Answer: Option (A) is the correct answer.
Explanation:
Force acting on a dam is as follows.
F = [tex]\frac{1}{2}\rho g\omega H^{2}[/tex] .......... (1)
Now, when we double the depth then it means H is increasing 2 times and then the above relation will be as follows.
F' = [tex]\frac{1}{2}\rho g\omega (2)^{2}[/tex]
F' = [tex]\frac{1}{2}\rho g\omega 4[/tex] ........... (2)
Now, dividing equation (1) by equation (2) as follows.
[tex]\frac{F}{F'}[/tex] = [tex]\frac{\frac{1}{2}\rho g\omega H^{2}}{\frac{1}{2}\rho g\omega 4}[/tex]
Cancelling the common terms we get the following.
[tex]\frac{F}{F'}[/tex] = [tex]\frac{1}{4}[/tex]
4F = F'
Thus, we can conclude that if doubled the depth of the dam the hydrostatic force will be 4F.
