CBSE physics

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Question
CBSEENPH12039331

A charge ‘q’ is moved from a point A above a dipole of dipole moment ‘p’ to a point B below the dipole in equatorial plane without acceleration. Find the work done in the process.

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Solution

Work done, W = q x VAB = q X 0 = 0

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Question
CBSEENPH12039332

In what way is the behaviour of a diamagnetic material different from that of a paramagnetic, when kept in an external magnetic field?

Solution

Difference in the behaviour of a diamagnetic material and paramagnetic material:

Diamagnetic Paramagnetic
1. Diamagnetic substance would move towards the weaker region of the magnetic field. 1. Paramagnetic substance move towards the stronger region of the magnetic field.
2. Diamagnetic substance is repelled by a magnet. 2. Paramagnetic substance moves towards the magnet.
3. Diamagnetic substance get aligned perpendicular to the field. 3. Paramagnetic substance get aligned along the magnetic field.

Question
CBSEENPH12039335

The plot of the variation of potential difference across a combination of three identical cells in series, versus current is shown below. What is the emf and internal resistance of each cell?

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Solution

Potential difference, E = 6/3 = 2 V
Internal resistance, r = straight r space equals space straight E over straight I equals 6 over 1 equals 6 space ohm
Given, three cells are connected in series,
r' = r/3 = 6/3 = 2 ohm

Question
CBSEENPH12039341

A battery of emf 12V and internal resistance 2 ohm is connected to a 4 ohm resistor as shown in the figure.

a) Show that a voltmeter when placed across the cell and across the resistor, in turn, gives the same reading.

b) To record the voltage and the current in the circuit, why is voltmeter placed in parallel and ammeter in series in the circuit?
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Solution

a) Emf, E = 12 V
Internal resistance, r = 2V
Now, using the formula,
E = V + Ir
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When the voltmeter is connected across the cell,
I = fraction numerator 12 over denominator 4 plus 2 end fraction space equals space 2 space A
V1 = 12 - 2(2) = 8 V
When the voltmeter is connected across the resistor,
V2 = IR
    = 2 x 4 = 8 V
That is, V1 = V2
Hence proved.
b) Voltmeter has very high resistance to ensure that it is connection does not alter the flow of current in the circuit. Current chooses the low resistance path. Therefore, voltmeter is connected in parallel to the load across which potential difference is to be measured.
Ammeter measures the value of current flowing through the circuit. Ammeter has a very low value of resistance to ensure that all the current flows through it. Hence, it should be connected in series.

Question
CBSEENPH12039342

Define an equipotential surface. Draw equipotential surfaces:

(i) in the case of a single point charge and in a constant electric field in Z-direction.

(ii) Why the equipotential surfaces about a single charge are not equidistant?

(iii) Can electric field exist tangential to an equipotential surface? Give reason.

Solution

i) Equipotential surface for a single point charge is:       

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Equipotential surface in a constant field in Z- direction.
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ii) The equipotential surface about a single charge is not equidistant because V is inversely proportional to r.
Also, the equipotential surfaces about a single charge are not equidistant because electric field due to a  single charge is not constant.
iii) Electric field cannot exist tangential to an equipotential surface because if the field lines are tangential, work will be done in moving a charge on the surface, which is against the theory of equipotential surface.