CBSE Set Qa1 Physics Sample Test Papers For Class 12th for students online

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Physics Class- XII CBSE)
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Q 1. What is the work done in moving a charge of 10 nC between two points on an equipotential surface ? (1 mark)
Ans 1. 
dw  = qdv
  on equipotential surface dv = 0
   
   dw = 0 i.e,no work is done in moving a change on an equipotential surface.

Q 2. Name the device used for measuring the internal resistance of a secondary cell ? (1 mark)
Ans 2.
  Internal resistance of a secondary cell can be measured by a potentiometer.

Q 3.  What is the nature of magnetic field in a moving coil galvanometer ? (1 mark)
Ans 3.  A radial  magnetic fields exists in a moving coil galvanometer.

Q 4.If a rate of change of current of 2 A/s induces an e.m.f.of 10mV in a solenoid, what is the self -inductance of the solenoid ? (1 mark)

Ans 4.  Given dI   = 2 A/S , 10mv = 10-3V
      dt     
      | | = L dI
                 dt
 
  
or  10 x 10

-3 = L x 2


       or L = 10-2 =  5 x 10-3 H = 5mH
2

Q 5. What type of magnetic material is used in making permanent magnets ? (1 mark)
Ans 5.
  A substance like steel which as a large retentivity and a large coercivity is used in making permanent magnets.

Q 6. Two metals A, B have work-functions 2eV, 4eV respectively. Which metal has lower threshold wavelength for photoelectric effect ? (1 mark)
Ans 6.  
Work function W = h v

0 = hc where  is threshold wavelength.

    
 1  
             W
 Metal B has higher value of W & hence lower threshold wavelength.

Q 7. Which have greater ionising power : alpha particles or beta particles ? (1 mark)
Ans 7.   Alpha particles have more ionising power than beta particles.

Q 8.  Which part of the electromagnetic spectrum is used in operating a RADAR ? (1 mark)
Ans 8.
  Microwaves are used in operating a RADAR.

Q 9. Which type of biasing gives a semiconductor diode very high resistance ? (1 mark)
Ans 9.
  Reverse biasing gives a semiconductor diode very high reistance.

Q 10.How is the luminosity of a star related to its radius ? (1 mark)
Ans 10.
 The luminosity 'L' is related to the radius 'r' by the equation.

      L = 4o T4

 where T is the absolute temperature of the star and 0 is the Stefan's constant.

Q 11. An electric dipole, when held at 300 with respect to a uniform electric field of 104N/C, experiences a torque of 9 x 10-26Nm. Calculate the dipole moment of the dipole. (2 marks)
Ans 11.
 Torque z = pE since
      Given z = 9 x 10-26 Nm , E = 104 N/C , O = 300,

      
9 x 10-26 = p x 104 x sin 30 = p x 104 x  1 
                                                                      2 

      or p = 9 x 10-26 x 2  =   18 x 10-30 c - m
                     104

Q 12.A set of n identical resistors, each of resistance R ohm, when connected in series have an effective resistance X ohm and when the resistors are connected in parellel, their effective resistance is Y ohm. Find the relation between R, X and Y. (2 marks)
Ans 12
When connected in series
       X = R + R + ............  n times = nR      ---(1)
       When connected in parellel
        1   =   1 +   1 + ........  n times   =    n
         Y        R      R                                  R
        or Y =  R   ---(2)
                   n
        Multiplying eqns (1) & (2)
        XY = nR x  R   =   R2
                         
n
        or R = XY

Q 13.State Kirchhoff's rules for eletrical networks. (2 marks)
Ans 13.
 Kirchoffs rules for elecrical networks :
1.  First law or Junction law : It states that the algebraic sum of the currents meeting at a juction in a closed circuit is zero.
     2.  Second law or loop rule : It states that in any closed path of an electrical circuit, the algebraic sum of  the emfs is equal to the algebraic sum of the resistance & the respective currents flowing through them.

Q 14.Show that Lenz's law is in accordance with the law of conservation of energy. (2 marks)

Ans 14.  Lenz's law is in accordance with the law of conservation of energy. For this, consider a coil connected to a galvanometer as shown.

        

When the N-pole of the magnet is moved towards the coil then according to Lenz's law the end of coil facing the magnet acquires north polarity.Therefore work has to be done against the force of repulsion in bringing the magnet closer to the coil. Similarily, when the N-pole of a magnet is moved away, south polarity develops on the end of the coil facing the agnet.Therefore work has to be done against the force of attraction in taking the magnet away. In both the cases, it is mechanical work done in moving the magnet wrt the coil that changes into electrical energy, producing induced current and the galvanometer shows deflection.Thus energy is only being transformed.
When we don't move the magnet, work done is zero.Hence no induced current is produced.

