A bucket is in the form of a frustum of a con | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given, volume of the frustum $= 28.49 \, 	ext{L} = 28.49 	imes 1000 \, 	ext{cm}^3 = 28490 \, 	ext{cm}^3$ (since $1 \, 	ext{L} = 1000 \, 	ext

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) Given, volume of the frustum $= 28.49 \, 	ext{L} = 28.49 	imes 1000 \, 	ext{cm}^3 = 28490 \, 	ext{cm}^3$ (since $1 \, 	ext{L} = 1000 \, 	ext{cm}^3$).Radius of the top $(r_1) = 28 \, 	ext{cm}$.Radius of the bottom $(r_2) = 21 \, 	ext{cm}$.Let the height of the bucket be $h \, 	ext{cm}$.The volume of a frustum of a cone is given by $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$.Substituting the values: $28490 = \frac{1}{3} 	imes \frac{22}{7} 	imes h 	imes (28^2 + 21^2 + 28 	imes 21)$.$28490 = \frac{22}{21} 	imes h 	imes (784 + 441 + 588)$.$28490 = \frac{22}{21} 	imes h 	imes 1813$.$h = \frac{28490 	imes 21}{22 	imes 1813}$.$h = \frac{598290}{39886} = 15 \, 	ext{cm}$.

$A$ bucket is in the form of a frustum of a cone and holds $28.490 \, \text{litres}$ of water. The radii of the top and bottom are $28 \, \text{cm}$ and $21 \, \text{cm}$, respectively. Find the height of the bucket. (in $\text{cm}$)

Similar Questions

The diameters of a frustum of a cone are $32 \,cm$ and $20 \,cm$. If its height is $8 \,cm$,find its curved surface area and volume. $(\pi = 3.14)$

$1\,m^3 = \dots \dots \dots cm^3$.

$16$ glass spheres,each of radius $2 \,cm$,are packed into a cuboidal box of internal dimensions $16 \,cm \times 8 \,cm \times 8 \,cm$. The box is then filled with water. Find the volume of water filled in the box (in $cm^3$).

The volume of a cone with radius $6 \, cm$ and height $21 \, cm$ is $\ldots \ldots \ldots \, cm^{3}$.

The volume of a sphere is $4.5 \pi \, cm^3$. Then,its diameter is $\ldots \ldots \ldots \ldots cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module