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The Needle of a Dip Circle Shows an Apparent Dip of 45° in a Particular Position and 53° When the Circle is Rotated Through 90°. Find the True Dip. - Physics

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प्रश्न

The needle of a dip circle shows an apparent dip of 45° in a particular position and 53° when the circle is rotated through 90°. Find the true dip.

बेरीज
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उत्तर

Given :
Apparent dip shown by the needle of the dip circle , `δ_1 = 45^circ`

Dip shown by the needle of the dip circle on rotating the circle, `δ_2 = 53^circ`

true dip (δ) is given by

`cot^2 δ = cot^2 δ_1 + cot^2 δ_2`

⇒`cot^2 δ = cot^2 45^circ + cot^2 53^circ`

⇒`cot^2 δ = 1.56`

⇒`cot^2 δ = 1.56`

⇒`δ = 38.6^circ ≈ 39^circ`

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पाठ 14: Permanent Magnets - Exercises [पृष्ठ २७८]

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एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
पाठ 14 Permanent Magnets
Exercises | Q 15 | पृष्ठ २७८

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