Q 15.Draw a labelled diagram of Hertz's experiment for producing electromagnetic waves. (2 marks)
Ans 15.
  Hertz's expt

Q 16.Give two differences between fringes formed in single slit diffraction and Young's double slit experiment. (2 marks)
Ans 16. Differences between fringes formed in single slit diffraction & Young's double slit expt :
(1) In Young's double slit experiment all the bright fringes are of the same intensity, whereas in single ;slit diffraction pattern all the bright fringes are not of same intensity.
 (2)  In Young's double slit experiment, the intensity of minima is generally zero or very small & there isgood contrast between bright and dark fringes. In single slit diffraction pattern the intensity at minima is never zero &  there is poor contrast between bright & dark fringes.

Q 17.Light of wavelength 600nm is incident on an aperture of size 2mm. Calculate the distance up to which the ray of light can travel such that its spread is less than the size of the aperture.  (2 marks)
Ans 17.
  The distance upto which a ray of light can travel that its spread is less than the size of the aperture is Fresnel distance which is given by Z


             F = a2
                    

        Given a = 2mm = 2 x 10-3m           

      = 600nm =600 x 10-9 m

 

                ZF = (2 x 10-3)   = 6.67 m
                               x    600 x 10-9


Q 18.Explain the principle used in Bunsen's grease photometer to compare the illuminating powers of two light sources. (2 marks)
Ans 18.  Bunsen's grease photometer is based on the principle of photometry which states that if the two sources produce equal illuminations on a surface, then the ratio of the luminious intensities of the two sources is equal to the ratio of the square of the respective distances of the two sources from that surface.
...............The two sources whose luminious intensities are to be compared are placed on the two sides of the screen of the Bunsen's photometer & their distances from the screen are so adjusted that  the grease spot becomes equally bright on its two sides. If E1 & E2 are the illuminations on the screen
due to the respective sources, then:

            E1 = E2
         
      


Q 19.A ray of light while travelling from a denser to a rarer medium undergoes total internal reflection. Derive the expression for the critical angle in terms of the speed of light in the respective media.
(2 marks)

Ans 19.  When the the angle of incidence is equal to the critical angle 'ic' then angle of reflection = 900
         =
sin i      =     sin ic        =    sin ic
                    sin r             sin90
         = Speed of light is denser medium
                   Speed of light is rarer medium

          sin ic = Speed of light is denser medium
                         Speed of light is rarer medium

Q 20.Draw a graph showing the variation of binding energy per nucleon with mass number of different nuclei. Mark the region where the nuclei are most stable. (2 marks)
Ans 20. 
                            

Q 21.Name the logic gate shown in the diagram and give its truth table. (2 marks)
Ans 21.  And Gate is shown in the given diagram
...............Its truth table is :

A
B
C
0
1
0
1
0
0
1
1
0
0
0
1

Q 22. State and explain Seebeck effect. (2 marks)
Ans 22.  Seebeck is the phenomenon of generation of electric current in a thermocouple by keeping its two junctions at different temperatures.
............... When two dissimilar metals are joined to form thermocouple, then due to diffusion of free electron from metal, with greater free electron density to lower free electron density, a potential difference is developed at the functions called contact potential. When there is temperature difference at the two
junctions, contact potentials at the two fuctions become different & there is an effective potential difference in the circuit called thermoems & current flows in the circuit.

Q 23.  Explain with the help of a circuit diagram how the value of an unknown resistance can be determined using a Wheatstone bridge. Give the formula used. (3 marks)
Ans 23

Slide wire bridge (Wheatstone bridge ) to field unknown resistence

...............In figure X is unknown resistance and R.B. is resistance box. After inserting the key K, jockey is moved on wire  AC till the galvanometer shows no deflection (Point B).
     If x is the resistance per unit length of wire AC.
         P = resistance of AB = xl
        
Q = resistance of BC = x(100 - l)
                                            R = P =          xl    
                                               X   Q        x(100-l)
        
        or                                           X = (100 - l )R
                                                                 l

 

Q 24 Earth receives an amount of heat radiaton 1.4 x 103 W/m2 from the sun.Assume that earth re-emits all the radiation recieved from the sun.Calculate the surface tempratueof the earth. (3 marks)
Ans 24 Amount of radiation recieved by earth per unit area per second = 1.4 x 103 w/m2
..........................................................  = solar constant
       Energy recieved per second by earth = 4 S           (1)
..............................where r is radius of earth.
       According to Stephans law,energy radiated per second per unit area = T4, where is Stephans constant.
       Energy radiated perb second by earth = 4 T4 (2)
       Equating   (1)  &  (2)
       4  T4     =  4S
                or T4    =     S 
...................................
                     T = (S)1/4 = ( 1.4 x 103)1/4 = 396.4 k
..........................              5.67 x 10-8
     
Q 25. A straight wire carries a current of 3A.Calulate the magnitude of the magnetic field at a point 10cm away from the wire.Draw diagram to show the direction of the magnetic field. (3 marks)
Ans 25  
I = 3A, r = 10cm = 0.1 m B = ?  

