Home
Class 12
PHYSICS
A dip circle lying initially in the mag...

A dip circle lying initially in the magnetic meridian ,the angle of dip is `45^(0)`. Now dip circle is rotated through `30^0` in the horizontal plane .The apparent angle of dip is

(A) `tan^(-1)``(2)/(sqrt(3))`

(B) `tan^(-1)``(sqrt(3))/(2)`

(C) `tan^(-1)``2`

(D)`tan^(-1)``(1)/(2)`

Promotional Banner

Similar Questions

Explore conceptually related problems

A dip circle lying initially in the magnetic meridian is rotated through angle theta in the horizontal plane. The ratio of tangent of apparent angle of dip to true angle of dip is

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

Evaluate: tan^(- 1)(-1/(sqrt(3)))+tan^(- 1)(-sqrt(3))+tan^(- 1)(sin(-pi/2))

A normal to parabola, whose inclination is 30^(@) , cuts it again at an angle of (a) tan^(-1)((sqrt(3))/(2)) (b) tan^(-1)((2)/(sqrt(3))) (c) tan^(-1)(2sqrt(3)) (d) tan^(-1)((1)/(2sqrt(3)))

The value of dip at a place is 45^@ . The plane of the dip circle is turned through 60^@ from the magnetic meridian. Find the apparent value of dip.

Find the value of expression: sin(2tan^(-1)(1/3))+cos(tan^(-1)2sqrt(2))

Find the value of expression: sin(2tan^(-1)(1/3))+cos(tan^(-1)2sqrt(2))

At a place the true value of angle of dip is 60^(@) . If dip circle is rotated by phi^(@) from magnetic meridian, the angle of dip is found to be tan^(-1)(2) . Then the value of phi is

A magnetic needle suspended in a vertical plane at 30^@ from the magnetic meridian makes an angle of 45^@ with the horizontal. Find the true angle of dip.

Find the value of expression: sin(2tan^(-1)1/3)+cos(tan^(-1)2sqrt(2))