State the underlying principle of a cyclotron. Write briefly how this machine is used to accelerate charged particles to high energies. from Physics Class 12 CBSE Year 2014 Free Solved Previous Year Papers

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Physics

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CBSE Class 12

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CBSE Physics 2014 Exam Questions

Short Answer Type

11.

The figure given below shows the block diagram of a generalized communication system. Identify the element labelled 'X' and write its function.



The element labelled 'X' is called 'channel'.

Channel connects transmitter and receiver. The signal from the transmitter is carried to the receiver by the communication channel.







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12.

Answer the following:

(a) Name the EM waves which are suitable for radar systems used in aircraft navigation. Write the range of frequency of these waves.

(b) If the earth did not have atmosphere, would its average surface temperature be higher or lower than what it is now? Explain.

(c) An EM wave exerts pressure on the surface on which it is incident. Justify.

(a) Microwaves are suitable for radar systems used in aircraft navigation. The range of frequency for these waves is 109 Hz to 1012 Hz.

(b) There would be no greenhouse effect on the surface of the Earth in the absence of atmosphere. As a result, the temperature of the Earth would decrease rapidly, making it difficult for human survival. 

(c) When the wave is incident on the metal surface, it is completely absorbed. Energy U and hence momentum (p = Energy divided by straight c right parenthesis is delivered to the surface of the earth. The momentum delivered becomes twice when, the wave is totally reflected because momentum is changed from p to –p. Thus, force and thereby pressure is exerted on the surface of the earth by EM waves.

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13.

(a) Obtain the expression for the energy stored per unit volume in a charged parallel plate capacitor.

(b) The electric field inside a parallel plate capacitor is E. Find the amount of work done in moving a charge q over a closed loop a b c d a. 



(a) Consider a parallel-plate capacitor of plate area A.




Let us say, charge Q is given to the capacitor. Now, in order to increase the separation between the plates, plate b is slowly pulled away from plate a.

Distance between the two capacitor plates = d

Force on plate b due to plate a is given by, F space equals fraction numerator Q squared over denominator 2 A epsilon subscript o end fraction
Work done inorder to displace the plate from its fixed position is, W=Fd 

     equals space fraction numerator straight Q squared straight d over denominator 2 Aε subscript straight o end fraction equals fraction numerator straight Q squared over denominator 2 straight C end fraction ; where C equals fraction numerator A epsilon subscript o over denominator d end fraction is the capacitance of the capacitor.
Work done is equal to increase in energy of the system.

therefore straight space straight U straight space equals straight space fraction numerator straight Q squared over denominator 2 straight C end fraction
Electric field is created in a volume which is given by, V = Ad

So, Energy stored per unit volume is given by, 


Error converting from MathML to accessible text.
where, E is the intensity of the electric field. 

(b) Work done, W= F.d ;

where, F is the force exerted on electrical charge and d is the displacement.

Since, the charge is moved along a closed path, net displacement is zero.

Therefore, work done= 0

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14.

Write the truth table for the combination of the gates shown. Name the gates used. 


From the logic gates given above, we can say that

R is the OR GATE

S is the AND GATE

  A

  B

  Y' = A+B

  Y = Y'.A

  0

  0

  0

  0

  0

  1

  1

  0

  1

  0

  1

  1

  1

  1

  1

  1

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15.

An electric dipole of length 4 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of 4  Nm. Calculate the potential energy of the dipole, if it has charge ±8 n C. 


Given,

Length of the dipole, l = 4 cm 

Torque, straight tau = 4square root of 3 N m

Using the formula for torque, we have

Torque, τ = PEsinθ = (Ql) Esinθ                                         ...(1)  


Potential energy, U= −PEcosθ= −(Ql)Ecosθ                    ...(2)

Dividing equation (2) by (1),

tau over U equals space fraction numerator left parenthesis Q l right parenthesis space E space sin space theta over denominator negative left parenthesis Q l right parenthesis space E space cos space theta end fraction space equals space minus tan space theta

U space equals negative space fraction numerator tau over denominator italic tan space theta end fraction equals negative space fraction numerator tau over denominator italic tan italic 60 to the power of o end fraction
U equals negative fraction numerator 4 square root of 3 over denominator square root of 3 end fraction equals space minus 4 space J

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16.

State the underlying principle of a cyclotron. Write briefly how this machine is used to accelerate charged particles to high energies.


Cyclotron works on the principle that an oscillating electric field can be used to accelerate a charge particle to high energy.

In a cyclotron, the charged particles across the gap between the two D-shaped magnetic field regions are accelerated by an electric field. The magnetic field is perpendicular to the paths of the charged particles that make them follow circular paths within the two Dee’s. An alternating voltage accelerates the charged particles each time they cross the Dee’s. The radius of each particle’s path increases with its speed. So, the accelerated particles spiral toward the outer wall of the cyclotron.

The accelerating electric field reverses just at the time the charge particle finishes its half circle so that it gets accelerated across the gap between the Dee’s. The particle gets accelerated again and again, and its velocity increases. Therefore, it attains high kinetic energy.

