In a certain position, the needle of a dip circle indicates an apparent dip of 45^ ^ ^ . When the circle is rotated by 90^ ^ ^ , the apparent dip shown is 53^ ^ ^ . Determine the true dip.

Correct answer: A. 39^ ^ ^ Let the true dip be . The apparent dips in two mutually perpendicular planes are _1 = 45^ ^ ^ and _2 = 53^ ^ ^ . The relation between true dip and apparent dips is given by: ^2 = ^2 _1 + ^2 _2 Substituting the given values: ^2 = ^2 45^ ^ ^ + ^2 53^ ^ ^ ^2 = (1)^2 + ((3)/(4))^2 ^2 = 1 + (9)/(16) = (25)/(16) = (5)/(4) = (4)/(5) = 0.8 = ^ -1 (0.8) 39^ ^ ^

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Permanent Magnets

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Permanent Magnets

In a certain position, the needle of a dip circle indicates an...

In a certain position, the needle of a dip circle indicates an apparent dip of

45^{}^{{}^{\circ}}

. When the circle is rotated by

90^{}^{{}^{\circ}}

, the apparent dip shown is

53^{}^{{}^{\circ}}

. Determine the true dip.

Options

A $39^{}^{{}^{\circ}}$
B $51^{}^{{}^{\circ}}$
C $49^{}^{{}^{\circ}}$
D $41^{}^{{}^{\circ}}$

Correct answer

A. $39^{}^{{}^{\circ}}$

Step-by-step solution

Let the true dip be

\delta

.

The apparent dips in two mutually perpendicular planes are

\delta_1 = 45^{}^{{}^{\circ}}

and

\delta_2 = 53^{}^{{}^{\circ}}

.

The relation between true dip and apparent dips is given by:

\cot^2 \delta = \cot^2 \delta_1 + \cot^2 \delta_2

Substituting the given values:

\cot^2 \delta = \cot^2 45^{}^{{}^{\circ}} + \cot^2 53^{}^{{}^{\circ}}

\cot^2 \delta = (1)^2 + \left(\dfrac{3}{4}\right)^2

\cot^2 \delta = 1 + \dfrac{9}{16} = \dfrac{25}{16}

\cot \delta = \dfrac{5}{4}

\tan \delta = \dfrac{4}{5} = 0.8

\delta = \tan^{-1}(0.8) \approx 39^{}^{{}^{\circ}}

Practice Permanent Magnets on Quantrex Academy →

More from Permanent Magnets