A) Depends on H
B) 1:1
C) 2:2
D) 1:2
Correct Answer: A
Solution :
Range is same for both holes \[\therefore \]\[2\sqrt{(H-{{h}_{1}}){{h}_{1}}}=2\sqrt{(H-{{h}_{2}}){{h}_{2}}}\] 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,\[H={{h}_{1}}+{{h}_{2}}\] Hence, the ratio of \[\frac{{{h}_{1}}}{{{h}_{2}}}\]depend on H.You need to login to perform this action.
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