Read the Passage
Camphor is a waxy, colourless solid with strong aroma that evaporates through the process of sublimation, if left in the open at room temperature.A cylindrical camphor tablet whose height is equal to its radius (r) evaporates when exposed to air such that that the rate of reduction of its volume is proportional to its total surface area. Thus, $\frac{dV}{dt} = kS$ is the differential equation, where V is the volume, S is the surface area and t is the time in hours.
(ii) Substituting $V = \pi r^3$ and $S = 2\pi r^2$, we get the differential equation $\frac{dr}{dt} = \frac{2}{3}k$. Solve it, given that $r(0) = 5$ mm.
(iii) (a) If it is given that $r = 3$ mm when $t = 1$ hour, find the value of k. Hence, find t for $r = 0$ mm.
OR
(iii) (b) If it is given that $r = 1$ mm when $t = 1$ hour, find the value of k. Hence, find t for $r = 0$ mm.
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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.
Correct Answer: (i) Order: 1, Degree: 1 (ii) r = (2/3)kt + 5 (iii) (a) k = -3, t = 2.5 hours OR (iii) (b) k = -6, t = 1.25 hours <\/strong>
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Bloom's Taxonomy: REMEMBER
Nature: COMPETENCY
Reasoning: This question assesses the student's ability to recall and apply the concepts of differential equations in a real-world context. It involves understanding the order and degree of a differential equation, solving a differential equation, and applying initial conditions to find specific solutions. The case study provides a context, making it competency-based.