles by using a
spring balance. If the weights are found to be
same, then h is (h << R, where R is the radius
of the earth)
$${{{R^2}{\omega ^2}} \over {2g}}$$
$${{{R^2}{\omega ^2}} \over g}$$
$${{{R^2}{\omega ^2}} \over {8g}}$$
$${{{R^2}{\omega ^2}} \over {4g}}$$
Explanation
At equator, g1 = g - R$${\omega ^2}$$
At height h, g2 = $$g\left( {1 - {{2h} \over R}} \right)$$ [as given h << R]
$$ \because $$ Weight same at poles and at h (so g1 = g2)
$$ \therefore $$ g - R$${\omega ^2}$$ = $$g\left( {1 - {{2h} \over R}} \right)$$
$$ \Rightarrow $$ $${R{\omega ^2} = {{2gh} \over R}}$$
$$ \Rightarrow $$ $${h = {{{R^2}{\omega ^2}} \over {2g}}}$$
At height h, g2 = $$g\left( {1 - {{2h} \over R}} \right)$$ [as given h << R]
$$ \because $$ Weight same at poles and at h (so g1 = g2)
$$ \therefore $$ g - R$${\omega ^2}$$ = $$g\left( {1 - {{2h} \over R}} \right)$$
$$ \Rightarrow $$ $${R{\omega ^2} = {{2gh} \over R}}$$
$$ \Rightarrow $$ $${h = {{{R^2}{\omega ^2}} \over {2g}}}$$
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