Question

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P of dip 30°. The number of oscillations per minute become 10 at another place Q of 60° dip. The ratio of the total magnetic field at the two places (B Q:_ BP_) is

• A. $\sqrt3:4$
• B. $4:\sqrt3$
• C. $\sqrt3:2$
• D. $2:\sqrt3$

Answer 1

Answer: (A)

$\sqrt3:4$

Answer 2

The correct answer is (A) : $\sqrt3:4$
$T∝\frac{1}{\sqrt{B\cosδ}}$
$⇒ \frac{T_1}{T_2} = \sqrt{\frac{B_2\cosδ_2}{B_1\cosδ_1}}$
$⇒ \frac{3s}{6s} = \sqrt{\frac{B_2}{B_1}×\frac{\frac{1}{2}}{\frac{\sqrt3}{2}}}$
$⇒ \frac{B_2}{B_1} = (\frac{1}{2})^2 × \sqrt3$
$= \frac{\sqrt3}{4}$