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  • The pressure acting on a submarine is Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be : (Assume that

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} )
 
Option: 1 \frac{200}{3}\; ^{o}/_{o}
Option: 2 \frac{200}{5}\; ^{o}/_{o}
Option: 3 \frac{5}{200}\; ^{o}/_{o}  
Option: 4 \frac{3}{200}\; ^{o}/_{o}

Answers (1)

best_answer

 \mathrm{P}_{1}=\rho \mathrm{g} \mathrm{d}+\mathrm{P}_{0}=3 \times 10^{5} \mathrm{~Pa}

As\ \mathrm{P}_{0}=1 \times 10^{5} \mathrm{~Pa}
\therefore \quad \rho \mathrm{g} \mathrm{d}=2 \times 10^{5} \mathrm{~Pa}

Now the depth is doubled

\mathrm{P}_{2}=2 \rho \mathrm{g} \mathrm{d}+\mathrm{P}_{0} =4 \times 10^{5}+10^{5}=5 \times 10^{5} \mathrm{~Pa} \\
 \text { \%increase }=\frac{\mathrm{P}_{2}-\mathrm{P}_{1}}{\mathrm{P}_{1}} \times 100 =\frac{5 \times 10^{5}-3 \times 10^{5}}{3 \times 10^{5}} \times 100=\frac{200}{3} \%

Posted by

avinash.dongre

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