If K1 and K2 are the thermal conductivities L 1 and L 2 are the lengths and A 1 and A 2 are the cross sectional areas of steel and copper rods respectively such that (K2/K1)=9, (A1/A2)=2, (L1/L2)=2. Then, for the arrangement as shown in the figure. The value of temperature T of the steel - copper junction in the steady state will be :

Question

If K1 and K2 are the thermal conductivities L 1 and L 2 are the lengths and A 1 and A 2 are the cross sectional areas of steel and copper rods respectively such that (K2/K1)=9, (A1/A2)=2, (L1/L2)=2. Then, for the arrangement as shown in the figure. The value of temperature T of the steel - copper junction in the steady state will be : (A) 18° C (B) 14° C (C) 45° C (D) 150° C

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/f9749b6cf9007bdbd65ba13de341a853-.png" /> ( d θ/ dt )=( K 1 A 1/ l 1)( T 1- T )=( K 2 A 2/ l 2)( T - T 2) ⇒ (450- T / T -0)=( K 2 A 2 I 1/ K 1 A 1 l 2)=9 × (1/2) × 2 ⇒ 450- T =9 T ⇒ T =45° C

$1 8^{\circ} C$

$1 4^{\circ} C$

$4 5^{\circ} C$

$15 0^{\circ} C$

$\frac{d θ}{d t} = \frac{K _{1} A _{1}}{l _{1}} (T_{1} - T) = \frac{K _{2} A _{2}}{l _{2}} (T - T_{2})$
$\Rightarrow \frac{450 - T}{T - 0} = \frac{K _{2} A _{2} I _{1}}{K _{1} A _{1} l _{2}} = 9 \times \frac{1}{2} \times 2$
$\Rightarrow 450 - T = 9 T \Rightarrow T = 4 5^{\circ} C$

18∘C

14∘C

45∘C

150∘C

dtdθ​=l1​K1​A1​​(T1​−T)=l2​K2​A2​​(T−T2​) ⇒T−0450−T​=K1​A1​l2​K2​A2​I1​​=9×21​×2 ⇒450−T=9T⇒T=45∘C

$1 8^{\circ} C$

$1 4^{\circ} C$

$4 5^{\circ} C$

$15 0^{\circ} C$

$\frac{d θ}{d t} = \frac{K _{1} A _{1}}{l _{1}} (T_{1} - T) = \frac{K _{2} A _{2}}{l _{2}} (T - T_{2})$
$\Rightarrow \frac{450 - T}{T - 0} = \frac{K _{2} A _{2} I _{1}}{K _{1} A _{1} l _{2}} = 9 \times \frac{1}{2} \times 2$
$\Rightarrow 450 - T = 9 T \Rightarrow T = 4 5^{\circ} C$