A disc of radius R=10 cm oscillates as a physical pendulum about an axis perpendicular to the plane of the disc at a distance r from its centre. If r=(R)/(4), the approximate period of oscillation is (Take g=10ms^(-2))

A disc of radius R=10 cm oscillates as a physical pendulum about an ax

1.	A disc of radius R=10 cm oscillates as a physical pendulum about an axis perpendicular to the plane of the disc at a distance r from its centre. If r=(R)/(4), the approximate period of oscillation is (Take g=10ms^(-2))
Answer» 0.84 s 0.94 s 1.26 s 1.42 s Solution : `T=2pisqrt((I)/(mgh))` Where I is the moment of inerrtia of the PENDULUM about an axiis through the pivot, m is the mass of the pendulum and h is the distance from the pivot to the centre of mass. In this case, a SOLID disc of R oscillates as a physical pendulum about an axiis perpendicular to the plane of the disc at a distance r from its centre. `thereforeI=(mR^(2))/(2)+mr^(2)=(mR^(2))/(2)+m((R)/(4))^(2)=(mR^(2))/(2)+(mR^(2))/(16)` `=(9mR^(2))/(16)""(becauser=(R)/(4))` Here, `R=10cm=0.1m,h=(R)/(4)` `thereforeT=2pisqrt(((9mR^(2))/(16))/((mgR)/(4)))=2pisqrt((9R)/(4g)` `=2pisqrt((9xx0.1)/(4xx10))=2pixx(3)/(2)xx(1)/(10)=0.94s`