The pressure acting on a submarine is $3 \times 10^{5}\;Pa$ 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 $1 \times 10^{5} \;Pa$ density of water is $10^{3}\, kg \,m ^{-3}, g =10 \,ms ^{-2}$ )
JEE MAIN 2021, Medium
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$P _{1}=\rho gd + P _{0}=3 \times 10^{5} Pa$
$\therefore \rho gd =2 \times 10^{5} Pa$
$P _{2}=2 \rho gd + P _{0}$
$=4 \times 10^{5}+10^{5}=5 \times 10^{5} Pa$
$\%$increase $=\frac{ P _{2}- P _{1}}{ P _{1}} \times 100$
$=\frac{5 \times 10^{5}-3 \times 10^{5}}{3 \times 10^{5}} \times 100=\frac{200}{3} \%$
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