A building is in the form of a cylinder surmo | vedclass

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

Question

(D) Let the radius of the hemispherical dome be $r$ meters and the total height of the building be $h$ meters.Since the base diameter of the dome is e

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) Let the radius of the hemispherical dome be $r$ meters and the total height of the building be $h$ meters.Since the base diameter of the dome is equal to $\frac{2}{3}$ of the total height,therefore $2r = \frac{2}{3}h$,which implies $r = \frac{h}{3}$.Let $H$ be the height of the cylindrical portion. Then $H = h - r = h - \frac{h}{3} = \frac{2}{3}h$ meters.The volume of air inside the building is the sum of the volume of the hemispherical dome and the volume of the cylindrical portion.Volume $= \frac{2}{3}\pi r^3 + \pi r^2 H = \frac{2}{3}\pi (\frac{h}{3})^3 + \pi (\frac{h}{3})^2 (\frac{2}{3}h) = \frac{2}{3}\pi (\frac{h^3}{27}) + \pi (\frac{h^2}{9}) (\frac{2}{3}h) = \frac{2\pi h^3}{81} + \frac{2\pi h^3}{27} = \frac{2\pi h^3 + 6\pi h^3}{81} = \frac{8\pi h^3}{81}$.Given the volume is $67 \frac{1}{21} = \frac{1408}{21} \, m^3$.So,$\frac{8}{81} 	imes \frac{22}{7} 	imes h^3 = \frac{1408}{21}$.$h^3 = \frac{1408}{21} 	imes \frac{81 	imes 7}{8 	imes 22} = \frac{1408}{21} 	imes \frac{567}{176} = 8 	imes 27 = 216$.$h = \sqrt[3]{216} = 6 \, m$.

$A$ building is in the form of a cylinder surmounted by a hemispherical dome (see figure). The base diameter of the dome is equal to $\frac{2}{3}$ of the total height of the building. Find the height of the building,if it contains $67 \frac{1}{21} \, m^3$ of air. (in $m$)

Similar Questions

The $CSA$ (Curved Surface Area) of a hemisphere is equal to $\ldots \ldots \ldots$

$A$ canal is $300 \,cm$ wide and $120 \,cm$ deep. The water in the canal is flowing with a speed of $20 \,km/h$. How much area will it irrigate in $20 \,minutes$ if $8 \,cm$ of standing water is desired? (in $hectares$)

$A$ toy is in the form of a cone surmounted on a hemisphere. The radius of the cone as well as the hemisphere is $3.5\, cm$ and the total height of the toy is $15.5\, cm$. Find its total surface area (in $cm^2$).

$A$ wall $24 \, m$ long,$0.4 \, m$ thick and $6 \, m$ high is constructed with bricks each of dimensions $25 \, cm \times 16 \, cm \times 10 \, cm$. If the mortar occupies $\frac{1}{10}$th of the volume of the wall,then find the number of bricks used in constructing the wall.

The volume of a hemisphere with diameter $1 \, cm$ is $\ldots \ldots \, cm^3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module