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NEET NEET Questions — Free MCQ Practice

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At a distance r, two equal charges are kept and they exert a force F on each other. What is the force acting on each charge, if the distance between them is doubled and charges are halved?

A.

F/4

B.

1.4 F

C.

F/16

D.

F/8

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A negatively charged plate has charge density of 2 × 10^–6 C/m². The initial distance of an electron which is moving toward plate but cannot strike the plate, if it is having energy of 200 eV

A.

1.77 mm

B.

3.51 mm

C.

1.77 cm

D.

3.51 cm

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Consider two points 1 and 2 in a region outside a charged sphere. Two points are not very far away from the sphere. If E and V represent the electric field vector and the electric potential, which of the following is not possible

A.

$| {\vec{E}}_{1} | = | {\vec{E}}_{2} |, V_{1} = V_{2}$

B.

${\vec{E}}_{1} \neq {\vec{E}}_{2}, V_{1} \neq V_{2}$

C.

${\vec{E}}_{1} \neq {\vec{E}}_{2}, V_{1} = V_{2}$

D.

$| {\vec{E}}_{1} | = | {\vec{E}}_{2} |, V_{1} \neq V_{2}$

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Explanation

Outside the charged sphere, (for equal distances from centre) if electric fields at two points are same then both points must be equipotential points.

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A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x = +1 cm and C be the point on the y-axis at y = +1 cm. Then the potentials at the points A, B and C satisfy

A.

V_A < V_B

B.

V_A > V_B

C.

V_A < V_C

D.

V_A > V_C

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Explanation

In a uniform electric field pointing in the positive x-direction, the electric potential increases as we move in the positive x-direction. Since point B (x = +1 cm) is farther along the positive x-axis than the origin A (x = 0), the potential at B is higher than the potential at A. Therefore, V_A < V_B.

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The electric potential at a point (x, y) in the x – y plane is given by V = –kxy. The field intensity at a distance r from the origin varies as

A.

r²

B.

r

C.

$\frac{1}{r}$

D.

$\frac{1}{r^{2}}$

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Two equal point charges are fixed at x = –a and x = +a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to

A.

x

B.

x²

C.

x³

D.

1/x

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Explanation

When a point charge Q is displaced by a small distance x along the x-axis, the change in its potential energy due to the two fixed charges (-q at -a and +q at +a) is approximately proportional to x^2. This is because the potential energy of a point charge in an electric field varies inversely with the distance from the source charges.

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An elementary particle of mass m and charge +e is projected with velocity v at a much more massive particle of charge Ze, where Z > 0. What is the closest possible approach of the incident particle ?

A.

$\frac{Z e^{2}}{2 π ε_{0} m v^{2}}$

B.

$\frac{Z e}{4 π ε_{0} m v^{2}}$

C.

$\frac{Z e^{2}}{8 π ε_{0} m v^{2}}$

D.

$\frac{Z e}{8 π ε_{0} m v^{2}}$

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Explanation

Suppose distance of closest approach is r, and according to energy conservation applied for elementary charge.

Energy at the time of projection = Energy at the distance of closest approach

⇒ $\frac{1}{2} m v^{2} = \frac{1}{4 π ε_{0}} . \frac{(Z e) . e}{r} \Rightarrow r = \frac{Z e^{2}}{2 π ε_{0} m v^{2}}$

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A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of –3Q, the new potential difference between the same two surfaces is

A.

V

B.

2V

C.

4V

D.

–2V

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Explanation

In case of a charged conducting sphere

$V_{inside} = V_{centre} = V_{surface} = \frac{1}{4 π ε_{o}} . \frac{q}{R}$ , $V_{outside} = \frac{1}{4 π ε_{0}} . \frac{q}{r}$

If a and b are the radii of sphere and spherical shell respectively, then potential at their surface will be

$V_{sphere} = \frac{1}{4 π ε_{0}} . \frac{Q}{a}$ and $V_{shell} = \frac{1}{4 π ε_{0}} . \frac{Q}{b}$

∴ $V = V_{sphere} - V_{shell} = \frac{1}{4 π ε_{0}} . [\frac{Q}{a} - \frac{Q}{b}]$

Now when the shell is given charge (–3Q), then the potential will be

$V'_{sphere} = \frac{1}{4 π ε_{0}} [\frac{Q}{a} + \frac{(- 3 Q)}{b}], V'_{shell} = \frac{1}{4 π ε_{0}} [\frac{Q}{b} + \frac{(- 3 Q)}{b}]$

∴ $V'_{sphere} - V'_{shell} = \frac{1}{4 π ε_{0}} [\frac{Q}{a} - \frac{Q}{b}] = V$

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A piece of cloud is having area 25 × 10⁶ m² and electric potential of 10⁵ volts. If the height of cloud is 0.75 km, then energy of electric field between earth and cloud will be

A.

250 J

B.

759 J

C.

1225 J

D.

1475 J

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Explanation

Energy $= \frac{1}{2} ε_{0} E^{2} \times (A \times d) = \frac{1}{2} ε_{0} (\frac{V^{2}}{d^{2}}) A d$

$= \frac{1}{2} \times \frac{8 .85 \times 10^{- 12} \times {(10^{5})}^{2} \times 25 \times 10^{6}}{0 .75 \times 10^{3}} = 1475 J$

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Capacitance of a capacitor made by a thin metal foil is 2 μF. If the foil is folded with paper of thickness 0.15 mm, dielectric constant of paper is 2.5 and width of paper is 400 mm, then length of foil will be

A.

0.34 m

B.

1.33 m

C.

13.4 m

D.

33.9 m

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Explanation

If length of the foil is l then $C = \frac{k ε_{0} (l \times b)}{d}$

⇒ $2 \times 10^{- 6} = \frac{2 .5 \times 8 .85 \times 10^{- 12} (l \times 400 \times 10^{- 3})}{0 .15 \times 10^{- 3}}$

⇒ l = 33.9 m

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