Question

A compass needle oscillates $20$ times per minute at a place where dip is ${45}^{\circ }$ and $30$ times per minute where dip is ${30}^{\circ }$. The ratio of horizontal component of earth's magnetic at second to first place is-

• A. $1.23$
• B. $1.63$
• C. $1.83$
• D. $1.51$

Answer 1

The correct option is C $1.83$

Given, ${\theta }_1={45}^{\circ };{\theta }_2={30}^{\circ }{T}_1=3\text{s};T_2=2\text{s}$

Time period of oscillation is,

$T=2\pi \sqrt{\frac{I}{M{B}_H}}$

$B_H\propto \frac{1}{T^2}[\because B_H=B\cos \theta ]$

$\Rightarrow \frac{(B_H{)}_2}{(B_H{)}_1}=\frac{T_1^2}{T_2^2}$

$\Rightarrow \frac{B_2\cos {\theta }_2}{B_1\cos {\theta }_1}=\frac{T_1^2}{T_2^2}$

$\frac{B_2}{B_1}=\frac{T_1^2}{T_2^2}\times \frac{\cos {\theta }_1}{\cos {\theta }_2}$

$={\left(\frac{3}{2}\right)}^2\times \frac{\cos {45}^{\circ }}{\cos {30}^{\circ }}=\frac{9}{4}\times \frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}}$

$=\frac{18}{4\sqrt{6}}=\frac{18}{9.8}$

$\therefore \frac{B_2}{B_1}=1.83$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.