Question

Water is flowing continuously from a tap having an internal diameter $8\times {10}^{-3}\text{m}$. The water velocity as it leaves the tap is $0.4\text{m/s}$. The diameter of the water stream at a distance $2\times {10}^{-1}\text{m}$ below the tap is close to

• A. $\frac{8}{\sqrt[4]{426}}\text{mm}.$
• B. $\frac{8}{\sqrt[4]{436}}\text{mm}$
• C. $\frac{6}{\sqrt[2]{228}}\text{mm}$
• D. $\frac{4}{\sqrt[3]{334}}\text{mm}$

Answer 1

The correct option is A $\frac{8}{\sqrt[4]{426}}\text{mm}.$

From Bernouili's theorem,
$\rho gh=\frac{1}{2}\rho (v_2^2-v_1^2)$
$\Rightarrow gh=\frac{1}{2}{v}_1^2({\left(\frac{v_2}{v_1}\right)}^2-1)$
$\Rightarrow gh=\frac{1}{2}{v}_1^2({\left(\frac{A_1}{A_2}\right)}^2-1)(\because A_1{v}_1=A_2{v}_2)$
$\Rightarrow {\left(\frac{A_1}{A_2}\right)}^2=1+\frac{2hg}{v_1^2}$
$\Rightarrow {\left(\frac{D_1}{D_2}\right)}^4=1+\frac{2hg}{v_1^2}$
Put the value,
$\Rightarrow D_2=\frac{D_1}{{(1+\frac{2gh}{v_1^2})}^{\frac{1}{4}}}$

$=\frac{8\times {10}^{-3}}{{(1+\frac{2\times 10\times 0.2}{(0.4{)}^2})}^{\frac{1}{4}}}$
$=\frac{8}{\sqrt[4]{426}}\text{mm}$
So, the diameter of the water stream is close to $\frac{8}{\sqrt[4]{426}}\text{mm}.$