A) 5%
B) 17%
C) 20%
D) 23%
Correct Answer: D
Solution :
[d] Volume of wire = \[\pi {{r}^{2}}h\] New radius of the wire = \[\frac{r\times 90}{100}=\frac{9r}{10}\] Let new length of the wire be L. \[\therefore \] Volume of new wire \[=\pi \,\,{{\left( \frac{9r}{10} \right)}^{2}}\times L=\frac{81}{100}\pi {{r}^{2}}L\] According to question, \[\pi {{r}^{2}}h=\frac{81}{100}\pi {{r}^{2}}L\Rightarrow L=\frac{100}{81}h\] Increase in length =\[\frac{100}{81}h-\,h=\frac{19}{81}h\] Percent increase = \[\frac{19/81h}{h}\times 100%=23.46%\] = 23% (approx.)
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