Book Store

Download books and chapters from book store.
Currently only available for.
CBSE

Subject

Physics

Class

CBSE Class 12
zigya logo

CBSE Physics 2016 Exam Questions

Short Answer Type

1.

a) Write the expression for the magnetic force acting on a charged particle moving with velocity v in the presence of magnetic field B. 

b) A neutron, an electron and an alpha particle, moving with equal velocities, enter a uniform magnetic field going into the plane of paper, as shown. Trace their paths in the field and justify your answer. 


a) 

Force acting on a charged particle q, which is moving with velocity v in a magnetic field B, is given by, 


                                          straight F with rightwards harpoon with barb upwards on top space equals space q space left parenthesis v with rightwards harpoon with barb upwards on top space x space B with rightwards harpoon with barb upwards on top space right parenthesis
The right-hand rule gives the direction of this force. The direction of the force is perpendicular to the plane containing velocity v and magnetic field B. 

b) 

A charged particle experiences a force when it enters the magnetic field. Due to the presence of magnetic field, the charged particle will move in a circular path. This is because the force is perpendicular to the velocity of the charged particle. 

Radius of the circular path in which the charged particle is moving is given by, 


mv squared over straight r space equals space q v B

space space space space space r space equals space fraction numerator m v over denominator q B end fraction

B space a n d space v space a r e space c o n s t a n t. space

W e space c a n space w r i t e comma space

r space proportional to space m over q 

Since the neutron has no charge, it will move along a straight line. 

The electron will follow a circular path which has a radius smaller than that of the alpha particle. This is because the mass to charge ratio of the alpha particle is more than that of the electron. 

Therefore, the electron will move in the clockwise direction and the electron will move in the anticlockwise direction as per the Right Hand Rule. 


2. Why are microwaves considered suitable for radar systems used in aircraft navigation?

Microwaves are considered suitable for radar systems used in aircraft navigation because they have a short wavelength range (10-3 m to 0.3 m), which makes them suitable for long range communication. 

3. Two cells of emfs 1.5 V and 2.0 V,  having internal resistances 0.2 Ω and 0.3 Ω, respectively, are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.

Given, 

EMF, E1 = 1.5 V ; E2 = 2 V 

Internal resistance, r = 0.2 straight capital omega space nd space 0.3 space straight capital omega

Effective emf of two cells connected in parallel is, 

Eeff =   fraction numerator straight E subscript 1 straight r subscript 2 space plus space straight E subscript 2 straight r subscript 1 over denominator straight r subscript 1 space plus space straight r subscript 2 end fraction

This implies, 

straight E subscript eff space equals space fraction numerator 1.5 space straight x space 0.3 space plus space 2.0 space straight x space 0.2 over denominator 0.5 end fraction space equals space 1.7 space straight V

The effective resistance can be calculated as:

Refffraction numerator straight r subscript 1 straight r subscript 2 over denominator straight r subscript 1 space plus space straight r subscript 2 end fraction

That is, 

Refffraction numerator 0.2 space straight x space 0.3 over denominator 0.5 end fraction space equals space 0.12 space capital omega




4.

A point charge +Q is placed at point O as shown in the figure. Is the potential
difference VA – VB is positive, negative or zero?


Electric potential at a distance r from the point charge +Q is given by, 

 V thin space left parenthesis straight r right parenthesis space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction space. space Q over r 

Potential at point A is, 


straight V space left parenthesis straight r subscript straight A space right parenthesis thin space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction space straight Q over straight r subscript straight A 

Similarly, potential at point B is given by, 

straight V space left parenthesis straight r subscript straight B space right parenthesis thin space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction space straight Q over straight r subscript straight B

Since rA < rB 


rightwards double arrow space straight V subscript straight A space greater than space straight V subscript straight B

therefore space straight V subscript straight A space minus space straight V subscript straight B space greater than thin space 0


5. Write the underlying principle of a moving coil galvanometer.

When a current-carrying coil is placed in a magnetic field, it experiences torque. This is the underlying principle of a moving coil galvanometer.

6. Two parallel-plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates, while Y contains a dielectric medium of straight epsilon subscript straight r space end subscript equals space 4



i) Calculate capacitance of each capacitor if the equivalent capacitance of the combination is 4 μF
ii) Calculate the potential difference between the plates x and Y

iii) Estimate the ratio of electrostatic energies stored in x and Y

i) 

Capacitance of a parallel plate capacitor is given by, 


straight C space equals space fraction numerator straight epsilon subscript straight o space straight epsilon subscript straight r space straight A over denominator straight d end fraction

Capacitance space of space plate space straight Y space is comma space

straight C subscript straight Y space equals space fraction numerator straight epsilon subscript straight o 4 straight A over denominator straight d end fraction

Capacitance space of space plate space straight X space is comma space

straight C subscript straight X space equals space fraction numerator straight epsilon subscript straight o straight A over denominator straight d end fraction

