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A dip needle vibrates in a vertical plane perpendicular to magnetic meridian. The time of
vibration is found to be 2s. The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be 2s, then the angle of dip is [UPCPMT 2012]

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0⁰

30⁰

45⁰

90⁰

answer is C.

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Detailed Solution

As, $T = 2 π \sqrt{\frac{I}{M B}} \Rightarrow T \propto \frac{1}{\sqrt{B}}, so T_{1} \propto \frac{1}{\sqrt{V}} and T_{2} \propto \frac{1}{\sqrt{H}}$

$V = H \tan δ \Rightarrow \frac{V}{H} = \tan δ$

But, $\frac{V}{H} = 1$ $[∵ T_{1} = T_{2} \Rightarrow V = H]$

$∴ \tan δ = 1 \Rightarrow δ = 45 °$

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