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Physics MCQs for NEET — Practice Questions with Answers

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Two charge +q and –q are situated at a certain distance. At the point exactly midway between them -

Electric field and potential both are zero

Electric field is zero but the potential is not zero

Electric field is not zero but the potential is zero

Neither electric field nor potential is zero

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Two insulated charged conducting spheres of radii 20 cm and 15 cm respectively and having an equal charge of 10 C are connected by a copper wire and then they are separated. Then -

Both the spheres will have the same charge of 10 C

Surface charge density on the 20 cm sphere will be greater than that on the 15 cm sphere

Surface charge density on the 15 cm sphere will be greater than that on the 20 cm sphere

Surface charge density on the two spheres will be equal

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Explanation

After redistribution, charges on them will be different, but they will acquire common potential

i.e. $k \frac{Q_{1}}{r_{1}} = k \frac{Q_{2}}{r_{2}}$ ⇒ $\frac{Q_{1}}{Q_{2}} = \frac{r_{1}}{r_{2}}$

As $σ = \frac{Q}{4 π r^{2}}$ ⇒ $\frac{σ_{1}}{σ_{2}} = \frac{Q_{1}}{Q_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}}$

⇒ $\frac{σ_{1}}{σ_{2}} = \frac{r_{2}}{r_{1}}$ ⇒ $σ \propto \frac{1}{r}$

i.e. surface charge density on smaller sphere will be more.

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Two equal charges q are placed at a distance of 2a and a third charge –2q is placed at the midpoint. The potential energy of the system is -

$\frac{q^{2}}{8 π ε_{0} a}$

$\frac{6 q^{2}}{8 π ε_{0} a}$

$- \frac{7 q^{2}}{8 π ε_{0} a}$

$\frac{9 q^{2}}{8 π ε_{0} a}$

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Explanation

$U_{s y s t e m} = \frac{1}{4 π ε_{0}} \frac{(q) (- 2 q)}{a} + \frac{1}{4 π ε_{0}} \frac{(- 2 q) (q)}{a} + \frac{1}{4 π ε_{0}} \frac{(q) (q)}{2 a}$

$U_{s y s t e m} = - \frac{7 q^{2}}{8 π ε_{0} a}$

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A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is -

$q E y^{2}$

$q E^{2} y$

$q E y$

$q^{2} E y$

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Explanation

Kinetic energy = Force × Displacement = qEy

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How much kinetic energy will be gained by an $α$ – particle in going from a point at 70 V to another point at 50 V ?

$40 e V$

$40 k e V$

$40 M e V$

$0 e V$

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Explanation

$K E = q (V_{1} - V_{2}) = 2 \times 1 .6 \times 10^{- 19} \times (70 - 50) = 40 e V$

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If a charged spherical conductor of radius 10 cm has potential V at a point distant 5 cm from its centre, then the potential at a point distant 15 cm from the centre will be -

$\frac{1}{3} V$

$\frac{2}{3} V$

$\frac{3}{2} V$

3 V

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Explanation

Potential inside the sphere will be same as that on its surface i.e. $V = V_{s u r f a c e} = \frac{K Q}{10} u n i t s$ , $V_{o u t} = \frac{K Q}{15} u n i t s$

∴ $\frac{V_{o u t}}{V} = \frac{2}{3}$ ⇒ $V_{o u t} = \frac{2}{3} V$

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What is the potential energy of the equal positive point charges of 1 μC each held 1 m apart in air ?

$9 \times 10^{- 3} J$

$9 \times 10^{- 3} e V$

$2 e V / m$

Zero

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Explanation

By using $U = 9 \times 10^{9} \frac{Q_{1} Q_{2}}{r}$

⇒ $U = 9 \times 10^{9} \times \frac{10^{- 6} \times 10^{- 6}}{1} = 9 \times 10^{- 3} J$

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An oil drop having charge 2e is kept stationary between two parallel horizontal plates 2.0 cm apart when a potential difference of 12000 volts is applied between them. If the density of oil is 900 kg/m³, the radius of the drop will be -

$2 .0 \times 10^{- 6} m$

$1 .7 \times 10^{- 6} m$

$1 .4 \times 10^{- 6} m$

$1 .1 \times 10^{- 6} m$

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Explanation

In equilibrium QE = mg ⇒ $Q . \frac{V}{d} = m g = (\frac{4}{3} π r^{3} ρ) g$

⇒ $2 \times 1 .6 \times 10^{- 19} \times \frac{12000}{2 \times 10^{- 2}} = \frac{4}{3} π r^{3} \times 900 \times 10$

⇒ r = 1.7 × 10^–6 m

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The ratio of momenta of an electron and an $α$ -particle which are accelerated from rest by a potential difference of 100 volt is

$\sqrt{\frac{2 m_{e}}{m_{α}}}$

$\sqrt{\frac{m_{e}}{m_{α}}}$

$\sqrt{\frac{m_{e}}{2 m_{α}}}$

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Explanation

Momentum $p = \sqrt{2 m K};$ where K = kinetic energy = Q.V

⇒ $p = \sqrt{2 m Q V}$ ⇒ $p \propto \sqrt{m Q}$

$\Rightarrow \frac{p_{e}}{p_{α}} = \sqrt{\frac{m_{e} Q_{e}}{m_{α} Q_{α}}} = \sqrt{\frac{m_{e}}{2 m_{α}}}$

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When a proton is accelerated through 1V, then its kinetic energy will be -

1840 eV

13.6 eV

1 eV

0.54 eV

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Explanation

$Δ K E = q V = e V = e \times 1 = 1 e V$

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