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QJEEM19Ph4294: JEE Main 2019 Physics - Class 11 JEEM19Ph4294 Solution: A circular disc of radius b has a hole of radius a at its…

A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc varies as (σ0r)\left( {{{{\sigma _0}} \over r}} \right) , then the radius of gyration of the disc about its axis passing through the centre is: JEE Main 2019 (Online) 12th April Morning Slot Physics - Rotational Motion Question 147 English

Multiple Choice Options

  • A a2+b2+ab2\sqrt {{{{a^2} + {b^2} + ab} \over 2}}
  • B a+b3\sqrt {{{a + b} \over 3}}
  • C a2+b2+ab3\sqrt {{{{a^2} + {b^2} + ab} \over 3}}
  • D a+b2\sqrt {{{a + b} \over 2}}

Step-by-Step Verified Solution

dI = (dm)r2
= ( σ\sigma dA)r2
= (σ0r2πdr)r2=σ02πr2dr\left( {{{{\sigma _0}} \over r}2\pi dr} \right){r^2} = {\sigma _0}2\pi {r^2}dr

I=∫DI=∫abσ02πr2drI = \int {DI} = \int\limits_a^b {{\sigma _0}2\pi {r^2}dr}
=σ02π(b3−a33)= {\sigma _0}2\pi \left( {{{{b^3} - {a^3}} \over 3}} \right)

m = ∫dm=∫σdA=σ02π∫abdr\int {dm = \int {\sigma dA = {\sigma _0}2\pi } } \int\limits_a^b {dr}

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