A) 5%
B) 17%
C) 20%
D) 23%
Correct Answer: D
Solution :
| \[\therefore \] Volume of wire \[=\pi {{r}^{2}}h\] |
| Now, new radius of the wire \[=\frac{r\times 90}{100}=\frac{9\,r}{10}\] |
| Let new length of the wire be L. |
| \[\therefore \]Volume of new wire \[=\pi {{\left( \frac{9r}{10} \right)}^{2}}\times L=\frac{81}{100}\pi {{r}^{2}}L\] |
| By given condition, \[\pi {{r}^{2}}h=\frac{81}{100}\pi {{r}^{2}}L\] |
| \[\Rightarrow \] \[L=\frac{100}{81}h\] |
| Increases in length \[=\frac{100}{81}h-h=\frac{19}{81}h\] |
| Percentage increase |
| \[=\frac{19/81\,\,h}{h}\times 100%=23.45%\] |
| = 23% (approx.) |
You need to login to perform this action.
You will be redirected in
3 sec
