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(B) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - R\omega^2 \cos^2 \lambda$,where $g$ is the accelerati

Vedclass · Accepted Answer

$1.68$

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(B) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - R\omega^2 \cos^2 \lambda$,where $g$ is the acceleration due to gravity if the earth were stationary,$R$ is the radius of the earth,and $\omega$ is the angular velocity of the earth.The increase in the value of $g$ when the earth stops rotating is $\Delta g = g - g' = R\omega^2 \cos^2 \lambda$.Given:$R = 6400 	imes 10^3 \ m = 6.4 	imes 10^6 \ m$$\lambda = 45^{\circ} \implies \cos^2 45^{\circ} = (1/\sqrt{2})^2 = 0.5$$\omega = \frac{2\pi}{T} = \frac{2 	imes 3.14}{24 	imes 3600} \approx 7.27 	imes 10^{-5} \ rad/sec$Substituting the values:$\Delta g = (6.4 	imes 10^6) 	imes (7.27 	imes 10^{-5})^2 	imes 0.5$$\Delta g = (6.4 	imes 10^6) 	imes (52.85 	imes 10^{-10}) 	imes 0.5$$\Delta g \approx 0.0169 \ m/sec^2$Converting to $C.G.S.$ units $(cm/sec^2)$:$1 \ m/sec^2 = 100 \ cm/sec^2$$\Delta g = 0.0169 	imes 100 = 1.69 \ cm/sec^2 \approx 1.68 \ cm/sec^2$.

If the earth were to cease rotating about its own axis,the increase in the value of $g$ in the $C.G.S.$ system at a place of latitude of $45^{\circ}$ will be ........ $cm/sec^{2}$.

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