The time period of oscillation of a bar magnet suspended horizontally along the magnetic meridian is $${T_0}.$$ If this magnet is replaced by another magnet of the same size and pole strength but with double the mass, the new time period will be

Solution :
$$\eqalign{ & {I_1} = \frac{{m{\ell ^2}}}{{12}}\,{\text{and}}\,{I_2} = \frac{{\left( {2m} \right){\ell ^2}}}{{12}} = 2{I_1}. \cr & {\text{Now,}}\,\,\frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{{I_1}}}{{{I_2}}}} \,or\,\frac{{{T_0}}}{{{T_2}}} = \sqrt {\frac{1}{2}} \cr & \therefore {T_2} = \sqrt 2 {T_0}. \cr} $$