An inverted tube barometer is kept on a lift with a moving downward with a deceleration $\alpha $ . The density of mercury is $\rho$ and acceleration due to gravity is $g$ . If the atmospheric pressure be $P_0$ then
AHeight of the mercury column in the lift will be $ \frac{{{P_0}}}{{\rho \left( {g + a} \right)}}$
BHeight of the mercury column in the lift will be $\frac{{{P_0}}}{{\rho \left( {g - a} \right)}}$
CHeight of the mercury column in the lift will be $\frac{{{P_0}}}{{\rho g}}$
DHeight of the mercury column in the lift will be $\frac{{{P_0}}}{{\rho a}}$
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AHeight of the mercury column in the lift will be $ \frac{{{P_0}}}{{\rho \left( {g + a} \right)}}$
a $P_{o}=\rho(g+a) h \Rightarrow h=\frac{P_{0}}{\rho(g+a)}$
$p(g+a)$
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