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A circular plate of radius \frac{R}{2} is cut from one edge of thin circular plate of radius R. The moment of inertia of remaining portion about an axis through ‘O’ and perpendicular to plane of plate is

Option: 1

\frac{11 MR^{2}}{24}


Option: 2

\frac{7 MR^{2}}{12}


Option: 3

\frac{13 MR^{2}}{32}

 


Option: 4

\frac{15 MR^{2}}{7}


Answers (1)

best_answer

Answer (3)

I_{remain}=I_{main}-I_{cut}

             =\frac{1}{2}MR^{2}-\left \{ \frac{1}{2}M'\frac{R^{2}}{4}+M'\frac{R^{2}}{4} \right \}

            =\frac{1}{2}MR^{2}-\frac{3}{2}\frac{M}{4}\frac{R^{2}}{4}=\frac{13 MR^{2}}{32}

Posted by

SANGALDEEP SINGH

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