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A magnetic compass needl oscillates 30 t...

A magnetic compass needl oscillates 30 times per minute at a place where the dip is `45^(@)` and 40 times per minute where the dip is `30^(@)` if `B_(1)` and `B_(2)` are respectively the total magnetic field due to the earth at the two places then the ratio `B_(1)//B_(2)` is best given by

A

0.7

B

2.2

C

3.6

D

1.8

Text Solution

Verified by Experts

The correct Answer is:
A
`T=2pisqrt((l)/(MH)) rArr H prop (1)/(T^(2))`
`H_(1)=B_(1)cos 45^(@)=(B_(1))/(sqrt2), H_(2)=B_(2)cos 30^(@) (sqrt3B_(2))/(2)`
`therefore" "(H_(1))/(H_(2))=((T_(2))/(T_(1)))^(2)=(B_(1))/(B_(2)).(sqrt2)/(sqrt3)=((30)/(40))^(2)" "therefore" "(B_(1))/(B_(2))=(9sqrt6)/(32)=0.7`
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