A) \[5\pi \]
B) \[10\pi \]
C) \[100\pi \]
D) \[50\pi \]
Correct Answer: B
Solution :
Let \[A=\pi {{r}^{2}}\] be area of metalic circular plate of \[r=50\] cm. Also, given\[\frac{dr}{dt}=1mm=\frac{1}{10}cm\] \[\therefore \]\[A=\pi {{r}^{2}}\] \[\Rightarrow \]\[\frac{dA}{dt}=2\pi r\frac{dr}{dt}=2\pi .50.\frac{1}{10}=10\pi \] Hence, area of plate increases in\[10\pi c{{m}^{2}}/\]hour.You need to login to perform this action.
You will be redirected in
3 sec