A) 24 s
B) \[3\,s\]
C) \[1.5\,s\]
D) \[48\,s\]
Correct Answer: B
Solution :
When the rods are joined in series (end to end), then equivalent Conductivity is K/2 \[{{Q}_{1}}=\frac{(k/2)A.\Delta \theta {{t}_{1}}}{\ell }\] If the rods are joined in parallel (lengthwise) the equivalent conductivity is 2 k Hence, \[{{Q}_{2}}=\frac{(2.k)A.\Delta \theta {{t}_{2}}}{\ell }\] Now \[{{Q}_{1}}={{Q}_{2}}(Given)\] \[\frac{(k/2)A.\Delta \theta {{t}_{1}}}{\ell }=\frac{(2.k)A.\Delta \theta {{t}_{2}}}{\ell }\] \[\Rightarrow \]\[{{t}_{2}}=\frac{{{t}_{1}}}{4}=\frac{12}{4}=3\,s.\] Hence, the correction option is (b).You need to login to perform this action.
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