The students of a Vidyalaya were asked to par | vedclass

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

Question

(D) The penholder is a cylinder with a base,so its surface area is the sum of the curved surface area and the area of the circular base.Radius $(r) =

Vedclass · Accepted Answer

$7920$

Vedclass · Answer

(D) The penholder is a cylinder with a base,so its surface area is the sum of the curved surface area and the area of the circular base.Radius $(r) = 3 \,cm$Height $(h) = 10.5 \,cm$Surface area of $1$ penholder $= 2\pi rh + \pi r^2 = \pi r(2h + r)$$= \frac{22}{7} 	imes 3 	imes (2 	imes 10.5 + 3) \,cm^2$$= \frac{22}{7} 	imes 3 	imes (21 + 3) \,cm^2$$= \frac{22}{7} 	imes 3 	imes 24 \,cm^2 = \frac{1584}{7} \,cm^2$Total cardboard required for $35$ competitors $= 35 	imes \left(\frac{1584}{7}ight) \,cm^2$$= 5 	imes 1584 \,cm^2 = 7920 \,cm^2$Thus,$7920 \,cm^2$ of cardboard is required.

Similar Questions

Find the volume of a sphere of radius $11.2 \, cm$. (in $, cm^3$)

$A$ dome of a building is in the form of a hemisphere. From inside,it was white-washed at the cost of $Rs. 4989.60$. If the cost of white-washing is $Rs. 20$ per square metre,find the inside surface area of the dome. (in $, m^2$)

The inner diameter of a circular well is $3.5 \, m$. It is $10 \, m$ deep. Find its inner curved surface area. [Assume $\pi = \frac{22}{7}$] (in $, m^2$)

$A$ right circular cylinder just encloses a sphere of radius $r$ (see figure). Find:
$(i)$ Surface area of the sphere,
$(ii)$ Curved surface area of the cylinder,
$(iii)$ Ratio of the areas obtained in $(i)$ and $(ii)$.

$A$ cuboidal water tank is $6 \,m$ long,$5 \,m$ wide and $4.5 \,m$ deep. How many litres of water can it hold (in $,l$)? $(1 \,m^3 = 1000 \,l)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module