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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 \left ( \frac{\sigma 0}{r} \right ), then the radius of gyration of the disc about its axis passing through the centre is :

  • Option 1)

    \frac{a+b}{2}

  • Option 2)

     

    \sqrt{\frac{a^{2}+b^{2}+ab}{3}}

  • Option 3)

    \sqrt{\frac{a^{2}+b^{2}+ab}{2}}

  • Option 4)

    \frac{a+b}{3}

Answers (1)

best_answer

 

 I=\int dm r^{2}

\int_{a}^{b}\frac{\sigma_0}{r}2\pi r dr.r^2=\sigma_0 2\pi [\frac{b^3-a^3}{3}]

I=mK^2 \\m=\int_{a}^{b}\frac{\sigma_0}{r}2\pi r dr=2\pi \sigma_0 (b-a) \\\therefore K=\sqrt{\frac{b^3-a^3}{3(b-a)}}=\sqrt{\frac{a^2+b^2+ab}{3}}

 


Option 1)

\frac{a+b}{2}

Option 2)

 

\sqrt{\frac{a^{2}+b^{2}+ab}{3}}

Option 3)

\sqrt{\frac{a^{2}+b^{2}+ab}{2}}

Option 4)

\frac{a+b}{3}

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