Question

The diameter and length of a roller are $120\text{cm}$ and $84\text{cm}$ respectively. To level the ground, $200$ rotations of the roller are required. Find the expenditure to level the ground at the rate of $\text{Rs.}10$ per sq. m.

Answer 1

Radius of the roller, $r=\frac{120}{2}=60\text{cm}$
Length of the roller, $h=84\text{cm}$

Area of the ground levelled in one rotation of the roller is equal to curved surface area of the roller.

$C.S.A=2\pi rh$

$=2\times \frac{22}{7}\times 60\times 84$

$=31680{\text{cm}}^2$

Area of the ground levelled in $200$ rotations of the roller is,
$200\times$ Area of the ground levelled in one rotation of the roller
$\Rightarrow 200\times 31680$
$\Rightarrow 6336000{\text{cm}}^2$

$\Rightarrow \frac{6336000}{10000}$ $(1{\text{m}}^2=10000{\text{cm}}^2)$
$\Rightarrow 633.6{\text{m}}^2$

Rate to level the ground $=\text{Rs.}10/{\text{m}}^2$
Expenditure (or total cost) to level the ground is equal to the total area of the ground multiplied by the rate to level the ground. So,
$\Rightarrow 633.6\times 10$
$\Rightarrow \text{Rs.}6,336$

Thus, the expenditure to level the ground is $\text{Rs.}6,336$.