Question ID - 88634 | SaraNextGen Top Answer

Consider a tank which initially holds . of brine that contains of salt. Another brine solution, containing of salt/ltr., is poured into the tank at the rate of ltr./min while, simultaneously, the well-stirred solution leaves the tank at the rate of . The problem is to find the amount of salt in the tank at any time

Let denote the amount of salt in the tank at any time. The time rate of change of , equals the rate at which salt enters the tank minus the rate at which salt leaves the tank. Salt enters the tank at the rate of . To determine the rate at which salt leaves the tank, we first calculate the volume of brine in the tank at any time , which is the initial volume plus the volume of brine added minus the volume of brine removed . Thus, the volume of brine at any time is

(a)

The concentration of salt in the tank at any time is , from which it follows that salt leaves the tank at the rate of

Thus, (b)

Or

A tank initially holds 100 ltr. Of a brine solution containing 20 lb of salt. At , fresh water is poured into the tank at the rate of 5 ltr./min, while the well-stirred mixture leaves the tank at the same rate. Then the amount of salt in the tank after 20 min

a)

b)

c)

d)

Question

Updated On 05-07-2025
By Lithanya Sara

1 Answer

127 votes

By Lithanya Sara · Accepted Answer

Answer Key / Explanation : (a) -

(a)

Here, , and . Hence

The solution of this linear equation is (1)

At , we are given that

Substituting these values into equation (1), we find that , so that equation (1) can be rewritten as

For

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