Question

The magnet of vibration magnetometer is heated so as to reduce its magnetic moment by $36\%$. By doing this the periodic time of the magnetometer will

• A. Decreases by $64\%$
• B. Increases by $36\%$
• C. Decreases by $25\%$
• D. Increases by $25\%$

Answer 1

The correct option is D

Increases by $25\%$

Period of oscillation in vibration magnetometer is given as,

$T=2\pi \sqrt{\frac{I}{M{B}_H}}$

$(I=$ moment of inertia of oscillating magnet, 𝑀= magnetic moment, 𝐵_𝐻= horizontal magnetic field of earth )

$T\propto \frac{1}{\sqrt{M}}$

$\frac{T_1}{T_2}=\sqrt{\frac{M_2}{M_1}}$

$M_2=M_1-36\%$ of $M_1$

$=M_1-\frac{36}{100}{M}_1=\frac{64}{100}{M}_1$

$\frac{T_1}{T_2}=\sqrt{\frac{64}{100}}=\frac{8}{10}$

$T_2=1.25T_1$

$\%$ change $=\frac{T_2-T_1}{T_1}\times 100$

$=\frac{1.25T_1-T_1}{T_1}\times 100=25$

Thus, time period of oscillation increased by $25\%$

Hence, the correct option is (b)