MHT CET · Physics · Motion In Two Dimensions
A wheel of radius 1 m rolls through \(180^{\circ}\) over a plane surface. The magnitude of the displacement of the point of the wheel initially in contact with the surface is.
- A \(2 \pi\)
- B \(\pi\)
- C \(\sqrt{\pi^2+4}\)
- D \(3 \pi\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\pi^2+4}\)
Step-by-step Solution
Detailed explanation

Distance travelled by the wheel in half revolution \(=\frac{\mathrm{C}}{2}=\frac{2 \pi \mathrm{r}}{2}=\pi \mathrm{r}\)
Where C is the circumference of the wheel.
\(\therefore\) From figure,
Displacement of initial point of contact after half revolution \(=\mathrm{AB}\)
\(\therefore A B^2=A D^2+\mathrm{DB}^2 \)
\( A B^2=(\pi r)^2+(2 r)^2=r^2\left(\pi^2+4\right) \)
\( \therefore A B=r \sqrt{\left(\pi^2+4\right)} \)
\( \therefore A B=\sqrt{\left(\pi^2+4\right)}\)
...(given \(r=1 \mathrm{~m}\) )
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