question_answer

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 :                                                   [JEE Main Held on 12-4-2019 Morning]

A) \[\frac{a+b}{2}\]                  

B) \[\frac{a+b}{3}\]

C) \[\sqrt{\frac{{{a}^{2}}+{{b}^{2}}+ab}{2}}\] 

D) \[\sqrt{\frac{{{a}^{2}}+{{b}^{2}}+ab}{3}}\]

Correct Answer: B

Solution :

\[dI=(dm){{r}^{2}}\] \[=(\sigma dA){{r}^{2}}\] \[=\left( \frac{{{\sigma }_{0}}}{r}2\pi rdr \right){{r}^{2}}\] \[=({{\sigma }_{0}}2\pi ){{r}^{2}}dr\] \[I=\int_{{}}^{{}}{dI}=\int\limits_{a}^{b}{{{\sigma }_{0}}2\pi {{r}^{2}}dr}\] \[={{\sigma }_{0}}2\pi \left( \frac{{{b}^{3}}-{{a}^{3}}}{3} \right)\] \[m=\int_{{}}^{{}}{dm}=\int_{{}}^{{}}{\sigma dA}\] \[={{\sigma }_{0}}2\pi \int\limits_{a}^{b}{dr}\]\[m={{\sigma }_{0}}2\pi \,(b-a)\] Radius of gyration \[k=\sqrt{\frac{I}{m}}=\sqrt{\frac{({{b}^{3}}-{{a}^{3}})}{3(b-a)}}\] \[=\sqrt{\left( \frac{{{a}^{3}}+{{b}^{3}}+ab}{3} \right)}\]