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# There is a hole in the bottom of the tank having water. If total pressure at bottom is 3 atm $\left( {1atm = {{10}^5}\dfrac{N}{{{m^2}}}} \right)$ then the velocity of water flowing from hole is:A. $\sqrt {400} \dfrac{m}{s}$ B. $\sqrt {600} \dfrac{m}{s}$ C. $\sqrt {60} \dfrac{m}{s}$ D.None of these

Last updated date: 14th Sep 2024
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Hint First, we will find the height of the liquid column using the difference of pressure at the bottom of the tank and at the top. After that we will use this height in Bernoulli's equation to find the velocity/efflux of the water.

Complete step-by-step solution
Total pressure at the bottom of the tank = 3 atm
${P_1} + \rho gh = {P_2}$ , where ${P_1}$ = pressure at the top of the tank (1 atm)
$\rho$ = density of water
h= height of the liquid column
${P_2}$ = pressure at the bottom of the tank (3 atm)
$\rho gh = 3 - 1 = 2atm$
Bernoulli’s theorem-
${P_1} + \rho g{h_1} + \dfrac{1}{2}\rho v_1^2 = {P_2} + \rho g{h_2} + \dfrac{1}{2}\rho v_2^2$
To find the velocity of water flowing out of the tank use
$\rho gh = \dfrac{1}{2}\rho {v^2}$
$v = \sqrt {2gh}$
$\rho$ of water= ${10^3}\dfrac{{kg}}{L}$
$v = \sqrt {\dfrac{4}{{{{10}^3}}}} \dfrac{m}{s}$
So, $v = \sqrt {400} \dfrac{m}{s}$

Option(A) $\sqrt {400} \dfrac{m}{s}$

Note You have to keep in mind that the value of the pressure of the tank at the top is not equal to zero, it’s equal to atmospheric pressure i.s. 1atm.
Height of the water column is found by the difference of both the pressure, not alone by pressure of the bottom of the tank, otherwise you will get the solution wrong.