A) Depends on H
B) \[1:1\]
C) \[2:2\]
D) \[1:2\]
Correct Answer: C
Solution :
[c] Range is same for both holes \[\therefore \,\,\,\,\,\,2\sqrt{(H-{{h}_{1}}){{h}_{1}}}=1\sqrt{(H-{{h}_{2}}){{h}_{2}}}\] Squaring both sides, \[4(H-{{h}_{1}}){{h}_{1}}=4(H-{{h}_{2}}){{h}_{2}}\] \[H{{h}_{1}}-h_{1}^{2}=H{{h}_{2}}-h_{2}^{2}\] On solving we get, \[H={{h}_{1}}+{{h}_{2}}\] (not possible) and \[{{h}_{1}}-{{h}_{2}}=0\] Hence, the ratio of \[\frac{{{h}_{1}}}{{{h}_{2}}}=1:1\]
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