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  •  A metallic ellipse in<img alt="\mathrm{ \frac{x^2}{100}+\frac{y^2}{91}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%5Cfrac%7Bx%5E2%7D%7B100%7D&plus;%5Cfrac%7By

 A metallic ellipse in\mathrm{ \frac{x^2}{100}+\frac{y^2}{91}=1} form is heated in such a  way that it coincides with its auxiliary circle in K seconds, then K=?
(Given that e change at the rate of \mathrm{\frac{10^{-1}}{671} }per second)

 

Option: 1

2020


Option: 2

2013


Option: 3

2040


Option: 4

2030


Answers (1)

best_answer

\mathrm{\text { Before heating, } e=\sqrt{1-\frac{91}{100}}=\frac{3}{10}}

Eccentricity of any circle is zero
 When ellipse coincides with auxiliary circle after
heating, e will change from \mathrm{\frac{3}{10} \text { to } 0 \text { in } k \text { secs. }}

\mathrm{\begin{aligned} & \therefore \frac{d e}{d t}=-\frac{10^{-1}}{671} \Rightarrow \int_{3 / 10}^0 d e=-\int_0^k \frac{10^{-1}}{671} \cdot d t \\ & \Rightarrow \quad 0-\frac{3}{10}=-\frac{10^{-1}}{671} \cdot(k-0) \\ & \Rightarrow \quad k=\frac{3}{10} \times \frac{671}{10^{-1}}=2013 \mathrm{secs} \end{aligned}}

Posted by

Irshad Anwar

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