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17.

A potentiometer wire of length 1 m has a resistance of 10 Ω. It is connected to a 6 V battery in series with a resistance of 5 Ω. Determine the emf of the primary cell which gives a balance point at 40 cm. 


Given,

Total length of the potentiometer wire, L = 1m

Resistance of the wire, R = 10 Ω

Voltage of the battery = 6 V

Resistance of the battery = 5 Ω

Therefore, total resistance of the circuit, R = (RAB + 5) Ω = 15 Ω



Using the figure given above, we have

Current in the circuit, I = V R = 6 x 15 A

Therefore, 

Voltage across ABVAB = i.RAB = 4 V

Emf of the cell, e = l over straight L straight V subscript 0                                ... (1) 

Here,

Balance point is obtained at, l = 40 cm

Total length, AB = L = 1 m = 100 cm

Putting the values in equation (1), we have

therefore straight space Emf comma straight space straight e straight space equals straight space 40 over 100 left parenthesis 4 right parenthesis straight space equals straight space 1.6 straight space straight V

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18.

For a single slit of width "a", the first minimum of the interference pattern of a monochromatic light of wavelength λ occurs at an angle of lambda over a . At the same angle of lambda over a  , we get a maximum for two narrow slits separated by a distance "a". Explain. 


Width of the slit = a

The path difference between two secondary wavelets is given by,

                       N λ = a sin θ.

Since, θ is very small sin  θ =  θ.

So, for the first order diffraction n = 1, the angle is λ/a.

Now we know that θ must be very small θ = 0 (nearly) because of which the diffraction pattern is minimum.

Now for interference case, for two interfering waves of intensity I1 and I2 we must have two slits separated by a distance.

We have the resultant intensity, I = straight I subscript 1 space plus space straight I subscript 2 space plus 2 square root of straight I subscript 1 straight I subscript 2 cosθ end root 
Since, θ = 0 (nearly) corresponding to angle λ/a so cos θ = 1 (nearly) 


So,

space space space space space space space space space straight I space equals straight I subscript 1 space plus space straight I subscript 2 space plus 2 square root of straight I subscript 1 straight I subscript 2 cosθ end root

rightwards double arrow space space space italic space straight I space equals space straight I subscript 1 space plus space straight I subscript 2 space plus 2 space square root of straight I subscript 1 straight I subscript 2 end root cos left parenthesis 0 right parenthesis

rightwards double arrow space space space space straight I space equals space straight I subscript 1 space plus space straight I subscript 2 space plus 2 space square root of straight I subscript 1 straight I subscript 2 end root italic space


We see the resultant intensity is sum of the two intensities, so there is a maxima corresponding to the angle λ/a.

This is why at the same angle of λ/a we get a maximum for two narrow slits separated by a distance "a".

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19.

A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens? 


Given, convex lens placed in contact with a plane mirror.

Image of the object coincides with the object.

So, the rays refracted from the first lens and then reflected by the plane mirror will be retracing their path.  This would happen when rays refracted by the convex lens fall normally on the mirror i.e., the refracted rays form a beam parallel to principal axis of the lens.

Hence, the object is at the focus of the convex lens.

Therefore, focal length, f = 20 cm

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20.

(a) State Ampere's circuital law, expressing it in the integral form.

(b) Two long coaxial insulated solenoids, S1 and S2 of equal lengths are wound one over the other as shown in the figure. A steady current "I" flow thought the inner solenoid S1 to the other end B, which is connected to the outer solenoid S2 through which the same current "I" flows in the opposite direction so as to come out at end A. If n1 and n2 are the number of turns per unit length, find the magnitude and direction of the net magnetic field at a point

(i) Inside on the axis and (ii) outside the combined system.        



(a) Ampere’s circuital law states that the line integral of magnetic induction  around a closed path in vacuum is equal to  times the total current I threading the closed path. 

 
In the above illustration, the Ampere’s Circuital Law can be written as follows: 

contour integral B ⃗. d l rightwards arrow equals mu subscript o space I semicolon space w h e r e comma space i equals vertical line i subscript 1 minus i subscript 2 vertical line.

(b) (i) The magnetic field due to a current carrying solenoid:

straight B with rightwards harpoon with barb upwards on top equals straight space straight mu subscript straight o straight space end subscript ni

where, n = number of turns per unit length


          i = current through the solenoid.

From the fig, we can say that, the direction of magnetic field due to solenoid S1 will be in the upward direction and the magnetic field due to S2 will be in the downward direction using right hand screw rule.

Therefore, 

straight B subscript net equals straight B subscript straight S 1 end subscript minus straight B subscript straight S 2 end subscript

rightwards double arrow straight B subscript net equals straight mu subscript 0 straight n subscript 1 straight I minus straight mu subscript 0 straight n subscript 2 straight I

space space space space space space space space space space space space equals straight mu subscript 0 straight I left parenthesis straight n subscript 1 minus straight n subscript 2 right parenthesis

Net magnetic field is in the upward direction.

(ii) Since, there is no current which is flowing outside the solenoid, the magnetic field is zero.

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