$\lambda  = {45^ \circ }\,;\,R = 6400 \times {10^3}m$

$\omega  = \frac{{2\pi }}{{24 \times 60 \times 60}}$

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}$