Question

[H] There are \( n+1 \) tanks each containing 100 litres and connected as shown. Throughout, the liquid in every tank is kept well-mixed. The 0-th tank contains 100 litres of pure water with 50 grams of salt dissolved in it. The remainder contain pure water. Pure water is pumped into tank 0 at 3 litres per minute and liquid leaves the system from tank \( n \) at 3 litres per minute. Let \( m_{k} \) denote the mass (in grams) of salt in tank \( k \). a) Show that \[ \frac{d m_{k}}{d t}=0.03\left(m_{k-1}-m_{k}\right), \quad k=1,2, \ldots, n \] and \[ \frac{d m_{0}}{d t}=-0.03 m_{0} \] b) Show that \[ m_{n}=\frac{50(0.03)^{n}}{n !} t^{n} e^{-0.03 t} \]

          [H] There are \( n+1 \) tanks each containing 100 litres and connected as shown.
Throughout, the liquid in every tank is kept well-mixed. The 0-th tank contains 100 litres of pure water with 50 grams of salt dissolved in it. The remainder contain pure water. Pure water is pumped into tank 0 at 3 litres per minute and liquid leaves the system from tank \( n \) at 3 litres per minute. Let \( m_{k} \) denote the mass (in grams) of salt in tank \( k \).
a) Show that
\[
\frac{d m_{k}}{d t}=0.03\left(m_{k-1}-m_{k}\right), \quad k=1,2, \ldots, n
\]
and
\[
\frac{d m_{0}}{d t}=-0.03 m_{0}
\]
b) Show that
\[
m_{n}=\frac{50(0.03)^{n}}{n !} t^{n} e^{-0.03 t}
\]

[H] There are n+1 tanks each containing 100 litres and connected as shown.
Throughout, the liquid in every tank is kept well-mixed. The 0-th tank contains 100 litres of pure water with 50 grams of salt dissolved in it. The remainder contain pure water. Pure water is pumped into tank 0 at 3 litres per minute and liquid leaves the system from tank n at 3 litres per minute. Let mk denote the mass (in grams) of salt in tank k.
a) Show that

(d mk)/(d t)=0.03(mk-1-mk), k=1,2, …, n

and

(d m0)/(d t)=-0.03 m0

b) Show that

mn=(50(0.03)^n)/(n !) t^n e^-0.03 t

Added by Ahsan N.

Question

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