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NEET NEET Questions — Free MCQ Practice

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An electric dipole is situated in an electric field of uniform intensity E whose dipole moment is p and moment of inertia is I. If the dipole is displaced slightly from the equilibrium position, then the angular frequency of its oscillations is

A.

${(\frac{p E}{I})}^{1 / 2}$

B.

${(\frac{p E}{I})}^{3 / 2}$

C.

${(\frac{I}{p E})}^{1 / 2}$

D.

${(\frac{p}{I E})}^{1 / 2}$

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Explanation

When dipole is given a small angular displacement θ about it's equilibrium position, the restoring torque will be

$τ = - p E \sin θ = - p E θ$ (as sinθ = θ)

or $I \frac{d^{2} θ}{d t^{2}} = - p E θ$ (as $τ = I α = I \frac{d^{2} θ}{d t^{2}}$ )

or $\frac{d^{2} θ}{d t^{2}} = - ω^{2} θ$ with $ω^{2} = \frac{p E}{I}$ ⇒ $ω = \sqrt{\frac{p E}{I}}$

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An infinite number of electric charges each equal to 5 nano-coulomb (magnitude) are placed along x-axis at x = 1 cm, x = 2 cm, x = 4 cm, x = 8 cm ………. and so on. In the setup if the consecutive charges have opposite sign, then the electric field in Newton/Coulomb at x = 0 is $(\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N - m^{2} / c^{2})$

A.

12 × 10⁴

B.

24 × 10⁴

C.

36 × 10⁴

D.

48 × 10⁴

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Explanation

$E = \frac{1}{4 π ε_{0}} . [\frac{5 \times 10^{- 9}}{{(1 \times 10^{- 2})}^{2}} - \frac{5 \times 10^{- 9}}{{(2 \times 10^{- 2})}^{2}} + \frac{5 \times 10^{- 9}}{{(4 \times 10^{- 2})}^{2}} - \frac{(5 \times 10^{- 9})}{{(8 \times 10^{- 2})}^{2}} + . ....]$

$\Rightarrow E = \frac{9 \times 10^{9} \times 5 \times 10^{- 9}}{10^{- 4}} [1 - \frac{1}{{(2)}^{2}} + \frac{1}{{(4)}^{2}} - \frac{1}{{(8)}^{2}} + . ..]$

$\Rightarrow E = 45 \times 10^{4} [1 + \frac{1}{{(4)}^{2}} + \frac{1}{{(16)}^{2}} + . ..] - 45 \times 10^{4} [\frac{1}{{(2)}^{2}} + \frac{1}{{(8)}^{2}} + \frac{1}{{(32)}^{2}} + . ..]$

$\Rightarrow E = 45 \times 10^{4} [\frac{1}{1 - \frac{1}{16}}] - \frac{45 \times 10^{4}}{(2)^{2}} [1 + \frac{1}{4^{2}} + \frac{1}{{(16)}^{2}} + . . .]$

E = 48 × 10⁴ – 12 × 10⁴ = 36 × 10⁴ N/C

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Two-point charges +q and –q are held fixed at (–d, 0) and (d, 0) respectively of a (x, y) coordinate system. Then

A.

E at all points on the y-axis is along $\hat{i}$

B.

The electric field $\vec{E}$ at all points on the x-axis has the same direction

C.

Dipole moment is 2qd directed along $\hat{i}$

D.

Work has to be done in bringing a test charge from infinity to the origin

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Suppose the charge of a proton and an electron differ slightly. One of them is -e and the other is $(e + ∆ e)$ . If the net of electrostatic force and gravitaional force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero,then $∆ e$ is of the order [Given mass of hydrogen, $m_{h}$ =1.67 $\times 10^{- 27}$ kg]

A.

$10^{- 20} C$

B.

$10^{- 23} C$

C.

$10^{- 37} C$

D.

