Question

what is the pressure on a swimmer 10 m below the surface of a lake

Answer 1

h=10m and p=1000 kgm^-3.

Take g=10ms^-2

Now,

P=Pa+pgh

= 1.01×10^5Pa+1000kgm^-3×10ms^-2×10m

=2.01×10^5 Pa

=2atn

Answer 2

Given:

• The depth of the swimmer below the surface of the lake = 10 m

To Find:

• The pressure on the swimmer.

solution:

We already have the formula to find the pressure which is given by,

P = $P_0$ + ρgh → {equation 1}

Where "$P_0$" is the atmospheric pressure is constant ( 1.05×$10^5$ Pa), "ρ" is the density of water which is constant( 1000 Kg/$m^3$), "g" is the gravitational force(9.8 m/$s^2$), and "h" is the depth.

On substituting the values we have in equation 1 we get,

⇒ P = 1.05×$10^5$+(1000×9.8×10)

Remove the brackets in the above formula and perform multiplication and addition operations. We get,

⇒ P = 1.99×$10^5$ Pa = 2 atm ( conversion of S.I. units)

∴ The pressure on the swimmer = 2 atm or 1.99×$10^5$ Pa.