The mass of a wrestler is 200 kg and the area of the sole of his foot is 0.01 m². If g = 10m/s^2, then what is the pressure(in atm) exerted on the floor when he is standing on one foot. (round off to the nearest integer).

We have, $\mathrm{Pressure}=\frac{\mathrm{Force}}{\mathrm{Area}}$
Mass of wrestler, m = 200 kg
Force, F = mg 

$=200\times 10$ 

$= 2000 \mathrm{N}$
When the wrestler is standing on one foot, the total area of foot, A = 0.01 m²
Pressure, $\mathrm{P}=\frac{2000 \mathrm{N}}{0.01\mathrm{ }{\mathrm{m}}^2}$ = 2,00,000 N/m² = 2,00,000 Pa
Since 1 atm = 101325 Pa

Pressure, P_atm = 1.97 atm

So, the pressure exerted on the floor when the wrestler is standing on one foot is approximately 2 atm.