A) 1.33
B) 1.5
C) 2
D) 0.414
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 \[dV=A\,(-dH)\] \[A\,(-dH)=a\sqrt{2gH}\,.\,\,dt\] Since, \[\int_{0}^{t}{dt}=\frac{A}{2\sqrt{2g}}\,\int_{H}^{0}{{{y}^{-1/2}}dy}\] ... (i) Now, substituting the proper limits in Eq. (i), derived in the theory, we have \[\int_{0}^{{{t}_{1}}}{dt=}\frac{A}{a\sqrt{2g}}\int_{H}^{H/2}{{{y}^{-1/2}}}\] \[\Rightarrow \] \[{{t}_{1}}=\frac{2A}{a\sqrt{2g}}[\sqrt{y}]_{H/2}^{H}\] \[\Rightarrow \] \[{{t}_{1}}=\frac{2A}{a\sqrt{2g}}\left[ \sqrt{H}-\sqrt{\frac{H}{2}} \right]\] \[\Rightarrow \] \[{{t}_{1}}=\frac{A}{a}\sqrt{\frac{H}{a}}\,(\sqrt{2}-1)\] ... (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), we get \[\frac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{2}-1\Rightarrow \frac{{{t}_{1}}}{{{t}_{2}}}=0.414\]
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