Search any question & find its solution
Question:
Answered & Verified by Expert
A tank of height $15 \mathrm{~m}$ and cross-section area $10 \mathrm{~m}^2$ is filled with water. There is a small hole of cross-section area $a$ which is much smaller than the container, located at a height of $12 \mathrm{~m}$ from the base of the container. How much force should be applied with a piston at the top level, so that the water coming out of the hole hits the ground at a distance of $16 \mathrm{~m}$ ? (Take, density of water $=1000 \mathrm{~kg} \mathrm{~m}^{-3}$ )
Options:
Solution:
1345 Upvotes
Verified Answer
The correct answer is:
$233 \mathrm{kN}$
As, range of water jet $=16 \mathrm{~m}$. If velocity of efflux is $v$, then
$v \times \sqrt{\frac{2 H}{g}}=16$, where $H=12 \mathrm{~m}$
Also, $v=\sqrt{2 g h+\frac{2\left(p-p_a\right)}{\rho}}, \rho=1000 \mathrm{~kg}-\mathrm{m}^{-3}$ and $h=3 \mathrm{~m}$
So combining these, we get
$$
\begin{aligned}
& \left(2 g h+\frac{2\left(p-p_a\right)}{\rho}\right)\left(\frac{2 H}{g}\right)=256 \\
& \Rightarrow \quad 2 \times 10 \times 3+\frac{2 \Delta p}{1000}=\frac{2560}{2 \times 12} \Rightarrow \Delta p=233 \mathrm{k} \frac{\mathrm{N}}{\mathrm{m}^2}
\end{aligned}
$$
So, force applied $=\Delta p \times A=233 \mathrm{kN}$
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.