Question

The pressure at a point in water is 10N /m square .the depth below this point where the pressure becomes double is (given density of water =10 raise to power three kg/m cube; g=10m/ s square)

Answer 1

Question #36954

The pressure at a point in water is $10\mathrm{N} / \mathrm{m}$ square. The depth below this point where the pressure becomes double is (given density of water = 10 raise to power three kg/m cube; g = 10 m/s square)

Solution:

Let

$$\begin{array}{l} P_{1} = 10 \, \mathrm{N/m^2} \quad ??? \\ \rho = 10^{3} \, \mathrm{Kg/m^3} \\ g = 10 \, \mathrm{m/s^2} \\ P_{2} = 2P_{1} \\ h = ? \end{array}$$

The pressure of water at the given point defined as

$$P_{1} = \rho g H$$

Were H is the water column height over the point

According to this

$$\begin{array}{l} P_{2} = \rho g (H + h) \\ P_{2} = 2P_{1} \\ \rho g (H + h) = 2P_{1} \\ H + h = \frac{2P_{1}}{\rho g} \\ h = \frac{2P_{1}}{\rho g} - H \\ h = \frac{2P_{1}}{\rho g} - \frac{P_{1}}{\rho g} \\ h = \frac{P_{1}}{\rho g} \\ h = \frac{10}{1000 \cdot 10} = 0.001 \, \mathrm{m} \\ \end{array}$$

Answer: 0.001 m.

The current numerical result is quite strange. It is possible that the pressure at a point is $10 \, \mathrm{N/cm^2}$. In this case:

$$\begin{array}{l} P_{1} = 10 \, \mathrm{N/cm^2} = 1 \cdot 10^{5} \, \mathrm{N/m^2} \\ h = \frac{1 \cdot 10^{5}}{1000 \cdot 10} = 10 \, \mathrm{m} \\ \end{array}$$

Answer: 10 m.