A) \[~\text{1}.\text{33}\]
B) \[\text{1}.\text{5}\]
C) \[\text{2}\]
D) \[0.\text{414}\]
Correct Answer: D
Solution :
The volume of liquid in time dt \[dV=av\times dt=a\sqrt{2gH}.dt\] (\[\because dH\]level falls in dt time ) The volume of liquid coming out of hole \[\text{dV}=\text{A}\left( -\text{dH} \right)\] \[A(-dH)=a\sqrt{2gH}.dt\] since- \[\int_{0}^{t}{dt=\frac{A}{a\sqrt{2g}}\int_{H}^{0}{{{y}^{-1/2}}dy}}\] ? (i) Now, substituting the proper limits in Eq. (i), derived in the theory, we have \[\int_{0}^{{{t}_{1}}}{dt=\frac{A}{a\sqrt{2g}}\int_{H}^{H/2}{{{y}^{-1/2}}dy}}\] \[\Rightarrow \] \[{{t}_{1}}=\frac{2A}{a\sqrt{2g}}[\sqrt{y}]_{H/2}^{H}\] \[\Rightarrow \] \[{{t}_{1}}=\frac{2A}{a\sqrt{2g}}\left[ \sqrt{H}-\sqrt{\frac{H}{2}} \right]\] \[\Rightarrow \] \[{{t}_{1}}=\frac{A}{a}\frac{\sqrt{H}}{g}(\sqrt{2}-1)\] ? (ii) Similarly, \[\,\int_{0}^{{{t}_{2}}}{dt}=-\frac{A}{a\sqrt{2g}}\int_{H/2}^{0}{{{y}^{-1/2}}dy}\] \[\Rightarrow \] \[{{t}_{2}}=\frac{A}{a}\sqrt{\frac{H}{g}}\] ? (iii) From Eqs.(ii) and (iii),weget \[\frac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{2}-1\Rightarrow \frac{{{t}_{1}}}{{{t}_{2}}}=0.414\]
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