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NEET PHYSICSMAGNETISM AND MATTERMedium

Question

If θ1\theta_1 and θ2\theta_2 be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip θ\theta is given by:

A

cot⁡2θ=cot⁡2θ1+cot⁡2θ2\cot^2 \theta = \cot^2 \theta_1 + \cot^2 \theta_2

B

tan⁡2θ=tan⁡2θ1+tan⁡2θ2\tan^2 \theta = \tan^2 \theta_1 + \tan^2 \theta_2

C

cot⁡2θ=cot⁡2θ1−tan⁡2θ2\cot^2 \theta = \cot^2 \theta_1 - \tan^2 \theta_2

D

tan⁡2θ=tan⁡2θ1−tan⁡2θ2\tan^2 \theta = \tan^2 \theta_1 - \tan^2 \theta_2

Step-by-Step Solution

  1. True Dip (θ\theta): The true angle of dip is related to the vertical component (VV) and horizontal component (HH) of Earth's magnetic field in the magnetic meridian by tan⁡θ=VH\tan \theta = \frac{V}{H} or cot⁡θ=HV\cot \theta = \frac{H}{V}.
  2. Apparent Dip: In a vertical plane making an angle α\alpha with the magnetic meridian, the effective horizontal component is H′=Hcos⁡αH' = H \cos \alpha, while the vertical component VV remains the same. The apparent dip θ1\theta_1 is given by: tan⁡θ1=VHcos⁡α⇒cot⁡θ1=Hcos⁡αV=cot⁡θcos⁡α\tan \theta_1 = \frac{V}{H \cos \alpha} \Rightarrow \cot \theta_1 = \frac{H \cos \alpha}{V} = \cot \theta \cos \alpha cos⁡α=cot⁡θ1cot⁡θ\cos \alpha = \frac{\cot \theta_1}{\cot \theta}
  3. Second Plane: Since the second plane is perpendicular to the first, it makes an angle (90∘−α)(90^\circ - \alpha) with the meridian. The apparent dip θ2\theta_2 is: tan⁡θ2=VHcos⁡(90∘−α)=VHsin⁡α⇒cot⁡θ2=Hsin⁡αV=cot⁡θsin⁡α\tan \theta_2 = \frac{V}{H \cos(90^\circ - \alpha)} = \frac{V}{H \sin \alpha} \Rightarrow \cot \theta_2 = \frac{H \sin \alpha}{V} = \cot \theta \sin \alpha sin⁡α=cot⁡θ2cot⁡θ\sin \alpha = \frac{\cot \theta_2}{\cot \theta}
  4. Derivation: Using the identity sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1: (cot⁡θ2cot⁡θ)2+(cot⁡θ1cot⁡θ)2=1\left( \frac{\cot \theta_2}{\cot \theta} \right)^2 + \left( \frac{\cot \theta_1}{\cot \theta} \right)^2 = 1 cot⁡2θ2+cot⁡2θ1cot⁡2θ=1\frac{\cot^2 \theta_2 + \cot^2 \theta_1}{\cot^2 \theta} = 1 cot⁡2θ=cot⁡2θ1+cot⁡2θ2\cot^2 \theta = \cot^2 \theta_1 + \cot^2 \theta_2

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MAGNETISM AND MATTER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMAGNETISM AND MATTERapparentanglesobservedverticalplanes

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