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, \[\Delta {{Q}_{1}}=\Delta {{Q}_{2}}\] \[\therefore \] \[{{t}_{2}}=\frac{{{t}_{1}}}{4}=\frac{12}{4}=3\,\min \]You need to login to perform this action.
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