A) 60
B) 68
C) 73
D) 75
Correct Answer: C
Solution :
Since, volume of cone \[{{V}_{1}}=\frac{4}{3}\pi {{r}^{2}}h\] After increasing by 20% of radius and height, then volume of cone \[{{V}_{2}}=\frac{4}{3}\pi {{\left( 1+\frac{1}{5} \right)}^{2}}r\left( 1+\frac{1}{5} \right)h\] \[=\frac{4}{3}\pi \times \frac{216}{125}{{r}^{2}}h\] \[\therefore \] Required percentage \[=\frac{{{V}_{2}}-{{V}_{1}}}{{{V}_{1}}}\times 100\] \[=\frac{\frac{4}{3}\pi {{r}^{2}}h\left( \frac{216}{125}-1 \right)}{\frac{4}{3}\pi {{r}^{2}}h}\times 100\] \[=\frac{91}{125}\times 100=72.8%\] \[=73%\] (approx..)
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