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2 separate reservation of oil A and B (capacity of $A=2 \times $ capacity of B) are filled completely with oil and oil is released simultaneously from both the reservoirs. The rate of flow of oil out of each reservoir at an instant of time is proportional to the quantity of oil in the reservoir at that time one hour after the oil is released the quantity of oil in reservoir A is 1.5 times the quantity of oil in reservoir B. If the reservoirs have same quantity of oil after $t= \large\frac{\log a}{\log b -\log c}$ then $a+b-c$

$(a)\;2 \\ (b)\;3 \\ (c)\;4 \\ (d)\;5 $

1 Answer

$\large\frac{dV_A}{\sigma t}$$=-K_1 V_A$
=> $\large\frac{dV_A}{V_A}=-k_1 dt$
$\log V_A=-K_1t+c$
at $t=0\quad V_A=V_0 A \quad \log V_{OA}=c$
$\log \bigg( \large\frac{VA}{V_{OA}}\bigg)$$=-K_1t$
$=> V_A=V_OA e^{-K_1t}$
Similarly for reservoir B $ VB =V_oB e^{-K_2t}$
$VOA=2 VOB$
$V_B=\large\frac{VOA}{2} e^{-K_2 t}$
at $t= 1 hours$
$V_A= \large\frac{3}{2} $$V_B$
$VOA e^{-K_1} =\large\frac{3}{2} \large\frac{V_o Ae^{-K_2}}{2}$
$e^{k_2-K_1}=\large\frac{3}{4}$
$K_1 -K_2= \log \bigg(\large\frac{4}{3}\bigg)$
at $t=t$
$V_A=V_B$
$VOAe^{-K1t}=\large\frac{VOA}{2} e^{-k_2t}$
$2= e^{(K_1-K_2)t}$
$e^{(K_2-K_1)t}= \large\frac{1}{2}$
$(K_2-K_1)t=-\log 2$
$t= \large\frac{\log 2}{K_1-K_2}$
$t= \large\frac{\log 2}{\log 4/3}$
$\quad=\large\frac{\log 2}{\log 4- \log 3}$
$a=2,b=4,c=3$
$a+b-c=3$
Hence b is the correct answer.
answered Feb 3, 2014 by meena.p
 

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