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A uniform quarter circular thin rod of mass M and radius R is pivoted at a point B on the floor. It can rotate freely in the vertical plane through B about a horizontal axis through B.
It is supported by a smooth vertical wall at its other free end A so that it remains at rest. Find the reaction force of wall on the rod.

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$(1 - \frac{2}{π}) M g$

$(1 - \frac{3}{π}) M g$

$(1 - \frac{4}{3 π}) M g$

$(1 - \frac{3}{4 π}) M g$

answer is D.

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Detailed Solution

Let the normal reaction of the wall be N. Note that the force Mg will be acting at COM of the rod which is located at a horizontal distance of $\frac{2 R}{π}$ from the centre
$N \cdot R = (R - \frac{2 R}{π}) Mg N = (1 - \frac{2}{π}) Mg$

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