Question

# If Earth is supposed to be a sphere of radius $$R$$, if $$g_{30^o}$$ is value of acceleration due to gravity at latitude of $$30^{\circ}$$ and $$g$$ at the equator, the value of $$g-g_{30^o}$$ is

A
14ω2R
B
34ω2R
C
ω2R
D
12ω2R

Solution

## The correct option is A $$\displaystyle \frac {3}{4} \omega^2 R$$Acceleration due to gravity at latitude '$$\lambda$$' is given by$$\displaystyle g_\lambda = g-R \omega^2 cos^2 \lambda$$At $$\displaystyle 30^{\circ}, g_{30}=g-R\omega^2 cos^2 30=g-\frac {3}{4}R\omega^2$$or, $$\displaystyle g-g_{30}=\frac {3}{4}R\omega^2$$Physics

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