The radius of a spherical balloon is increasing at the rate of 0.5 cm/s. At the instant when the radius is 4 cm, match List-I with List-II
| List-I (Quantity) | List-II (Rate of increase, in cm/s, cm2/s or cm3/s) | ||
| (A) | Volume of the balloon | (I) | \(4\pi\) |
| (B) | Surface area of the balloon | (II) | \(32\pi\) |
| (C) | Circumference of a great circle of the balloon | (III) | \(\pi\) |
| (D) | Area of a great circle of the balloon | (IV) | \(16\pi\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
Each quantity is a function of the radius r, so by the chain rule its rate is (derivative with respect to r) × \(\frac{dr}{dt}\), with r = 4 and \(\frac{dr}{dt} = 0.5\). A great circle has the same radius as the sphere.
(A) Volume of the balloon
This matches (II) \(32\pi\).
\( \displaystyle V = \frac{4}{3}\pi r^{3} \)
\( \displaystyle \frac{dV}{dt} = 4\pi r^{2}\frac{dr}{dt} \)
\( = 4\pi \times 16 \times 0.5 = 32\pi \)
(B) Surface area of the balloon
This matches (IV) \(16\pi\).
\( S = 4\pi r^{2} \)
\( \displaystyle \frac{dS}{dt} = 8\pi r\frac{dr}{dt} \)
\( = 8\pi \times 4 \times 0.5 = 16\pi \)
(C) Circumference of a great circle of the balloon
This matches (III) π.
\( C = 2\pi r \)
\( \displaystyle \frac{dC}{dt} = 2\pi\frac{dr}{dt} \)
\( = 2\pi \times 0.5 = \pi \)
(D) Area of a great circle of the balloon
This matches (I) \(4\pi\).
\( A = \pi r^{2} \)
\( \displaystyle \frac{dA}{dt} = 2\pi r\frac{dr}{dt} \)
\( = 2\pi \times 4 \times 0.5 = 4\pi \)