Home
Class 12
PHYSICS
A dip needle vibrates in the vertical pl...

A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be 2 sec. 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

A

`0^(@)`

B

`30^(@)`

C

`45^(@)`

D

`90^(@)`

Text Solution

Verified by Experts

The correct Answer is:
C
`As " " T_(1)=2pi sqrt((I)/(MV'))T_(2)=2pi sqrt((I)/(MH))`
` " " V=H tan delta Rightarrow (V)/(H)=tan delta`
`But " " (V)/(H)=1`
`therefore " " tan delta=1 Rightarrow=45^(@)`
Promotional Banner

Similar Questions

Explore conceptually related problems

A dip needle in a plane perpendicular to magnetic meridian will remain

The time of vibration of a dip needle vibration in the vertical plane in the magnetic needle is made to vibrate in the horizontal plane, the time of vibration is 3sqrt2 s . Then angle of dip will be-

A dip needle vibrates in a vertical plane with time period of 3 second . The same needle is suspended horizontally and made to vibrate is a horizontal plane . The time period is again 3 sec. The angle of dip at that place is .

A dip circle is taken to geomagnetic equator. The needle is allowed to move in a vertical plane perpendicular to the magnetic meridian. The needle will stay

A dip needle lies initially in the magnetic merdian when it shows an angle of dip theta at a place. The dip circle is rotated through an angle x in the horizontal plane and then it shows an angle of dip theta^(') . Then tantheta^(')/tantheta is

The number of oscillations per second of a vibrating objects is called its time period