Question

7. A cylindrical garden roller is 75 cm long and its radius is 14 cm. How many revolutions will it make to
level an area of 9,900cm²​

Answer 1

$\boxed {\huge {\underline{ \mathfrak{ \red{Answer:-}}}}}$

The total number of Revolution will be 1.5

${ \large{\underline {\underline { \tt{Given:}}}}}$

Length of roller = 75cm

Radius of roller = 14cm

Garden area = 9900 cm²

So,

Curved surface area of the roller

= 2 × π × r × h

=2 ×22/7 × 14 ×75

=2 × 22 × 2 × 75

=88 ×75

${\large {\boxed{\fbox{= 6600cm²}}}}$

Number of revolutions =9900/6600=1.5