A wheel of radius 1m rolls …
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Sia ? 6 years, 2 months ago
Horizontal distance covered by the wheel in half revolution
= {tex}\pi{/tex}R.
So the displacement of the point which was initially in contact with ground = AA' =
{tex}\begin{array}{l}{\sqrt{(\pi R)^{2}+(2 R)^{2}}} \\ {(A s R=1 m)}\end{array} {/tex}{tex}=R \sqrt{\pi^{2}+4}=\sqrt{\pi^{2}+4}{/tex}
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