The capacity of a tank,in the form of a cylin | vedclass

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

Question

(A) The radius of the base of the cylinder is $r = \frac{28}{2} = 14 \, m$.Let $h$ be the depth (height) of the tank.The capacity (volume) of the tank

Vedclass · Accepted Answer

$2464$

Vedclass · Answer

(A) The radius of the base of the cylinder is $r = \frac{28}{2} = 14 \, m$.Let $h$ be the depth (height) of the tank.The capacity (volume) of the tank is given by $V = \pi r^2 h = 6160 \, m^3$.Substituting the values: $\frac{22}{7} 	imes 14 	imes 14 	imes h = 6160$.$22 	imes 2 	imes 14 	imes h = 6160 \implies 616 	imes h = 6160 \implies h = 10 \, m$.The inner curved surface area is $A = 2 \pi r h = 2 	imes \frac{22}{7} 	imes 14 	imes 10 = 880 \, m^2$.The cost of painting the inner curved surface at the rate of ₹ $2.8$ per $m^2$ is $880 	imes 2.8 = ₹ 2464$.

The capacity of a tank,in the form of a cylinder,is $6160 \, m^3$. If the diameter of its base is $28 \, m$,find out the cost of painting its inner curved surface at the rate of ₹ $2.8$ per $m^2$. (Use $\pi = \frac{22}{7}$)

Similar Questions

$A$ $1 \, m^3$ of metal weighing $90 \, Kg$ is rolled into a square bar $9 \, m$ long. An exact cube is cut off from the bar. How much does it weigh? (in $kg$)

Find the curved surface area (in $cm^{2}$) of a hemisphere of diameter $28\, cm$.

The radius and height of a right circular cone are in the ratio $3: 4$. If its volume is $96\pi\, cm^3$,what is its slant height (in $cm$)?

The base of a right prism is an equilateral triangle. If the lateral surface area and volume are $120 \, cm^2$ and $40 \sqrt{3} \, cm^3$,respectively,then the side of the base of the prism is ...... $cm$.

The base of a right pyramid is an equilateral triangle of side $4 \text{ cm}$. The height of the pyramid is half of its slant height. Its volume is ........ $\text{cm}^3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module