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(C) The sphere is accelerating horizontally with $a = 2g$. The effective acceleration $g_{eff}$ acting on the liquid in the frame of the sphere is the

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$P_0 + \rho gR\sqrt{5}$

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(C) The sphere is accelerating horizontally with $a = 2g$. The effective acceleration $g_{eff}$ acting on the liquid in the frame of the sphere is the vector sum of the gravitational acceleration $g$ (downwards) and the pseudo-acceleration $a = 2g$ (backwards).$g_{eff} = \sqrt{g^2 + (2g)^2} = \sqrt{g^2 + 4g^2} = \sqrt{5g^2} = g\sqrt{5}$.The pressure variation in the liquid is given by $\Delta P = \rho g_{eff} h$,where $h$ is the displacement along the direction of $g_{eff}$.The minimum pressure $P_0$ occurs at the point furthest in the direction of $g_{eff}$ (the 'top' relative to the effective gravity vector),and the pressure increases as we move in the direction of $g_{eff}$.The distance from the point of minimum pressure to the centre of the sphere along the direction of $g_{eff}$ is $R$.Therefore,the pressure at the centre is $P_{centre} = P_0 + \rho g_{eff} R$.Substituting $g_{eff} = g\sqrt{5}$,we get $P_{centre} = P_0 + \rho (g\sqrt{5}) R = P_0 + \rho gR\sqrt{5}$.

$A$ hollow sphere of radius $R$ is filled completely with an ideal liquid of density $\rho$. The sphere is moving horizontally with an acceleration $2g$,where $g$ is the acceleration due to gravity. If the minimum pressure of the liquid is $P_0$,then find the pressure at the centre of the sphere.

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