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Physics

Class

CBSE Class 12

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CBSE Physics 2011 Exam Questions

Short Answer Type

11.

A current is induced in coil C1 due to the motion of current carrying coil C2.

(a) Write any two ways by which a large deflection can be obtained in the galvanometer G.

(b) Suggest an alternative device to demonstrate the induced current in place of galvanometer. 



a) Two ways by which we can obtain deflection in the galvanometer is by:

i) Moving coil C2 towards C1 with high speed.

ii) By placing a soft iron laminated core at the centre of coil C1.

b) By replacing galvanometer with the torch bulb in coil C1, one can demonstrate the induced current. The bulb begins to glow as a result of the induced current.

2306 Views

12.

State Biot-Savart law, giving the mathematical expression for it.

Use this law to derive the expression for the magnetic field due to a circular coil carrying current at a point along its axis.

How does a circular loop carrying current behave as a magnet? 

Biot-Savart law states that the magnetic field strength (dB) produced due to a current element of current I and length dl at a point having position vector  to current element is given by,

                                       stack d B with rightwards harpoon with barb upwards on top space equals space fraction numerator mu subscript o over denominator 4 pi end fraction fraction numerator I space stack d l with rightwards harpoon with barb upwards on top space cross times space r with rightwards harpoon with barb upwards on top over denominator r cubed end fraction 

where, mu subscript o  is permeability of free space.

The magnitude of magnetic field is given by, 

d B space equals space fraction numerator mu subscript o over denominator 4 pi end fraction fraction numerator I space d l space sin straight space theta over denominator r squared end fractiontheta spaceis the angle between the current element and position vector.

Magnetic field at the axis of a circular loop:

Consider a circular loop of radius R carrying current I. Let, P be a point on the axis of the circular loop at a distance x from its centre O.

Let,  be a small current element at point A. 

                            

Magnitude of magnetic induction dB at point P due to this current element is given by,

 delta B space equals space fraction numerator mu subscript o over denominator 4 pi end fraction fraction numerator I space delta l space sin alpha over denominator r squared end fraction italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space... space left parenthesis 1 right parenthesis

The direction of stack delta B with rightwards harpoon with barb upwards on top is perpendicular to the plane containing stack delta l with rightwards harpoon with barb upwards on top space a n d space r with rightwards harpoon with barb upwards on top
Angle between stack delta l with rightwards harpoon with barb upwards on top space a n d space r with rightwards harpoon with barb upwards on top space i s space 90 to the power of o

Therefore, 

stack delta B with rightwards harpoon with barb upwards on top space equals space fraction numerator mu subscript o I over denominator 4 pi end fraction fraction numerator delta l space sin 90 to the power of 0 over denominator r squared end fraction equals fraction numerator mu subscript o I delta l space over denominator 4 pi r squared end fraction                           ... (2)

The magnetic induction stack delta B with rightwards harpoon with barb upwards on top can also be resolved into two components, PM and PN’ along the axis and perpendicular to the axis respectively. Thus if we consider the magnetic induction produced by the whole of the circular coil, then by symmetry the components of magnetic induction perpendicular to the axis will be cancelled out, while those parallel to the axis will be added up. 

Thus, resultant magnetic induction B with rightwards harpoon with barb upwards on top at axial point P is given by, 
T h e space c o m p o n e n t space o f space stack delta B with rightwards harpoon with barb upwards on top space a l o n g space t h e space a x i s comma space delta B subscript x space equals space fraction numerator mu subscript o over denominator 4 pi end fraction fraction numerator I space delta l space space s i n alpha over denominator r squared end fraction
Error converting from MathML to accessible text. 

Therefore the magnitude of resultant magnetic induction at axial point P due to the whole circular coil is given by, 

B = contour integral fraction numerator mu subscript o I R over denominator 4 pi left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction. d l space equals space fraction numerator mu subscript o I R over denominator 4 pi left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction contour integral d l space
contour integral d l space equals l e n g t h space o f space t h e space l o o p space equals space 2 pi R

italic space T h e r e f o r e italic comma italic space
B space equals space fraction numerator mu subscript o I R over denominator italic 4 pi italic left parenthesis R to the power of italic 2 italic plus x to the power of italic 2 italic right parenthesis to the power of italic 3 italic divided by italic 2 end exponent end fraction left parenthesis 2 pi R right parenthesis
space space space equals space fraction numerator mu subscript o I R squared over denominator 2 left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction space T e s l a

If the coil contains N turns, then B equals space fraction numerator mu subscript o N I R squared over denominator 2 left parenthesis R squared plus x squared right parenthesis to the power of 3 divided by 2 end exponent end fraction space T e s l a is the required magnetic field.

A magnetic needle placed at the center and axis of a circular coil shows deflection. This implies that a circular coil behaves as a magnet. 

4278 Views

13.

