A) \[\frac{x\,\pi \,{{R}^{2}}\,{{n}_{1}}}{{{n}_{2}}}\]
B) \[\frac{x\,\pi \,{{R}^{2}}\,{{n}_{2}}}{{{n}_{1}}}\]
C) \[\frac{2\pi \,\,R\,{{n}_{1}}}{{{n}_{2}}}\]
D) \[\pi {{R}^{2}}x\]
Correct Answer: B
Solution :
Let actual height of water of the tank be h, then \[_{1}{{n}_{2}}=\frac{actual\,\,depth}{apparent\,\,depth}\] Also \[_{1}{{n}_{2}}=\frac{{{n}_{2}}}{{{n}_{1}}}\] \[\therefore \,\,\,\,\,\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{h}{x}\] where x is an apparent depth. \where x is an apparent depth. \[\therefore \,\,\,\,\,\,\,\,\,\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{\frac{dh}{dt}}{\frac{dx}{dt}}\] \[\therefore \,\,\,\,\,\,\,\,\,\frac{dh}{dt}=\frac{{{n}_{2}}}{{{n}_{1}}}\times \frac{dx}{dt}\] or change in actual rate of flow \[=\,\,\,\frac{{{n}_{2}}}{{{n}_{1}}}\times change\] in apparent rate of flow \[\frac{dh}{dt}=\frac{{{n}_{2}}}{{{n}_{1}}}\times x\,\,cm/min\] Multiplying both sides by it \[\pi {{R}^{2}}\], we have \[\frac{dh}{dt}\times \pi {{R}^{2}}=\frac{{{n}_{2}}}{{{n}_{1}}}\times x\times \pi {{R}^{2}}\] \[\therefore \,\,\,\,\,Amount\,\,of\,\,water\,\,drained=x\,\pi \,{{R}^{2}}\frac{{{n}_{2}}}{{{n}_{1}}}\]You need to login to perform this action.
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