Question

The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by 19%. By doing this the periodic time of the magnetometer will

• A. Increase by 19%
• B. Increase by 11%
• C. Decrease by 19%
• D. Decrease by 21%

Answer 1

Answer: (B)

Increase by 11%

Answer 2

$T = 2 \pi \sqrt { \frac { I } { M B _ { H } } } \Rightarrow T \propto \frac { 1 } { \sqrt { M } } \Rightarrow \frac { T _ { 1 } } { T _ { 2 } } = \sqrt { \frac { M _ { 2 } } { M _ { 1 } } }$If $M _ { 1 } = 100$then $M _ { 2 } = ( 100 - 19 ) = 81$. So,

$\frac { T _ { 1 } } { T _ { 2 } } = \sqrt { \frac { 81 } { 100 } } = \frac { 9 } { 10 } \Rightarrow T _ { 2 } = \frac { 10 } { 9 } T _ { 1 } = 11 \% T _ { 1 }$