Question
For diver's safety, a $10 \mathrm{~m}$ platform diving pool should be $5 \mathrm{~m}$ deep. However, with an excellent dive, a diver usually reaches a maximum depth of $2.5 \mathrm{~m}$.
(i) Calculate the pressure due to the weight of the water at the depth of $2.5 \mathrm{~m}$.
(ii) Calculate the depth below the surface of water at which the pressure due to the weight of the water equals $1.0 \mathrm{~atm}$. [Density of water $=10^3 \mathrm{~kg} / \mathrm{m}^3, 1 \mathrm{~atm}=101.3 \mathrm{kPa}$ ]

Answer


Data : $\mathrm{h}=250 \mathrm{~m}, \rho=1000 \mathrm{~kg} / \mathrm{m}^3, \mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^2, 1 \mathrm{~atm}=101.3 \mathrm{kPa}$
(i) $\rho=h p g=(250)(1000)(9.8)=2.45 \mathrm{mPa}$ $=\frac{2.45 \times 10^6}{1.013 \times 10^5}=24.18 \mathrm{~atm}$
This gives the pressure at a depth of $250 \mathrm{~m}$.
(ii) $\mathrm{h}=\frac{p}{\rho g}=\frac{1.013 \times 10^5}{10^3 \times 9.8}=10.34 \mathrm{~m}$
This gives the required depth.

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