Question

A dip needle vibrates in a vertical plane perpendicular to magnetic meridian. The time of vibration is found to be $2 \,s$, the same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be $2 \,s$. 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
$T_{1}=2\pi\sqrt{\frac{I}{MB_{V}}}$
In second case
$T_{2}=2\pi\sqrt{\frac{I}{MB_{H}}}$
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 $T_{2}=T_{1}=2\,s$
$\therefore tan\delta=1$
$\delta=tan^{-1}\left(1\right)=45^{\circ}$