$A$ torque of $1.732 \times 10^{-5} \text{ Nm}$ is required to hold a magnet at $90^{\circ}$ with the horizontal component of the Earth's magnetic field. The torque required to hold it at $60^{\circ}$ will be $\left[\sin 90^{\circ}=1, \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right] [\sqrt{3} = 1.732]$

• A
  $1.5 \times 10^{-5} \text{ Nm}$
• B
  $1 \times 10^{-5} \text{ Nm}$
• C
  $1.732 \times 10^{-5} \text{ Nm}$
• D
  $0.5 \times 10^{-5} \text{ Nm}$