Answer
$0.97\;atm$
Work Step by Step
Applying the Bernoulli’s equation, we obtain
$P_2+\frac{1}{2}\rho v_2^2+\rho hh_2=P_1+\frac{1}{2}\rho v_1^2+\rho hh_1$
Here, $h_1=h_2$ and $P_2=P_{air}=1.01\times10^{5}\;Pa$
Therefore,
$P_1=P_{air}+\frac{1}{2}\rho v_1^2-\frac{1}{2}\rho v_2^2$
Substituting the given values
$P_1=1.01\times10^{5}+\frac{1}{2}\times1000\times( 15^2-5.4^2)$
or, $P_1=198920\;Pa$
or $P_1=1.97\;atm$
Thus, the gauge pressure is $(1.97 –1.00)\; atm= 0.97\;atm$
