If there is no torsion in the suspension thread, then the time period of a magnet executing SHM is

Question

If there is no torsion in the suspension thread, then the time period of a magnet executing SHM is (A) T=(1/2π) √ (MB/I) (B) T=(1/2π) √ (I/MB) (C) T=2π √ (I/MB) (D) T=2π √ (MB/I)

Tardigrade Admin · Accepted Answer

If there is no torsion in the suspension thread, then the time period of a magnet executing SHM is T=2 π √(I/M B) where, I is moment of inertia of short bar magnet, B is Earth's magnetic field and M is magnetic moment.

Q. If there is no torsion in the suspension thread, then the time period of a magnet executing $S H M$ is

$T = \frac{1}{2 π} \frac{MB}{I}$

$T = \frac{1}{2 π} \frac{I}{MB}$

$T = 2 π \frac{I}{MB}$

$T = 2 π \frac{MB}{I}$

If there is no torsion in the suspension thread, then the time period of a magnet executing SHM is $T = 2 π \frac{I}{MB}$ where, $I$ is moment of inertia of short bar magnet, $B$ is Earth's magnetic field and $M$ is magnetic moment.

Q. If there is no torsion in the suspension thread, then the time period of a magnet executing SHM is

T=2π1​IMB​​

T=2π1​MBI​​

T=2πMBI​​

T=2πIMB​​