The force acting on a window of are 50`xx`50 cm of a submarine at a depth of 2000 m in an ocean ,the interior of which is maintained at sea level atmospheric pressure is (density of sea water = `10 ^(3) kg m^(-3)` ,g =`10 m s^(-2)` )

Options

A. `5 xx10^(5)N`
B. `25 xx10^(5)` N
C. `5 xx 10^(6)`N
D. `25 xx 10^(6)`N

Answer

C

Explanation

Here, h=2000 m, `rho=10^(3) kg m^(-3), g=10 m s^(-2)` The pressure outside the submarine is
`P=P_(a)+rhogh`
where `P_(a)` is the atmospheric pressure.
Pressure inside the submarine is `P_(a)` .
Hence, net pressure acting on the window is gauge pressure.
Gauge pressure, `P_(g)=P-P_(a)=rhogh`
`=10^(3) kg m^(-3) xx 10 m s^(-2)xx 2000 m=2xx10^(7)` Pa
Area of a window is
`A=50 cm xx50 cm = 2500 xx 10^(-4) m^(2)`
Force acting on the window is
`F=P_(g)A`
`=2xx10^(7)Pa xx 2500xx10^(-4) m^(2)=5xx10^(6)N`