A reservoir contains 50 L of pure water initially. Salted water flows into the reservoir at the rate of 2 L per min. It contains 2g of salt per liter. The mixture is kept uniform by stirring and it flows out of the bottom of the reservoir at the same rate. The time taken for the quantity of salt in the reservoir to increase from 40g to 80 g is a $\log_{e} b$ min, then the value of a+b is equal to ….

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answer is 28.

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Detailed Solution

Let at any moment concentration of solution be xg/L.
Let the change be dx in dt time.
$∴ d x = 4 d t - \frac{2 x}{50} d t \Rightarrow 25 d x = (100 - x) d t \Rightarrow 25 (\frac{d x}{100 - x}) = d t \Rightarrow 25 \int_{40}^{80} \frac{d x}{100 - x} = \int_{0}^{t} d t \Rightarrow 25 {[\log (100 - x)]}_{40}^{80} = {[t]}_{0}^{t} \Rightarrow 25 \log \frac{60}{20} = t \Rightarrow t = 25 \log_{e} 3 ∴ a = 25 a n d b = 3 \Rightarrow a + b = 28$