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

Physics Class- XII CBSE)
You are on Set no 1 Answer 1 to 12

Q1) Name the physical quantity whose SI unit is Coulomb/Volt.  (Marks 1)
Ans1)  Capacitance

Q2) Write the frequency limit of visible range of electromagnetic spectrum in kHz.  (Marks 1)
Ans2) 8 x 10¹¹ HHz to 4 x 10¹¹ KHz

Q3) How does the conductance of a semi-conducting material change with rise in temperature?  (Marks 1)
Ans3) Increases.

Q4) The force experienced by a particle of charge e moving with velocity in a magnetic field is given by. = e( x ) Of these, name the pairs of vectors which are always at right angles to each other.  (Marks 1)
Ans4)  = e( x )
  |  and   | 

Q5) Two wires A and B are of same metal, have the same area of cross-section and have their lengths in the ratio 2 : 1. What will be the ratio of currents flowing through them respectively when the same potential difference is applied across length of each of them? 
(Marks 1)
Ans5) Resistance length
R_A/R_B = 1/2
Current is inversely proportional to resistance
.^.. I_A/I_B = 1/2

Q6) Calculate the rms value of the alternating current shown in the figure.  (Marks 1)

Ans6) Irms = i² = ((2² + (-2)² +2²)/3) = (12/3) = 2A

Q7) The image of an object formed by a lens on the screen is not in sharp focus. Suggest a method to get clear focusing of the image on the screen without disturbing the position of the object the lens or the screen.  (Marks 1)
Ans7) The reason for such image is spherical aberration. Clearer focusing can be got by using the other central position of the lens & blocking the inter portion or using other proton & blocking central portion.

Q8) Two points A and B are placed between two parallel plates having a potential difference V as shown in the figure.

Will these protons experience equal or unequal force?  (Marks 1)
Ans8) Both A and B will experience equal force as field between the plates is uniform.

Q9) Define the terms 'threshold frequency' and 'stopping potential' for photo-electric effect. Show graphically how the stopping potential, for a given metal, varies with frequency of the incident radiations. Mark threshold frequency on this graph.  (Marks 2)
Ans9) Threshold frequency: It is the maximum frequency of the incident radiation below which no emission of photoelectrons takes place.
Stopping potential : It is the maximum negative potential required to stop the fastest photoelectrons i e, photoelectric current becomes zero.

Q10) Draw labelled diagram of a Leclanche cell. Write the function of charcoal powder and manganese dioxide used in its porous pot.  (Marks 2)
Ans10)  

Charcoal powder makes manganese dioxide conducting and manganese dioxide acts as depolarizer.

Q11) How does the mutual inductance of a pair of coils change when :
(i) the distance between the coils is increased?
(ii) the number of turns in each coil is decreased?
Justify your answer in each case.  (Marks 2)
Ans11) (i) Mutual inductance decrease with the increase of the distance between the coils. This is due to lower coupling between the coils.
(ii) With the decreases in the no of turns in each coil, the mutual inductance of a pair of coils is given by
_nN_sN_pA/l

Q12) Define the terms magnetic inclination and horizontal component of Earth's magnetic field at a place. Establish the relationship between the two with the help of the diagram.  (Marks 2)
Ans12)  Magnetic inclination : at a place is defined as the angle when the direction of total intensity of earth's magnetic field makes with a horizontal line in the magnetic median.
Horizontal component of earth's magnetic field : if the component of total intensity of earth's magnetic field in the horizontal direction in magnetic meridian.

: magnetic inclination H : Horizontal component of earth's magnetic field

Q13) An electron in an atoms revolves around the nucleus in an orbit of radius 0.53 A^o . Calculate the equivalent magnetic moment if the frequency of revolution of electron is 6.8 x 10⁹ MHz.  (Marks 2)
Ans13) r = 0.53A^o = 0.53 x 10^-10 m
=6.8 x 10⁹ MHz = 6.8 x 10¹⁵ Hz
e = 1.6 x 10^-19 c
Now I = e
& magnetic moment M = IA = er²
= 6.8 x 10¹⁵ x 1.6 x 10^-19 x 3.14 (0.53 x 10^-10)²
= 9.596 x 10^-24 Am 

Q14) Write the function of base region of a transistor. Why is the region made thin and slightly doped?  (Marks 2)
Ans14) The base provides an interaction between emitter & collection of transistor. The base is thin & lightly doped to reduce recombination of holes & electrons in this region & so there is appreciable collector current.

