A) x p R2 n1/n2
B) x p R2 n2/n1
C) 2 p R n1/n2
D) p R2 x
Correct Answer: B
Solution :
Apparent depth \[h'=\frac{h}{_{air}{{\mu }_{liquid}}}\] Þ \[\frac{dh'}{dt}=\frac{1}{_{a}{{\mu }_{w}}}=\frac{1}{_{a}{{\mu }_{w}}}\frac{dh}{dt}\]Þ\[x=\frac{1}{_{a}{{\mu }_{w}}}\frac{dh}{dt}\]Þ \[\frac{dh}{dt}={{\,}_{a}}{{\mu }_{w}}\,x\] Now volume of water \[V=\pi {{R}^{2}}h\] Þ \[\frac{dV}{dt}\] \[=\pi {{R}^{2}}\frac{dh}{dt}\] \[=\pi {{R}^{2}}.{{\,}_{a}}{{\mu }_{w}}\,x\] \[={{\,}_{a}}{{\mu }_{w}}\pi {{R}^{2}}x\] \[=\frac{{{\mu }_{w}}}{{{\mu }_{a}}}\pi {{R}^{2}}x\]\[=\left( \frac{{{n}_{2}}}{{{n}_{1}}} \right)\,\pi {{R}^{2}}x\]
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