A uniform circular disc of radius a is taken. A circular portion of radius b has been…
Physics · JEE Main · NTA Exams — Laws of Motion
A uniform circular disc of radius a is taken. A circular portion of radius b has been removed from its as shown in the figure. If the centre of hole is at a distance c from the centre of the disc, the distance \(\mathbf{x}_{2}\) of the centre of mass of the remaining part from the initial centre of mass O is given by :
- \(\frac{\pi b^{2}}{\left(a^{2}-c^{2}\right)}\)
- \(\frac{c b^{2}}{\left(a^{2}-b^{2}\right)}\)
- \(\frac{\pi c^{2}}{\left(a^{2}-b^{2}\right)}\)
- \(\frac{c a^{2}}{\left(c^{2}-b^{2}\right)}\)
Answer
(B) c b^ 2 (a^ 2 -b^ 2 )
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