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Physics MCQs for NEET — Practice Questions with Answers

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Inside a hollow charged spherical conductor, the potential -

Is constant

Varies directly as the distance from the centre

Varies inversely as the distance from the centre

Varies inversely as the square of the distance from the centre

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Explanation

Inside the hollow sphere, at any point the potential is constant.

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Two small spheres each carrying a charge q are placed r meter apart. If one of the spheres is taken around the other one in a circular path of radius r, the work done will be equal to

Force between them × r

Force between them × 2πr

Force between them / 2πr

Zero

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Explanation

The force is perpendicular to the displacement.

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Two charged spheres of radii 10 cm and 15 cm are connected by a thin wire. No current will flow, if they have -

The same charge on each

The same potential

The same energy

The same field on their surfaces

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Explanation

Because current flows from higher potential to lower potential.

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The electric field inside a spherical shell of uniform surface charge density is -

Zero

Constant, less than zero

Directly proportional to the distance from the centre

None of the above

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Explanation

All charge resides on the outer surface so that according to Gauss law, electric field inside a shell is zero.

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The electric potential V at any point O (x, y, z all in metres) in space is given by $V = 4 x^{2} v o l t$ . The electric field at the point $(1 m, 0, 2 m)$ in volt/metre is -

8 along negative x-axis

8 along positive x-axis

16 along negative x-axis

16 along positive z-axis

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Explanation

The electric potential $V (x, y, z) = 4 x^{2} v o l t$

Now $\vec{E} = - (\hat{i} \frac{\partial V}{\partial x} + \hat{j} \frac{\partial V}{\partial y} + \hat{k} \frac{\partial V}{\partial z})$

Now $\frac{\partial V}{\partial x} = 8 x, \frac{\partial V}{\partial y} = 0$ and $\frac{\partial V}{\partial z} = 0$

Hence $\vec{E} = - 8 x \hat{i}$ , so at point (1m, 0, 2m)

$\vec{E} = - 8 \hat{i} v o l t / m e t r e$ or 8 volt/metre along negative X-axis.

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A hollow metal sphere of radius 5 cm is charged so that the potential on its surface is 10 V. The potential at the centre of the sphere is -

0 V

10 V

Same as at point 5 cm away from the surface

Same as at point 25 cm away from the surface

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Explanation

Since potential inside the hollow sphere is same as that on the surface.

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If a unit positive charge is taken from one point to another over an equipotential surface, then -

Work is done on the charge

Work is done by the charge

Work done is constant

No work is done

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Explanation

On the equipotential surface, electric field is normal to the charged surface (where potential exists) so that no work will be done.

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A conductor with a positive charge

Is always at +ve potential

Is always at zero potential

Is always at negative potential

May be at +ve, zero or –ve potential

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Explanation

May be at positive, zero or negative potential, it is according to the way one defines the zero potential.

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Two spheres A and B of radius 4 cm and 6 cm are given charges of 80 μc and 40μc respectively. If they are connected by a fine wire, the amount of charge flowing from one to the other is -

20μc from A to B

16μc from A to B

32μc from B to A

32μc from A to B

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Explanation

Total charge $Q = 80 + 40 = 120 μ C$ . By using the formula $Q_{1} ' = Q [\frac{r_{1}}{r_{1} + r_{2}}]$ .

A new charge on sphere A is $Q_{A}^{'} = Q [\frac{r_{A}}{r_{A} + r_{B}}] = 120 [\frac{4}{4 + 6}] = 48 μ C$ .

Initially it was 80 μC i.e., (80-48)= 32 μC charge flows from A to B.

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On rotating a point charge having a charge q around a charge Q in a circle of radius r. The work done will be

$q \times 2 π r$

$\frac{q \times 2 π Q}{r}$

Zero

$\frac{Q}{2 ε_{0} r}$

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Explanation

Since charge Q moving on equipotential surface so work done is zero.

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