Question

Sea water of density 1020 kgm$^{3}$ exerts a pressure of 1.02 × 10$^{7}$ Pa at the bottom of sea. Find the depth of sea. (g = 10 ms$^{–2}$​)

Answer 1

Given:

Density of sea water (d) = 1020 kgm$^{3}$

Acceleration due to gravity (g) = 10 ms$^{–2}$

To find:

Depth of sea.

Solution:

In order to find depth of sea water first we need to find pressure of sea water.

Pressure of sea water (P)

= 1.02 × 10$^{7}$

= 102,00,000 Nm$^{–2}$

We know that,

P = h × d × g

$\therefore$ h = P/dg

$= \frac{10200000}{1020 \times 10} \\ \\ = 1000 \: m \\ \\ = 1 \: km.$

Hence, depth of sea is 1 km.