At present the acceleration due to gravity at latitude `45^(@)` on earth is `9.803ms^(-2)`. If earth stops rotating, the acceleration due to gravity at the same place would be`" "(Romega^(2)=0.034ms^(-2))`

Options

A. `9.837ms^(-2)`
B. `9.82ms^(-2)`
C. `9.81ms^(-2)`
D. `9.786ms^(-2)`

Answer

b

Explanation

`g_(phi)=g-Romega^(2)cos^(2)phi`
`=g-Romega^(2)((1)/(sqrt2))^(2)=g-(Romega^(2))/(2)`
`g=g_(phi)+(Romega^(2))/(2)=9.803+(0.034)/(2)` ltbRgt `=9.82ms^(-2)`.