Three rods of identical area of cross-section andmade from the same metal form the sides of an isosceles triangle ABC - right angled at B- The points A and B are maintained at temperatures T and -2 T respectively- In the steady state the temperature of the point C is T c- Assuming that only heat conduction takes place- T C - T is equal toA- 1-2-1B- 3-2-1C- 1-2-2-1D- 1-3-2-1

The correct option is B ∵ T_b nbsp; gt;T_A ⇒ Heat nbsp;will flow B to A via two paths nbsp; (i) B to A (ii) and along BCA as shown. nbsp; Rate of fl ...

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Byju's Answer

Standard XII

Physics

Thermal Resistance

Three rods of...

Question

Three rods of identical area of cross-section and

made from the same metal form the sides of an

isosceles triangle ABC , right angled at B. The

points A and B are maintained at temperatures

T and $\sqrt{2}$ T respectively. In the steady state the

temperature of the point C is $T_{c}$ . Assuming that

only heat conduction takes place, $\frac{T_{C}}{T}$ is equal to

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Solution

The correct option is B

∵ $T_{b} > T_{A} \Rightarrow$ Heat will flow B to A via two paths

(i) B to A (ii) and along BCA as shown.

Rate of flow of heat in path BCA will be same

i.e ${(\frac{Q}{t})}_{B C} = {(\frac{Q}{T})}_{C A}$

$\Rightarrow \frac{k (\sqrt{2} T - T_{C}) A}{a} = \frac{k (T_{c} - T) A}{\sqrt{2} a}$

$\Rightarrow \frac{T_{c}}{T} = \frac{3}{1 + \sqrt{2}}$

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The correct option is B ∵ Tb>TA⇒ Heat will flow B to A via two paths (i) B to A (ii) and along BCA as shown. Rate of flow of heat in path BCA will be same i.e (Qt)BC = (QT)CA ⇒ k(√2T − TC)Aa = k(Tc − T)A√2a ⇒TcT = 31 + √2