Q15) Derive an expression for the energy stored in a charged parallel plate capacitor with air as the medium between its plates.  (Marks 2)
Ans15) Suppose the capacitor is charged gradually at any stage the charge on the capacitor is q
Potential = q/c
small amount of work done in giving an additional charge dq is
dW = q/c dq 
Total work done in giving a charge Q
= Energy stored
For a parallel plate capacitor 
Energy stored

Q16) In the diagram given below for the stationary orbit of the hydrogen atom, mark the transitions representing the Balmer and Lyman series. (Marks 2)

Ans16)  

Q17) The given figure shows an inductor L and resistor R connected in parallel to a battery B 

through a switch S. The resistance of R is the same as that of the coil that makes L. Two identical bulbs, P and Q are put in each arm of the circuit as shown. When S is closed, which of the two bulbs will light up earlier? Justify your answer.  (Marks 2)
Ans17) P will light up earlier in the induced I, will be induced current which will oppose the growth of current through it & hence reduce the current 

Q18) Two points electric charges of value q and 2q are kept at a distance d apart from each other in air. A third charge Q is to be kept along the same line in such a way that the net force acting on q and 2q is zero. Calculate the position of charge Q in terms of q and d.  (Marks 2)
Ans18)  

For equilibrium of charge q Force bet Q and q should be equal & apposite to force between q & 2q
i.e. (1/4o)(Qq/x²) = (1/4o)(2qq/d²)  - (i)
For equilibrium of charge 2q, for a between 2q & Q should be equal & apposite to force between 2q and 2q.
i.e. (1/4o)(2qQ/(d - x)²) = (1/4o)(2qq/d²)  - (ii)
Equating (i) & (ii)
2x² = (d - x)²
or 2x = (d - x) or x = d / (2 +1)
i.e. , at a distance d/(2 + 1) from charge q

Q19) Explain, with the help of a circuit diagram, the use of potentiometer for determination of internal resistance of a primary cell. Derive the necessary mathematical expression.  (Marks 3)
Ans19) To find internal resistance of a cell E, a current is set up as shown below.

Close key K and maintain suitable constant current in the potentiometer wire with the help of rheostat Rh. Adjust the position of jockey at different points of wire and find a point J on the wire where if jockey is pressed, galvanometer shows no deflection. Note the length AJ (= l₁) of the potentiometer wire. Now e.m.f of the cell, E = potential difference across the length of the potentiometer wire.
or E = Kl₁  - (i)
Where K is the potential gradient across the wire.
Close Key K₁ so that the resistance R is introduced in the cell circuit. Again find the position of the jockey on the potentiometer wire where galvanometer shows no deflection. Let it be at J₁. Note the length of the wire AJ₁ (= l₂ say). 
Then, Potential difference between two poles of the cell, V = potential difference across the length l₂ of the potentiometer wire 
i.e. V = Kl₂  - (ii)
Dividing (i) by (ii) we have
E/V = l₁/l₂  - (iii)
We know that the internal resistance r₁ of a cell of e.m.f. E, when a resistance R is connected in its circuit is given by
r₁ = (E - V)/V x R = (E/V - 1) R  - (iv)
Putting the value from (iii) in (iv) we get
r₁ = (l₁/l₂ - 1) R = (l₁ - l₂)/l₂ x R
Thus, knowing the values of l₁, l₂ and R the internal resistance r₁ of the cell can be determined. 

Q20) Sketch the wave-fronts corresponding to converging rays. Verify Snell's law of refraction using Huygens' wave theory.  (Marks 3)
Ans20)  

Snell's law of refraction using Huygen's wave Theory:
Let XY be a plane surface separating two media 1 and 2, the velocity of light being c₁ in the first medium and c₂ in the second medium. Suppose a plane wave front AB is incident on surface XY. It first strikes at A and then the successive points towards c. Let i be the angle of incidence. According to Huyghen's principle, from each point on AC, the secondary wavelets start growing in the second medium with speed c₂. Let the wave disturbance take time t to travel from B to C, then BC = c₁t. During the time, the disturbance from B reaches the point C, the secondary wavelets from point A must have spread over a hemisphere of radius AD = c₂t in the second medium. The tangent plane CD drawn from point C over this hemisphere of radius c₂t will be the new refracted wave front.

