A tank, as shown in the figure below, formed using a combination of a cylinder and a cone, offers better drainage as compared to a flat bottomed tank. A tap is connected to such a tank whose conical part is full of water. Water is dripping out from a tap at the bottom at the uniform rate of 2\ cm³/s. The semi-vertical angle of the conical tank is 45°. On the basis of given information, answer the following questions : (i) Find the volume of water in the tank in terms of its radius r. (ii) Find rate of change of radius at an instant when r = 2√(2) cm. (iii)(a) Find the rate at which the wet surface of the conical tank is decreasing at an instant when radius r = 2√(2) cm. (iii)(b) Find the rate of change of height 'h' at an instant when slant height is 4 cm.

(i) V = 13π r³; (ii) -(1)/(4π) cm/s; (iii)(a) decreasing at 2 cm²/s; (iii)(b) -(1)/(4π) cm/s

MathsCBSE Class 12 · Board exam 2023, set 65/1/1 · 4 marks · Case-based

CBSE Class 12 Mathematics 2023: question 37

Set in the CBSE Class 12 board exam of 2023, paper 65/1/1, 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-basedApplication of Derivatives

A tank, as shown in the figure below, formed using a combination of a cylinder and a cone, offers better drainage as compared to a flat bottomed tank. A tap is connected to such a tank whose conical part is full of water. Water is dripping out from a tap at the bottom at the uniform rate of $2 cm^{3} / s$ . The semi-vertical angle of the conical tank is $4 5^{\circ}$ .

On the basis of given information, answer the following questions :

(i)
Find the volume of water in the tank in terms of its radius r.
1 mark
(ii)
Find rate of change of radius at an instant when $r = 22$ cm.
1 mark
(iii)(a)
Find the rate at which the wet surface of the conical tank is decreasing at an instant when radius $r = 22$ cm.
2 marks
OR
(iii)(b)
Find the rate of change of height 'h' at an instant when slant height is 4 cm.
2 marks

Figure in the paper: Photo of a storage tank: a cylindrical upper part with a conical bottom, supported on a metal frame with legs. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 65/1/1.

Hints

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

Hint 1

What does a $4 5^{\circ}$ semi-vertical angle tell you about the radius and the height?

Hint 2

Differentiate the volume with respect to t and use $d V / d t = - 2$ , which is negative because water is leaving.

Hint 3

The wet surface is the curved surface $π r l$ with $l = 2 r$ . For (iii)(b), find r and h when $l = 4$ .

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) Semi-vertical angle $4 5^{\circ}$ gives $tan 4 5^{\circ} = \frac{r}{h} = 1$ , so $h = r$ . Then $V = \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{3}$ .
1 mark
(ii) $\frac{d V}{d t} = π r^{2} \frac{d r}{d t}$ , and $\frac{d V}{d t} = - 2$ cm³/s because the volume is decreasing.
½ mark
(ii) At $r = 22$ : $- 2 = π (8) \frac{d r}{d t}$ , so $\frac{d r}{d t} = - \frac{1}{4 π}$ cm/s. The radius decreases at $\frac{1}{4 π}$ cm/s.
½ mark
(iii)(a) Slant height $l = r^{2} + h^{2} = 2 r$ . The wet (curved) surface is $C = π r l = 2 π r^{2}$ .
1 mark
(iii)(a) $\frac{d C}{d t} = 22 π r \frac{d r}{d t}$ .
½ mark
(iii)(a) At $r = 22$ : $\frac{d C}{d t} = 22 π (22) (- \frac{1}{4 π}) = - 2$ cm²/s. The wet surface is decreasing at $2$ cm²/s.
½ mark
(iii)(b) $l^{2} = h^{2} + r^{2} = 2 h^{2}$ since $r = h$ . When $l = 4$ : $h^{2} = 8$ , so $h = r = 22$ cm.
1 mark
(iii)(b) Since $h = r$ at all times, $\frac{d h}{d t} = \frac{d r}{d t}$ . By (ii), at $r = 22$ this is $- \frac{1}{4 π}$ cm/s. The height decreases at $\frac{1}{4 π}$ cm/s.
1 mark

Answer: (i) $V = \frac{1}{3} π r^{3}$ ; (ii) $- \frac{1}{4 π}$ cm/s; (iii)(a) decreasing at 2 cm²/s; (iii)(b) $- \frac{1}{4 π}$ cm/s

Where the marks are

(i) Uses $h = r$ to get $V = \frac{1}{3} π r^{3}$ (1 mark)
(ii) Differentiates: $d V / d t = π r^{2} d r / d t$ (½ mark)
(ii) $d r / d t = - 1/ (4 π)$ cm/s (½ mark)
(iii)(a) $C = 2 π r^{2}$ (1 mark)
(iii)(a) $d C / d t = 22 π r d r / d t$ (½ mark)
(iii)(a) $- 2$ cm²/s (½ mark)
(iii)(b) Finds $r = h = 22$ when $l = 4$ (1 mark)
(iii)(b) $d h / d t = d r / d t = - 1/ (4 π)$ cm/s (1 mark)

Revise the chapter

Chapter: Application of Derivatives (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Application of Derivatives, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 65/1/1: every question, by section.

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CBSE Class 12 Mathematics 2023: question 37

Hints

Step-by-step solution

Revise the chapter

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