JEE PYQ: Mechanical Properties Of Fluids Question 27
Question 27 - 2019 (10 Apr 2019 Shift 2)
A submarine experiences a pressure of $5.05 \times 10^6$ Pa at depth of $d_1$ in a sea. When it goes further to a depth of $d_2$, it experiences a pressure of $8.08 \times 10^6$ Pa. Then $d_1 - d_1$ is approximately (density of water = $10^3$ kg/m$^2$ and acceleration due to gravity = 10 ms$^{-2}$):
(1) 300 m
(2) 400 m
(3) 600 m
(4) 500 m
Show Answer
Answer: (1) 300 m
Solution
$P_1 = P_0 + \rho g d_1$ $P_2 = P_0 + \rho g d_2$ $\Delta P = P_2 - P_1 = \rho g \Delta d$ $3.03 \times 10^6 = 10^3 \times 10 \times \Delta d$ $\Rightarrow \Delta d = 300$ m
BETA
