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NEET Practice Questions (MCQs) with Answers & Solutions

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A Galvanometer of resistance$ 50 \Omega $ is connected to a battery of 3 volt along with a resistance of $2950 \Omega $ in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

A. $ 6050 \Omega $ B. $ 4450 \Omega $ C. $ 5050 \Omega $ D. $ 5550 \Omega $

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Explanation

total initial resistance = G + R = 50 + 2950 = $3000 \Omega $ $ \therefore = { V \over G+R} = { 3 \over 3000} = 0.001 Amp $ Let x be the effective resistance of the circuit $ 3 volt = 3000 \times 0.001 = x \times 20 /30 \times 0.001 $ $ x =4500 \Omega $ $ \therefore resistance to be added = 4500 - 50 = 4450 \Omega $

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A Galvanometer coil has a resistance of $15 \Omega $ and gives full scale deflection for a current of 4 mA. To convert it to an ammeter of range 0 to 6 Amp.....................

A. $ 10 m \Omega $ resistance is to be connected in parallel to the galvanometer. B. $ 10 m \Omega $ resistance is to be connected in series with the galvanometer. C. $ 0.1 \Omega $ resistance is to be connected in parallel to the galvanometer. D. $ 0.1 \Omega $ resistance is to be connected in series with the galvanometer.

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Explanation

$ S = { G . I_G \over -I_G } = { 4 \times 10^{-3} \times 15 \over 6 - ( 4 \times 10^{-3 }) }= 10 m \Omega $ above shunt resistance should be connected in parallel

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The deflection in moving coil Galvanometer is reduced to half when it is shunted with a $40 \Omega $ coil. The resistance of the Galvanometer is ............................

A. $ 60 \Omega $ B. $ 10 \Omega $ C. $ 40 \Omega $ D. $ 20 \Omega $

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Explanation

$ I_G .G = ( I- I_G). S $ $ G = { (I -I_G ) .S \over I_G} = S = 40 \Omega $

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A conducting circular loop of radius a carries a constant current I. It is placed in a uniformmagnetic field $ \vec B$ , such that $\vec B $ is perpendicular to the plane of the Loop. The magnetic force acting on the Loop is ...................

A. $ \vec B I r $ B. $ \vec B I \pi r^2 $ C. 0 D. $ BI ( 2 \pi r ) $

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Explanation

Net force on a current carrying closed Loop is always zero if it is placed in a uniform mag. field.

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Two thin long parallel wires separated by a distance Y are carrying a current I Amp each. The magnitude of the force per unit length exerted by one wire on their is

A. $ { \mu_0 I^2 \over Y^2 } $ B. $ { \mu_0 I^2 \over 2 \pi Y^2 } $ C. $ { \mu_0 \over 2 \pi } { 1 \over Y } $ D. $ { \mu_0 \over 2 \pi } { 1 \over Y^2 } $

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Explanation

$ F = { \mu_o \over 2 \pi } { I_1 I_2 l \over y } $ $ { F \over l } = { \mu_o \over 2 \pi } { I .I \over y } = { \mu _ o \over 2 \pi } { I^2 \over y } $

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If two streams of protons move parallel to each other in the same direction, then they

A. Do not exert any force on each other B. Repel each other C. Attract each other D. Get rotated to be perpendicular to each other.

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Explanation

For charge particles, if they are moving freely in space, electrostatic force is dominant over mag. force between them. Hence due to ele. force they repel each other.

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A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent magnet such that $ \vec B $ is in plane of the coil. If due to a current I in the triangle a torque $ \tau $ acts on it, the side l of the triangle is

A. $ { 2 \over \sqrt 3 } \left( \tau \over BI ) \right) ^{1/2} $ B. $ { 2 \over \sqrt 3 } \left( \tau \over BI ) \right) $ C. $ { 2 \over \sqrt 3 } \left( \tau \over BI ) \right) ^{1/2} $ D. $ \left( 1 \over \sqrt 3 ) \right) \left( \tau \over BI \right) $

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Explanation

$ In \triangle CAD $ $ AC^2 = AD^2 + DC^2 $ $ l^2 = { l^2 \over 4 } + DC^2 $ $ DC = { \sqrt 3 \over 2 } l $ $ Area of \triangle ABC $ $ A = {1 \over 2 } (l) \left( {\sqrt 2 \over 2 }l \right) $ $ A = { 1 \over 4 } \sqrt 3 l^2 $ $ torques acting on \triangle ABC is $ $ \tau = IAB sin \theta $ $ = I \left( {1 \over 4 } \sqrt 3 l^2 \right) B sin ^\circ $ $ \theta = 90 ^\circ $ $ \tau = { \sqrt 3 \over 4 } I l^2 B $ $ \therefore l^2 = - { 4 \tau \over \sqrt 3 I B } $ $ \therefore l^2 = { 4 \tau \over \sqrt 3 I B } $ $ \therefore l = 2 \left ({ \tau \over \sqrt I B } \right) ^{1/2 } $

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In a moving coil galvanometer, the deflection of the coil $\theta $ is related to ele. Current I by the relation.

A. $ I \alpha tan \theta $ B. $ I \alpha \theta $ C. $ I \alpha \sqrt^2 $ D. $ I \alpha \sqrt \theta $

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Explanation

$ \tau = NIAB and \tau = k \theta $ $ \therefore NIAB = k \theta $ $ I = \left({ k \over NAB} \right) \theta $ $ \therefore I \alpha \theta $

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The unit of ele. current "AMPEAR" is the current whichwhen flowing through each of two parallel wires spaced 1 meter apart in vaccum and of infinite length will give rise to a force between them equal to N/m

A. 1 B. $ 2 \times 10^{-7} $ C. $ 1 \times 10^{-2} $ D. $ 4 \pi \times 10^{-7} $

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Explanation

$ F = { \mu_0 \over 2 \pi } { I_1 I_2 l \over y } $ $ { F \over l } = 2 \times 10^{-7} N/m $

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A coil having N turns is wound tightly in the formof a spiral with inner and outer radii "a" and "b" respectively. When a current I passes through the coil, the magnetic field at the centre is ...............

A. $ { \mu_o NI \over b }$ B. $ { 2 \mu_o NI \over b }$ C. $ { \mu _o NI \over 2 (b-a) } ln \left( b \over a \right) $ D. $ { \mu _o \over 2NI (b-a) } ln (ab) $

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Explanation

$ No. of turns per unit width = { N \over b-a} $ $ \therefore the no.of turns in thickness dx is dN = \left( N \over b-a\right) dx $ $ \therefore mag. field at the centre is dB = dN \left( \mu_o I \over 2x \right) $ $ dB = \left( {N \over b-a} \right) { \mu_o I \over 2x} .dx $ $ \therefore mag.field B = \int dB$ $ = { \mu_o NI \over 2 (b-a) } n \left( { n \over a } \right) $

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