Search any question & find its solution
Question:
Answered & Verified by Expert
Water level is maintained in a cylindrical vessel upto a fixed height $H$. The vessel is kept on a horizontal plane. At what height above the bottom should a hole be made in the vessel, so that the water stream coming out of the hole strikes the horizontal plane of the greatest distance from the vessel?
Options:
Solution:
2280 Upvotes
Verified Answer
The correct answer is:
$h=\frac{H}{2}$
The velocity with which water will came out
$v=\sqrt{2 g(H-h)}$
The time of flight of water ejected from the hole.
$\begin{aligned} t & =\sqrt{\frac{2 h}{g}}=\text { time of flight } \\ \therefore \quad x & =v \times t \\ & =\sqrt{2 g(H-h)} \times \sqrt{\frac{2 h}{g}}=\sqrt{4\left(H h-h^2\right)}\end{aligned}$
For $x$ to be maximum
$\begin{aligned} & \frac{d x}{d h}=0 \\ \Rightarrow & d \frac{\sqrt{4\left(H h-h^2\right)}}{d h}=0 \\ \Rightarrow & 0=H-2 h \Rightarrow h=\frac{H}{2}\end{aligned}$
$v=\sqrt{2 g(H-h)}$
The time of flight of water ejected from the hole.
$\begin{aligned} t & =\sqrt{\frac{2 h}{g}}=\text { time of flight } \\ \therefore \quad x & =v \times t \\ & =\sqrt{2 g(H-h)} \times \sqrt{\frac{2 h}{g}}=\sqrt{4\left(H h-h^2\right)}\end{aligned}$
For $x$ to be maximum
$\begin{aligned} & \frac{d x}{d h}=0 \\ \Rightarrow & d \frac{\sqrt{4\left(H h-h^2\right)}}{d h}=0 \\ \Rightarrow & 0=H-2 h \Rightarrow h=\frac{H}{2}\end{aligned}$
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.