A) the rate of increase in internal energy is\[\frac{5}{2}R(\alpha +\beta t)\]
B) the current flowing in the element is \[\sqrt{\frac{5}{2r}R(\alpha +\beta t)}\]
C) the piston moves upwards with constant acceleration
D) the piston moves upwards with constant speed
Correct Answer: A
Solution :
We know that Internal energy, \[U=\frac{nfRT}{2}=\frac{5R}{2}\left( \alpha t+\frac{1}{2}\beta {{t}^{2}} \right)\] Differentiate with respect to t \[\frac{dU}{dt}=\frac{5R}{2}[\alpha +\beta t]\] But \[dQ=n{{C}_{p}}dT\] \[\therefore \] \[\frac{dQ}{dt}=n{{C}_{p}}\times \frac{dT}{dt}\Rightarrow {{i}^{2}}r=\frac{7}{2}R\times [\alpha +\beta t]\] \[i=\sqrt{\frac{7R}{2r}(\alpha +\beta t)}\] \[pV=nRT\] or \[V=\frac{nRT}{p}=\frac{nR}{p}\left( \alpha t+\frac{1}{2}\beta {{t}^{2}} \right)\] \[X=\frac{nR}{pA}\left( \alpha t+\frac{1}{2}\beta {{t}^{2}} \right)\] \[v=\frac{nR}{pA}(\alpha +\beta t)\] and acceleration \[=\frac{nR}{pA}\times \beta \] So, the rate of increase in internal energy is \[\frac{5}{2}R(\alpha +\beta t)\]and the piston moves upwards with constant acceleration.You need to login to perform this action.
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