Search any question & find its solution
Question:
Answered & Verified by Expert
Water is flowing continuously from a tap having an internal diameter $8 \times 10^{-3} \mathrm{~m}$. The water velocity as it leaves the tap is $0.4 \mathrm{~ms}^{-1}$. The diameter of the water stream at a distance $2 \times 10^{-1} \mathrm{~m}$ below the lap is close to :
Options:
Solution:
1230 Upvotes
Verified Answer
The correct answer is:
$3.6 \times 10^{-3} \mathrm{~m}$
$3.6 \times 10^{-3} \mathrm{~m}$
Equation of continuity
$$
\begin{aligned}
& \Rightarrow(\mathrm{a} \times \mathrm{v}) \text { top }=(\mathrm{a} \times \mathrm{v}) \text { bottom } \\
& \mathrm{v}_{\mathrm{b}}^2-(0.4)^2=2 \times 9.8 \times 0.2\left[\mathrm{v}^2-\mathrm{u}^2=2 \mathrm{gh} \text { is used }\right] \\
& \mathrm{v}_{\mathrm{b}}=2 \mathrm{~m} / \mathrm{s} \text { (nearly) } \\
& \pi\left[8 \times 10^{-3}\right] \times 0.4=\pi \mathrm{d}^2 \times 4 \\
& \mathrm{~d} \approx 3.6 \times 10^{-3} \mathrm{~m}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow(\mathrm{a} \times \mathrm{v}) \text { top }=(\mathrm{a} \times \mathrm{v}) \text { bottom } \\
& \mathrm{v}_{\mathrm{b}}^2-(0.4)^2=2 \times 9.8 \times 0.2\left[\mathrm{v}^2-\mathrm{u}^2=2 \mathrm{gh} \text { is used }\right] \\
& \mathrm{v}_{\mathrm{b}}=2 \mathrm{~m} / \mathrm{s} \text { (nearly) } \\
& \pi\left[8 \times 10^{-3}\right] \times 0.4=\pi \mathrm{d}^2 \times 4 \\
& \mathrm{~d} \approx 3.6 \times 10^{-3} \mathrm{~m}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.