The diameters of the front and the rear wheels of a tractor are 63 cm and 1.54 m respectively. The rear wheel is rotating at $24\dfrac{6}{11}$ revolutions per minute. Find :

(i) the revolutions per minute made by the front wheel.

(ii) the distance travelled by the tractor in 40 minutes.

(i) Given:

Diameter of the rear wheel = 1.54 m

Diameter of the front wheel = 0.63 m

Radius of the rear wheel = $\dfrac{d}{2}$ = $\dfrac{1.54}{2}$ = 0.77 m

Radius of the front wheel = $\dfrac{d}{2}$ = $\dfrac{0.63}{2}$ = 0.315 m

Distance travelled by the tractor in 1 revolution of the rear wheel = Circumference of the rear wheel

= 2πr

$= 2 \times \dfrac{22}{7} \times 0.77\\[1em] = \dfrac{44}{7} \times 0.77\\[1em] = 44 \times 0.11\\[1em] = 4.84 \text{ m}$

The number of revolutions per minute for the rear wheel is = $24\dfrac{6}{11}$ = $\dfrac{270}{11}$

So, the distance traveled by the rear wheel in one minute is = 4.84 x $\dfrac{270}{11}$

= 0.44 x 270

= 118.8 m

Let x be the number of revolutions made by the front wheel.

The total distance travelled by tractor in 1 min = Number of revolutions made by the front wheel in 1 min x Circumference of wheel

$⇒ x \times 2 \times \dfrac{22}{7} \times 0.315 = 118.8\\[1em] ⇒ x \times 44 \times 0.045 = 118.8\\[1em] ⇒ x \times 1.98 = 118.8\\[1em] ⇒ x = \dfrac{118.8}{1.98}\\[1em] ⇒ x = 60$

Hence, the number of revolutions made by the front wheel is 60.

(ii) Distance travelled by the tractor in 40 minutes = Number of revolutions made by the rear wheel in 40 min x Circumference of the rear wheel

$= \dfrac{270}{11} \times 40 \times 4.84 \\[1em] = 270 \times 40 \times 0.44 \\[1em] = 4,752 \text{ m}$

Hence, the distance travelled by the tractor in 40 minutes is 4,752 m = 4.752 km.