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(D) Given: $a = g/2$. Let the liquid level in the left limb rise by $x$ and in the right limb fall by $x$. The pressure at point $A$ (left corner) is

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Given: $a = g/2$. Let the liquid level in the left limb rise by $x$ and in the right limb fall by $x$. The pressure at point $A$ (left corner) is $P_A = P_0 + \rho g h + 2\rho g(h-x) = P_0 + 3\rho g h - 2\rho g x$.The pressure at point $B$ (right corner) is $P_B = P_0 + \rho g x$.The pressure difference between $A$ and $B$ due to horizontal acceleration is $P_A - P_B = \rho_{eff} \cdot L \cdot a$,where $L = 2h$ is the length of the horizontal arm.Considering the effective mass distribution,the pressure difference is $P_A - P_B = [2\rho(h+x) + \rho(h-x)] \cdot a$ is incorrect; the correct approach is using the pressure gradient: $P_A - P_B = (2\rho) \cdot (2h) \cdot a$.$P_A - P_B = (2\rho)(2h)(g/2) = 2\rho g h$.Substituting the expressions for $P_A$ and $P_B$:$(P_0 + 3\rho g h - 2\rho g x) - (P_0 + \rho g x) = 2\rho g h$.$3\rho g h - 3\rho g x = 2\rho g h$.$\rho g h = 3\rho g x \implies x = h/3$.The difference in heights is $\Delta H = (h+x) - (h-x) = 2x = 2h/3$.Since $2h/3$ is not in the options,the correct answer is $D$.

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