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All Physics Chemistry Botany Zoology

Language English हिंदी

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Two equal negative charge – q is fixed at the fixed points (0, a) and (0, –a) on the Y-axis. A positive charge Q is released from rest at the point (2a, 0) on the X-axis. The charge Q will

Execute simple harmonic motion about the origin

Move to the origin and remain at rest

Move to infinity

Execute oscillatory but not simple harmonic motion

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Explanation

Two equal negative charges -q are fixed on the Y-axis, and a positive charge Q is released from rest on the X-axis. The net force on Q will be attractive towards the origin, but the force will not be along the line joining Q and the origin, causing oscillatory but not simple harmonic motion.

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An electric line of force in the xy plane is given by equation x² + y² = 1. A particle with unit positive charge, initially at rest at the point x = 1, y = 0 in the xy plane will -

Not move at all

Will move along straight line

Will move along the circular line of force

Information is insufficient to draw any conclusion

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Explanation

Charge will move along the circular line of force because x² + y² = 1 is the equation of circle in xy-plane.

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A positively charged ball hangs from a silk thread. We put a positive test charge q₀ at a point and measure F/q₀, then it can be predicted that the electric field strength E

> F/q₀

= F/q₀

< F/q₀

Cannot be estimated

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Explanation

Because of the presence of positive test charge q₀ in front of positively charged ball, charge on the ball will be redistributed, less charge on the front half surface and more charge on the back half surface. As a result of this net force F between ball and point charge will decrease i.e. actual electric field will be greater than F/q₀.

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A solid metallic sphere has a charge +3Q. Concentric with this sphere is a conducting spherical shell having charge –Q. The radius of the sphere is a and that of the spherical shell is b (b > a). What is the electric field at a distance R(a < R < b) from the centre

$\frac{Q}{2 π ε_{0} R}$

$\frac{3 Q}{2 π ε_{0} R}$

$\frac{3 Q}{4 π ε_{0} R^{2}}$

$\frac{4 Q}{4 π ε_{0} R^{2}}$

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Explanation

Electric field at a distance R is only due to sphere because electric field due to shell inside it is always zero. Hence electric field = $\frac{1}{4 π ε_{0}} . \frac{3 Q}{R^{2}}$

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A point charge q is placed at a distance a/2 directly above the centre of a square of side a. The electric flux through the square is

$\frac{q}{ε_{0}}$

$\frac{q}{π ε_{0}}$

$\frac{q}{4 ε_{0}}$

$\frac{q}{6 ε_{0}}$

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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 $(K = \frac{1}{4 π ε_{0}})$

$K \frac{2 λ_{1} λ_{2}}{R^{2}}$

$K \frac{2 λ_{1} λ_{2}}{R}$

$K \frac{λ_{1} λ_{2}}{R^{2}}$

$K \frac{λ_{1} λ_{2}}{R}$

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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.

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The charge on 500 cc of water due to protons will be

6.0 × 10²⁷ C

2.67 × 10⁷ C

6 × 10²³ C

1.67 × 10²³ C

If both the assertion and the reason are true and the reason is a correct explanation of the assertion
If both the assertion and reason are true but the reason is not a correct explanation of the assertion
If the assertion is true but the reason is false
If both the assertion and reason are false

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Explanation

Q = ne; where n = number of moles × 6.02 × 10²³ × no. of protons in one atom of water

(no. of protons in an atom of water = protons in hydrogen x 2 + protons in oxygen = 1 x 2 + 8 = 10)

⇒ $Q = \frac{500}{18} \times 6 .02 \times 10^{23} \times 10 \times 1 .6 \times 10^{- 19} = 2 .67 \times 10^{7} C$

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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

$\frac{1}{4 π ε_{0}} A γ_{0}^{3}$

$4 π ε_{0} A γ_{0}^{3}$

$\frac{4 π ε_{0} A}{γ_{0}}$

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

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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.

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

$\frac{q}{2 π^{2} ε_{0} R^{2}}$

$\frac{q}{4 π^{2} ε_{0} R^{2}}$

$\frac{q}{4 π ε_{0} R^{2}}$

$\frac{q}{2 π ε_{0} R^{2}}$

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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

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

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

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

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

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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.

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

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