Question

Three rods of material x and two rods of material y are connected as shown in figure. All the rods are of identical length and cross - sectional area. The end A is maintained at 100${}^{\circ }$C and the junction C at 0${}^{\circ }$C. It is given that resistance of rod of material x is R. Further, $K_x$ = ${2K}_y$ .

Match the entries of column I and column II

$$\begin{array}{cc}column1& column2\\ A.TemperatureofjunctionB& P.Doesnotbehavelikeabalancedwheatstone{s}^{\prime }sbridge\\ B.TemperatureofjunctionC& Q.Istwiceoftheother\\ C.Thisnetworkis& R.{\frac{400}{7}}^{\circ }C\\ D.Resistanceofrodofmaterialy& S.{\frac{300}{7}}^{\circ }C\end{array}$$

• A. A-P B-R C-P D-R
• B. A-P B-Q C-P D-R
• C. A-R B-S C-P D-Q
• D. A-Q B-Q C-Q D-P

Answer 1

The correct option is C

A-R B-S C-P D-Q

$\text{Thermal resistance R}=\frac{1}{KA}If{K}_x=2K_y,R_x=\frac{Ry}{2}\left(or\right)R_y=2R_x=2R\text{So, final}\text{By the logic of current, for point B,}\frac{T_1-100}{R}+\frac{T_1-T_2}{R}+\frac{T_1-0}{2R}=0\Rightarrow 2T_1-200+2T_1-2T_2+T_1=0\Rightarrow 5T_{1}2T_{2}2=200\to \left(1\right)\text{For point C}\frac{T_2-100}{2R}+\frac{(t_2-T_2-0)}{R}=0\Rightarrow 5T_2-2T_1=100\to \left(2\right)\text{Solving (1) and (2), we get,}{T}_1=\frac{{400}^{0}C}{7}and{T}_2=\frac{{380}^{0}C}{7}\text{Thermal resistance between B and D is},\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{3R}+\frac{1}{3R}=\frac{1}{R}+\frac{2}{3R}\Rightarrow R_{eq}=\frac{3R}{5}$