When a freely suspended bar magnet is heated, its magnetic dipole moment decreases by 36 %. Then its periodic time would

Options

A. increase by 36 %
B. increase by 25 %
C. decrease by 25 %
D. decrease by 64 %

Answer

B

Explanation

Initial periodic time of oscillation,
` T= 2pi sqrt( (I)/( mB_h))`
(Where `B_h=` horizontal component of Earth.s magnetic field)
Here I and `B_h` remain constant and so,
`T prop (1)/( sqrt(m))`
` therefore (T_1)/( T_2) = sqrt((m_2)/( m_1))`
`therefore (T_1)/( T_2) = sqrt((64)/( 1000)) (because ` If `m_1 =100` unit, then `m_2=100-36 =64` unit)
`therefore (T_1)/( T_2) = (8)/(10)`
`therefore T_2 = (10)/(8) T_(1) = 1.25 T_(1)`
Percentage increase in periodic time,
`(T_2 - T_1)/( T_1) xx 100% = (0.25 T_1)/( T_1) xx 100% = 25%`