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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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27.8 m/s

41.0 m/s

9.6 m/s

19.7 m/s

answer is B.

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Detailed Solution

According to Bernoulli's theorem,

$P_{B} + h ρ g = P_{A} + \frac{1}{2} ρ v_{A}^{2}$

$v_{A} > > v_{B})$

$3.10 P + 53 \times 660 \times 10 = P + \frac{1}{2} \times 660 v_{A}^{2}$

$\Rightarrow 2.1 \times 1.01 \times 10^{5} + 3.498 \times 10^{5} = \frac{1}{2} \times 660 \times v_{A}^{2}$

$\Rightarrow 5.619 \times 10^{5} = \frac{1}{2} \times 660 \times v_{A}^{2}$

$∴ v_{A} = \sqrt{\frac{2 \times 5.619 \times 10^{5}}{660}} = 41 {ms}^{- 1}$

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