Competency-Based
Assessment Made Simple.

Step-by-Step Solution

(i) Order and Degree:

The given differential equation is $\frac{dV}{dt} = kS$. Since it involves the first derivative of V with respect to t, the order is 1. The power of the highest order derivative is also 1, so the degree is 1.

(ii) Solving the differential equation:

Given $V = \pi r^3$ and $S = 2\pi r^2$, we have $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = 3\pi r^2 \frac{dr}{dt}$.

Substituting into the given differential equation: $3\pi r^2 \frac{dr}{dt} = k(2\pi r^2)$.

Simplifying, we get $\frac{dr}{dt} = \frac{2}{3}k$.

Integrating both sides with respect to t: $\int dr = \int \frac{2}{3}k dt$.

This gives $r = \frac{2}{3}kt + C$, where C is the constant of integration.

Given $r(0) = 5$ mm, we substitute $t = 0$ and $r = 5$ into the equation: $5 = \frac{2}{3}k(0) + C$, so $C = 5$.

Thus, the solution is $r = \frac{2}{3}kt + 5$.

(iii) (a) Finding k and t for r = 0:

Given $r = 3$ mm when $t = 1$ hour, we substitute these values into the equation: $3 = \frac{2}{3}k(1) + 5$.

Solving for k: $\frac{2}{3}k = -2$, so $k = -3$ mm/hour.

Now, we want to find t when $r = 0$: $0 = \frac{2}{3}(-3)t + 5$, which simplifies to $0 = -2t + 5$.

Solving for t: $2t = 5$, so $t = \frac{5}{2} = 2.5$ hours.

OR (iii) (b) Finding k and t for r = 0:

Given $r = 1$ mm when $t = 1$ hour, we substitute these values into the equation: $1 = \frac{2}{3}k(1) + 5$.

Solving for k: $\frac{2}{3}k = -4$, so $k = -6$ mm/hour.

Now, we want to find t when $r = 0$: $0 = \frac{2}{3}(-6)t + 5$, which simplifies to $0 = -4t + 5$.

Solving for t: $4t = 5$, so $t = \frac{5}{4} = 1.25$ hours.