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The Diameter of a Circle is Increasing at the Rate of 1 Cm/Sec. When Its Radius is π, the Rate of Increase of Its Area is (A) π Cm2/Sec - Mathematics

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प्रश्न

The diameter of a circle is increasing at the rate of 1 cm/sec. When its radius is π, the rate of increase of its area is

विकल्प

  •  π cm2/sec

  •  2π cm2/sec

  •  π2 cm2/sec

  • 2 cm2/sec2

MCQ
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उत्तर

 π2 cm2/sec

\[\text { Let D be the diameter and A be the area of the circle at any time t. Then },\]

\[A = \pi r^2 \left( \text { where r is the radius of the cicle } \right)\]

\[ \Rightarrow A=\pi\frac{D^2}{4}\left[ \because r = \frac{D}{2} \right]\]

\[ \Rightarrow \frac{dA}{dt} = 2\pi\frac{D}{4}\frac{dD}{dt}\]

\[ \Rightarrow \frac{dA}{dt} = \frac{\pi}{2} \times 2\pi \times 1 \left[ \because \frac{dD}{dt} = 1 cm/\sec \right]\]

\[ \Rightarrow \frac{dA}{dt} = \pi^2 {cm}^2 /\sec\]

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अध्याय 13: Derivative as a Rate Measurer - Exercise 13.4 [पृष्ठ २६]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 13 Derivative as a Rate Measurer
Exercise 13.4 | Q 21 | पृष्ठ २६

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