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)

$45^{\circ}$

Answer 2

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 $T_{2}=T_{1}=2\,s$ $\therefore tan\delta=1$ $\delta=tan^{-1}\left(1\right)=45^{\circ}$