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JEE MATHEMATICS MCQ: RATE MEASURE

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Application +4 | -1

A solid box is formed by placing a cylinder, having equal height and radius on top of a cube such that, the circular base of cylinder is the inscribed circle for square top of the cube. If the radius of cylinder is changing at the rate

\frac{1}{2 \pi+16}

\mathrm{cm} / \mathrm{sec}

, then the rate of change of volume of the box when radius is

2 \mathrm{~cm}

, is (Assuming that box always remain in the given shape)

View answer and explanation

Correct option: A

6

Option A (6) is correct because the rate of change of volume is calculated by differentiating the volume equation with respect to time and substituting the given values.

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