- Home
- Standard 11
- Physics
If the earth were to cease rotating about its own axis. The increase in the value of $g$ in $C.G.S.$ system at a place of latitude of $45^o$ will be ........ $cm/sec^{2}$.
$2.68$
$1.68$
$3.36$
$0.34$
Solution
$\lambda = {45^ \circ }\,;\,R = 6400 \times {10^3}m$
$\omega = \frac{{2\pi }}{{24 \times 60 \times 60}}$
The value of acceleration due to gravity with latitude $\lambda $
due to rotation of earth is,
$g' = g – R{\omega ^2}{\cos ^2}\lambda $
$g – g' = R{\omega ^2}{\cos ^2}\lambda $
$ = \frac{{6400 \times {{10}^3}}}{2} \times {\left( {\frac{{2 \times 3.14}}{{24 \times 60 \times 60}}} \right)^2}$
$ = \frac{{6400 \times {{10}^3} \times 4 \times 3.14 \times 3.14}}{{2 \times 24 \times 60 \times 60 \times 24 \times 60 \times 60}}$
$ = 16.89 \times {10^{ – 3}}m/{\sec ^2}$
$ = 16.89 \times {10^{ – 1}}cm/{\sec ^2}$
$ = 1.68cm/{\sec ^2}$
