JEE Main 16 March 2021 Shift 1 Solved Paper

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