Question

A tap is attached to a water tank as shown in figure. Water level above the tap is maintained at $h=0.2\text{m}$. The cross-sectional area of the tap is $\sqrt{2}\times {10}^{-4}{\text{m}}^2$. Assuming constant pressure throughout the stream of water, find the velocity & cross-sectional area of the stream $0.2\text{m}$ below the opening of the tap :-
$[\text{Assume}g=10{\text{m/s}}^2]$

• A. $2\sqrt{2}\text{m/s};{10}^{-4}{\text{m}}^2$
• B. $1\text{m/s};{10}^{-4}{\text{m}}^2$
• C. $2\text{m/s};{10}^{-4}{\text{m}}^2$
• D. $2\text{m/s};{10}^{-5}{\text{m}}^2$

Answer 1

The correct option is A $2\sqrt{2}\text{m/s};{10}^{-4}{\text{m}}^2$

Velocity of water at tap outlet $\left(v_1\right)=\sqrt{2gh}$
$=\sqrt{2\times 10\times \frac{2}{10}}=\sqrt{4}=2\text{m/s}$
Area of cross-section of tap $\left(A_1\right)=\sqrt{2}\times {10}^{-4}{\text{m}}^2$
Volume flow rate of water $\left(V\right)=A_1{v}_1$
$=2\sqrt{2}\times {10}^{-4}{\text{m}}^3/s$
Using Bernoulli's theorem (i) between the surface of tank & opening of the tap:
$\Delta P=\rho gH={10}^3\times 10\times 0.2$
$=2000\text{Pa}=2\text{kPa}...\left(1\right)$

(ii) Between the opening of the tap & point $2\text{m}$ below it-
$\rho gH=\frac{1}{2}\rho (v_2^2-v_1^2)$
$\Rightarrow 2000=\frac{1}{2}\times {10}^3\times (v_2^2-2^2)$
[using $\left(1\right)$]
$\Rightarrow 2=\frac{1}{2}\times (v_2^2-4)$
$\Rightarrow v_2^2=8$
or $v_2=2\sqrt{2}\text{m/s}$

Since volume flow rate for water stream coming out of tap will be constant,
$V=v_1{A}_1=v_2{A}_2$
$\Rightarrow 2\sqrt{2}\times {10}^{-4}=2\sqrt{2}\times A_2$
$\therefore A_2={10}^{-4}{\text{m}}^2$
Thus, option (a) is the correct answer.