MathsCBSE Class 12 · Board exam 2025, set 65/7/1 · 4 marks · Case-based

CBSE Class 12 Mathematics 2025: question 37

Set in the CBSE Class 12 board exam of 2025, paper 65/7/1, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 4 marks · Case-basedCompetency-basedDifferential Equations

Case Study – 2

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 (r) evaporates when exposed to air such that the rate of reduction of its volume is proportional to its total surface area. Thus, $\frac{d V}{d t} = k S$ is the differential equation, where V is the volume, S is the surface area and 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.
1 mark
(ii)
Substituting $V = π r^{3}$ and $S = 2 π r^{2}$ , we get the differential equation $\frac{d r}{d t} = \frac{2}{3} k$ . Solve it, given that $r (0) = 5$ mm.
1 mark
(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.
2 marks
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.
2 marks

Figure in the paper: Photograph of cylindrical-shaped camphor tablets. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 65/7/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Order is the highest derivative present; degree is its power when the equation is polynomial in derivatives.

Hint 2

$\frac{d r}{d t}$ is a constant, so integrate directly and use $r (0) = 5$ to find the constant.

Hint 3

Substitute the given $r$ and $t$ into your solution to get $k$ , then set $r = 0$ and solve for $t$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) In $\frac{d V}{d t} = k S$ the highest-order derivative is $\frac{d V}{d t}$ (first order) and it appears with power 1. Order $= 1$ , Degree $= 1$ .
1 mark
(ii) Separating variables and integrating: $\int d r = \frac{2 k}{3} \int d t \Rightarrow r = \frac{2}{3} k t + C$ .
½ mark
(ii) Using $r (0) = 5$ : $C = 5$ , so $r = \frac{2}{3} k t + 5$ .
½ mark
(iii)(a) Put $r = 3$ , $t = 1$ : $3 = \frac{2}{3} k + 5 \Rightarrow \frac{2}{3} k = - 2 \Rightarrow k = - 3$ .
1 mark
(iii)(a) Then $r = - 2 t + 5$ . For $r = 0$ : $t = \frac{5}{2}$ hours $= 2.5$ hours.
1 mark
(iii)(b) Put $r = 1$ , $t = 1$ : $1 = \frac{2}{3} k + 5 \Rightarrow \frac{2}{3} k = - 4 \Rightarrow k = - 6$ .
1 mark
(iii)(b) Then $r = - 4 t + 5$ . For $r = 0$ : $t = \frac{5}{4}$ hours $= 1.25$ hours.
1 mark

Answer: (i) Order 1, degree 1; (ii) $r = \frac{2}{3} k t + 5$ ; (iii)(a) $k = - 3$ , $t = 2.5$ h; (iii)(b) $k = - 6$ , $t = 1.25$ h

Where the marks are

(i) Order = 1 (½ mark), Degree = 1 (½ mark)
(ii) Integrates to $r = \frac{2}{3} k t + C$ (½ mark); uses $r (0) = 5$ to get $r = \frac{2}{3} k t + 5$ (½ mark)
(iii)(a) Finds $k = - 3$ (1 mark); $r = - 2 t + 5$ and $t = \frac{5}{2}$ h (1 mark)
(iii)(b) Finds $k = - 6$ (1 mark); $r = - 4 t + 5$ and $t = \frac{5}{4}$ h (1 mark)

Revise the chapter

Chapter: Differential Equations (notes, formulas, where marks are lost and every past question)
Topic: Variables Separable Method

Practise this chapterBoard-style questions from Differential Equations, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 65/7/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Mathematics 2025: question 37

Hints

Step-by-step solution

Revise the chapter

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