MHT CET · Maths · Differential Equations
If the radius of a circle increases at the rate of \(7 \mathrm{~cm} / \mathrm{sec}\), then the rate of increase of its area after 10 minutes is
- A \(1,84,800 \mathrm{~cm}^{2} / \mathrm{sec}\)
- B \(1,64,800 \mathrm{~cm}^{2} / \mathrm{sec}\)
- C \(1,88,400 \mathrm{~cm}^{2} / \mathrm{sec}\)
- D \(1,68,400 \mathrm{~cm}^{2} / \mathrm{sec}\)
Answer & Solution
Correct Answer
(A) \(1,84,800 \mathrm{~cm}^{2} / \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
\(\frac{d \varepsilon}{d t} =7 \mathrm{~cm} / \mathrm{sec} . \)
\( A =\pi \varepsilon^{2} \)
\( \frac{d A}{d t} =2 \pi \varepsilon \frac{d \varepsilon}{d t}-(1) \)
\( \text { Radius after } 10 \text { minutes } ; \varepsilon=70 \mathrm{~cm} \times 60 \mathrm{~cm} \)
\( \frac{d A}{d t} =2 \times \frac{22}{7} \times 70 \times 7 \times 60[\mathrm{Put} \text { in (1)... }\)
\( =1,84,800 \mathrm{~cm}^{2} / \mathrm{sec}.\)
\( A =\pi \varepsilon^{2} \)
\( \frac{d A}{d t} =2 \pi \varepsilon \frac{d \varepsilon}{d t}-(1) \)
\( \text { Radius after } 10 \text { minutes } ; \varepsilon=70 \mathrm{~cm} \times 60 \mathrm{~cm} \)
\( \frac{d A}{d t} =2 \times \frac{22}{7} \times 70 \times 7 \times 60[\mathrm{Put} \text { in (1)... }\)
\( =1,84,800 \mathrm{~cm}^{2} / \mathrm{sec}.\)
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