                          B =     10-7 x 2 x 3
        0.1

Q 26. Explain how a vibration magnetometer can be used to determine the magnetic moment of a bar magnet. (3 marks)
Ans 26. A vibration magnetometer consists of wooden box with glass slides.A plane mirror is placed at base & index line is marked along its length.A long glass tube is fixed vertically at the center of the box.It has a torsion head T from which hangs an unspun silk fibre.A stirrup of non-magnetic material like brass is attached to the free of  the fibre,so that it is just above the base of the box.
 Setting : Magnetometer is set in magnet meridian.
 This is done by placing a compass needle on the index line & rotating the box till the compass needle is parallel to the index line.Magnet is placed on the stirrup so that the north-pole of the magnet can be set into vibrations  of small amplitude using another magnet from outside & its time period can be determined.                                        

    

   Knowing T, BH & I( I = mass ) ,where l and b are the length & breadth of the

    magnet,the magnetic moment M can be determined.


VIBRATION MAGNETOMETER

Q 27. Deduce lens maker's formula for a thin biconvex lens. 
(3 marks)
Ans 27. 
Consider a point object O lying on the principal axis of the lens.A ray of light starting from O and incient normally on the surface XP1Y along OP1 passes straight.Another ray incident on XP1Y  along OA is refracted along AB.If the lens material were continuous and there were no boundry/second surface XP2Y of the lens,the refracted ray AB would go straight meeting the first refracted ray at I1.Therefore,I1 would have been a real image  of O formed after refraction at XP1Y.
                  ....................(1)

       Actually,the lens material is not continious.Therefore,the refracted ray AB sufers furthes refraction at B and emerges along BI,meeting atually the principal axis at I.Therefore I is the final real image of O,formed after refraction through the convex lens.
                        For refraction at the second surface XP2Y,we can regard I1 as a virtual object,whose real image is formed at I.
        Therefore,
                       

         
Let R2 be radius of curvature of second surface of the lens.
                                     As refraction is now taking place from denser to rarer medium,therefore, using          eqn.(21),we get

             
              
              ..............(3)

             Put reflective index of material of lens w.r.t. surrounding medium.
          When object on the left of lens is at image is formed at the principal focus of the lens.
                                       when u = ,v = f = focal length of the lens.
                                          putting this in eqn. 3
                                          1    (M-1) ( 1 - 1 )
                                          f                  R1 R2

Q 28. Explain the origin of spectral lines of hydrogen using Bohr's theory. (3 marks)
Ans 28.Bohr's postulates of hydrogen atom :
1. There is a positively charged nucleus at the center of the atom around which electrons resolve in circular orbits.
...............The centripetal force is provided by coulomb's force of attraction exerted by nucleus.
                           Kze2  = mv2
                              r2          r
                           Kze2  = mv2
                              
r
2. Electrons revolve in certain discrete,non-radiating orbits called statimary orbits,for which the angular momentum is an integral multiple of      h 
..............................
.......................... 
3. 
When electron jumps from one stationary orbit of higher energy E2 to another lower energy E1 it emits energy of frequency to where
...............E2-E1  =   hv.

...............From eqns of (1) & (2)
............... r =      x2h2
...............  mke2   

............... Total energy of electron = K.E + P.E
                                              1 mv2 + (-ke2)
                                              2                r
............... using equation (1)      
         Total energy =  1  (kze2) - kze2      =  kze2
..............................
. 2    r           r               2r
...............substituting r from eqn (3)
...............Total energy =

When electron jumps from outer level (energy E2) to inner level (energy E1), then from Bohr's theory:
       h v =  E2-E1

        =
         hv =

               v =   C 
                          
hv = hc    =
   


Depending m values of n1 & n2,radiations of different wave lengths are emitted.

