Question

As per the given figure, two plates A and B of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of plates are 4.0 cm and 2.5 cm, respectively and the area of cross-section is 120 cm² for each plate. The equivalent thermal conductivity of the compound plate is \((1+\frac{5}{α})K,\) then the value of α will be ___.

Answer 1

Correct Answer: 21

 <div>Given hyperbola : \(x^2 - y^2 = 1\)</div> 

<div>Eccentricity of hyperbola</div>

<div><span class="math-tex">\(\begin{array}{l}
e_H = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{2}
\end{array}\)</span></div>

<div>Given : eccentricity of ellipse is reciprocal</div>

<div><span class="math-tex">\(\begin{array}{l}
e_E = \frac{1}{\sqrt{2}}
\end{array}\)</span></div>

<div>For ellipse</div>

<div><span class="math-tex">\(\begin{array}{l}
e_E = \sqrt{1-\frac{b^2}{a^2}}
\end{array}\)</span></div>

<div><span class="math-tex">\(\begin{array}{l}
\sqrt{1-\frac{b^2}{a^2}} = \frac{1}{\sqrt{2}}
\end{array}\)</span></div>

<div><span class="math-tex">\(\begin{array}{l}
1-\frac{b^2}{a^2} = \frac{1}{2}
\end{array}\)</span></div>

<div><span class="math-tex">\(\begin{array}{l}
\frac{b^2}{a^2} = \frac{1}{2} \Rightarrow b^2 = \frac{a^2}{2}
\end{array}\)</span></div>

<div>Equation of tangent to ellipse : \(y = mx \pm \sqrt{a^2m^2 - b^2}\)</div>

<div>Equation of tangent to hyperbola : \(y = mx \pm \sqrt{m^2 - 1}\)</div>

<div>For common tangent, constants must be equal</div>

<div><span class="math-tex">\(\begin{array}{l}
\sqrt{a^2m^2 - b^2} = \sqrt{m^2 - 1}
\end{array}\)</span></div>

<div><span class="math-tex">\(\begin{array}{l}
a^2m^2 - b^2 = m^2 - 1
\end{array}\)</span></div>

<div>Given slope \(m = \sqrt{\frac{5}{2}}\)</div>

<div><span class="math-tex">\(\begin{array}{l}
a^2\cdot\frac{5}{2} - \frac{a^2}{2} = \frac{5}{2} - 1
\end{array}\)</span></div>

<div><span class="math-tex">\(\begin{array}{l}
\frac{5a^2 - a^2}{2} = \frac{3}{2}
\end{array}\)</span></div>

<div><span class="math-tex">\(\begin{array}{l}
\frac{4a^2}{2} = \frac{3}{2} \Rightarrow 2a^2 = \frac{3}{2}
\end{array}\)</span></div>

<div><span class="math-tex">\(\begin{array}{l}
a^2 = \frac{3}{4}
\end{array}\)</span></div>

<div><span class="math-tex">\(\begin{array}{l}
b^2 = \frac{a^2}{2} = \frac{3}{8}
\end{array}\)</span></div>

<div><span class="math-tex">\(\begin{array}{l}
4(a^2 + b^2) = 4\left(\frac{3}{4} + \frac{3}{8}\right)
\end{array}\)</span></div>

<div><span class="math-tex">\(\begin{array}{l}
= 4\cdot \frac{9}{8} = \frac{36}{8} = \frac{9}{2}
\end{array}\)</span></div>

<div>Hence, answer = <b>\(\frac{9}{2}\)</b></div>