A right circular cylinder always has its height equal to its radius r. The radius is increasing at the rate of 1 cm/s. Match List-I with List-II, where List-II gives each rate at the instant when r = 3 cm (in cm2/s or cm3/s).
| List-I (Rate of change of) | List-II (Value) | ||
| (A) | Volume | (I) | \(6\pi\) |
| (B) | Curved surface area | (II) | \(12\pi\) |
| (C) | Total surface area | (III) | \(24\pi\) |
| (D) | Area of the base | (IV) | \(27\pi\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
Since h = r, each measure is a function of r alone. By the chain rule, each rate is its derivative with respect to r times \(\displaystyle\frac{dr}{dt} = 1\), taken at r = 3.
(A) Volume
This matches (IV) \(27\pi\).
\(V = \pi r^2 h = \pi r^3\)
\(\displaystyle\frac{dV}{dt} = 3\pi r^2 \times 1 = 27\pi\)
(B) Curved surface area
This matches (II) \(12\pi\).
\(S = 2\pi r h = 2\pi r^2\)
\(\displaystyle\frac{dS}{dt} = 4\pi r \times 1 = 12\pi\)
(C) Total surface area
This matches (III) \(24\pi\).
\(T = 2\pi r h + 2\pi r^2 = 4\pi r^2\)
\(\displaystyle\frac{dT}{dt} = 8\pi r \times 1 = 24\pi\)
(D) Area of the base
This matches (I) \(6\pi\).
\(B = \pi r^2\)
\(\displaystyle\frac{dB}{dt} = 2\pi r \times 1 = 6\pi\)