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Question: A sniper fires a rifle bullet into a gasoline tank making a hole 53.0 m below the surface of gasolin...

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

A

(a) 27.8ms−127.8ms^{- 1}

A

(b) 41.0ms−141.0ms^{- 1}

A

(c) 9.6ms−19.6ms^{- 1}

A

(d) 19.7ms−119.7ms^{- 1}

Explanation

Solution

(b)

Sol.

According to Bernoulli's theorem,

PB+hρg=PA+12ρvA2P_{B} + h\rho g = P_{A} + \frac{1}{2}\rho v_{A}^{2} (As vA>>vB)3.10P+53×660×10=P+12×660vA2(\text{As }v_{A} > > v_{B})3.10P + 53 \times 660 \times 10 = P + \frac{1}{2} \times 660v_{A}^{2}

⇒ 2.1×1.01×105+3.498×105=12×660×vA22.1 \times 1.01 \times 10^{5} + 3.498 \times 10^{5} = \frac{1}{2} \times 660 \times v_{A}^{2}

⇒ 5.619×105=12×660×vA25.619 \times 10^{5} = \frac{1}{2} \times 660 \times v_{A}^{2}

∴ vA=2×5.619×105660v_{A} = \sqrt{\frac{2 \times 5.619 \times 10^{5}}{660}}= 41 m/s