A submarine experiences a pressure of `5.05xx10^(6)Pa` at a depth of `d_(1)` in a sea When it goes futher to a depth of `d_(2)`. It experiences a pressure of `8.08xx10^(6)Pa`. The `d_(2)-d_(1)` is approximately (density of water `=10^(3)kg//m^(3)` and acceleration due to gravity `=10ms^(-2))`

Options

A. 500 m
B. 400 m
C. 300 m
D. 600 m

Answer

C

Explanation

`P_(2)-P_(1)=(d_(2)-d_(1))grho=3.03xx10^(6)," "d_(2)-d_(1)=(3.03xx10^(6))/(10^(3)xx10)=303m~~300m`