JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Elasticity & SHM

A compass needle oscillates 20 times per minute at a place where the dip is

3 0^{\circ}

and 30 times per minute where the dip is

6 0^{\circ}

. The ratio of total magnetic field due to the earth at two place respectively is

\frac{4}{x}

. The value of

x

Answer: 243

Solution

The period of oscillation of a compass needle is

T \propto \frac{1}{B _{H}}

, where

B_{H} = B cos θ

is the horizontal component.

Thus

T \propto \frac{1}{B c o s θ}

.

Given:

T_{1} = \frac{60}{20} = 3

T_{2} = \frac{60}{30} = 2

θ_{1} = 3 0^{\circ}

θ_{2} = 6 0^{\circ}

\frac{T _{1}}{T _{2}} = \frac{B _{2} cos θ _{2}}{B _{1} cos θ _{1}} ⟹ \frac{3}{2} = \frac{B _{2} \cdot \frac{1}{2}}{B _{1} \cdot \frac{3}{2}}

Squaring:

\frac{9}{4} = \frac{B _{2}}{3 B _{1}} ⟹ \frac{B _{1}}{B _{2}} = \frac{4}{9 3} = \frac{4}{243}

.

Therefore,

x = 243

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Integer answer