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NEET NEET Questions — Free MCQ Practice

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An electron of mass m_e initially at rest moves through a certain distance in a uniform electric field in time t₁. A proton of mass m_p also initially at rest takes time t₂ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio of t₂/ t₁ is nearly equal to

A.

1

B.

${(m_{p} / m_{e})}^{1 / 2}$

C.

${(m_{e} / m_{p})}^{1 / 2}$

D.

1836

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Explanation

For electron $s = \frac{1}{2} \frac{e E}{m_{e}} \times t_{1}^{2}$ , For proton $s = \frac{1}{2} \frac{e E}{m_{p}} \times t_{2}^{2}$

∴ $\frac{t_{2}^{2}}{t_{1}^{2}} = \frac{m_{p}}{m_{e}} \Rightarrow \frac{t_{2}}{t_{1}} = \sqrt{\frac{m_{p}}{m_{e}}} = {(\frac{m_{p}}{m_{e}})}^{1 / 2}$

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Point charges +4q, –q and +4q are kept on the x-axis at points x = 0, x = a and x = 2a respectively, then

A.

Only -q is in stable equilibrium

B.

None of the charges are in equilibrium

C.

All the charges are in unstable equilibrium

D.

All the charges are in stable equilibrium

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Explanation

Force on each charge is zero. But if any of the charge is displaced, the net force starts acting on all of them.

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An electron enters an electric field with its velocity in the direction of the electric lines of force. Then

A.

The path of the electron will be a circle

B.

The path of the electron will be a parabola

C.

The velocity of the electron will decrease

D.

The velocity of the electron will increase

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Explanation

Because electric field applies the force on electron in the direction opposite to it’s motion.

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The dimension of (1/2) $ε_{0} E^{2} (ε_{0}$ : permittivity of free space; E: electric field) is

A.

MLT^–1

B.

ML²L^–2

C.

ML^–1T^–2

D.

ML²T^–1

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Explanation

Energy density $= \frac{Energy}{Volume}$

so it’s dimensions are $\frac{M L^{2} T^{- 2}}{L^{3}} = [M L^{- 1} T^{- 2}]$

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A table tennis ball that has been covered with conducting paint is suspended by a silk thread so that it hangs between two plates, out of which one is earthed and other is connected to a high voltage generator. This ball

A.

Is attracted towards high voltage plate and stays there

B.

Hangs without moving

C.

Swing backward and forward hitting each plate in turn

D.

Is attracted to the earthed plate and stays there

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Explanation

Firstly being a conductor it is attracted by the high voltage plate, when charge is shared, ball is repelled until it goes to other plate and whole of the charge is transferred to the earth and the process is repeated.

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If an insulated non-conducting sphere of radius R has charge density ρ. The electric field at a distance r from the centre of sphere (r < R) will be

A.

$\frac{ρ R}{3 ε_{0}}$

B.

$\frac{ρ r}{ε_{0}}$

C.

$\frac{ρ r}{3 ε_{0}}$

D.

$\frac{3 ρ R}{ε_{0}}$

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Explanation

For non-conducting sphere $E_{i n} = \frac{k . Q r}{R^{3}} = \frac{ρ r}{3 ε_{0}}$

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Infinite charges of magnitude q each are lying at x =1, 2, 4, 8... meter on X-axis. The value of the intensity of the electric field at point x = 0 due to these charges will be

A.

12 × 10⁹q N/C

B.

Zero

C.

6 × 10⁹q N/C

D.

4 × 10⁹q N/C

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Explanation

Net field at origin $E = \frac{q}{4 π ε_{0}} [\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{4^{2}} + . ... \infty]$

$= \frac{q}{4 π ε_{0}} [1 + \frac{1}{4} + \frac{1}{16} + . .... \infty]$

Using formula for infinite Geometric progression (G.P.) ,

if Sum = $a + \frac{a}{r} + \frac{a}{r^{2}} + \frac{a}{r^{3}} . . . . . . . . . . . . . . . . . . . . . . . . \infty$

and r<1

then ,sum = $\frac{a}{1 - r}$

In question, a =1 and r= $\frac{1}{4}$

$∴ S u m = \frac{1}{1 - \frac{1}{4}} = \frac{4}{3}$

$= \frac{q}{4 π ε_{0}} [\frac{1}{1 - \frac{1}{4}}] = 12 \times 10^{9} q N / C$

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Two infinitely long parallel conducting plates having surface charge densities +σ and –σ respectively, are separated by a small distance. The medium between the plates is a vacuum. If ε₀ is the dielectric permittivity of vacuum, then the electric field in the region between the plates is

A.

$0 v o l t s / m e t e r$

B.

$\frac{σ}{2 ε_{o}} v o l t s / m e t e r$

C.

$\frac{σ}{ε_{o}} v o l t s / m e t e r$

D.

$\frac{2 σ}{ε_{o}} v o l t s / m e t e r$

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Explanation

In the given scenario of two infinitely long parallel conducting plates with opposite surface charge densities +Ïƒ and -Ïƒ separated by a small distance in vacuum, the electric field between the plates is given by E = Ïƒ/Îµ0, where Îµ0 is the permittivity of free space.

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A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The total flux for the surface of the cylinder is given by

A.

$2 π R^{2} E$

B.

$π R^{2} / E$

C.

$(π R^{2} - π R) / E$

D.

Zero

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Explanation

For a cylinder placed in a uniform electric field parallel to its axis, the total electric flux through the curved surface is zero as the field lines are parallel to the surface. Hence, the total flux for the surface of the cylinder is zero.

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Electric field at a point varies as r° for

A.

An electric dipole

B.

A point charge

C.

A plane infinite sheet of charge

D.

A line charge of infinite length

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Explanation

E = σ / (2ε₀)

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