A deflection magnetometer is placed with its arm along the east-west direction and a bar magnet is placed along the arm of the magnetometer. Due to the magnet, the deflection observed is $\mathrm{\theta }$ and the period of oscillation of the needle in the magnetometer is $\mathrm{T}$. When the magnet is removed, the period of oscillation is ${\mathrm{T}}_0$. The relation between $\mathrm{T}$ and ${\mathrm{T}}_0$ is

• Correct answer
  A

  ${\mathrm{T}}^2={\mathrm{T}}_0^2\cos \mathrm{\theta }$
• B

  $\mathrm{T}={\mathrm{T}}_0\cos \mathrm{\theta }$
• C

  $\mathrm{T}=\frac{{\mathrm{T}}_0}{\cos \mathrm{\theta }}$
• D

  ${\mathrm{T}}^2=\frac{{\mathrm{T}}_0^2}{\cos \mathrm{\theta }}$

Hence, the correct answer is option: 'A'.