Question
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 height $$H$$ 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 $${h_2}$$
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 height $$H$$ 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 $${h_2}$$
A.
Depends on $$H$$
B.
$$1:1$$
C.
$$2:2$$
D.
$$1:2$$
Answer :
$$1:1$$
Solution :
Range is same for both holes
$$\therefore 2\sqrt {\left( {H - {h_1}} \right){h_1}} = 2\sqrt {\left( {H - {h_2}} \right){h_2}} $$
Squaring both sides,
$$\eqalign{ & 4\left( {H - {h_1}} \right){h_1} = 4\left( {H - {h_2}} \right){h_2} \cr & H{h_1} - h_1^2 = H{h_2} - h_2^2 \cr} $$
On solving we get,
$$H = {h_1} + {h_2}\left( {{\text{not}}\,{\text{possible}}} \right)\,{\text{and}}\,{h_1} - {h_2} = 0$$
Hence, the ratio of $$\frac{{{h_1}}}{{{h_2}}} = 1:1.$$
Range is same for both holes
$$\therefore 2\sqrt {\left( {H - {h_1}} \right){h_1}} = 2\sqrt {\left( {H - {h_2}} \right){h_2}} $$
Squaring both sides,
$$\eqalign{ & 4\left( {H - {h_1}} \right){h_1} = 4\left( {H - {h_2}} \right){h_2} \cr & H{h_1} - h_1^2 = H{h_2} - h_2^2 \cr} $$
On solving we get,
$$H = {h_1} + {h_2}\left( {{\text{not}}\,{\text{possible}}} \right)\,{\text{and}}\,{h_1} - {h_2} = 0$$
Hence, the ratio of $$\frac{{{h_1}}}{{{h_2}}} = 1:1.$$
