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CBSE Class 12 Physics Solved Question Paper 2011

Short Answer Type

A resistance R is connected across a cell of emf $element of$ 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 $straight epsilon$ , V and R.

The expression for internal resistance is given by,

$Error converting from MathML to accessible text.$

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Two uniformly large parallel thin plates having charge densities + $straight sigma$ and – $sigma$ 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

$rightwards double arrow space E space equals space fraction numerator m g over denominator q end fraction$ , is acting vertically downwards (along –Y axis).

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

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

i) V_A - V_Bis positive when charge Q is positive.

ii) V_A- V_B is negative.

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Define the term 'wattless current'.

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

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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.

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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 $straight mu space straight C$

Charge on each capacitor, $straight q subscript 1 space equals straight q subscript 2 space equals space CV space equals space left parenthesis 1 μF right parenthesis cross times left parenthesis 6 straight V right parenthesis space equals space 6 μC$

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

$C subscript 1 apostrophe space equals space C subscript 2 apostrophe space equals space 3 cross times 1 space mu F space equals space 3 space mu F$

i)

Now, charge on capacitor C_{1 ,}q₁’ = C₁’V_{1 = $3 cross times 6 space equals space 18 space mu C$

Charge on capacitor C₂ remains the same, i.e., 6 $straight mu space straight C$}
ii)

Potential difference across C₁ remains the same.

_{Potential difference across C₂is, $V subscript 2 apostrophe space equals space fraction numerator q subscript 2 over denominator C subscript 2 apostrophe end fraction equals 6 over 3 equals space 2 space V$}

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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.

2064 Views

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,

theta italic space italic equals italic space italic 60 to the power of o

Horizontal component,

straight H space equals space 0.4 space straight G space equals space 0.4 space cross times space 10 to the power of negative 4 end exponent space straight T

Let, B_e be the earth’s magnetic field, then

$Error converting from MathML to accessible text.$

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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., $mu subscript r space less than space 1$ .

So, magnetic material is diamagnetic.

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10.

Two small identical electrical dipoles AB and CD, each of dipole moment 'p' are kept at an angle of 120^o as shown in the figure. What is the resultant dipole moment of this combination? If this system is subjected to electric field ( $straight E with rightwards harpoon with barb upwards on top$ ) 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 120^o to each other.

Resultant magnetic dipole moment is given by,

$Error converting from MathML to accessible text.$

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

Now, torque is given by,

$tau space equals p with rightwards harpoon with barb upwards on top space cross times space E with rightwards harpoon with barb upwards on top italic space italic space equals space p E space sin theta italic space italic space equals p E sin 30 to the power of o space space equals space 1 half pE space$