The pressure acting on a submarine is 3 times | vedclass

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Question

(A) The total pressure $P$ at a depth $d$ is given by $P = P_{0} + ho gd$,where $P_{0}$ is the atmospheric pressure.Given $P_{1} = 3 	imes 10^{5} \

Vedclass · Accepted Answer

$\frac{200}{3} \%$

Vedclass · Answer

(A) The total pressure $P$ at a depth $d$ is given by $P = P_{0} + ho gd$,where $P_{0}$ is the atmospheric pressure.Given $P_{1} = 3 	imes 10^{5} \; Pa$ and $P_{0} = 1 	imes 10^{5} \; Pa$.So,$ho gd = P_{1} - P_{0} = 3 	imes 10^{5} - 1 	imes 10^{5} = 2 	imes 10^{5} \; Pa$.If the depth is doubled $(d' = 2d)$,the new pressure $P_{2}$ is $P_{2} = P_{0} + ho g(2d) = P_{0} + 2(ho gd)$.Substituting the values: $P_{2} = 1 	imes 10^{5} + 2(2 	imes 10^{5}) = 1 	imes 10^{5} + 4 	imes 10^{5} = 5 	imes 10^{5} \; Pa$.The percentage increase in pressure is given by $\frac{P_{2} - P_{1}}{P_{1}} 	imes 100$.Percentage increase $= \frac{5 	imes 10^{5} - 3 	imes 10^{5}}{3 	imes 10^{5}} 	imes 100 = \frac{2 	imes 10^{5}}{3 	imes 10^{5}} 	imes 100 = \frac{200}{3} \%$.

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