A sniper fires a rifle bullet into a gasoline tank making a hole 53.0 m below the surface of gasoline. The tank was sealed at 3.10 atm. The stored gasoline has a density of 660 kgm –3 . The velocity with which gasoline begins to shoot out of the hole is
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According to Bernoulli's theorem,
$P_B + h \rho g = P_A + \frac{1}{2} \rho v_A^2 \quad (As \ v_A \gg v_B)$ $3.10P + 53 \times 660 \times 10 = P + \frac{1}{2} \times 660 v_A^2$ $\Rightarrow 2.1 \times 1.01 \times 10^5 + 3.498 \times 10^5 = \frac{1}{2} \times 660 \times v_A^2$ $\Rightarrow 5.619 \times 10^5 = \frac{1}{2} \times 660 \times v_A^2$ $\therefore v_A = \sqrt{\frac{2 \times 5.619 \times 10^5}{660}} = 41 \ \mathrm{m/s}$
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