Two identical cylindrical tanks are connected by a narrow tube with a cork at its middle (fig). The radius of tank is $R = 20.0 c m,$ the radius of the tube is $r = 1.0 m m$ . The length of the tube is $L = 1.5 m .$ Water $(ρ = 1000 k g / m^{3})$ is poured into one of the tanks to a height of $h = 50 c m,$ while the second tank is empty. At the instant $t = 0,$ the cork is opened. [Both the tanks are open at top]

$(Viscosity of water η = 1 \times 10^{- 3} P a - s)$
Given poiseuille’s equation for volume flow rate of fluid in a tube $\frac{d V}{d t} = \frac{Δ P π r^{4}}{8 η L}$
$Δ P \to$ pressure difference across ends of the tube
$r \to$ radius of tube
$L \to$ length of tube

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Total mechanical energy of the fluid is conserved during the process

Water level in the two containers cannot become equal in a finite time

The difference between the levels of the water in the tanks decreases to e times in time $2.4 \times 10^{4} s .$

The difference between the levels of the water in the tanks decreases to e times in time $1.6 \times 10^{4} s .$

answer is B, D.

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Detailed Solution

$\frac{ρ g (h_{1} - h_{2}) π r^{4}}{8 η L} = \frac{A d h_{2}}{d t} = \frac{- A d h_{1}}{d t} = k d h_{1} = \frac{- k d t}{A}, d h_{2} = \frac{+ k d t}{A} d (h_{1} - h_{2}) = \frac{- 2 k d t}{A} \Rightarrow k = - \frac{A}{2 d t} d (h_{1} - h_{2}) \frac{ρ g (h_{1} - h_{2}) π r^{4}}{8 η L} = \frac{- A d (h_{1} - h_{2})}{2 d t} \frac{d (Δ h)}{d t} = \frac{- ρ g π r^{4} 2 Δ h}{8 η L A} \int_{Δ h}^{Δ h_{0}} \frac{d (Δ h)}{Δ h} = - \frac{ρ g π r^{4}}{4 η L A} \int_{0}^{t} d t Δ h = Δ h_{o} e^{- \frac{ρ g π r^{4}}{4 η L A} t} Δ h = \frac{Δ h_{o}}{e} i n a t i m e t_{0} = \frac{4 η L A}{ρ g π r^{4}} = 2.4 \times 10^{4} s a n d Δ h \to 0 a s t \to α$