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A small magnetic needle performs 10 osci...

A small magnetic needle performs 10 oscillations/minute in the earth's horizontal magnetic field . When a bar magnet is placed near the small magnet in same position , frequency of oscillations becomes `10sqrt2` oscillations/minute . If the bar magnet be turned around end to end , the rate of oscillation of small magnet will become

A

2 vibrations/min

B

4 vibrations/min

C

10 vibrations/min

D

14 vibrations/min

Text Solution

Verified by Experts

The correct Answer is:
A
`(60)/(10) = 2pi sqrt((l)/(MB_(H)))`
`(60)/(14) = 2pi sqrt((l)/(M (B_(H) + B)))`
`therefore (7)/(5) = sqrt((B_(H) + B)/(B_(H)))` or `B = (24)/(25) B_(H)`
Hence ,
`(60)/(t) = 2pi sqrt((l)/(M (B_(H) - B))) = 2pi sqrt((l)/(MB (1- 24//25)))`
` = 5 xx 2pi sqrt((l)/(MB)) = 5 xx (60)/(10) = 30`
`t = (60)/(30) = 2` vibrations/ min
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