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NEET Practice Questions (MCQs) with Answers & Solutions

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What is the angle between the electric dipole moment and the electric field strength due to it on the equatorial line

$0 °$

$90 °$

$180 °$

None of these

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The electrostatic field due to a charged conductor just outside the conductor is

zero and parallel to the surface at every point inside the conductor

zero and is normal to the surface at every point inside the conductor

parallel to the surface at every point and zero inside the conductor

normal to the surface at every point and zero inside the conductor

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Explanation

The electric field is zero inside a conductor. Just outside a conductor, the electric field lines are perpendicular to its surface.

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The law, governing the force between electric charges is known as

Ampere's law

Ohm's law

Faraday's law

Coulomb's law

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Explanation

Coulomb's law is used to calculate the force between charges.

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F_g and F_e represents gravitational and electrostatic force respectively between electrons situated at a distance 10 cm. The ratio of F_g/ F_e is of the order of

10⁴²

10^–43

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Explanation

Gravitational force between electrons $F_{G} = \frac{G {(m_{e})}^{2}}{r^{2}}$

Electrostatics force between electrons $F_{e} = k . \frac{e^{2}}{r^{2}}$

$\frac{F_{G}}{F_{e}} = \frac{G {(m_{e})}^{2}}{k . e^{2}} = \frac{6 .67 \times 10^{- 11} \times {(9 .1 \times 10^{- 31})}^{2}}{9 \times 10^{9} \times {(1 .6 \times 10^{- 19})}^{2}} = 2 .39 \times 10^{- 43}$

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A soap bubble is given a negative charge, then its radius

Decreases

Increases

Remains unchanged

Nothing can be predicted as information is insufficient

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Explanation

Due to mutual repulsion of charges distributed on the surface of bubble.

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In the absence of other conductors, the surface charge density

Is proportional to the charge on the conductor and its surface area

Inversely proportional to the charge and directly proportional to the surface area

Directly proportional to the charge and inversely proportional to the surface area

Inversely proportional to the charge and the surface area

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Explanation

Surface charge density $σ = \frac{q}{A}$

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Out of gravitational, electromagnetic, Vander Waals, electrostatic and nuclear forces; which two are able to provide an attractive force between two neutrons

Electrostatic and gravitational

Electrostatic and nuclear

Gravitational and nuclear

Some other forces like Vander Waals

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Explanation

Gravitational force and nuclear force both are attractive in nature.

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A total charge Q is broken in two parts Q₁ and Q₂ and they are placed at a distance R from each other. The maximum force of repulsion between them will occur, when

$Q_{2} = \frac{Q}{R}, Q_{1} = Q - \frac{Q}{R}$

$Q_{2} = \frac{Q}{4}, Q_{1} = Q - \frac{2 Q}{3}$

$Q_{2} = \frac{Q}{4}, Q_{1} = \frac{3 Q}{4}$

$Q_{1} = \frac{Q}{2}, Q_{2} = \frac{Q}{2}$

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Explanation

$Q_{1} + Q_{2} = Q$ ..... (i) and $F = k \frac{Q_{1} Q_{2}}{r^{2}}$ .....(ii)

From (i) and (ii) $F = \frac{k Q_{1} (Q - Q_{1})}{r^{2}}$

For F to be maximum $\frac{d F}{d Q_{1}} = 0$ ⇒ $Q_{1} = Q_{2} = \frac{Q}{2}$

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Three charges 4q, Q, and q are in a straight line in the position of 0, l/2, and l respectively. The resultant force on q will be zero if Q =

– q

–2q

$- \frac{q}{2}$

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Explanation

The force between 4q and q; $F_{1} = \frac{1}{4 π ε_{0}} \cdot \frac{4 q \times q}{l^{2}}$

The force between Q and q; $F_{2} = \frac{1}{4 π ε_{0}} \cdot \frac{Q \times q}{{(l / 2)}^{2}}$

We want $F_{1} + F_{2} = 0$ or $\frac{4 q^{2}}{l^{2}} = - \frac{4 Q q}{l^{2}}$

⇒ Q = –q

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Two small spheres each having the charge +Q are suspended by insulating threads of length L from a hook. This arrangement is taken in space where there is no gravitational effect, then the angle between the two suspensions and the tension in each will be

$180^{o}, \frac{1}{4 π ε_{0}} \frac{Q^{2}}{{(2 L)}^{2}}$

$90^{o}, \frac{1}{4 π ε_{0}} \frac{Q^{2}}{L^{2}}$

$180^{o}, \frac{1}{4 π ε_{0}} \frac{Q^{2}}{2 L^{2}}$

$180^{o}, \frac{1}{4 π ε_{0}} \frac{Q^{2}}{L^{2}}$

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