A magnetic moment m oscillating freely in earth's horizontal magnetic field makes a oscillation per minute. If the magnetic moment is quadrupled and the earth's field doubled, the number of oscillation made per minute would becomes

\frac{n}{2 \sqrt{2}}

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\frac{n}{\sqrt{2}}

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2 \sqrt{2} n

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\sqrt{2} n

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Solution

The correct option is C $2 \sqrt{2} n$
No. of oscillation per min= $\frac{1}{2 π} \sqrt{\frac{M B_{H}}{I}}$

$n α \sqrt{M B_{H}}$

$M \to 4 t i m e s$

$B_{H} = 2 t i m e s$

$v \to \sqrt{8} t i m e s$

$v^{1} = \sqrt{8} V$

$= 2 \sqrt{2} n$

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A magnetic moment m oscillating freely in earth's horizontal magnetic field makes a oscillation per minute. If the magnetic moment is quadrupled and the earth's field doubled, the number of oscillation made per minute would becomes

The correct option is C 2√2nNo. of oscillation per min=12π√MBHI nα√MBH M→4times BH=2times v→√8times v1=√8V =2√2n

The correct option is C $2 \sqrt{2} n$
No. of oscillation per min= $\frac{1}{2 π} \sqrt{\frac{M B_{H}}{I}}$

$n α \sqrt{M B_{H}}$

$M \to 4 t i m e s$

$B_{H} = 2 t i m e s$

$v \to \sqrt{8} t i m e s$

$v^{1} = \sqrt{8} V$

$= 2 \sqrt{2} n$