Q 29. A beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields E and B respectively.If the electric field is cut-off ,the electron beam moves in a circular
path of radius 'r'.Derive the expression for e/m of electrons in terms of  r,E and B. (3 marks)
Ans 29.  In the first eam when the electron beam goes undeflected
...............evb = eE
...............or V = E          (1)
  .......................B
.When electric field is cut off ,only magnetic force acts on it which provides thecentripetal force.
        eB = m2
                   
r
         or   e  =  m 2
               m     rB
substituting   from eqn (1)
                   e     =   E2
                  
m           rB3

Q 30. Draw a labelled diagram of Van de Graff generator.Give its principle and explain its working.
Ans 30 Van the Graft generator Principle.
This generator is based on : (i) The actions of sharp points i.e. the phenomenon of corona discharge. (ii) The property that charge given to hollow conductor is  transferred to outer surface and is distrbuted uniformily over it. Working :The spray comb is given a positive potential (~104volt) w.r.t. the earth by high tension source H.T.Due discharging action of sharp points, a positively charged electric wind is set up, which sprays positive charge on belt (corona discharge). As the belt moves,and reaches the sphere,a negative charge is induced on the sharp ends of the collecting comb B2 and an equal positive charge is induced on the farther end of B2. This positive charge shifts immediately to the outer surface of S.Due to discharging action of sharp points of B2,a negatively charged electric wind is set up.This neutralises the positive charge on the belt.The uncharged belt returns down,collects the positive charge from B1 ,which in turn is collected by B2. This is repeated .Thus thepositive charge on S goes on accumalating.

Now the capacity of spherical shell = where R is radius of the shell. As V =  Q           Q    
............ C..........
       

Hence the potential V of the spherical shell goes on increasing with increase in Q.
...............OR

The Capacitance of a capacitor is defined as the ratio of charge on the plates.Its S.I. unit is Farad.
Electrostatic energy stored in a capacitor :
Let the capacitor be charged gradually.At any stage,change on it is q.
          Therefore P.D between the plates of capacitor = q
                                                                                C
          small amount of work done in giving an additional charge dq is
               dw = q  dq
                       C
          Total work done in giving a charge Q to the capacitor
          
          But Q = CV
           Energy stored = W =  Q2 = 1 CV2
                                                     2C     2
           Energy density = Total energy = 1/2 CV2
............... ............... ........volume ............Ad
           where d : distance between plates
                        A : Coin - sectioned area of plate
                  C :
     Energy density (u) =

                        V =Ed
          u =

Q 31. Prove that the power dissipated in an ideal resistor connected to an a.c. source is V2eff/R.Acapacitor,a resistor  and a 40 mH inductor are connected in series to an a.c. source of frequency 60 Hz.Calculate the capacitance of the capacitor,if the current is in phase with the voltage. (2+3 marks)
Ans 31.  let V = V0 sin wt
       In an ideal resistor connected to an a.c source, current & voltage are in phase.
        I = I0 sin wt
        small work done oin time dt
        dw =IV dt = I0 V0 sin2 wt dt
       Total work done =  dw   =    I0V0 sin2 wt dt
                                               =    I0V0  T 
                                                                2
        
                               
        Numerical : L = 40 MH = 40 x 10 -3H
                         
 v = 60 Hz.
                If current is in phase with the voltage
                           WL =  1 
                                    WC
               or C =       1            =         1    
                              W2L            2 v2L
                                 1                                      =        176 x 10-6F
                     
4 x (3.14)2  x (60) x 40 x 10 -3  
                                                                         =        176 MF

Q 32. Explain with the help of a circuit diagram the working principle of a transistor as an amplifier in the common emmitter configuration .Derive the expression for the voltage gain of the amplifier.(5 marks)

Ans 32.
           


...The input (emitter base) circuit is forward biased with battery voltage VEB, and the output(collector-emitter) circuit is reversed biased with battery voltage Vce.Due to which the resistance of input circuit is low and that of output circuit is high.RL is a load resistance conected in collector circuit.The low input a.c voltage sgnal is applied across base-emitter  circuit and the amplified a.c. voltage signal (i.e. output) is obtained as the change in collector volta.
....When no a.c. signal voltage is applied to input circuit but emitter base circuit is closed let us consider,that Ie,Ib and Ic be the emitter current,base current and collector current  respectively.Then according to;

        Kirchhoff's first law
                      Ie = Ib + IC                                                    ---------(i)
        
If VC is collector voltage,then
                  
VCE = VC + IcRL
                     
or Vc =VCE - IcRL                                         ----------(ii)
  
When the input signal is fed to the emitter base circuit,it will change the emitter voltage and hence the emitter current,which in turn will change the collector current.Due to this the collector voltage Vc will vary in accordance with relation (ii).These variations in the collector volage appear as an amplified output.
...............To find voltage gain :
...............change in output  =   Vc = Vce - IcRL
............... ............... ........ = - IcRL ( Vce = 0).
...............Voltage gain = Av =  V 0     = V c
............... ............... ............  VL          VL
............... ...................... .. 
............... ......................Av = -13ac  R L 
............... ...................... ...............RL  


Now RL RL & 13 ac has very large value, there is large voltage gain