Question

the diameter of a roller is 42cm and its length is 120 cm. it takes 500 revolutions to move once to land a playground. find the area of the ground

Answer 1

Diameter=42cm

Radius=42÷2=21

Area of roller=CSA of cyclinder

=2πrh

=2×22÷7×21×120

=15840

Area of 1 revolution=15840cm

Area for 500 revolution=15840×500

=7920000cm

Answer 2

Given,

Diameter of a roller = 42 cm

Length of the roller = 120 cm

Number of revolutions to move once to land a playground = 500

To find,

Area of the ground

Solution,

We know that radius is half the diameter.

So, $r = \frac{d}{2}$

$r = \frac{42}{2} = 21$

Radius of the roller = 21 cm

Area swept by the roller in one revolution = Curved surface area of the roller

Roller is in cylindrical shape, so consider curved surface area of the cylinder.

Curved surface area of the cylinder = 2πrh

Height is nothing but the length of the roller.

2πrh=      $2*\frac{22}{7}*21*120 = 15840cm^{2}$

Therefore , area of 1 revolution=15840 cm.

Area of 500 revolutions = $15840 * 500 =7920000 cm^{2}$

Area of 500 revolutions = area of the ground, as it takes 500 revolutions to move the whole land in the ground for once.

Hence, the area of the ground is $7920000 cm^{2}$ .

 

=7920000cm