Question

A dip needle vibrates in a vertical plane perpendicular to magnetic meridian. The time of vibration is found to be $2s$, the same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be $2s$. Then, the angle of dip is

• A. $0^{\circ }$
• B. $30^{\circ }$
• C. $45^{\circ }$
• D. $90^{\circ }$

Answer 1

Answer: (C)

Time of vibration is
$T=2\pi \sqrt{\frac{I}{MB}}$
In first case
$T_1=2\pi \sqrt{\frac{I}{M{B}_V}}$
In second case
$T_2=2\pi \sqrt{\frac{I}{M{B}_H}}$
Divide $\left(ii\right)$ by $\left(i\right)$, we get
$\therefore \frac{T_2}{T_1}=\sqrt{\frac{B_V}{B_H}}$
As $tan\delta =\frac{B_V}{B_H}$ where $\delta$ is the angle of dip
$\therefore \frac{T_2}{T_1}=\sqrt{tan\delta }$
As $T_2=T_1=2s$
$\therefore tan\delta =1$
$\delta =ta{n}^{-1}\left(1\right)=45^{\circ }$