A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is T', the ratio (T' /T) is

Question

A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is T', the ratio (T' /T) is (A) (1/2√2) (B) 1/2 (C) 2 (D) 1/4

Tardigrade Admin · Accepted Answer

When the magnet is divided into 2 equal parts, the magnetic dipole movement
M′ = pole strength × length

= \frac{M}{2}

and moment of inertia

I^{'} = \frac{1}{12} \times ma ss (l e n g t h)^{2}

= \frac{1}{12} \times \frac{m}{2} (\frac{ℓ}{2})^{2}

\Rightarrow I^{'} = \frac{I}{8}

Time period

= 2 π I \frac{^{'}}{M ^{'} B} = 2 π \frac{I /8}{\frac{M}{2} B}

T^{'} = \frac{T}{2}

\Rightarrow \frac{T ^{'}}{T} = \frac{1}{2}

$\frac{1}{2 2}$

$1/2$

$2$

$1/4$

Q. A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is T′, the ratio TT′​ is

22​1​

1/2

2

1/4

When the magnet is divided into 2 equal parts, the magnetic dipole movement M′ = pole strength × length =2M​ and moment of inertia I′=121​×mass(length)2 =121​×2m​(2ℓ​)2 ⇒I′=8I​ Time period =2πIM′B′​​=2π2M​BI/8​​ T′=2T​ ⇒TT′​=21​

$\frac{1}{2 2}$

$1/2$

$2$

$1/4$