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(C) In a frame of reference accelerating with $a = 2g$ horizontally,the effective acceleration $\vec{g}_{eff}$ acting on the liquid is $\vec{g}_{eff}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) In a frame of reference accelerating with $a = 2g$ horizontally,the effective acceleration $\vec{g}_{eff}$ acting on the liquid is $\vec{g}_{eff} = \vec{g} - \vec{a}$.Since $\vec{g} = -g\hat{j}$ and $\vec{a} = 2g\hat{i}$,we have $\vec{g}_{eff} = -2g\hat{i} - g\hat{j}$.The magnitude of the effective acceleration is $g_{eff} = \sqrt{(-2g)^2 + (-g)^2} = \sqrt{4g^2 + g^2} = g\sqrt{5}$.The pressure variation in the liquid is given by $dP = \rho \vec{g}_{eff} \cdot d\vec{r}$.The pressure is minimum at the point furthest in the direction of $\vec{g}_{eff}$. Let this point be $A$. The pressure increases as we move opposite to the direction of $\vec{g}_{eff}$.The pressure at any point at a distance $r$ from the point of minimum pressure along the direction of $\vec{g}_{eff}$ is $P = P_0 + \rho g_{eff} r$.The centre of the sphere is at a distance $R$ from the point of minimum pressure along the direction of $\vec{g}_{eff}$.Therefore,the pressure at the centre is $P_c = P_0 + \rho g_{eff} R = P_0 + \rho (g\sqrt{5}) R = P_0 + \rho gR\sqrt{5}$.

$A$ 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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