MHT CET202020 Oct 2020Morning ShiftPhysicsMagnetic Properties of MatterActual
A torque of 1 732 10⁻⁵ Nm is required to hold a magnet at 90^ with the horizontal component of earth's magnetic field. The torque required to hold it at 60^ will be [ 2 =1, 3 = 3 2 ][ 3 =1 732]
Options
- A1 5 10⁻⁵ Nm
- B1 10⁻⁵ Nm
- C1 732 10⁻⁵ Nm
- D0 5 10⁻⁵ Nm
Correct answer
A. 1 5 10⁻⁵ Nm
Step-by-step solution
array ll ₁=1.732 10⁻⁵ Nm & =90^ ₂=? & =60^ array ₁= BM = BM 90^ = BM ₂= BM 60^ = BM 3 2 =1.732 10⁻⁵ 1.732 2 =1.4999 10⁻⁵=1.5 10⁻⁵ Nm