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A magnet is heated so as to reduce its magnetic moment by 19%. Periodic Time of the magnetometer will be

  • Option 1)

    \uparrow 19 \%

  • Option 2)

    \uparrow 11 \%

  • Option 3)

    \downarrow 19 \%

  • Option 4)

    \downarrow 10 \%

 

Answers (1)

best_answer

 

\\* T = 2\pi\sqrt{\frac{I}{mB}} \Rightarrow T\;\alpha\;\frac{1}{\sqrt{M}} \\* \frac{T_{1}}{T_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{81}{100}} = \frac{9}{10} \\* \Rightarrow T_{2} = \frac{10}{9}T_{1} \Rightarrow 11\%

SHM of Bar Magnet in a Magnetic Field -

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

I \rightarrow Moment of Inertia of magnet

M \rightarrow Magnetic moment of magnet

B \rightarrow Magnetic field

- wherein

* In uniform Magnetic field Bar magnets performs SHM.

* Formula

 

 


Option 1)

\uparrow 19 \%

This is incorrect.

Option 2)

\uparrow 11 \%

This is correct.

Option 3)

\downarrow 19 \%

This is incorrect.

Option 4)

\downarrow 10 \%

This is incorrect.

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Aadil

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