A compass needle oscillates 20 times per minute at a place where the dip is 30° and 30 times per minute where the dip is 60°. The ratio of total magnetic field due to the earth at two places respectively is 4/√(x). The value of x is Numerical answer type. Work it out, then open the answer.

Answer: 243. Frequency n ∝ √(Bcos δ), so n² ∝ Bcos δ. (20²)/(30²) = (B₁cos 30°)/(B₂cos 60°) ⇒ 4/9 = (B₁)/(B₂)·√(3) (B₁)/(B₂) = (4)/(9√(3)) = (4)/(√(81× 3)) = (4)/(√(243)) So x = 243.

A compass needle oscillates 20 times per minute at a place where the… | JEE Main 2023 (12 Apr, Shift 1)

Q8JEE Main 2023 · 12 APR · SHIFT 1Difficult

Vibration Magnetometer

A compass needle oscillates 20 times per minute at a place where the dip is $30^\circ$ and 30 times per minute where the dip is $60^\circ$ . The ratio of total magnetic field due to the earth at two places respectively is $\frac{4}{\sqrt{x}}$ . The value of $x$ is

Numerical answer type. Work it out, then open the answer.

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Answer: 243

Frequency $n \propto \sqrt{B\cos\delta}$ , so $n^2 \propto B\cos\delta$ .
$\dfrac{20^2}{30^2} = \dfrac{B_1\cos 30^\circ}{B_2\cos 60^\circ} \Rightarrow \dfrac{4}{9} = \dfrac{B_1}{B_2}\cdot\sqrt{3}$
$\dfrac{B_1}{B_2} = \dfrac{4}{9\sqrt{3}} = \dfrac{4}{\sqrt{81\times 3}} = \dfrac{4}{\sqrt{243}}$
So $x = 243$ .

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