Question

Water from a tap emerges vertically downwards with an initial speed of $1.0m{{s}^{-1}}$. The cross-sectional area of the tap is ${{10}^{-4}}{{m}^{2}}$. Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-sectional area of the stream 0.15m below the tap is $\left( g=10m{{s}^{-1}} \right)$
$\begin{align}
  & \left( A \right)5.0\times {{10}^{-4}}{{m}^{2}} \\
 & \left( B \right)1.0\times {{10}^{-5}}{{m}^{2}} \\
 & \left( C \right)5.0\times {{10}^{-5}}{{m}^{2}} \\
 & \left( D \right)2.0\times {{10}^{-5}}{{m}^{2}} \\
\end{align}$

Answer 1

Hint: The pressure difference is the product of the density of liquid, acceleration due to gravity and height of the liquid. And also the pressure difference is half the product of the density of the liquid and the square of change in its velocity. Thus by equating both the equations we will get the value of velocity. Thus by applying the continuity equation we will get the cross sectional area.

Formula used:
The continuity equation is given by,
${{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}$
where, ${{A}_{1}}\And {{A}_{2}}$ are the area of cross section
${{v}_{1}}\And {{v}_{2}}$ are the velocity of liquid
The pressure difference is given by,
$\vartriangle P=\rho gh$
where, $\rho $ is the density of the liquid
g is the acceleration due to gravity
h is the height of the liquid

Complete answer:
The volume of flow rate $={{10}^{-4}}\dfrac{{{m}^{3}}}{\sec }$
The pressure difference is given by the product of the density of liquid, acceleration due to gravity and height of the liquid. That is,
$\vartriangle P=\rho gh$
Substituting the values we get,
$\vartriangle P={{10}^{3}}\times 10\times 0.15$
$\Rightarrow \vartriangle P=1500Pa$ ………….(1)
We also know that,
$\vartriangle P=\dfrac{1}{2}\rho \left( v_{2}^{2}-v_{1}^{2} \right)$
Thus substituting the values we get,
$\vartriangle P=\dfrac{1}{2}\times {{10}^{3}}\left( {{v}^{2}}-1 \right)$ ……………(2)
By equating equation (1) and (2) we get,
$1500=\dfrac{1}{2}\times {{10}^{3}}\left( {{v}^{2}}-1 \right)$
By rearranging the equation we get,
$3000={{10}^{3}}\left( {{v}^{2}}-1 \right)$
$\Rightarrow 3={{v}^{2}}-1$
$\Rightarrow {{v}^{2}}=4$
$\therefore v=2\dfrac{m}{s}$
By applying the equation of continuity we get,
${{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}$
Rearranging the equation for ${{A}_{2}}$ we get,
${{A}_{2}}=\dfrac{{{A}_{1}}{{v}_{1}}}{{{v}_{2}}}$
$\Rightarrow {{A}_{2}}=\dfrac{{{10}^{-4}}\times 1}{2}$
$\therefore {{A}_{2}}=5\times {{10}^{-5}}{{m}^{2}}$

Thus option (C ) is correct.

Note:
The continuity equation is a statement of the conservation of mass. Bernoulli's theorem says the principle conservation of energy for ideal fluids in a streamline flow. Bernoulli’s theorem can be applied in airplane wings, flying discs etc. Thus Bernoulli’s theorem can otherwise be defined as the increase in speed creates decrease in pressure.