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In a cylindrical water tank, there are two small holes $A$ and $B$ on the wall at a depth of $h_1$, from the surface of water and at a height of $h_2$ from the bottom of water tank. Surface of water is at height of $h_2$ from the bottom of water tank. Surface of water is at heigh $H$ from the bottom of water tank. Water coming out from both holes strikes the ground at the same point $S$. Find the ratio of $h_1$ and $h_2$
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Verified Answer
The correct answer is:
Depends on $H$
Depends on $H$
Range is same for both holes
$$
\therefore 2 \sqrt{\left(\mathrm{H}-h_1\right) h_1}=2 \sqrt{\left(\mathrm{H}-h_2\right) h_2}
$$
Squaring both sides,
$$
\begin{aligned}
& 4\left(H-h_1\right) \mathrm{h}_1=4\left(H-h_2\right) h_2 \\
& H h_1-h_1^2=H h_2-h_2^2
\end{aligned}
$$
On solving we get,
$$
H=h_1+h_2
$$
Hence, the ratio of $\frac{h_1}{h_2}$ depends on $H$.
$$
\therefore 2 \sqrt{\left(\mathrm{H}-h_1\right) h_1}=2 \sqrt{\left(\mathrm{H}-h_2\right) h_2}
$$
Squaring both sides,
$$
\begin{aligned}
& 4\left(H-h_1\right) \mathrm{h}_1=4\left(H-h_2\right) h_2 \\
& H h_1-h_1^2=H h_2-h_2^2
\end{aligned}
$$
On solving we get,
$$
H=h_1+h_2
$$
Hence, the ratio of $\frac{h_1}{h_2}$ depends on $H$.
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