Question

If the earth was to cease rotating about its own axis, the increase in the value of $g$ in $\text{C.G.S.}$ system at a place of latitude of ${45}^{\circ }$ will be

• A. $2.68$
• B. $1.68$
• C. $3.36$
• D. $0.34$

Answer 1

The correct option is B $1.68$

Given:
$\theta ={45}^{\circ };R=6400\times {10}^3\text{m}$

Angular velocity $\left(\omega \right)$ of the earth is given by

$\omega =\frac{2\pi }{T}$

$\Rightarrow \omega =\frac{2\pi }{24\times 60\times 60}(\because T=24\text{hr})$

The value of acceleration due to gravity with latitude $\theta$ due to rotation of earth is,

$g^{\prime }=g-R{\omega }^2{\cos }^2\theta \Rightarrow (g-g^{\prime })=R{\omega }^2{\cos }^2\theta$

$\Rightarrow g-g^{\prime }=6400\times {10}^3\times {\left(\frac{2\times 3.14}{2\times 60\times 60}\right)}^2\times {\left(\frac{1}{\sqrt{2}}\right)}^2$

$\Rightarrow g-g^{\prime }=16.89\times {10}^{-3}{\text{m/sec}}^2=16.89\times {10}^{-1}{\text{cm/sec}}^2$

$\Rightarrow g-g^{\prime }=1.68{\text{cm/sec}}^2$

Hence, option $\left(b\right)$ is correct.

Why this question? To familiarize students with the acceleration due to gravity at a point not on the equator and the effect of earth's rotation on it. The acceleration due to gravity latitude $\theta$ is given by $g^{\prime }=g-{\omega }^{2}R{\cos }^2\theta$