Thus comma space

space space space space space space straight C subscript straight Y space equals space 4 space straight C subscript straight X
rightwards double arrow space straight C subscript eq space equals space fraction numerator straight C subscript straight X space straight C subscript straight Y over denominator straight C subscript straight X space plus straight C subscript straight Y space end fraction space equals space 4 space μF

rightwards double arrow space fraction numerator straight C subscript straight X space left parenthesis 4 straight C subscript straight X right parenthesis over denominator 5 space straight C subscript straight X end fraction space equals space 4 space straight mu space straight F

rightwards double arrow space space straight C subscript straight X space equals space 5 space straight mu space straight F

rightwards double arrow space space straight C subscript straight Y space equals space 20 space μF 

ii) 

Potential difference between plates X and Y can be calculated as, 

            Q = CV 

rightwards double arrow space straight C subscript straight X space straight V subscript straight X space equals space straight C subscript straight Y straight V subscript straight Y space space

rightwards double arrow space straight V subscript straight X over straight V subscript straight Y space equals space straight C subscript straight Y over straight C subscript straight X space equals space 4

rightwards double arrow space straight V subscript straight X space equals space 4 space straight V subscript straight Y space

Also comma space straight V subscript straight Y space plus space straight V subscript straight X space equals space 15 space

rightwards double arrow space straight V subscript straight Y space equals space 3 space straight V space

rightwards double arrow space straight V subscript straight X space equals space 12 space straight V 

iii) 

The ratio of electrostatic energies can be calculated as: 


straight E space equals space fraction numerator straight Q squared over denominator 2 straight C end fraction

rightwards double arrow space straight E subscript straight X over straight E subscript straight Y space equals space straight C subscript straight Y over straight C subscript straight X space equals space 4 space

rightwards double arrow space straight E subscript straight x over straight E subscript straight Y space equals space 4 over 1


7.

i) Define mutual Inductance. 

ii) A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes 20 A in 0.5 s, what is the change of flux linkage with the other coil?


i) Mutual Induction: It is the phenomenon in which a change of current in one coil induces an emf in another coil places near it. The coil in which the current changes is called the primary coil and the coil in which the emf is induced is called the secondary coil. 

ii) 

EMF induced is, e = - M . dI over dt
straight e space equals space minus 1.5 space straight x space fraction numerator 20 space minus space 0 over denominator 0.5 end fraction
straight e space equals space minus 60 space straight V

So, the flux linked with the other coil is given by, 


increment straight ϕ space equals space straight e space straight x space increment straight t space equals space minus 60 space straight x space 0.5 space equals space minus space 30 space Wb


8. Define 'quality factor' of resonance in a series LCR circuit. What is its SI unit?

Quality factor relates the maximum or peak energy stored in the circuit (the reactance) to the energy dissipated (the resistance) during each cycle of oscillation. 

Quality factor is expressed as, 

straight Q space equals space fraction numerator straight omega subscript straight o straight L over denominator straight R end fraction

Q-factor is a dimensionless quantity. 


9. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?

According to Gauss's law, flux through a closed surface is given by,

                                 straight ϕ space equals space straight q over straight epsilon subscript straight o

Here, q is the charge enclosed by the gaussian surface. 

That is, on increasing the radius of the gaussian surface, charge q remains unchanged.

So, flux through the gaussian surface will not be affected when its radius is increased. 

10. A charge is distributed uniformly over a ring of radius 'a'. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence, show that for points at large distance from the ring, it behaves like a point charge.

Consider a ring of radius 'a' which carries uniformly distributed positive total charge Q. 




To find: electric field due to a ring at a point P lying at a distance x from its centre along the central axis perpendicular to the plane of the ring.
 

As the charge is distributed uniformly over the ring, the charge density over the ring is, 

                                  straight lambda space equals fraction numerator straight Q over denominator 2 πa end fraction
The perpendicular component of electric field due to charge on the ring along the x-axis cancels each other out.

As there is same charge on both sides of the ring, the magnitude of the electric field at P due to the segment of charge dQ is given by, 

dE = ke dQ over straight r squared
Exintegral subscript ring space k space fraction numerator d Q over denominator r squared end fraction. space c o s space theta 

   space equals space integral subscript 0 superscript 2 πa end superscript space k space fraction numerator lambda d l over denominator r squared end fraction space x over r

space equals k space lambda space x over r cubed integral subscript 0 superscript 2 pi a end superscript space d l

equals space k lambda space x over r cubed space 2 pi a space

equals space k space fraction numerator Q over denominator 2 pi a end fraction space x over r cubed space. space space 2 pi a space

equals space kQ space straight x over straight r cubed

equals space kQ space fraction numerator straight x over denominator square root of open parentheses straight x squared plus straight a squared close parentheses cubed end root end fraction

1. At the centre (X = 0) , electric field is zero. 

2. When x>> a, a can be neglected in the denominator. 

Therefore, 

 straight E space equals space kQ space bevelled fraction numerator straight x over denominator square root of open parentheses straight x squared plus straight a squared close parentheses cubed end root end fraction

straight E space equals space straight k space straight Q over straight x squared space

1.png
curious learner
Do a good deed today
Refer a friend to Zigya