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 \frac{T'}{T}  is

  • Option 1)

    \frac{1}{2\sqrt{2}}

  • Option 2)

    \; \; \; \frac{1}{2}\;

  • Option 3)

    \; \; 2\; \;

  • Option 4)

    \; \frac{1}{4}

 

Answers (1)

As we learnt in

Time Period of Freely Suspended Magnet -

T= 2\pi \sqrt{\frac{I}{MB}}

- wherein

I - Moment of Inertia

 

 

 

 

For an oscillating magnet, T=2\pi \sqrt{\frac{I}{MB}}

where\; I=ml^{2}/12,M=xl,x=pole \; strength

When the magnet is divided into 2 equal parts, the magnetic dipole moment

M'=Pole \; strength \times length=\frac{x\times l}{2}=\frac{M}{2}..............(i)

I'=\frac{Mass\times (length)^{2}}{12}

=\frac{(m/2)(l/2)^{2}}{12}=\frac{ml^{2}}{12\times 8}=\frac{I}{8}...............(ii)     

\therefore \: \; Time\; period\; T'=2\pi \sqrt{\frac{I'}{M'B}}

\therefore \: \; \frac{T'}{T}=\sqrt{\frac{I'}{M'}\times \frac{M}{I}}=\sqrt{\frac{I'}{I}\times \frac{M}{M'}}..............(iii)

\therefore \: \; \frac{T'}{T}=\sqrt{\frac{1}{8}\times \frac{2}{1}}=\frac{1}{2}

 


Option 1)

\frac{1}{2\sqrt{2}}

Incorrect

Option 2)

\; \; \; \frac{1}{2}\;

Correct

Option 3)

\; \; 2\; \;

Incorrect

Option 4)

\; \frac{1}{4}

Incorrect

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