A magnetic needle oscillates in a horizontal plane with a period T at a place where the angle of dip is $60^{0}$ . When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be

\frac{T}{\sqrt{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{2 T}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 T

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{T}{\sqrt{2}}$
$T_{1} \times \frac{1}{\sqrt{B_{H_{1}}}}$

Now, ${B_{H})}_{2} = B_{H_{1}} cos 60^{0}$

$\frac{T_{1}}{T_{2}} = \sqrt{\frac{B_{H_{2}}}{B_{H_{1}}}}$

$\frac{T_{1}}{T_{2}} = \sqrt{\frac{B_{H_{1}} cos 60^{0}}{B_{H_{2}}}}$

$T_{2} = \sqrt{2} T_{1}$

Suggest Corrections

Similar questions

Q. The angle of the dip, if a dip needle oscillating in a vertical plane makes 40 oscillations per minute in a magnetic meridian and 30 oscillations per minute in a vertical plane at right angle to the magnetic meridian, is:

Q. A magnetic needle freed to rotate around a fixed vertical plane stay at an angle

60^{0}

with the horizontal . If the dip at that place is

37^{0}

, find the angle of the fixed vertical plane with the meridian.

A magnetic needle is free to rotate in a vertical plane which makes an angle of $60^{0}$ with the magnetic meridian. If the needle stays in a direction making an angle of tan $^{- 1} (\frac{2}{\sqrt{3}})$ with the horizontal, what would be the dip at that place ?

Q. A thin magnetic needle oscillates in a horizontal plane with a period

T

. It is broken into

n

equal parts by cutting perpendicular to the length. Find the time period of each part.

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

Join BYJU'S Learning Program

A magnetic needle oscillates in a horizontal plane with a period T at a place where the angle of dip is 600. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be

The correct option is A T√2T1×1√BH1 Now, BH)2=BH1cos600 T1T2=√BH2BH1 T1T2= ⎷BH1cos600BH2 T2=√2T1

A magnetic needle oscillates in a horizontal plane with a period T at a place where the angle of dip is $60^{0}$ . When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be

The correct option is A $\frac{T}{\sqrt{2}}$
$T_{1} \times \frac{1}{\sqrt{B_{H_{1}}}}$

Now, ${B_{H})}_{2} = B_{H_{1}} cos 60^{0}$

$\frac{T_{1}}{T_{2}} = \sqrt{\frac{B_{H_{2}}}{B_{H_{1}}}}$

$\frac{T_{1}}{T_{2}} = \sqrt{\frac{B_{H_{1}} cos 60^{0}}{B_{H_{2}}}}$

$T_{2} = \sqrt{2} T_{1}$