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# CBSE Physics 2011 Exam Questions

1.

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at 60° with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth's magnetic field at the place.

Given,

Angle of dip,
Horizontal component,
Let, Be be the earth’s magnetic field, then

2.

Define the term 'wattless current'.

Current flowing in an ac circuit without any net dissipation of power is called wattless current.

3.

Two uniformly large parallel thin plates having charge densities + and –are kept in the X-Z plane at a distance 'd' apart. Sketch an equipotential surface due to electric field between the plates. If a particle of mass m and charge '-q' remains stationary between the plates, what is the magnitude and direction of this field?

Equipotential surface between the plates due to the electric field is given by the figure shown below. We can see that, the equipotential surface is at a distance d/2 from either plate in X-Z plane.

Given, a particle of mass ‘m’ and charge ‘-q’ remains stationary in between the plates. For the negative charge,

i) weight mg acts, vertically downward

ii) electric force qE acts vertically upward.

So, mg = qE

, is acting vertically downwards (along –Y axis).

4.

How are infrared waves produced? Why are these referred to as 'heat waves’? Write their one important use.

Infrared waves are produced by hot bodies and molecules.

They are referred as heat waves because they are readily absorbed by water molecules in most materials, which increase their thermal motion, so they heat up the material.

Infrared waves are used for therapeutic purpose and long distance photography.

5.

The permeability of a magnetic material is 0.9983. Name the type of magnetic materials it represents.

Here, permeability is less than 1 i.e.,  .

So, magnetic material is diamagnetic.

6.

A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the direction of electric and magnetic field vectors?

The direction of electric field vectors is along X-axis.

Magnetic field vector is along Y-axis.

7.

Figure shows two identical capacitors, C1 and C2, each of 1 μF capacitance connected to a battery of 6 V. Initially switch 'S' is closed. After sometime 'S' is left open and dielectric slabs of dielectric constant K = 3 are inserted to fill completely the space between the plates of the two capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted?

When switch S is closed, potential difference across capacitor is 6V

Potential across the battery, V1 = V2 = 6V

Capacitance of the capacitor, C1 = C2 = 1

Charge on each capacitor,

When switch S is opened, the potential difference across C1 remains 6 V, while the charge on capacitor C2 remains 6 . After insertion of dielectric between the plates of each capacitor, the new capacitance of each capacitor becomes

i)

Now, charge on capacitor C1 ,q1’ = C1’V1 =

Charge on capacitor C2 remains the same, i.e., 6

ii)

Potential difference across C1 remains the same.

Potential difference across Cis,

8.

Two small identical electrical dipoles AB and CD, each of dipole moment 'p' are kept at an angle of 120o as shown in the figure. What is the resultant dipole moment of this combination? If this system is subjected to electric field () directed along + X direction, what will be the magnitude and direction of the torque acting on this?

Given, AB and CD are dipoles kept at an angle of 120o to each other.

Resultant magnetic dipole moment is given by,

Resultant magnetic dipole makes an angle 60o with Y-axis or 30o with x-axis.

Now, torque is given by,

Direction of torque is along negative Z-direction.

9.

A resistance R is connected across a cell of emf  and internal resistance r. A potentiometer now measures the potential difference between the terminals of the cell as V. Write the expression for 'r' in terms of  , V and R.

The expression for internal resistance is given by,

10.

A point charge Q is placed at point O as shown in the figure. Is the potential difference VA VB positive, negative, or zero, if Q is (i) positive (ii) negative?

With increase in distance, potential due to point charge decreases. So,

i) VA - Vis positive when charge Q is positive.

ii) VA - VB is negative.

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