The pressure acting on a submarine is $3 \times 10^{5}$ Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be:
(Assume that atmosphere pressure is $1 \times 10^{5}$ Pa; density of water is $10^{3}$ kg $m^{- 3}$ , g = 10 $m s^{- 2}$ )

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\frac{200}{5} %$

$\frac{5}{200} %$

$\frac{3}{200} %$

$\frac{200}{3} %$

answer is D.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$P_{o} + ρ {gh=3×10}^{5} P a$

$If depth is doubled,new pressure on submarine will be,$

$\Rightarrow P' = P_{0} + ρ {g2h=10}^{5} + (4 \times 10^{5}) =5 \times 10^{5} P a$

$⇒ \frac{Δ P}{P} \times 100 = \frac{(P' - P)}{P} \times 100 = \frac{(5 - 3)}{3} \times 100 = \frac{200}{3} %$

$\Rightarrow ρ gh = 2 \times 10^{5} P a$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test