Question

53. Three rods (lengths 2L,L,L) made of the same material and having the same area of cross-section
are joined as shown in figure. The end points A, B and C are maintained at constant temperatures
100°C, 50°C and 0°C, respectively. Assuming that there is no loss of heat from the surface of
the rods, find the temperature that the junction P ultimately reaches
A 50°C

B. 40°C

C 30°C

D 20°C​

Answer 1

The given arrangement is,

$\setlength{\unitlength}{1mm}\begin{picture}\thicklines\put(-50,0){\line(1,0){50}}\put(0,0){\line(1,1){17.6777}}\put(0,0){\line(1,-1){17.6777}}\put(-53,-5){\sf{A}}\put(-3,-5){\sf{P}}\put(18,20){\sf{B}}\put(18,-22){\sf{C}}\put(-25,-5){\sf{2L}}\put(6,10){\sf{L}}\put(6,-12){\sf{L}}\multiput(-50,0)(50,0){2}{\circle*{1}}\multiput(17.6777,17.6777)(0,-35.3554){2}{\circle*{1}}\end{picture}$

which can be arranged as an electrical circuit where,

• resistance of rod is taken as resistance of current carrying wire.

• thermal conductivity of the rod is taken as electrical conductivity of the wire.

• heat current is taken as electrical current.

• temperature difference is taken as potential difference.

like,

$\setlength{\unitlength}{1mm}\begin{picture}\thicklines\put(-50,0){\line(1,0){15}}\put(-15,0){\line(1,0){15}}\put(0,0){\line(1,1){5}}\qbezier(15,15)(16.33885,16.33885)(17.6777,17.6777)\put(0,0){\line(1,-1){5}}\qbezier(15,-15)(16.33885,-16.33885)(17.6777,-17.6777)\put(-53,-5){\sf{A}}\put(-3,-5){\sf{P}}\put(18,20){\sf{B}}\put(18,-22){\sf{C}}\put(-27,-5){\sf{2L}}\put(6,13){\sf{L}}\put(6,-14){\sf{L}}\multiput(-50,0)(50,0){2}{\circle*{1}}\multiput(17.6777,17.6777)(0,-35.3554){2}{\circle*{1}}\multiput(-35,0)(4,0){5}{\qbezier(0,0)(1,1)(2,2)\qbezier(2,2)(3,1)(4,0)}\multiput(5,5)(2,2){5}{\qbezier(0,0)(0,1)(0,2)\qbezier(0,2)(1,2),(2,2)}\multiput(5,-5)(2,-2){5}{\qbezier(0,0)(0,-1)(0,-2)\qbezier(0,-2)(1,-2),(2,-2)}\put(-5,0){\vector(1,0){0}}\put(4,4){\vector(1,1){0}}\put(4,-4){\vector(1,-1){0}}\put(-8,3){ \sf{i_1} }\put(0,5){ \sf{i_2} }\put(5,-3){ \sf{i_3} }\end{picture}$

Now the potential at point,

• A, $\sf{V_A=100\ V}$

• B, $\sf{V_B=50\ V}$

• C, $\sf{V_C=0\ V}$

Here the three wires having same cross sectional area $\sf{A}$ are of same material so they have same resistivity $\sf{\rho.}$

We know the resistance of a wire is given by,

• $\sf{R=\dfrac{\rho L}{A}}$

The resistance of the wire AP,

• $\sf{R_1=\dfrac{\rho(2L)}{A}=\dfrac{2\rho L}{A}}$

The resistance of the wire PB,

• $\sf{R_2=\dfrac{\rho L}{A}}$

The resistance of the wire PC,

• $\sf{R_3=\dfrac{\rho L}{A}}$

By Ohm's Law, the current through the wire AP,

$\sf{\longrightarrow i_1=\dfrac{V_P-V_A}{R_1}}$

$\sf{\longrightarrow i_1=\dfrac{(V_P-100)A}{2\rho L}}$

The current through the wire PB,

$\sf{\longrightarrow i_2=\dfrac{V_B-V_P}{R_2}}$

$\sf{\longrightarrow i_2=\dfrac{(50-V_P)A}{\rho L}}$

The current through the wire PC,

$\sf{\longrightarrow i_3=\dfrac{V_C-V_P}{R_3}}$

$\sf{\longrightarrow i_3=-\dfrac{V_PA}{\rho L}}$

Applying Junction Rule at the junction P,

$\sf{\longrightarrow i_1=i_2+i_3}$

$\sf{\longrightarrow\dfrac{(V_P-100)A}{2\rho L}=\dfrac{(50-V_P)A}{\rho L}-\dfrac{V_PA}{\rho L}}$

$\sf{\longrightarrow\dfrac{V_P-100}{2}=50-2V_P}$

$\sf{\longrightarrow V_P=40\ V}$

So in the actual question, the temperature at the junction P will be,

$\sf{\longrightarrow\underline{\underline{T_P=40^oC}}}$

Hence (B) is the answer.