The acceleration due to gravity on the earth’s surface at the poles is g
Question:

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)

  1. (1) $\frac{R^{2} \omega^{2}}{2 g}$

  2. (2) $\frac{R^{2} \omega^{2}}{g}$

  3. (3) $\frac{R^{2} \omega^{2}}{4 g}$

  4. (4) $\frac{R^{2} \omega^{2}}{8 g}$


Correct Option: , 2

Solution:

(2) Value of $g$ at equator, $g_{A}=g \cdot-R \omega^{2}$

Value of $g$ at height $h$ above the pole,

$g_{B}=g \cdot\left(1-\frac{2 h}{R}\right)$

As object is weighed equally at the equator and poles, it means $g$ is same at these places.

$g_{A}=g_{B}$

$\Rightarrow g-R \omega^{2}=g\left(1-\frac{2 h}{R}\right)$

$\Rightarrow R \omega^{2}=\frac{2 g h}{R} \Rightarrow h=\frac{R^{2} \omega^{2}}{2 g}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.