COMEDK · Maths · 27. Application of Derivatives
An open hemispherical storage tank has radius 13 m. Oil flows into the tank such that the depth '\(h\)' of oil in the tank changes at the rate of 3 m / hr. When the depth \(h = 1\)m, the rate of change of the area of the top surface of the oil is
- A \(26\pi\) m\(^2\) / hr
- B \(75\pi\) m\(^2\) / hr
- C \(72\pi\) m\(^2\) / hr
- D \(24\pi\) m\(^2\) / hr
Answer & Solution
Correct Answer
(C) \(72\pi\) m\(^2\) / hr
Step-by-step Solution
Detailed explanation
Let \(R\) be the radius of the hemispherical tank, so \(R = 13\) m. Let \(r\) be the radius of the top surface of the oil at depth \(h\). The distance from the center of the hemisphere to the oil surface is \(R - h\). Using the Pythagorean theorem in the cross-section of the…
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