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Physics MCQs for NEET — Practice Questions with Answers

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Point charge q₁ = 2 μC and q₂ = –1 μC are kept at points x = 0 and x = 6 respectively. Electrical potential will be zero at points

x = 2 and x = 9

x = 1 and x = 5

x = 4 and x = 12

x = –2 and x = 2

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Equipotential surfaces associated with an electric field which is increasing in magnitude along the x-direction are

Planes parallel to yz-plane

Planes parallel to xy-plane

Planes parallel to xz-plane

Coaxial cylinders of increasing radii around the x-axis

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A bullet of mass 2 gm is having a charge of 2 μC. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of 10 m/s ?

5 kV

50 kV

5 V

50 V

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Explanation

By using $\frac{1}{2} m v^{2} = Q V$

⇒ $\frac{1}{2} \times 2 \times 10^{- 6} \times {(10)}^{2} = 2 \times 10^{- 6} V$ ⇒ V = 50 kV

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In a certain charge distribution, all points having zero potential can be joined by a circle S. Points inside S have positive potential, and points outside S have a negative potential. A positive charge, which is free to move, is placed inside S .

It will remain in equilibrium

It can move inside S, but it cannot cross S

It must cross S at some time

It may move, but will ultimately return to its starting point

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Explanation

A free positive charge move from higher (positive) potential to lower (negative) potential. Hence, it must cross S at some time.

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A square of side ‘a’ has charge Q at its centre and charge ‘q’ at one of the corners. The work required to be done in moving the charge ‘q’ from the corner to the diagonally opposite corner is -

Zero

$\frac{Q q}{4 π \in_{0} a}$

$\frac{Q q \sqrt{2}}{4 π \in_{0} a}$

$\frac{Q q}{2 π \in_{0} a}$

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Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are +q and –q. The potential difference between the centres of the two rings is -

Zero

$\frac{Q}{4 π ε_{0}} [\frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}}]$

$Q R / 4 π ε_{0} d^{2}$

$\frac{Q}{2 π ε_{0}} [\frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}}]$

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Explanation

The potential difference between the centers of two thin wire rings of radius R, placed at a distance d apart with charges +q and -q, is given by (q/2Ï€Îµâ‚€)[1/R - 1/âˆš(RÂ² + dÂ²)]. This is obtained by considering the potential due to each ring at the center of the other ring and taking their difference.

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A hollow metallic sphere of radius R is given a charge Q. Then the potential at the centre is -

Zero

$\frac{1}{4 π ε_{0}} . \frac{Q}{R}$

$\frac{1}{4 π ε_{0}} . \frac{2 Q}{R}$

$\frac{1}{4 π ε_{0}} . \frac{Q}{2 R}$

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Explanation

Potential V any where inside the hollow sphere, including the centre is $V = \frac{1}{4 π ε_{0}} \frac{Q}{r}$ .

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A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in -

Reduction of charge on the plates and increase of potential difference across the plates

Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates

Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates

None of the above

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Explanation

Battery is disconnected so Q will be constant as C ∝ K. So with introduction of dielectric slab capacitance will increase using Q = CV, V will decrease and using $U = \frac{Q^{2}}{2 C}$ , energy will decrease.

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The capacity of a condenser is 4 × 10^–6 farad and its potential is 100 volts. The energy released on discharging it fully will be -

0.02 Joule

0.04 Joule

0.025 Joule

0.05 Joule

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Explanation

$The energy released = the energy stored in the capacitor . Energy stored in the capacitor - U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 4 \times 10^{- 6} \times {(100)}^{2} = 0 .02 J$

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The insulated spheres of radii R₁ and R₂ having charges Q₁and Q₂ respectively are connected to each other. There is -

No change in the energy of the system

An increase in the energy of the system

Always a decrease in the energy of the system

A decrease in the energy of the system unless Q₁R₂ = Q₂R₁

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Explanation

When $\frac{Q_{1}}{R_{1}} \neq \frac{Q_{2}}{R_{2}};$ current will flow in connecting wire so that energy decreases in the form of heat through the connecting wire.

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