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CBSE

Subject

Physics

Class

CBSE Class 12

CBSE Physics 2011 Exam Questions

Short Answer Type

1.

Where on the surface of Earth is the angle of dip 90°? 


At poles, angle of dip is 90°

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2.

A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. What is the potential at the centre of the sphere?


Potential at centre of sphere is equal to 10 V.

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3.

Define electric dipole moment. Write its S.I. unit. 


Electric dipole moment is defined as the product of charge and the distance between the charges, and is directed from negative to positive charge.

The SI unit of electric dipole moment is coulomb metre (Cm). 

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4.

Two bar magnets are quickly moved towards a metallic loop connected across a capacitor ‘C’ as shown in the figure. Predict the polarity of the capacitor. 



Left face of the coil will act as North Pole and right face as South Pole so as to oppose the current induced in the coil. As seen from left, the direction of the current in the coil will be anti-clockwise. 



That is, plate of the capacitor A will be positive and plate B will be negative.

Upper plate is positive and lower plate is negative.

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5.

In the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine the potential at point B.


Current in 2  resistor is 1 A. 

Now, applying Kirchhoff’s law ACDB,

VA + 1 + 2 x 1 - 2 = VB

As, VA = 0

Therefore, 

VB = 1 + 2 - 2 = 1 V

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6.

A thin straight infinitely long conducting wire having charge density straight lambda is enclosed by a cylindrical surface of radius r and length l, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder. 


Given, charge density = straight lambda 

Radius and length of the cylindrical surface are r and l respectively.

Charge enclosed by the cylindrical surface = straight lambda l

Now, using Gauss theorem,

Electric flux, straight ϕ 

             Error converting from MathML to accessible text. 

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7.

Plot a graph showing the variation of coulomb force (F) versus 1 over straight r squared, where r is the distance between the two charges of each pair of charges: (1 straight muC, 2 straight muC) and (2 straight muC – 3 straight muC). Interpret the graphs obtained. 


Force between charged particles is given by, 


The graph between F and 1 over straight r squared is a straight line of slope  passing through the origin.



Magnitude of slope is more for attraction, therefore attractive force is greater than the repulsive force. 

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8.

How are radio waves produced?


Radio waves are produced by the accelerated motion of charges in conducting wires.

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9.

What are eddy currents? Write any two applications of eddy currents.


Eddy currents are the currents induced in a metallic plate when it is kept in a time varying magnetic field. Magnetic flux linked with the plate changes and so the induced current is set up. Eddy currents are sometimes so strong, that metallic plate becomes red hot.

Applications:

  1. In induction furnace, the metal to be heated is placed in a rapidly varying magnetic field produced by high frequency alternating current. Strong eddy currents are set up in the metal produce so much heat that the metal melts. This process is used in extracting a metal from its ore. The arrangement of heating the metal by means of strong induced currents is called the induction furnace. 

  2. Induction Motor: The eddy currents may be used to rotate the rotor. When a metallic cylinder (or rotor) is placed in a rotating magnetic field, eddy currents are produced in it. According to Lenz’s law, these currents tend to reduce to relative motion between the cylinder and the field. The cylinder, therefore, begins to rotate in the direction of the field. This is the principle of induction motion. 
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10.

Write the expression for Lorentz magnetic force on a particle of charge ‘q’ moving with velocity v in a magnetic field B  . Show that no work is done by this force on the charged particle.

OR 

A steady current (I1) flows through a long straight wire. Another wire carrying steady current (I2) in the same direction is kept close and parallel to the first wire. Show with the help of a diagram how the magnetic field due to the current I1 exerts a magnetic force on the second wire. Write the expression for this force.

Lorentz magnetic force is given by, 

Therefore, work done is, 


As, 

So, Work done, W = 0

                                                            OR

Consider two long straight conductors PQ and RS placed parallel to each other carrying currents I1 and I2 respectively. Conductors experience an attractive force when the conductors move in the same direction while, repulsive force is experienced by the conductors when currents move in opposite direction.

PQ and RS which are placed at a distance r, carry currents I1 and I2 in the same direction.

Suppose, a current element ‘ab’ of length  of wire RS.

The magnetic field produced by current-carrying conductor PQ at the location of other wire RS,


So, total force on the conductor of length L is given by, 

Force acting per unit length of conductor is given by,  

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