Q. The top surface of an incompressible liquid is open to the atmosphere. The pressure at a depth $P_{1}.$ How does the pressure $P_{2}\ $at depth $h_{2\ } = 2h_{1}$ compare with $P_{1}?$
A) $P_{2} > 2P_{1}$
B) $P_{2} = 2P_{1}$
C) $P_{2} < 2P_{1}$
D) $P_{2} = P_{1}$
Explanation
(c) $p_{1} = p_{0} + \rho gh_{1}$
$p_{2} = p_{0} + \rho gh_{2} = p_{0} + 2\rho gh_{1}$
$= 2\left( p_{0} + \rho gh_{1} \right) - p_{0} = 2p_{1} - p_{0}(p_{2} < 2p_{1})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.