CSA of a cylindrical tank with diameter 2.1 , | vedclass

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

Question

(D) Given: Diameter of the cylindrical tank $= 2.1 \, m$.Radius $r = \frac{	ext{diameter}}{2} = \frac{2.1}{2} = 1.05 \, m$ or $\frac{21}{20} \, m$.He

Vedclass · Accepted Answer

$33$

Vedclass · Answer

(D) Given: Diameter of the cylindrical tank $= 2.1 \, m$.Radius $r = \frac{	ext{diameter}}{2} = \frac{2.1}{2} = 1.05 \, m$ or $\frac{21}{20} \, m$.Height $h = 5 \, m$.The formula for the Curved Surface Area $(CSA)$ of a cylinder is $2 \pi r h$.Substituting the values: $CSA = 2 	imes \frac{22}{7} 	imes \frac{21}{20} 	imes 5$.$CSA = 2 	imes \frac{22}{7} 	imes \frac{21}{20} 	imes 5 = 2 	imes 22 	imes \frac{3}{20} 	imes 5 = 44 	imes \frac{15}{20} = 44 	imes 0.75 = 33 \, m^2$.Therefore,the $CSA$ is $33 \, m^2$.

$CSA$ of a cylindrical tank with diameter $2.1 \, m$ and height $5 \, m$ is $\ldots \ldots \ldots \ldots \, m^2$.

Similar Questions

The radii of a frustum of a cone are $6 \, cm$ and $4 \, cm$. If its height is $9 \, cm$,its volume is $\ldots \ldots \ldots \pi \, cm^3$.

Curved surface area of a hemisphere $= \ldots \ldots \ldots . .$

Total surface area of a cone exceeds its curved surface area by $\ldots \ldots \ldots . . .$

$A$ cylinder is closed at both the ends by $30 \, cm$ high cones. The radius of the cylinder is $21 \, cm$ and the total height of the solid is $140 \, cm$. Find the volume of this solid (in $cm^3$).

$A$ metallic sphere with radius $6 \, cm$ is melted and recast into a wire of radius $0.4 \, cm$. Find the length of the wire in metres.

Vedclass Test Series

Exam Paper Generator

Online Exam Module