$10^{- 47} C$

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Explanation

(c) Net charge on one H-atom

=-e+e+ $∆ e$

Net electrostatic repulsive force between two H-atoms

$F_{e} = \frac{K {(∆ e)}^{2}}{d^{2}}$

$F_{G} = \frac{G m_{1}^{2}}{d^{2}}$

It is given that

$F_{e} - F_{G} = 0$

$\Rightarrow \frac{K {(∆ e)}^{2}}{d^{2}} - \frac{G m_{1}^{2}}{d^{2}} = 0$

$\Rightarrow {(∆ e)}^{2} = \frac{(6.67 \times 10^{- 11}) {(1.67 \times 10^{- 27})}^{2}}{9 \times 10^{9}}$

$\Rightarrow ∆ e = 1.437 \times 10^{- 37} C$

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An electric dipole is place at an angle of $30^{\circ}$ with an electric field intensity 2 $\times 10^{5}$ N/C. It experiences a torque equal to 4 Nm. The charge on the dipole, if the dipole length is 2 cm, is

A.

8 mC

B.

2 mC

C.

5 mC

D.

7 $μ$ C

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Explanation

(b) $∵$ Torque on an electric dipole in an electric field,

$τ$ =p $\times E \Rightarrow |τ| = p E \sin θ$

where $θ$ is the angle between E and p

$\Rightarrow 4 = ρ \times 2 \times 10^{5} \times \frac{1}{2} \Rightarrow p = 4 \times 10^{- 5} c m \Rightarrow p = q 2 l ∴ q 2 l = 4 \times 10^{- 5}$

Where 2l=2cm= $2 \times 10^{- 2} m$

$q = \frac{4 \times 10^{- 5}}{2 \times 10^{- 2}}$

$\Rightarrow 2 \times 10^{- 3} C = 2 m C$

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What is the flux through a cube of side a if a point charge of q is a one of its corner?

A.

$\frac{2 q}{ε_{0}}$

B.

$\frac{q}{8 ε_{0}}$

C.

$\frac{q}{ε_{0}}$

D.

$\frac{q}{2 ε_{0}} 6 a^{2}$

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Explanation

Charge enclosed=q/8

Therefore,flux $ϕ = \frac{q_{e n c l o s e d}}{ε_{0}}$

$ϕ = \frac{q}{8 ε_{0}}$

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Two parallel metal plates having charges +Q and -Q faces each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will

A.

become zero

B.

increase

C.

decrease

D.

remain same

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Explanation

When the parallel metal plates are dipped in kerosene oil, the electric field between the plates decreases. This is because the kerosene oil has a higher permittivity than vacuum or air, which reduces the effective electric field according to the relation E = Eâ‚€/Îº, where Îº is the relative permittivity.

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The mean free path of electrons in a metal is $4 \times 10^{- 8} m .$ The electric field which can give on an average 2 eV energy to an electron in the metal will be in a unit of $V m^{- 1}$ :

A.

$8 \times 10^{7}$

B.

$5 \times 10^{- 11}$

C.

$8 \times 10^{- 11}$

D.

$5 \times 10^{7}$

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Explanation

Energy=2eV

$e V_{0} = 2 e V$

$\Rightarrow V_{0} = 2$

Now, electric field E= $\frac{2}{4 \times 10^{- 8}}$

= $0.5 \times 10^{8}$

= $5 \times 10^{7} V m^{- 1}$

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Four particles each having charge q are placed at the vertices of a square of side a. The value of the electric potential at the midpoint of one of the side will be

A.

0

B.

$\frac{1}{4 π ϵ_{0}} \frac{2 q}{a} (2 + \frac{2}{\sqrt{5}})$

C.

$\frac{1}{4 π ϵ_{0}} \frac{2 q}{a} (2 - \frac{2}{\sqrt{5}})$

D.

$\frac{1}{4 π ϵ_{0}} \frac{2 q}{a} (1 + \frac{1}{\sqrt{5}})$

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If W be the amount of heat produced in the process of charging an uncharged capacitor then the amount of energy stored in it is

A.

2W

B.

$\frac{W}{2}$

C.

W

D.

zero

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Explanation

Energy stored in capacitor = Heat produced in process of charging

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