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A cylindrical vessel of height $50 \mathrm{~cm}$ is filled with water and rests on a table. A small hole is made at the height $h$ from the bottom of the vessel so that the water jet could hit the table surface at a maximum distance $x_{\max }$ from the vessel as shown in the figure. The value of $x_{\max }$ will be (Neglect the viscosity of water.)
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Verified Answer
The correct answer is:
$50 \mathrm{~cm}$
Given, height of cylinder, $H=50 \mathrm{~cm}$
The given situation is shown in the figure.
Range (maximum distance) is given by
$$
x=2 \sqrt{h(H-h)}
$$
$x$ will be maximum only when $\frac{d x}{d h}=0$
$$
2.1[h(-1)+(H-h) 1]=0
$$
$$
\begin{array}{cc}
& 2 \sqrt{h(H-h)}-h+(H-h)=0 \\
& H-2 h=0 \\
\therefore & h=\frac{H}{2}
\end{array}
$$
Hence, $x$ will be maximum, when $h=\frac{H}{2}$
$\therefore$ From Eqs. (i), we get
$$
\begin{aligned}
x_{\max } & =2 \sqrt{\frac{H}{2}\left(H-\frac{H}{2}\right)} \\
& =2 \sqrt{\frac{H}{2} \cdot \frac{H}{2}}=H \\
x_{\max } & =50 \mathrm{~cm}
\end{aligned}
$$
The given situation is shown in the figure.
Range (maximum distance) is given by
$$
x=2 \sqrt{h(H-h)}
$$
$x$ will be maximum only when $\frac{d x}{d h}=0$
$$
2.1[h(-1)+(H-h) 1]=0
$$
$$
\begin{array}{cc}
& 2 \sqrt{h(H-h)}-h+(H-h)=0 \\
& H-2 h=0 \\
\therefore & h=\frac{H}{2}
\end{array}
$$
Hence, $x$ will be maximum, when $h=\frac{H}{2}$
$\therefore$ From Eqs. (i), we get
$$
\begin{aligned}
x_{\max } & =2 \sqrt{\frac{H}{2}\left(H-\frac{H}{2}\right)} \\
& =2 \sqrt{\frac{H}{2} \cdot \frac{H}{2}}=H \\
x_{\max } & =50 \mathrm{~cm}
\end{aligned}
$$
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