Question

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

Answer 1

Radius of roller
$= \frac{diameter}{2} \\ = \frac{120}{2} = 60cm$
Area of Roller = curved surface area of cylinder
= 2πrh cm^2
here π = 22/7
r = radius
h = height
$= 2 \times ( \frac{22}{7} ) \times 60 \times 84 \\ = \frac{221760}{7} \\ = 31680 \: {cm}^{2}$
In 200 rotations of roller area covered =
$200 \times 31680 \: {cm}^{2} \\ = 6336000 \: {cm}^{2}$
because in one rotation roller covers 31680 cm^2
In 200 rotations of roller area covered =
$= \frac{6336000}{100 \times 100} \\ = 633.6 \: {m}^{2}$
per sq m expenditure is 10 Rs
so for 633.6 m^2, total expenditure is
$= 633.6 \times 10 \\ = 6336 \: Rs$

Answer 2

Answer:

Given :

$radius = r = \frac{diameter}{2} \\ \: \: \: \: \: \: \: = \frac{120}{2} = 60cm \\ length = height = h = 84cm \\ no.of \: rotating \: = n = 200 \\ rate \: of \: levelling \: = rs \: 10per \: sq.cm \:$

To find :

The expenditure to travel the ground

Formula :

curved surface area of cylinder =2πrh

expenditure =area of travelled *rate of travelling

Area levelled by roller in 1 rotation =curved surface area of roller.

Area of ground levelled by roller in 'N'Djamena rotation =n* C. S. A of roller (cylinder)

=n*2πrh

$n \times 2\pi \: rh \: \\ = 200 \times 2 \times \frac{22}{7} \times 60 \times 84 {cm}^{2} \\ = 400 \times 1320 \times 12 \\ = 6336000 {cm}^{2} \\ \frac{6336000}{100 \times 100} {m}^{2} {1 {cm}^{2} = \frac{1}{1000} {m}^{2} } \\ = 633.6 {m}^{2} \\ expenditure \: = area \: of \: ground \: travelled \: \times rate \: of \: levelling \: \\ = 633.6 \times 10 \\ = 6336 \\ \\ expenditure \: to \: travel \: the \: ground \: is \: 6336.$