question_answer

Two rods of same material have same length and area. The heat\[\Delta Q\]flows through them for 12 min when they are joint side by side. If now both the rods are joined in parallel, then the same amount of heat\[\Delta Q\]will flow in:

A)  24 min                                 

B)  3 min

C)  12 min                                 

D)  6 min

Correct Answer: B

Solution :

                  When two rods are joined, then the rate of flow of heat is given by \[Q=KA\frac{({{\theta }_{1}}-{{\theta }_{2}})}{l}t\] where K is coefficient of thermal conductivity, A is area and I is length when rods are joined in series. \[\Delta {{Q}_{1}}=\frac{A({{T}_{1}}-{{T}_{2}}){{t}_{1}}}{\frac{{{l}_{1}}}{{{K}_{1}}}+\frac{{{l}_{2}}}{{{K}_{2}}}}\]    Given, \[{{l}_{1}}={{l}_{2}}=l,{{K}_{1}}={{K}_{2}}=K,\]we have                 \[\Delta {{Q}_{1}}=\frac{A({{T}_{1}}-{{T}_{2}}){{t}_{1}}}{\frac{l}{{{K}_{1}}}+\frac{l}{{{K}_{2}}}}\]                 \[=\frac{A({{T}_{1}}-{{T}_{2}}){{t}_{1}}}{l}\frac{K}{2}\] when rods are joined in parallel                 \[\Delta {{Q}_{2}}=({{K}_{1}}A+{{K}_{2}}A)\frac{({{T}_{1}}-{{T}_{2}}){{t}_{2}}}{l}\]                 \[=2\frac{KA({{T}_{1}}-{{T}_{2}}){{t}_{2}}}{l}\] Given, \[\Delta {{Q}_{1}}=\Delta {{Q}_{2}}\] \[\therefore \]  \[{{t}_{2}}=\frac{{{t}_{1}}}{4}=\frac{12}{4}=3\,\min \]