Water in filled in a cylindrical vessel of height H. A hole is made at height z from the bottom, as shown in the figure. The value of of z for which the range (R) of the emerging water through the hole will be maximum for

• A

  $z=\frac{H}{4}$
• Correct answer
  B

  $\quad z=\frac{\mathrm{H}}{2}$
• C

  $z=\frac{\mathrm{H}}{8}$
• D

  $\quad z=\frac{H}{3}$

Solution

\(x=2 \sqrt{y(H-y)}\) If hole is made at depth \(y=H-y\), then \(\begin{aligned} & X=2 \sqrt{(H-y)[H-(H-y)]} \\ & =2 \sqrt{y(H-y)} \\ & =x \end{aligned}\) Hence, range will remain same. Now, \(\frac{d X}{d y}=\frac{1}{2}\left(y H-y^2\right)^{-1 / 2}(H-2 y)\) For maximum value of \(X, \frac{d X}{d y}=0\) \(\Rightarrow \frac{H-2 y}{2\left(y H-y^2\right)^{1 / 2}}=0\) [Since denominator can't be zero, \(y \neq 0 \& y \neq H\) ] \(\Rightarrow y=\frac{H}{2}\)

Hence, the correct answer is option: 'B'.