A) 24 min
B) 3 min
C) 12 min
D) 6 min
Correct Answer: B
Solution :
When rods are joined in series then heat flow \[\Delta {{Q}_{1}}=\frac{A({{T}_{1}}-{{T}_{2}}){{t}_{1}}}{\frac{{{l}_{1}}}{{{K}_{1}}}+\frac{{{l}_{1}}}{{{K}_{2}}}}\] \[=\frac{A({{T}_{1}}-{{T}_{2}}){{t}_{1}}}{\frac{l}{{{K}_{1}}}+\frac{l}{{{K}_{2}}}}=\frac{A({{T}_{1}}-{{T}_{2}}){{t}_{1}}}{l}=\frac{K}{2}\] When rods are joined in parallel \[\Delta {{Q}_{2}}=({{K}_{1}}A+{{K}_{2}}A)\frac{({{T}_{1}}-{{T}_{2}}){{t}_{2}}}{l}\] \[=\frac{2KA({{T}_{1}}-{{T}_{2}}){{t}_{2}}}{l}\] \[\because \] \[\Delta {{Q}_{1}}=\Delta {{Q}_{2}}\] (given) \[\therefore \] \[{{t}_{2}}=\frac{{{t}_{1}}}{4}=\frac{12}{4}=3\min \]You need to login to perform this action.
You will be redirected in
3 sec