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Question: The time period of oscillation of a bar magnet suspended horizontally along the magnetic meridian is...

The time period of oscillation of a bar magnet suspended horizontally along the magnetic meridian is T0. 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

A

T02\frac { T _ { 0 } } { 2 }

B

T02\frac { T _ { 0 } } { \sqrt { 2 } }

C

2T0\sqrt { 2 } T _ { 0 }

D

2T02 T _ { 0 }

Answer

2T0\sqrt { 2 } T _ { 0 }

Explanation

Solution

T=2πIMBH⇒T∝I∝w⇒T′=2T0T = 2 \pi \sqrt { \frac { I } { M B _ { H } } } \Rightarrow T \propto \sqrt { I } \propto \sqrt { \mathrm { w } } \Rightarrow T ^ { \prime } = \sqrt { 2 } T _ { 0 }