A wheel of circumference 2 m rolls on a circular path of radius 80 m. What is the angle made by the wheel about the centre of the path, if it rotates 64 times on its own axis?
[amp_mcq option1=”1.6 rad” option2=”1.4 rad” option3=”1.2 rad” option4=”0.8 rad” correct=”option1″]
This question was previously asked in
UPSC CBI DSP LDCE – 2023
The wheel has a circumference of 2 m. When the wheel rotates once on its own axis, the distance it covers rolling on a surface is equal to its circumference.
The wheel rotates 64 times on its own axis.
Total distance covered by the wheel rolling on the circular path = Number of rotations $\times$ Circumference of the wheel
Total distance = $64 \times 2$ m = 128 m.
This distance is an arc length on the circular path. The formula relating arc length (s), radius (R), and the angle ($\theta$) subtended at the center (in radians) is $s = R \times \theta$.
Here, the arc length $s = 128$ m and the radius of the circular path $R = 80$ m.
So, $128 = 80 \times \theta$.
Solve for $\theta$:
$\theta = \frac{128}{80}$ radians.
Simplify the fraction:
$\theta = \frac{128 \div 16}{80 \div 16} = \frac{8}{5}$ radians.
$\frac{8}{5} = 1.6$ radians.
When a wheel rolls, the distance covered in one rotation is equal to its circumference. The arc length ($s$) of a sector in a circle is given by the formula $s = R\theta$, where $R$ is the radius and $\theta$ is the angle subtended at the center *in radians*.