JEE PYQ: Mechanical Properties Of Fluids Question 1
Question 1 - 2021 (16 Mar 2021 Shift 1)
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 atmospheric pressure is $1 \times 10^5$ Pa, density of water is $10^3$ kg m$^{-3}$, $g = 10$ ms$^{-2}$)
(1) $\frac{200}{3}%$
(2) $\frac{200}{5}%$
(3) $\frac{5}{200}%$
(4) $\frac{3}{200}%$
Show Answer
Answer: (1) $\frac{200}{3}%$
Solution
$P_1 = \rho g d + P_0 = 3 \times 10^5$ Pa $\therefore \rho g d = 2 \times 10^5$ $P_2 = 2\rho g d + P_0 = 4 \times 10^5 + 10^5 = 5 \times 10^5$ Pa $%\text{increase} = \frac{P_2 - P_1}{P_1} \times 100 = \frac{5 \times 10^5 - 3 \times 10^5}{3 \times 10^5} \times 100 = \frac{200}{3}%$
BETA
