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A magnetic needle suspended in a vertica...

A magnetic needle suspended in a vertical plane at `30^@` from the magnetic meridian makes an angle of `45^@` with the horizontal. Find the true angle of dip.

A

approx45^(@) `

B

`30^(@) `

C

`approx41^(@) `

D

approx51^(@) `

Text Solution

Verified by Experts

The correct Answer is:
C
In a vertical plane at `30^(@)` from the magnetic meridian, the horizontal component is,

While vertical component is still. Thereofre, apparent dip will be given by `tan theta'=(V)/(H)=(V)/(H cos30^(@))`
`but " " (V)/(H)=tan theta' " " (where theta= "true angle of dip")`
`therefore " " tan theta'=(tan theta)/(cos 30^(@))`
`therefore " " theta=tan^(-1)[tan theta'cos30^(@)] `
`therefore " " a=tan^(-1)[(tan45^(@))(cos30^(@))]approx41^(@) `
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