JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2019
- Hard
- MCQ
in circular plate of mass \(M\) and radius \(R\) has its density varying as \(p\left( r \right) = {p_0}\,r\) with \(P_0\) as constant and \(r\) is the distance from its center. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is \(I = aMR^2\) . The value of the coefficient \(a\) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\frac{8}{5}\)
Step-by-step Solution
Detailed explanation
\(M = \int\limits_0^R {{\rho _0}r\left( {2\pi rdr} \right) = \frac{{{\rho _0} \times 2\pi \times {R^3}}}{3}} \)
\(\,\,\,\mathop {{I_0}}\limits_{\left( {MOl\,about\,COM} \right)} = \int\limits_0^R {{\rho _0}r} \left( {2\pi rdr} \right) \times {r^2} = \frac{{{\rho _0} \times 2\pi {R^5}}}{5}\)
By parallel axis theorem
\(I = {I_0} + M{R^2}\)
\( = \frac{{{\rho _0} \times 2\pi {R^5}}}{5} + \frac{{{\rho _0} \times 2\pi {R^3}}}{3} \times {R^2} = {\rho _0}2\pi {R^5} \times \frac{8}{{15}}\)
\( = M{R^2} \times \frac{8}{5}\)
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