Question

the magnet is turned
ar magnet produces a neutral point at P. If
net is turned through 180° , the needle
makes 10v2 oscillations/ min. If the
et is removed all together frequency of
oscillation of the needle at P is
1) 20 oscillations / min
magnet is remd
2) 0
3) 10 oscillations / min
4) 30 oscillations / min​

Answer 1

Answer:-

Option C , is correct option, 10 oscillation/ minute.

The Explanation :-

At P, the magnetic field of the magnet B is equal and opposite to the horizontal component of earth's magnetic field BH.When the magnet is turned through 180∘, B and BH will add up and the magnetic field at P will be equal to 2BH.

In the absence of the magnet, magnetic field at P will be equal to BH.

In the absence of the magnet, magnetic field at P will be equal to BH.Now we know that frequency of oscillation,

$= > f = \sqrt{b}$

$= > \frac{f1}{f2} = \sqrt{ \frac{b1}{b2} }$

$= > f1 = \frac{10}{2} \: per \: min$

=> B1 =2 BH , B2= BH

After putting, the value in the above equation ,we get f2=10/ minute.

Oscillation :-

Oscillation is the variation, typically in time, of some measure as seen, for example, in a swinging pendulum.

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