Question

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

• A. $5.0 \times 10^{-3} \, m$
• B. $7.5 \times 10^{-3} \, m$
• C. $9.6 \times 10^{-3} \, m$
• D. $3.6 \times 10^{-3} \, m$

Answer 1

Answer: (D)

Here, $d_{1}= 8 \times10^{-3} m$,
$v_{1}=0.4 \,m\,s^{-1}, h=0.2\, m$
According to equation of motion,

$v_{2}=\sqrt{v_{1}^{2}+2gh}=\sqrt{\left(0.4\right)^{2}+2\times10\times0.2}$
$\approx 2\, m\, s^{-1}$
According to equation of continuity, $A_{1}v_{1}=A_{2}v_{2}$
$\therefore \pi\times\left(\frac{8\times10^{-3}}{2}\right)^{2} \times0.4=\pi \times\left(\frac{d_{2}}{2}\right)^{2}\times2$
or $d_{2}=3.6\times10^{-3}\,m$