JEE Mains · Physics · STD 11 - 2. motion in straight line
A disc is rolling without slipping on a surface. The radius of the disc is \(R\). At \(t=0\), the top most point on the disc is \(A\) as shown in figure. When the disc completes half of its rotation, the displacement of point \(A\) from its initial position is
- A \(R \sqrt{\left(\pi^2+4\right)}\)
- B \(R \sqrt{\left(\pi^2+1\right)}\)
- C \(2 R\)
- D \(2 R \sqrt{\left(1+4 \pi^2\right)}\)
Answer & Solution
Correct Answer
(A) \(R \sqrt{\left(\pi^2+4\right)}\)
Step-by-step Solution
Detailed explanation
Displacement \(=\sqrt{(2 R)^2+(\pi R)^2}=R \sqrt{4+\pi^2}\)
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