Answer
$9.24\times10^{4}\;Pa$
Work Step by Step
Applying the Bernoulli’s equation between the points B and C, we obtain
$p_B+\frac{1}{2}\rho v_B^2+\rho gh_B=p_C+\frac{1}{2}\rho v_C^2+\rho gh_C$
Here, $p_C=p_0$, $v_C=3.19\;m/s$ $h_C=0$, $h_B=(h_1+d+h_2)$ and the speed of the liquid at B can be assumed to be equal to $v_C$
Substituting the above values, we obtain
$p_B+\frac{1}{2}\rho v_C^2+\rho g(h_1+d+h_2)=p_0+\frac{1}{2}\rho (0)^2+\rho g(0)$
or, $p_B=p_0-\rho g(h_1+d+h_2)$
or, $p_B=[1.0\times10^5-1000\times9.81\times(0.25+0.12+0.40)]\;Pa$
or, $\boxed{p_B=9.24\times10^{4}\;Pa}$
Therefore, the pressure in the liquid at the topmost point B is $9.24\times10^{4}\;Pa$
