The perpendicular height of a conical tent is | vedclass

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

Question

(B) Given: Perpendicular height $h = 4 \frac{2}{3} 	ext{ m} = \frac{14}{3} 	ext{ m}$.Base diameter $d = 6 	ext{ m}$,so radius $r = \frac{d}{2} = 3

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given: Perpendicular height $h = 4 \frac{2}{3} 	ext{ m} = \frac{14}{3} 	ext{ m}$.Base diameter $d = 6 	ext{ m}$,so radius $r = \frac{d}{2} = 3 	ext{ m}$.The volume of a conical tent is given by $V = \frac{1}{3} \pi r^2 h$.Substituting the values: $V = \frac{1}{3} 	imes \frac{22}{7} 	imes (3)^2 	imes \frac{14}{3}$.$V = \frac{1}{3} 	imes \frac{22}{7} 	imes 9 	imes \frac{14}{3} = 22 	imes 2 = 44 	ext{ m}^3$.Since $11$ persons sleep in the tent,the average volume of air per person is $\frac{	ext{Total Volume}}{	ext{Number of persons}} = \frac{44}{11} = 4 	ext{ m}^3$.

The perpendicular height of a conical tent is $4 \frac{2}{3} \text{ m}$ and the diameter of its base is $6 \text{ m}$. If $11$ persons can sleep in this tent,find how many average cubic meters $(m^3)$ of air each person gets?

Similar Questions

The volumes of two cubes are in the ratio of $8: 125$. Find the ratio of their edges and surface areas.

$A$ cylindrical tower is $5 \, m$ in diameter and $14 \, m$ high. The cost of white washing its curved surface at $50$ paise per $m^{2}$ is (in $Rs.$)

The length of a room is $12\, m$,width $8\, m$,and height $6\, m$. How many boxes will it hold if each box occupies $1.5\, m^3$ of space?

The radius of the base of a solid cylinder is $r \text{ cm}$ and its height is $3 \text{ cm}$. It is recast into a cone of the same radius. The height of the cone will be (in $\text{cm}$):

$A$ rectangular tin sheet is $12 \, cm$ long and $5 \, cm$ broad. It is rolled along its length to form a cylinder by making the opposite edges just touch each other. Then the volume of the cylinder is

Vedclass Test Series

Exam Paper Generator

Online Exam Module