JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
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 \(10^{3}\, kg \,m ^{-3}, g =10 \,ms ^{-2}\) )
- A \(\frac{200}{3} \%\)
- B \(\frac{200}{5} \%\)
- C \(\frac{5}{200} \%\)
- D \(\frac{3}{200} \%\)
Answer & Solution
Correct Answer
(A) \(\frac{200}{3} \%\)
Step-by-step Solution
Detailed explanation
\(P _{1}=\rho gd + P _{0}=3 \times 10^{5} Pa\) \(\therefore \rho gd =2 \times 10^{5} Pa\) \(P _{2}=2 \rho gd + P _{0}\) \(=4 \times 10^{5}+10^{5}=5 \times 10^{5} Pa\) \(\%\)increase \(=\frac{ P _{2}- P _{1}}{ P _{1}} \times 100\)…
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