Question 287 of 308

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$2R$B$R\sqrt{\left({\pi }^{2}+4\right)}$C$R\sqrt{\left({\pi }^{2}+1\right)}$D$2R\sqrt{\left(1+4{\pi }^{2}\right)}$

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Answer: (B) $R\sqrt{\left({\pi }^{2}+4\right)}$

The displacement is the hypotenuse of the diameter and the distance travelled for half the rotation $\pi R$.
By using Pythagoras theorem,
Displacement of $A=\sqrt{(2R{)}^{2}+(\pi R{)}^{2}}$
$=R\sqrt{{\pi }^{2}+4}$