Verified Solution • 2025–26 Session moderate

QJEEM19Ph7a6b: JEE Main 2019 Physics - Class 11 JEEM19Ph7a6b Solution: A thin circular plate of mass M and radius R has its density…

A thin circular plate of mass M and radius R has its density varying as ρ\rho (r) = ρ\rho 0r with ρ\rho 0 as constant and r is the distance from its centre. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is I = aMR2. The value of the coefficient a is :

Multiple Choice Options

  • A 12{1 \over 2}
  • B 32{3 \over 2}
  • C 85{8 \over 5}
  • D 35{3 \over 5}

Step-by-Step Verified Solution

M=∫0Rρ0r×2πrdr=2πρ0R33M = \int\limits_0^R {{\rho _0}r \times 2\pi rdr = {{2\pi {\rho _0}{R^3}} \over 3}}

IC=∫0Rρ0r×2πrdr×r2=2πρ0R53{I_C} = \int\limits_0^R {{\rho _0}r \times 2\pi rdr \times {r^2} = {{2\pi {\rho _0}{R^5}} \over 3}}

∴\therefore I=IC+MR2=2πρ0R5(13+15)I = {I_C} + M{R^2} = 2\pi {\rho _0}{R^5}\left( {{1 \over 3} + {1 \over 5}} \right)

⇒\Rightarrow 16πρ0R515=85[23πρ0R3]R2=85MR2{{16\pi {\rho _0}{R^5}} \over {15}} = {8 \over 5}\left[ {{2 \over 3}\pi {\rho _0}{R^3}} \right]{R^2} = {8 \over 5}M{R^2}

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