"The pressure acting on a submarine is 3 $\times$ 105 Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be : (Assume that atmospheric pressure is 1 $\times$ 105 Pa density of water is 103 kg m$-$3, g = 10 ms$-$2)"

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Accepted Answer

"P = P₀ + h$\rho$g = 3 $\times$ 10⁵ Pa

$\Rightarrow$ h$\rho$g = 3 $\times$ 10⁵ $-$ 1 $\times$ 10⁵

$\Rightarrow$ h$\rho$g = 2 $\times$ 10⁵

$\therefore$ 2h$\rho$g = 4 $\times$ 10⁵

$\therefore$ P' = P₀ + 4 $\times$ 10⁵

$\therefore$ P' = 5 $\times$ 10⁵ Pa

$\therefore$ % increase in pressure = ${{P' - P} \over P} \times 100$

$= {{(5 - 3) \times {{10}^5}} \over {3 \times {{10}^5}}} \times 100$

$= {{200} \over 3}$%"

The pressure acting on a submarine is 3 $\times$ 10⁵ Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be :

(Assume that atmospheric pressure is 1 $\times$ 10⁵ Pa density of water is 10³ kg m^$-$3, g = 10 ms^$-$2)

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The pressure acting on a submarine is 3 $\times$ 105 Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be : (Assume that atmospheric pressure is 1 $\times$ 105 Pa density of water is 103 kg m$-$3, g = 10 ms$-$2)

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More Properties of Solids and Liquids questions

Drill 25 more like these. Every day. Free.

The pressure acting on a submarine is 3 $\times$ 10⁵ Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be :

(Assume that atmospheric pressure is 1 $\times$ 10⁵ Pa density of water is 10³ kg m^$-$3, g = 10 ms^$-$2)