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Thermodynamics

# Three conducting rods of same material & area of cross - section are figured as shown . Then , $\;\large\frac{l_{1}}{l_{2}}=?$

$(a)\;\large\frac{1}{3}\qquad(b)\;\large\frac{1}{2}\qquad(c)\;\large\frac{1}{6}\qquad(d)\;1$

Answer : (c) $\;\large\frac{1}{6}$
Explanation :
Let the temperature of point O be
So , $\;\large\frac{(T_{A}-T_{O})}{l_{1}}=\large\frac{(T_{O}-T_{B})}{l_{2}}+\large\frac{(T_{O}-T_{C})}{l_{3}}$
$\large\frac{10}{l_{1}}=\large\frac{50}{l_{2}}+\large\frac{20}{2l_{2}}$
$\large\frac{10}{l_{1}}=\large\frac{50}{l_{2}}+\large\frac{10}{l_{2}}=\large\frac{60}{l_{2}}$
$\large\frac{l_{1}}{l_{2}}=\large\frac{1}{6}$