Are these Physics questions free to attempt?

Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

Do I need an account to practise Physics MCQs?

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

Are these previous year NEET questions?

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.

Physics NEET Questions — Free MCQ Practice

● ● ● ● ●

Two infinitely long parallel wires having linear charge densities λ₁ and λ₂ respectively are placed at a distance of R meters. The force per unit length on either wire will be $(K = \frac{1}{4 π ε_{0}})$

A.

$K \frac{2 λ_{1} λ_{2}}{R^{2}}$

B.

$K \frac{2 λ_{1} λ_{2}}{R}$

C.

$K \frac{λ_{1} λ_{2}}{R^{2}}$

D.

$K \frac{λ_{1} λ_{2}}{R}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

When two infinitely long parallel wires with linear charge densities Î»1 and Î»2 are placed at a distance R, the force per unit length on either wire is given by the expression (2kÎ»1Î»2)/R, where k = 1/(4Ï€Îµ0) is the Coulomb constant. This formula arises from the application of Coulomb's law and the principle of linear superposition.

● ● ● ● ●

The electric field in a region is radially outward with magnitude $E = A γ_{0}$ . The charge contained in a sphere of radius $γ_{0}$ centered at the origin is

A.

$\frac{1}{4 π ε_{0}} A γ_{0}^{3}$

B.

$4 π ε_{0} A γ_{0}^{3}$

C.

$\frac{4 π ε_{0} A}{γ_{0}}$

D.

$\frac{1}{4 π ε_{0}} \frac{A}{γ_{0}^{3}}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Flux linked with the given sphere $φ = \frac{Q}{ε_{o}};$

where Q = Charge enclosed by the sphere.

Hence Q = φε₀ = (EA)ε₀

⇒ Q = 4π (γ₀)² × Aγ₀ε₀ = 4πε₀Aγ₀³.

● ● ● ● ●

Charge q is uniformly distributed over a thin half-ring of radius R. The electric field at the centre of the ring is

A.

$\frac{q}{2 π^{2} ε_{0} R^{2}}$

B.

$\frac{q}{4 π^{2} ε_{0} R^{2}}$

C.

$\frac{q}{4 π ε_{0} R^{2}}$

D.

$\frac{q}{2 π ε_{0} R^{2}}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

Two equal charges are separated by a distance d. A third charge placed on a perpendicular bisector at x distance will experience maximum coulomb force when

A.

$x = \frac{d}{\sqrt{2}}$

B.

$x = \frac{d}{2}$

C.

$x = \frac{d}{2 \sqrt{2}}$

D.

$x = \frac{d}{2 \sqrt{3}}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

When two equal charges are separated by a distance d, the maximum Coulomb force experienced by a third charge placed on the perpendicular bisector occurs when the third charge is at a distance x = d/(2âˆš2) from the midpoint of the line joining the two charges. This position maximizes the net force due to the two equal charges.

● ● ● ● ●

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.

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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 $ω^{2} = \frac{p E}{I}$ ⇒ $ω = \sqrt{\frac{p E}{I}}$

● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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

● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

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.

$10^{- 47} C$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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

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

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

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

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

● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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= $2 \times 10^{- 2} m$

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

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

● ● ● ● ●

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.

$\frac{q}{2 ε_{0}} 6 a^{2}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Charge enclosed=q/8

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

$ϕ = \frac{q}{8 ε_{0}}$

Physics MCQs for NEET — Practice Questions with Answers

Frequently Asked Questions