Question

Pressure at the bottom of sea at some particular place is 9,00,0000 Pa. If density of sea water is $1.0{\mathrm{gcm}}^{-3}$, then what is the depth of sea (Neglect the pressure of the atmosphere and take $\mathrm{g}=10{\mathrm{ms}}^{-2}$.

• A. $865\mathrm{m}$
• B. $8600\mathrm{km}$
• C. $86000\mathrm{m}$
• D. $8.6\times {10}^7\mathrm{m}$

Answer 1

The correct option is A

$865\mathrm{m}$

Given :pressure at sea bottom at some particular place$=9,00,0000\mathrm{Pa}$.
density of sea water=$1.0{\mathrm{gcm}}^{-3}$
to find the depth of sea
We know that the pressure exerted by any fluid at certain depth is given by
$$\begin{gathered} \mathrm{P}=\mathrm{\rho gh}(\mathrm{where}\mathrm{h}\mathrm{is}\mathrm{the}\mathrm{depth}\mathrm{of}\mathrm{the}\mathrm{sea},\mathrm{\rho }\mathrm{is}\mathrm{density},\mathrm{g}-\mathrm{acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity}\mathrm{and}\mathrm{p}-\mathrm{pressure}) \\ 9000000=1040\times 9.8\times \mathrm{h}(\mathrm{as}1.04\mathrm{g}/{\mathrm{cm}}^{-3}=1040\mathrm{kg}/{\mathrm{m}}^{-3}) \\ \Rightarrow \mathrm{h}=\frac{9000000}{1040\times 9.8}=833.6\mathrm{m} \\ \mathrm{hence}\mathrm{option}\left(\mathrm{a}\right)\mathrm{is}\mathrm{the}\mathrm{correct}\mathrm{answer}. \end{gathered}$$