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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

Suppose there is a uniform circular disc of mass M kg and radius r m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by \(\frac{ x }{256} Mr ^2\). The value of x is ___________.
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  1. A 100
  2. B 109
  3. C 128
  4. D 156
Verified Solution

Answer & Solution

Correct Answer

(B) 109

Step-by-step Solution

Detailed explanation

\(M =\sigma \pi R ^2\) \(\sigma \pi R ^2=16 m\) \(m =\frac{\sigma \pi R ^2}{16}\) \(I _{\text {system }}=\frac{ MR ^2}{2}-2\left(\frac{ mR ^2}{2 \times 16}+\frac{9 mR ^2}{16}\right)\) \(=\frac{ MR ^2}{2}-2 \times \frac{19 mR ^2}{32}\) \(=\frac{ MR ^2}{2}-\frac{19}{16} mR ^2\)…
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