A) 0%
B) 25%
C) 62.5%
D) 75%
Correct Answer: C
Solution :
| [c] Let, original radius = r and original height = h Original volume\[=\pi {{r}^{2}}h\] New radius = 50% of r\[=\left( \frac{50}{100}\times r \right)=\frac{r}{2}\] New height = 150% of h \[=\left( \frac{50}{100}\times h \right)=\frac{3h}{2}\] New volume \[=\left\{ \left. \pi \,\times {{\left( \frac{r}{2} \right)}^{2}}\times \frac{3h}{2} \right\} \right.=\frac{3\pi {{r}^{2}}h}{8}\] Volume decreased\[=\left( \pi {{r}^{2}}h-\frac{3\pi {{r}^{2}}h}{8} \right)=\frac{5\pi {{r}^{2}}h}{8}\] Decrease percentage \[=\left( \frac{5\pi {{r}^{2}}h}{8}\times \frac{1}{\pi {{r}^{2}}h}\times 100 \right)%\] \[=\frac{125}{2}%=62.5%\] |
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