Question

Two identical rods $AB$ and $CD$ are connected across the end points of a diameter of a ring as shown in the figure. Radius of ring is equal to length of rod $AB$ (of $CD$) and area of cross-section is same everywhere in the assembly. Rods and ring also have same material. If point $A$ maintained at ${100}^{o}C$ and point $D$ is maintained at $0^{o}C$, then temperature of point $C$ will be

• A. ${62}^{o}C$
• B. ${37}^{o}C$
• C. ${28}^{o}C$
• D. ${45}^{o}C$

Answer 1

The correct option is C ${28}^{o}C$

Given that,

Point A maintained $={100}^{0}C$

Point D maintained $=0^{0}C$

Now, according to figure

$\frac{k(100-{\theta }_B)a}{l}=\frac{2k({\theta }_B-{\theta }_C)a}{\pi l}=\frac{k({\theta }_C-0)a}{l}$

$100-{\theta }_B=\frac{2}{\pi }({\theta }_B-{\theta }_C)={\theta }_C$

$2{\theta }_B-2{\theta }_C=\pi {\theta }_C$

${\theta }_B=\left(\frac{\pi +2}{2}\right){\theta }_C={\theta }_C$

$100=[1+\left(\frac{\pi +2}{2}\right)]{\theta }_C$

$100=\frac{\pi +4}{2}\times {\theta }_C$

${\theta }_C=\frac{200}{\pi +4}$

${\theta }_C=28.01$

${\theta }_C={28}^0$

Hence, the temperature of point C will be ${28}^0$