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........ $ms^{-1}$
Diffcult
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(b)According to Bernoulli's theorem,
${P_B} + h\rho g = {P_A} + \frac{1}{2}\rho v_A^2$$({\rm{As }}{v_A} > > {v_B})$
$3.10P + 53 \times 660 \times 10 = P + \frac{1}{2} \times 660v_A^2$
==> $2.1 \times 1.01 \times {10^5} + 3.498 \times {10^5}$$ = \frac{1}{2} \times 660 \times v_A^2$
==> $5.619 \times {10^5} = \frac{1}{2} \times 660 \times v_A^2$
${v_A} = \sqrt {\frac{{2 \times 5.619 \times {{10}^5}}}{{660}}} = 41 m/s$
${P_B} + h\rho g = {P_A} + \frac{1}{2}\rho v_A^2$$({\rm{As }}{v_A} > > {v_B})$
$3.10P + 53 \times 660 \times 10 = P + \frac{1}{2} \times 660v_A^2$
==> $2.1 \times 1.01 \times {10^5} + 3.498 \times {10^5}$$ = \frac{1}{2} \times 660 \times v_A^2$
==> $5.619 \times {10^5} = \frac{1}{2} \times 660 \times v_A^2$
${v_A} = \sqrt {\frac{{2 \times 5.619 \times {{10}^5}}}{{660}}} = 41 m/s$
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