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CUET
Physics
Magnetism and Matter
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Angle of dip $\delta = 45^0$
$\Rightarrow \delta' = \frac{tan\delta}{cos\theta}= \frac{tan 45^0}{cos 30^0} = \frac{1}{\sqrt 3 /2} = \frac{2}{\sqrt 3}$
$\delta'= tan^{-1}\frac{2}{\sqrt 3}$
