Question

A wheel with a rubber tyre has an outside diameter of 25 cm. When the radius has been decreased a quarter of a centimeter, the number of revolutions of the wheel in one metre will:
A. Be increased about 2%.
B. Be increased about 1%.
C. Be increased about 20%.
D. Be increased $\dfrac{1}{2}$ %
E. Remain the same

Answer 1

Hint: Find the number of revolutions when the radius is unchanged and find the number of revolutions when the radius is changed. Then find the percentage change in the revolution.

Complete step by step solution:
It is given that the outside diameter of the wheel with the rubber tyre is $25$cm. Thus,
Diameter of the tyre$\left( D \right) = 25$cm
We know that the radius is just half of the diameter. That is,
Radius$\left( r \right) = \dfrac{D}{2}$
Substitute 25 as the value of $D$ in the above formula:
Radius$\left( r \right) = \dfrac{{25}}{2} = 12.5$cm
It can be noticed that the distance travelled by the wheel in one revolution is equal to the circumference of the tyre. That is,
Distance covered in one revolution$ = $ Circumference of the tyre
The circumference of the tyre is given as$2\pi r$ , where $r$ is the radius. So, the distance in one revolution is given as:
Distance in one revolution$ = $ $2\pi r$
Substitute 3.14 as the value of $\pi $ and $12.5$ as the value of $r$ in the above expression:
Distance in one revolution$ = $ $2 \times 3.14 \times 12.5$
Distance in one revolution$ = $ $78.5$cm
Thus, the distance travelled by the tyre in one revolution is $78.5$cm.
The question is asking for the number of revolutions, when the covered distance is one meter, which is equal to$100$cm.
The number of revolution in $100{\text{ cm}} = \dfrac{{100}}{{78.5}} = 1.27$
It means that when the radius of the tyre is $12.5$cm, then there are $1.27$ revolution to cover $100$ cm of distance.
Now, it is given in the problem that the radius is decreased by one-fourth of the centimeter. It means that:
New radius$\left( {{r_1}} \right) = 12.5 - \left( {\dfrac{1}{4}} \right)$
New radius$\left( {{r_1}} \right)$ $ = 12.5 - 0.25 = 12.25$cm
Therefore, the number of revolutions in 100cm with a new radius$ = \dfrac{{100}}{{2\pi {r_1}}}$
Substitute 3.14 as the value of $\pi $ and $12.25$ as the value of ${r_1}$ in the above expression:
Number of revolutions in 100cm with new radius$ = \dfrac{{100}}{{2 \times 3.14 \times 12.25}} = 1.29$cm
It means that when the radius of the tyre is $12.25$cm, then there is a $1.29$ revolution to cover $100$ cm of distance.
Now, we have to find the percentage increase in the number of revolutions. Use the formula:
Percentage increase in number of revolution$ = \dfrac{{{\text{Increase in number of revolution}}}}{{{\text{original number of revolution}}}} \times 100$
Substitute the values:
Percentage Increase in number of revolution $ = \dfrac{{1.29 - 1.27}}{{1.29}} \times 100 \approx 2\% $
Therefore, when the radius has been decreased a quarter of a centimeter then the number of revolution is increased by$2\% $.
Therefore the option A is correct.

Note:It has to be noticed that the circumference of the circle is the linear distance around the circle. We can also say that if the circle is opened and form a straight line then the length of that line is the circumference of the circle. Thus, the distance covered by the tyre in one revolution is equal to the circumference of the circle.