NEET Practice Questions (MCQs) with Answers & Solutions

$ tan \phi ' = { tan \phi \over cos \beta } $ $where \phi' = Apparent angle of dip $ $ = { tan 60 ^ \circ \over cos 60 ^ \circ } $ $ \phi = True angle of dip $ $ tan \phi' = 2 $ $ \beta = Angle made by vertical plane with magnetic meridian $ $ \phi' = tan^{-1} (2) $

The tangent of the angle of dip ($\theta$) initially is given as $\tan \theta$. When the dip circle is rotated by an angle $x$ in the horizontal plane, the new angle of dip is $\theta'$. According to the trigonometric relationship in spherical coordinates, the new tangent of the angle of dip can be represented as $\tan \theta' = \frac{\tan \theta}{\cos x}$. Therefore, $\frac{\tan \theta'}{\tan \theta} = \frac{1}{\cos x}$.

$ in vertical plane T = \sqrt [2 \pi] { I \over MB_V} .......(1)$ $ in horizontal plane T = \sqrt [2 \pi] { I \over MB_H} ........(2)$ but in both the cause T = 2 sec $ \sqrt { I \over MB_V } = \sqrt { I \over MB_H} $ $ { 1 \over B_V} = { 1 \over B_H} $ $ 1 = { B_V \over B_H } $ $ 1 = tan \phi $ $ \phi = 45 ^ \circ $

$ B_{ 1 axis } = { \mu_o \over 4 \pi } { 2M \over d^3 } ........(1) $ $B_{ 2 equator } = { \mu_o \over 4 \pi } { M \over d^3 }.......(2) $ At point P $ B_{ resultant} = \sqrt { B_1^2 + B_2 ^2 } $ $ = \sqrt 5 { \mu_o \over 4 \pi } { M \over d^3 } $

$ Magentic suspectibility x_m \alpha { 1 \over T } $ $ { x_{m2} \over x_{m1}} = { T_1 \over T_2 } $ $ { x_{m2} \over 0.0060} = { 273 - 73 \over 273 - 173 } $ $ { X_{m2} \over 0.0060} = {200 \over 100 } $ $ X_{m2} = 0.0120 $

Ferro magnetic substances, magnetised strongly in the direction of magnetic field. Para magnetic substances magnetised weaklyin the direction of magnetic field. Diamagnetic substances is magnetised weaklyin opposite direction of magnetic field.

$ {x_{m2} \over x_{m1} }= { T_1 \over T_2 } $ $ { 1.8 \times 10^{-5} \over 1.2 \times 10^{-5} } = { 300 \over T_2 } $ $ T_2 = 200 Kelvin $

$ Baxis = { 2M \over x^3 } = 9 (In CGS ) $ $ = { M \over x^3 } = { 9 \over 2 } $ $ B_{equator} = { M \over \left( x \over 2 \right)^3} $ $ = 8 \left( { M \over x^3 } \right) $ $ = 8 \left( 9 \over 2 \right) $ $ = 36 Gauss $

$ Volume of the domain = ( 1 \times 10^{-6 })^3 m^3 = 10^{-18} m^3 $ $ \therefore New dipole moment mnet = Nm $ $ = 8 \times 10^{10} \times 9 \times 10^{-24} $ $ = 72 \times 10^{-14} A.m^2 $ $ \therefore Magnetization M = { m_{net} \over vol} $ $ = { 72 \times 10^{-14} \over 10^{-18 }}{A.m^2 \over m^3 } $ $ = 72 \times 10^{+4} $ $ = 7.2 \times 10^5 Amp.meter^{-1} $

Diamagentic materials