A) 110 m
B) 55 m
C) 220 m
D) 230 m
Correct Answer: C
Solution :
\[\therefore \]\[\frac{2\pi {{r}_{2}}}{2\pi {{r}_{1}}}=\frac{23}{22}\] \[\Rightarrow \] \[\frac{{{r}_{2}}}{{{r}_{1}}}=\frac{23}{22}\] Let \[{{r}_{1}}=22\,x\,\,m\] and \[{{r}_{2}}=23\,\,x\,\,m\] \[{{r}_{1}}-{{r}_{2}}=5\therefore x=5\,\,m\] \[\therefore \]Inner radius \[=22\times 5=110\,\,m\] \[\therefore \]Required diameter \[=110\times 2=220\,\,m\]You need to login to perform this action.
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