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TS EAMCET · Maths · Differential Equations

The volume of a spherical balloon is increasing at the rate of 2 cm3/sec. When its radius is 4 cm, the rate of change of its surface area (in cm/2sec) is

  1. A 1
  2. B 2
  3. C 3
  4. D 4
Verified Solution

Answer & Solution

Correct Answer

(A) 1

Step-by-step Solution

Detailed explanation

Let V be the volume & S be the surface area of spherical balloon having radius r. Given:dVdt=2cm3/sec To find:dSdtr=4 As, volume V=43πr3 ⇒dVdt=433πr2drdt ⇒2=433πr2drdt ∴drdt=12πr2 Now, S=4πr2 ⇒dSdt=4π×2rdrdt…
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