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 in cylindrical tank be h. \[_{1}{{n}_{2}}=\frac{\text{actual}\,\text{depth}}{\text{apparent}\,\text{depth}}\]or\[\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{h}{x}\]or\[\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{\frac{dh}{dt}}{\frac{dx}{dt}}\] \[\therefore \]\[\frac{dh}{dt}=\frac{{{n}_{2}}}{{{n}_{1}}}\times \frac{dx}{dt}\]or Change in actual rate of flow \[=\frac{{{n}_{2}}}{{{n}_{1}}}\times \]change in apparent rate of flow. or\[\frac{dh}{dt}=\frac{{{n}_{2}}}{{{n}_{1}}}\times xcm\,{{\min }^{-1}}\] Thus, amount of water drained in cc per min \[=\frac{dh}{dt}\times \pi {{R}^{2}}\] \[=\frac{{{n}_{2}}}{{{n}_{1}}}\times x\times \pi {{R}^{2}}\] \[=\frac{x\pi {{R}^{2}}{{n}_{2}}}{{{n}_{1}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
