Answer
$4.34\times10^{4}$
Work Step by Step
According to the Bernoulli’s equation:
$p_1+\frac{1}{2}\rho v_1^2+\rho gh_1=p_2+\frac{1}{2}\rho v_2^2+\rho gh_2$
To find the minimum pressure difference, we have to put $v_1=v_2$
Therefore,
$(p_1-p_2)_{min}=\rho g(h_2-h_1)$
Substituting the given values
$(p_1-p_2)_{min}=1000\times9.81\times(6.59-2.16)\;Pa$
or, $(p_1-p_2)_{min}=4.34\times10^{4}\;Pa$
