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Mathematics · 2025 · 4 marks
CBSE 2025 · Region 7 · Set 2 · Q38
Camphor is a waxy, colourless solid with strong aroma that evaporates through the process of sublimation, if left in the open at room temperature.…
Camphor is a waxy, colourless solid with strong aroma that evaporates through the process of sublimation, if left in the open at room temperature.
(Cylindrical-shaped Camphor tablets) A cylindrical camphor tablet whose height is equal to its radius ( $\displaystyle \mathbf{r}$ ) evaporates when exposed to air such that the rate of reduction of its volume is proportional to its total surface area. Thus, $\displaystyle \frac{\mathrm{dV}}{\mathrm{dt}}=\mathrm{kS}$ is the differential equation, where V is the volume, S is the surface area and $\displaystyle \mathrm{t}$ is the time in hours. Based upon the above information, answer the following questions :
(i)
Write the order and degree of the given differential equation.
(ii)
Substituting $\displaystyle \mathrm{V}=\pi \mathrm{r}^{3}$ and $\displaystyle \mathrm{S}=2 \pi \mathrm{r}^{2}$, we get the differential equation $\displaystyle \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{2}{3} \mathrm{k}$. Solve it, given that $\displaystyle \mathrm{r}(0)=5 \mathrm{~mm}$.
(iii)
(a)
If it is given that $\displaystyle \mathrm{r}=3 \mathrm{~mm}$ when $\displaystyle \mathrm{t}=1$ hour, find the value of k . Hence, find t for $\displaystyle \mathrm{r}=0 \mathrm{~mm}$.
OR
(b)
If it is given that $\displaystyle \mathrm{r}=1 \mathrm{~mm}$ when $\displaystyle \mathrm{t}=1$ hour, find the value of k . Hence, find t for $\displaystyle \mathrm{r}=0 \mathrm{~mm}$.
Marking-scheme solution
(i)
Order = $\displaystyle 1$, Degree = $\displaystyle 1$
(ii)
Separating the variable and integrating, $\displaystyle \int dr = \dfrac{2k}{3}\int dt \Rightarrow r = \dfrac{2}{3}kt + C$
Putting $\displaystyle t = 0, r = 5$, we get $\displaystyle C = 5$
\[r = \frac{2}{3}kt + 5 \]
(iii)
(a)
Putting $\displaystyle r = 3, t = 1$, $\displaystyle 3 = \dfrac{2}{3}k(1) + 5 \Rightarrow k = -3$
$\displaystyle r = -2t + 5$, For $\displaystyle r = 0$, $\displaystyle t = \dfrac{5}{2}$ hrs or $\displaystyle 2.5$ hrs
OR
(b)
Putting $\displaystyle r = 1, t = 1$, $\displaystyle 1 = \dfrac{2}{3}k + 5 \Rightarrow k = -6$
$\displaystyle \therefore\; r = -4t + 5$, For $\displaystyle r = 0$, $\displaystyle t = \dfrac{5}{4}$ hrs or $\displaystyle 1.25$ hrs
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