AIIMS2017Physics-Magnetism

AIIMS 2017 Physics Magnetic Dip MCQ Question

Type: MCQ-numerical-Medium-Class 12

The angle of dip if dip needle oscillating in vertical plane makes 40 oscillations per minute in a magnetic meridian and 30 oscillations per minute in vertical plane at right angle to the magnetic meridian is

A

θ = sin⁻¹(0.5625)

B

θ = sin⁻¹(0.325)

C

θ = sin⁻¹(0.425)

D

θ = sin⁻¹(0.235)

Correct Answer

Option A

Detailed Explanation

The angle of dip (θ) can be determined using the formula tan(θ)=N12N222N22\tan(θ) = \frac{N_1^2 - N_2^2}{2N_2^2}, where N1N_1 is the number of oscillations in the magnetic meridian (40 oscillations/min) and N2N_2 is the number of oscillations at right angles to it (30 oscillations/min). Substituting these values gives tan(θ)=4023022×302=16009001800=7001800=718\tan(θ) = \frac{40^2 - 30^2}{2 \times 30^2} = \frac{1600 - 900}{1800} = \frac{700}{1800} = \frac{7}{18}, leading to θ=sin1(0.5625)θ = \sin^{-1}(0.5625).

Other options are incorrect because they correspond to different values of tan(θ)\tan(θ) that do not match the calculated ratio, thus not representing the angle of dip based on the observed oscillation frequencies.

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