Show graphically, the variation of the de-Broglie wavelength (straight lambda) with the potential (V) through which an electron is accelerated from rest.


Wavelength is inversely proportional to the potential. 

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14.

In the circuit shown, R1 = 4 Ω, R2 = R3 = 15 Ω, R4 = 30 Ω and E = 10 V. Calculate the equivalent resistance of the circuit and the current in each resistor.



Given, R1 = 4 Ω, R2 = R3 = 15 Ω, R4 = 30 Ω and E = 10 V

Here, R2 , R3 and R4 are connected parallel to each other.

Therefore, equivalent resistance is given by,

1 over straight R equals straight space 1 over straight R subscript 2 plus 1 over straight R subscript 3 plus 1 over straight R subscript 4

space space space space space equals straight space 1 over 15 plus 1 over 15 plus 1 over 30

rightwards double arrow    straight space straight R straight space equals straight space 6 straight capital omega space is space the space effective space resistance.

Now, R1 is in series with R. So, equivalent resistance is given by, 

R subscript e q end subscript space equals space R space plus R subscript 1

italic space italic space italic space italic space italic space italic space italic equals italic space italic 6 italic plus italic space italic 4 italic space italic space

italic space italic space italic space italic space italic space italic space equals space 10 space capital omega

Current I1 is given by, I subscript 1 space equals space E over R subscript e q end subscript equals 10 over 10 equals space 1 space A                  ... (1)

This current is divided at A into three parts I2, I3, I4 .

therefore space I subscript 2 plus I subscript 3 plus I subscript 4 space equals space 1                                                            ... (2)
Also, 

A l s o italic comma italic space

I subscript 2 R subscript 2 space equals space I subscript 3 R subscript 3 space equals space I subscript 4 R subscript 4

rightwards double arrow space I subscript 2 space cross times 15 equals I subscript 3 space cross times 15 space equals I subscript 4 space cross times 30

rightwards double arrow space I subscript 2 equals I subscript 3 equals 2 I subscript 4 italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic. italic.. space left parenthesis 3 right parenthesis 

Now, putting the values of I subscript 2 and straight I subscript 3 in (2), we get

Error converting from MathML to accessible text.

Thus, I1 = 1A, I2 = I3 = 0.4 A and I4 = 0.2 A.

1557 Views

15.

With the help of a labelled diagram, state the underlying principle of a cyclotron. Explain clearly how it works to accelerate the charged particles.

Show that cyclotron frequency is independent of energy of the particle. Is there an upper limit on the energy acquired by the particle? Give reason.   

Underlying principle of cyclotron: When a charged particle is kept in a magnetic field it experiences a force and the perpendicular magnetic field causes the particle to rotate.

Working:

Consider the figure which is shown. 

A Dee which is at a negative potential accelerates the positive ions which are produced from the source S. Magnetic field which is perpendicular will move the ions in a circular motion inside the Dees. The magnetic field and the frequency of the applied voltages are so chosen that as the ion comes out of a Dee, the Dees change its polarity (positive becoming negative and vice-versa) and the ion is further accelerated.

  

Now, the ions move with higher velocity along a circular path of greater radius. The phenomenon is continued till the ion reaches at the periphery of the Dees where an auxiliary negative electrode (deflecting plate) deflects the accelerated ion on the target to be bombarded.

Expression for cyclotron frequency:

Let the velocity of the positive ion having charge ‘q’ be ‘v’.

Then, 

                                    q v B space equals space fraction numerator m v squared over denominator r end fraction
rightwards double arrow space r space equals space fraction numerator m v over denominator q B end fraction ;m is the mass of ion, r the radius of the path of ion in the dee and B is the strength of the magnetic field.

Angular velocity of the ion is given by, 

straight omega equals straight v divided by straight r equals qB divided by straight m space

Now, time taken by the ion in describing a semi-circle i.e., turning through an angle  is given by, 

t space equals space pi over omega equals fraction numerator pi m over denominator q B end fraction

The applied alternating potential should also have the same semi-periodic time (T/2) as that taken by the ion to cross either Dee.

That is, 
 
                         T over 2 equals t space equals space fraction numerator pi m over denominator q B end fraction 

rightwards double arrow space T equals space fraction numerator 2 pi m over denominator q B end fraction, is the expression for period of revolution. 
So, Frequency of revolution of particle is given by, f space equals space 1 over T equals space fraction numerator q B over denominator 2 pi m end fraction

This frequency is called the cyclotron frequency which is independent of the speed of the particle. 

1544 Views

16.

In a transistor, doping level in base is increased slightly. How will it affect (i) collector current and (ii) base current?    


When doping level is increased in base,

i) collector current decreases

ii) base current increases slightly.

1277 Views

17.

Using Gauss's law obtains the expression for the electric field due to a uniformly charged thin spherical shell of radius R at a point outside the shell. Draw a graph showing the variation of electric field with r, for r > R and r < R.     


