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QJEEM20Pha5e9: JEE Main 2020 Physics - Class 11 JEEM20Pha5e9 Solution: Mass per unit area of a circular disc of radius depends on the…

Mass per unit area of a circular disc of radius aa depends on the distance r from its centre as σ(r)\sigma \left( r \right) = A + Br . The moment of inertia of the disc about the axis, perpendicular to the plane and assing through its centre is:

Multiple Choice Options

  • A 2πa4(A4+aB5)2\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)
  • B πa4(A4+aB5)\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)
  • C 2πa4(aA4+B5)2\pi {a^4}\left( {{{aA} \over 4} + {B \over 5}} \right)
  • D 2πa4(A4+B5)2\pi {a^4}\left( {{A \over 4} + {B \over 5}} \right)

Step-by-Step Verified Solution

JEE Main 2020 (Online) 7th January Evening Slot Physics - Rotational Motion Question 142 English Explanation

dI = dm(r2)

and dm = σ\sigma 2 π\pi rdr

∴\therefore dI = σ\sigma 2 π\pi rdr(r2) = σ\sigma 2 π\pi r3dr

Given σ(r)\sigma \left( r \right) = A + Br

∴\therefore dI = 2 π\pi (A + Br)r3dr

∫dI=∫0a2πr3(A+Br)dr\int {dI = \int\limits_0^a {2\pi {r^3}\left( {A + Br} \right)dr} }

⇒\Rightarrow I = 2π[Ar44+Br55]2\pi \left[ {{{A{r^4}} \over 4} + {{B{r^5}} \over 5}} \right]

= 2πa4(A4+aB5)2\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)

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