The circular ends of a bucket are of radii 35 cm and 14 cm and the height of the bucket is 40 cm. Its volume is
(a) 60060 cm3
(b) 80080 cm3
(c) 70040 cm3
(d) 80160 cm3
(b) 80080 cm3
Let R and r be the radii of the top and base of the bucket, respectively, and let h be its height.
Then,
$R=35 \mathrm{~cm}, r=14 \mathrm{~cm}, h=40 \mathrm{~cm}$
Volume of the bucket = Volume of the frustum of the cone
$=\frac{1}{3} \pi h\left[R^{2}+r^{2}+R r\right] \mathrm{cm}^{3}$
$=\frac{1}{3} \times \frac{22}{7} \times 40 \times\left[(35)^{2}+(14)^{2}+(35 \times 14)\right] \mathrm{cm}^{3}$
$=\left(\frac{880}{21} \times 1911\right) \mathrm{cm}^{3}$
$=80080 \mathrm{~cm}^{3}$
Hence, the volume of the bucket is $80080 \mathrm{~cm}^{3}$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.