Three rods of identical cross-sectional area made from the same metal form the sides of an isosceles triangle A B C right angled at B. The points A and B are maintained at temperatures T and √2 T respectively in the steady state. Assuming that only heat conduction takes place, temperature of point C will be

Question

Three rods of identical cross-sectional area made from the same metal form the sides of an isosceles triangle A B C right angled at B. The points A and B are maintained at temperatures T and √2 T respectively in the steady state. Assuming that only heat conduction takes place, temperature of point C will be (A) (3 T/√2+1) (B) (T/√2+1) (C) (T/√3(√2-1)) (D) (T/√2-1)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/3966c98c7f79bb63afba245e26078e79-.png" /> As TB >TA, heat flows from B to A through both paths B A and B C A. Rate of heat flow in B C= Rate of heat flow in C A ∴ (K A(√2 T-TC)/L)=(K A(TC-T)/√2 L) Solving this, we get TC=(3 T/√2+1)

$\frac{3 T}{2 + 1}$

$\frac{T}{2 + 1}$

$\frac{T}{3 ( 2 - 1 )}$

$\frac{T}{2 - 1}$

As $T_{B} > T_{A}$ ,
heat flows from $B$ to $A$ through both paths $B A$ and $BC A$ .
Rate of heat flow in $BC =$ Rate of heat flow in $C A$
$∴ \frac{K A ( 2 T - T _{C} )}{L} = \frac{K A ( T _{C} - T )}{2 L}$
Solving this, we get $T_{C} = \frac{3 T}{2 + 1}$

2​+13T​

2​+1T​

3​(2​−1)T​

2​−1T​

As TB​>TA​, heat flows from B to A through both paths BA and BCA. Rate of heat flow in BC= Rate of heat flow in CA ∴LKA(2​T−TC​)​=2​LKA(TC​−T)​ Solving this, we get TC​=2​+13T​

$\frac{3 T}{2 + 1}$

$\frac{T}{2 + 1}$

$\frac{T}{3 ( 2 - 1 )}$

$\frac{T}{2 - 1}$

As $T_{B} > T_{A}$ ,
heat flows from $B$ to $A$ through both paths $B A$ and $BC A$ .
Rate of heat flow in $BC =$ Rate of heat flow in $C A$
$∴ \frac{K A ( 2 T - T _{C} )}{L} = \frac{K A ( T _{C} - T )}{2 L}$
Solving this, we get $T_{C} = \frac{3 T}{2 + 1}$