A)
B)
C)
D)
Correct Answer: B
Solution :
\[\varphi =BA\]\[=B\times \pi {{r}^{2}}\] \[\therefore \varphi \propto {{r}^{2}}\Rightarrow \varphi =k{{r}^{2}}\] (k = constant) \[\therefore e=\frac{d\varphi }{dt}=k.2r\frac{dr}{dt}\] From 0 ? 1, r is constant, \\[\frac{dr}{dt}=0\] hence, e = 0 From 1 ? 2, \[r=\alpha t,\]\\[\frac{dr}{dt}=\alpha \] hence \[e\propto r\] Þ \[e\propto t\] From 2 ? 3, again r is constant, \\[\frac{dr}{dt}=0\]hence e = 0
You need to login to perform this action.
You will be redirected in
3 sec
