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NEET Practice Questions (MCQs) with Answers & Solutions

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A 100 W bulb designed to operate on 100 V is to be connected across a 500 V source. Find the resistance to be put in series so that bulb consumes 100 W only.

100 $Ω$

500 $Ω$

400 $Ω$

300 $Ω$

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A neutron of mass m enters a uniform magnetic field B with the velocity v perpendicular to it. The radius of the circular path made by a neutron is :

$\frac{mv}{B}$

$\frac{mv}{B^{2}}$

$Infinity$

$Zero$

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Explanation

$r = \frac{mv}{qB} = \frac{mv}{0} = \infty As neutron is electrically neutral$

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Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents flowing in them are 3I and 4I respectively. The resultant magnetic field induction at the center is :

$\frac{μ_{0} I}{π}$

$\frac{5 μ_{0} I}{2 R}$

$\frac{7 μ_{0} I}{2 R}$

$Zero$

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Explanation

$(2) B_{1} = \frac{3 μ_{0} I}{2 x R} B_{2} = \frac{4 μ_{0} I}{2 x R} since B_{1} perpendicular to B_{2}, so resultant magnetic field B = \sqrt{B_{1}^{2} + B_{2}^{2}} = \frac{5 μ_{0} I}{2 R}$

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A uniform electric field and a uniform magnetic field are acting in the same direction in a certain region. If a proton is projected in this region such that its velocity is pointed along the direction of fields, then the proton :

will turn towards the left of the direction of motion

will turn towards the right of the direction of motion

will speed up

will slow down

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Explanation

$(3) Magnetic force \overset{⇀}{Fm} = q (\overset{⇀}{V} x \overset{⇀}{B}) Here \overset{⇀}{V} ∥ \overset{⇀}{B} i . e ., no megnetic force act on the particle . Electric force \overset{⇀}{F} = q \overset{⇀}{E} . Electric force is parallel to the electric field because charge is + ve so particle be accelerated .$

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When a proton is released from rest in a room, it starts with an initial acceleration $a_{0}$ towards the east. When it is projected towards the north with a speed $v_{0}$ , it moves with initial acceleration $3 a_{0}$ towards east. The electric and magnetic fields in the room are -

$\frac{{Ma}_{0}}{e} west, \frac{{Ma}_{0}}{{ev}_{0}} up$

$\frac{{Ma}_{0}}{e} west, \frac{2 {Ma}_{0}}{{ev}_{0}} down$

$\frac{{Ma}_{0}}{e} east, \frac{2 {Ma}_{0}}{{ev}_{0}} d o w n$

$\frac{{Ma}_{0}}{e} east, \frac{3 {Ma}_{0}}{{ev}_{0}} u p$

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Explanation

When a proton is released from rest, it experiences an initial acceleration a0 towards the east due to the electric field. When it is projected with a velocity v0 towards the north, its acceleration due to the magnetic field is 2a0 towards the east, making the total acceleration 3a0 towards the east. Hence, the electric and magnetic fields are (a0e/e) east and (3a0/ev0) up, respectively.

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$A positively charged particle enters a magnetic field of B \hat{i} with a velocity V \hat{j} . The particle will move$

In a circle

In y-z plane

With constant speed

All of these are correct

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Explanation

When charged particle moves perpendicular to uniform magnetic field, then path of the particle is circular and plane perpendicular to the direction of the magnetic field.

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If the planes of two identical concentric coils are the perpendicular and magnetic moment of each coil is M, then the resultant magnetic moment of the two coil system is

$\sqrt{2} M$

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Explanation

$(2) The resultant magnetic moment M' = \sqrt{M^{2} + M^{2} + 2 MM \cos 90 °} = M \sqrt{2}$

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A voltage has a resistance G and it can measure a maximum voltage of V. To convert it in a voltmeter of range nV, the value of resistance required is :

(n - 1)G in series

(n - 1)G in parallel

$\frac{G}{n - 1}$ in parallel

$\frac{G}{n - 1}$ in series

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A current wire is hidden in a wall. Its position can be located with the help of a

Moving coil galvanometer

Voltmeter

Hotwire ammeter

Magnetic needle

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Explanation

Magnetic needle deflect in the presence of a magnetic field produced by the wire.

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When an electron enters perpendicularly in a magnetic field with velocity v, time period of its revolution is T. If it enters in the same magnetic field with a velocity 2v, then its time period will be:

$\frac{T}{2}$

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Explanation

$(4) T = \frac{2 πm}{qB} T \neq f (v)$

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