Electric field intensity at a point outside a uniformly charged thin spherical shell:

Consider a uniformly charged thin spherical shell of radius R carrying charge Q. Let us assume a spherical Gaussian surface of radius r (>R), concentric with the given shell inorder to find the electric field outside the shell.

If  is the electric field outside the shell, then by symmetry electric field strength has same magnitude Eo on the Gaussian surface and is directed radially outward. Also, the direction of normal at each point is radially outward. So, angle between stack straight E subscript straight i with rightwards harpoon with barb upwards on top space a n d space stack d s with rightwards harpoon with barb upwards on top space i s space z e r o space a t space e a c h space p o i n t.
Therefore, electric flux through Gaussian surface, stretchy contour integral subscript s stack E subscript o with rightwards harpoon with barb upwards on top space. space stack d s with rightwards harpoon with barb upwards on top 
                                                                      space italic equals contour integral stack E subscript o with rightwards harpoon with barb upwards on top space. space stack d s with rightwards harpoon with barb upwards on top cos space 0

space equals space stack E subscript o with rightwards harpoon with barb upwards on top.4 pi r squared

Now, charge enclosed by the Gaussian surface is Q [Gaussian surface is outside the given charged shell].

Therefore, using Gauss theorem, 

space space space space space stretchy contour integral subscript s stack E subscript o with rightwards harpoon with barb upwards on top. space stack d E with rightwards harpoon with barb upwards on top space equals 1 over epsilon subscript o cross times c h a r g e space e n c l o s e d italic space

rightwards double arrow space space space E subscript o space 4 pi r to the power of italic 2 space equals italic 1 over epsilon subscript omicron cross times Q space

italic rightwards double arrow italic space italic space italic space italic space italic space italic space italic space italic space italic space E subscript o space equals space fraction numerator italic 1 over denominator italic 4 pi epsilon subscript omicron end fraction Q over r to the power of italic 2

This is the required electric field outside a given thin charged shell.

If straight sigma is the surface charge density of the spherical shell, then

phi space equals space 4 pi R squared sigma C
therefore space E subscript o space equals space fraction numerator italic 1 over denominator italic 4 pi epsilon subscript o end fraction fraction numerator italic 4 pi R to the power of italic 2 sigma over denominator r to the power of italic 2 end fraction space equals space fraction numerator R to the power of italic 2 sigma over denominator epsilon subscript o r to the power of italic 2 end fraction 

Electric field inside the shell:

If is the electric field inside the shell, then by symmetry electric field strength has the same magnitude Ei on the Gaussian surface and is directed radially outward. Also the directions of normal at each point are radially outward. So angle between Ei and  is zero at each point.

Thus, electric flux through Gaussian surface is 0 because of the absence of the charge.

The below graph shows the variation of electric field with r, for r >R and r < R.

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18.

Define the terms (i) drift velocity, (ii) relaxation time.

A conductor of length L is connected to a dc source of emf ε. If this conductor is replaced by another conductor of same material and same area of cross-section but of length 3L, how will the drift velocity change?

i) Drift velocity: The electrons drift towards the direction of positive potential, whenever a potential difference is applied across the conductor. The small average velocity of free electrons along the direction of positive potential is called the drift velocity.

Drift velocity is denoted by vd.

ii) Relaxation time: The time for which an electron moves freely between two successive collisions of electron with lattice ions/atoms is called the relaxation time.

 

Drift velocity is given by, 

straight V subscript straight d space equals space eτ over straight m straight epsilon over straight L

straight i. straight e. comma space straight V subscript straight d space italic proportional to italic 1 over L

where L is the length of the conductor,

 is the emf of the DC source,

L is the length of the conductor.

When, the conductor is replaced by another conductor of length 3L, drift- velocity will become one-third because both of them are inversely proportional to each other.

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19.

When monochromatic light travels from one medium to another its wavelength changes but frequency remains the same, Explain.      


When monochromatic light travels from one medium to another it’s wavelength and speed both changes and the frequency remains unchanged.

                                             nu subscript 1 over lambda subscript 1 equals nu subscript 2 over lambda subscript 2

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20.

Two convex lenses of same focal length but of aperture A1 and A2 (A2 < A1), are used as the objective lenses in two astronomical telescopes having identical eyepieces. What is the ratio of their resolving power? Which telescope will you prefer and why? Give reason. 


Resolving power is given by, R = fraction numerator A over denominator 122 space lambda end fraction ; where A is the aperture. 
therefore space space space space space space space space R subscript 1 over R subscript 2 equals A subscript 1 over A subscript 2

Magnification of telescope is given by, m = f subscript o over f subscript e

m is same for both. 

Inorder to view the finer details of the object we use telescope of high resolving power.

Therefore, telescope having convex lens aperture A1 is used

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