# Water is filled in a beaker upto a height of 3 m: Bernoulli's Equation

**Problem:**
Water is filled in a beaker upto a height of 3 m. An orifice (hole) is at a height of 52.5 cm from the bottom of beaker (see figure). The ratio of cross-sectional area of the orifice and the beaker is 0.1. The square of the speed of the liquid coming out from the orifice is, (Take g = 10 m/s^{2})
(IIT JEE 2005)

- 50 (m/s)
^{2}
- 50.5 (m/s)
^{2}
- 51 (m/s)
^{2}
- 52 (m/s)
^{2}

**Answer:** The answer is A i.e., the square of the speed of the liquid coming out from the orfice is 50 (m/s)^{2}.

**Solution:**
Let $A_1$ be the cross-sectional area of the beaker and $v_1$ be the flow velocity at a height $h_1={3}\;\mathrm{m}$. Let $A_2$ be the cross-sectional area of the orifice and $v_2$ be the flow velocity at the orifice located at a height $h_2={0.525}\;\mathrm{m}$. Using continuity equation on an imaginary streamline from top of the beaker to the orifice,
\begin{align}
A_1v_1=A_2v_2,
\end{align}
we get,
\begin{align}
\label{lkb:eqn:1}
v_1=(A_2/A_1)v_2.
\end{align}
Bernoulli's equation,
\begin{align}
P_0+\rho g h_1+\frac{1}{2}\rho v_1^2=P_0+\rho g h_2+\frac{1}{2}\rho v_2^2,
\end{align}
gives,
\begin{align}
\label{lkb:eqn:2}
v_2^2=v_1^2+2g(h_1-h_2).
\end{align}
Eliminate $v_1$ from above equations to get,
\begin{align}
v_2^2&=\frac{2g(h_1-h_2)}{1-({A_2}/{A_1})^2}\\
&=\frac{2(10)(3-0.525)}{1-0.01}\\
&={50}\;\mathrm{m^2/s^2}.\nonumber
\end{align}

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