A ship sinks in the bottom of a \(250 m\) deep lake. The atmospheric pressure over the lake is \(103 \times 10\) s Pa. Taking the density of water in the lake to be \(1000 kg / m ^{3}\), calculate the pressure exerted on the boat. [Acceleration due to gravity \(\left.=10 m / s ^{2}\right]\)
A. \(2.60 \times 10^{6} Pa\) B. \(2.50 \times 10^{6} Pa\) C. \(2.60 \times 105 Pa\) D. \(1.63 \times 105 Pa\)
Correct Answer: A
Explanation
Pressure exerted on the boat \(=P_{\text {atm }}+P_{\text {water }}\). $$ \begin{aligned} &\begin{aligned} P _{\text {water }} &=\rho h g \\ &=(250 \times 1000 \times 10) \\ &=2.5 \times 106 Nm ^{-2} \end{aligned} \\ &\begin{aligned}5 \times 10 Nm ^{-2} &=2.5 \times 106 Pa \\ & P_{\text {atm }}=1.03 \times 10^{5} Pa \\ \text { Pressure exerted on the boat } \\ &=\left(1.03 \times 10^{5}+2.5 \times 10^{6}\right) Pa \\ &=2.6 \times 10^{6} Pa \end{aligned} \end{aligned} $$
