A) \[{{\text{M}}^{\text{-1}}}{{\text{L}}^{\text{-2}}}{{\text{T}}^{\text{3}}}\text{ }\!\!\theta\!\!\text{ }\]
B) \[{{\text{M}}^{\text{-1}}}{{\text{L}}^{\text{-2}}}{{\text{T}}^{\text{3}}}\text{ }\!\!\theta\!\!\text{ }\]
C) \[\text{M}{{\text{L}}^{\text{-2}}}{{\text{T}}^{2}}\text{ }\!\!\theta\!\!\text{ }\]
D) \[\text{M}{{\text{L}}^{\text{-2}}}{{\text{T}}^{-2}}{{\text{ }\!\!\theta\!\!\text{ }}^{-1}}\]
Correct Answer: A
Solution :
When thermal conductivity is K, thermal resistivity is \[\frac{1}{K}\], then thermal resistance \[R=\frac{1\times L}{KA}\] Now heat conducted is given by \[\frac{{{Q}_{1}}}{t}={{\theta }_{1}}\frac{kA({{\theta }_{1}}-{{\theta }_{2}})}{L}=\frac{{{\theta }_{1}}-{{\theta }_{2}}}{R}\] Therefore, dimensions of R = dimensions of \[\frac{({{\theta }_{1}}-{{\theta }_{2}})t}{Q}\] \[=\frac{\theta T}{M\times ({{L}^{2}}{{T}^{-2}}{{\theta }^{-1}})}={{M}^{-1}}{{L}^{-2}}{{T}^{3}}\theta \]You need to login to perform this action.
You will be redirected in
3 sec