Differential equation for the variation of amount of salt 'x' in a tank is given by : (dx/dt)+(x/20) = 10... time at which the amount of salt increases to 100kg
- A. 100 ln 2
- B. 50 ln 2
- C. 20 ln 2
- D. 10 ln 2
Answer: C. 20 ln 2
Explanation: Linear ODE. Solution x(t) = 200 (1 - e^(-t/20)). 100 = 200 (1 - e^(-t/20)). 0.5 = 1 - e^(-t/20). e^(-t/20) = 0.5. -t/20 = ln(0.5) = -ln2. t = 20 ln 2.