In covering a distance $s$ metres,a circular wheel of radius $r$ metres makes $\frac{s}{2 \pi r}$ revolutions. Is this statement true? Why?
(A) The statement is true.
The distance covered by a wheel in one complete revolution is equal to its circumference,which is given by the formula $2 \pi r$ metres.
To find the total number of revolutions made in covering a distance of $s$ metres,we divide the total distance by the distance covered in one revolution:
$\text{Number of revolutions} = \frac{\text{Total distance}}{\text{Distance in one revolution}} = \frac{s}{2 \pi r}$.
The distance covered by a wheel in one complete revolution is equal to its circumference,which is given by the formula $2 \pi r$ metres.
To find the total number of revolutions made in covering a distance of $s$ metres,we divide the total distance by the distance covered in one revolution:
$\text{Number of revolutions} = \frac{\text{Total distance}}{\text{Distance in one revolution}} = \frac{s}{2 \pi r}$.
Explore More
Similar Questions
With the vertices $A, B$ and $C$ of a triangle $ABC$ as centres,arcs are drawn with radii $5 \, cm$ each as shown in the figure. If $AB = 14 \, cm, BC = 48 \, cm$ and $CA = 50 \, cm$,then find the area of the shaded region. (Use $\pi = 3.14$) (in $cm^2$)
Difficult
View SolutionThe length of the diagonals of a square garden $ABCD$ is $120\, m$. As shown in the figure,there are flower beds on two opposite sides of the garden in the shape of minor segments,the center of which is the point of intersection of the diagonals. Find the area of these flower beds. $(\pi=3.14)$ (in $m^2$)
Difficult
View SolutionThe radius of a circular ground is $35 \, m$. Inside it,a $3.5 \, m$ broad road runs around its boundary. $A$ part of the road between two radii forming an angle of measure $72^{\circ}$ at the centre is to be repaired. Find the cost of repairing at the rate of ₹ $80 / m^{2}$. (in ₹)
Difficult
View SolutionIf the sum of the circumferences of two circles with radii $R_{1}$ and $R_{2}$ is equal to the circumference of a circle of radius $R$,then
Easy
View SolutionIn the given diagram,$\Delta ABC$ is a right-angled triangle in which $m \angle B = 90^{\circ}$ and $AB = BC = 14 \text{ cm}$. $A$ minor sector $BAPC$ is a sector of a circle with center $B$ and radius $BA$. $A$ semicircle arc $\widehat{AQC}$ is drawn on diameter $\overline{AC}$. Find the area of the shaded region in $\text{cm}^2$.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo