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.) |
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