Question

the diameter and length of a roller are 70 cm and 100 cm respectively if it moves over to level a playground of area 4400 m square find the number of revolutions the roller will take to level 2 playground

Answer 1

d=70
r=35
csa=2*22/7*35*100=22000 sqcm
no.of revolution=2*4400*1000/22000=400 revolution ans

Answer 2

Answer: There are 4000 revolutions required to level 2 playground.

Step-by-step explanation:

Since we have given that

Diameter of a roller = 70 cm

Radius of a roller is given by

$\frac{70}{2}=35\ cm$

Length of a roller = 100 cm

As we know the formula for "Curved surface area of cylinder":

$C.S.A=2\pi rh\\\\C.S.A=2\times \frac{22}{7}\times 35\times 100\\\\C.S.A=2\times 22\times 5\times 100\\\\C.S.A=22000\ cm^2$

As we know that

$1\ m=100\ cm\\\\1\ m^2=10000\ cm^2\\\\1\ cm^2=\frac{1}{10000}\ m^2\\\\22000\ cm^2=\frac{22000}{10000}\ m^2=2.2\ m^2$

Area of playground = 4400 m²

As we have mentioned that the roller will take to level 2 playground which is given by

$4400\times 2\\\\=8800\ m^2$

So, Number of revolution the roller will take is given by

$\frac{\text{ Area of playground}}{\text{ Area of roller }}\\\\=\frac{8800}{2.2}\\\\=4000$

Hence, there are 4000 revolutions required to level 2 playground.