Question

A solid is composed of a cylinder with hemispherical ends. If the whole length of the solids is $104$ cm and the radius of each of the hemispherical ends is $7$ cm, find the cost of polishing its surface at the rate of Rs.$10$ per $d{m}^2$.

Answer 1

Solution:

As per the parameters given in the question, we have

Radius of the hemispherical ends (say r) =$7$ cm

Height of the solid =$(h+2r)$ = $104$ cm

The curved surface area of the cylinder (say S) = $2\pi rh$

$S=2\pi \left(7\right)h$ …….1

$\Rightarrow$ $h+2r=104$

$\Rightarrow$ $h+2\times 7=104$

$\therefore h=90cm$

Put the value of $h$ in (1), we will get

$S=2\pi \left(7\right)\left(90\right)$

$S=3958,40$ cm²

So, the curved surface area of the cylinder = $S$ = $3958.40$ cm²

Curved surface area of the two hemisphere (say SA) = $2$ (2 πr²)

$SA$= $2$(2$\times$π$\times$ 7${}^2$)

SA = 615.75 cm²

Therefore, the total curved surface area of the solid = Curved surface area of the cylinder + Curved surface area of the two hemisphere = TSA

$TSA=S+SA$

$TSA=3958.40+615.75$

$TSA$ = $4575.75$ cm² = $45.757$ dm²

The cost of polishing the 1 dm² surface of the solid is Rs. $10$

So, the cost of polishing the $45.757$ dm² surface of the solid = $10\times$45.757= Rs. 457.57

Hence,

The cost of polishing the whole surface of the solid is Rs. $457.57$