Verified Solution • 2025–26 Session moderate

QJEEM19Ph0a15: JEE Main 2019 Physics - Class 11 JEEM19Ph0a15 Solution: A thin disc of mass M and radius R has mass per unit area (r) = …

A thin disc of mass M and radius R has mass per unit area σ\sigma (r) = kr2 where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is :

Multiple Choice Options

  • A MR23{{M{R^2}} \over 3}
  • B MR26{{M{R^2}} \over 6}
  • C 2MR23{{2M{R^2}} \over 3}
  • D MR22{{M{R^2}} \over 2}

Step-by-Step Verified Solution

IDisc=∫0R(dm)r2⇒IDisc=∫0R(σ2πrdr)r2{I_{Disc}} = \int\limits_0^R {\left( {dm} \right)} {r^2} \Rightarrow {I_{Disc}} = \int\limits_0^R {\left( {\sigma 2\pi rdr} \right)} {r^2}

IDisc=∫0R(kr22πrdr)r2{I_{Disc}} = \int\limits_0^R {\left( {k{r^2}2\pi rdr} \right)} {r^2}     Mass of Disc

IDisc=2πk∫0Rr2dr    M=∫0R2πrdr kr2{I_{Disc}} = 2\pi k\int\limits_0^R {{r^2}dr} \,\,\,\,M = \int\limits_0^R {2\pi rdr\,k{r^2}}

IDisc=2πk(r66)0R    M=2πk∫0Rr3dr{I_{Disc}} = 2\pi k\left( {{{{r^6}} \over 6}} \right)_0^R\,\,\,\,M = 2\pi k\int\limits_0^R {{r^3}dr}

IDisc=2πkR66    M=2πkr44∣0R{I_{Disc}} = 2\pi k{{{R^6}} \over 6} \,\,\,\,M = 2\pi k\left. {{{{r^4}} \over 4}} \right|_0^R

IDisc=πkR63=(πkR42)R223   M=2πkr44∣0R{I_{Disc}} = {{\pi k{R^6}} \over 3} = \left( {{{\pi k{R^4}} \over 2}} \right){{{R^2}2} \over 3}\,\,\,M = 2\pi k\left. {{{{r^4}} \over 4}} \right|_0^R

IDisc=M2R23;IDisc=23MR2{I_{Disc}} = {{M2{R^2}} \over 3};{I_{Disc}} = {2 \over 3}M{R^2}

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