Physics MCQs for NEET — Practice Questions with Answers

Mag. moment of circular Loop carrying current is $ M = IA = I ( \pi R^2 ) = I \pi \left ( { L \over 2 \pi } \right) ^2 $ $ M = { IL^2 \over 4 \pi } \Rightarrow L = \sqrt { 4 \pi M \over I } $

$ L = 10 \times 10^{-2} m $ $ r = 15 \times 10^{-2} m $ $ OP = \sqrt { 225 -25 } $ $ = \sqrt 200 cm $ since, at theneutral point, magnetic field due to the magnetic equal to $B_H$ $ B_H = { \mu o \over 4 \pi } { M \over (OP^2 + AO^2 ) ^{3/2} }$ = 1.35 Amp. meter

$ 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 $