A) 20%
B) 40%
C) 60%
D) 72.8%
Correct Answer: D
Solution :
Let the height and radius of a cone are r and h respectively. \[\therefore \] \[{{V}_{1}}=\frac{1}{3}\pi {{r}^{2}}h\] New radius \[=\frac{120}{100}r=\frac{6r}{5}\] And New height \[=\frac{120}{100}h=\frac{6h}{5}\] \[\therefore \] New Volume, \[{{V}_{2}}=\frac{1}{3}\pi {{\left( \frac{6r}{5} \right)}^{2}}\left( \frac{6h}{5} \right)\] \[=\frac{216}{125}\left( \frac{1}{3}\pi {{r}^{2}}h \right)\] Increase in volume \[=\left( \frac{216}{125}-1 \right)\left( \frac{1}{3}\pi {{r}^{2}}h \right)\] \[=\frac{91}{125}\left( \frac{1}{3}\pi {{r}^{2}}h \right)\] % increase \[=\frac{91}{125}\times \frac{\frac{1}{3}\pi {{r}^{2}}h}{\frac{1}{3}\pi {{r}^{2}}h}\times 100%\] \[=\frac{91\times 100}{125}%\] \[=72.8%\]
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