Given, pressure on submarine at a certain depth, pi=3×105Pa Since, we know that at a certain depth or below the surface of a liquid of density ρ, pressure is given as pi=p0+ρgh ⇒3×105=p0+ρgh ⇒3×105=1×105+ρgh(∵p0=1×105Pa) ⇒3×105−1×105=ρgh ⇒2×105=ρgh....(i) When the depth is doubled, then final pressure will be pf=p0+ρg2h=1×105+2(ρgh) =1×105+2(2×105) [using Eq. (i)] =1×105+4×105 ⇒pf=5×105Pa ∴ Percentage increase in pressure