Consider a ray POQ normal to both the incident and the refracted wave fronts. Let the angle of incidence and refraction be i and r, which can be taken as the angles made respectively by the incident wave front AB and refracted wave front CD with the surface of separation XY. For CD to be the true refracted wave front, all the wave disturbances must take the same time in traveling from the incidence wave front AB and wave front CD.
Now the total time taken by the ray to travel from P and Q
= PO/c₁ + OQ/c₂
= (AO sin i)/c₁ + (OC sin r)/c₂
= (AO sin i)/c₁ + ((AC - AO) sin r)/c₂
= (AC sin r)/c₂ + AO[(sin i)/c₁- (sin r)/c₂]
Different rays from the incident wave front have different values of AO. Since the time taken by these rays to reach the wavefront CD is same, the coefficient of AO must vanish i.e.,
sin i/c₁ = sin r/c₂
or sin i/sin r = c₁/c₂ = Constant, n₂₁
This proves Snell's law of refraction. The constant n₂₁ is called the refractive index of second medium with respect to the first medium. 
Further, since the incident ray SA, the normal AN and refracted ray AD are respectively perpendicular to the incident wave front AB, the dividing surface XY and the refracted wavefront CD (all perpendicular to the plane of the paper), therefore, they all lie in the plane of the paper, i.e., in the same plane. This proves another law of refraction.

Q21) An electric dipole is held in a uniform electric field.
(i) Show that no translatory force acts on it.
(ii) Derive an expression for the torque acting on it.  (Marks 3)
Ans21)  

Consider an electric dipole consisting of two equal and opposite point charges -q and +q separated by a small distance 2a, having dipole moment |p| = q x 2a Let this dipole be held in a uniform external electric field at an angle with the direction of . Force on charge +q  = q, along the direction of Force on charge -q = q, in a direction opposite to .

As the force on the two charged of the dipole are equal and opposite, therefore, net force on the electric dipole is zero.
These forces being equal, unlike and parallel form a couple, which rotates the dipole is clock-wise direction, lending to align its axis along the direction of the field.
Draw AC  |  and BC ||
.^.. perpendicular distance between the forces = arm of couple = AC 
As Torque = moment of the couple 
= force x arm of couple
.^.. = F x AC = F x AB sin
 = F x 2a sin
 = (q x 2a) E sin
  = p E sin
.^.. in the vector form, we can rewrite this equation. As
= x

Q22) Derive an expression for the width of the central maxima for diffraction of light at a single slit. How does this width change with increase in width of the slit?  (Marks 3)
Ans22)  

The set of parallel rays falling on the slit form a plane wave from WW'. According to Huyghen's  principle, each point on the unblocked portion of plane wave front AB sends out secondary wavelets in all the direction.
The secondary waves, from points equidistant from the center C of the slit lying in the portion CA and CB of wave front travel the same distance in reaching O, and hence the path difference between them is zero. These secondary waves reinforce each other, resulting in the maximum intensity at point O.
Consider the secondary waves traveling in a direction making and angle with CO. All the secondary waves traveling in this direction reach a point P on the screen. The intensity at P will depend on the path difference between the secondary waves emitted from the corresponding points of the wave front. Draw AN perpendicular to BK. Path difference between the secondary waves reaching P from A and B = BN = AB sin = a sin
If this path difference is ,(the wavelength of light used), then P will be point of maximum intensity. This is because the whole wavefront can be considered to be divided into two equal halves CA and CB and if the path difference between the secondary waves from A and B is , then the path difference between the secondary waves from A and C reaching P will be /2 and path difference between the secondary waves from B and C reaching P will again be /2. Also for every point in the upper half AC, there is a corresponding point in the lower half CB for which the path difference between the secondary waves, reaching P is /2 Thus, destructive interference takes place at P and therefore, P is a point of first secondary minimum. Thus for I minima
a sin = = sin = /a  - (i)
The width of central maximum is the distance between first secondary minimum on either side of O
Let I minima be at a distance x 
If Q is small, sin ~ =x/D
From (i)
x/D = /a or x =D/a
Width of central maximum = 2x =2D/a 

Q23) A capacitor of capacitance 100F and a coil of resistance 50 and inductance 0.5 H are connected in series with a 110 V, 50 Hz source. Calculate the rms. value of the current in the circuit.  (Marks 3)
Ans23) Given C =100 F = 100 x 10^-6 F = 10^-4 F
R = 50 , L = 0.5 H, V_rms = 110V
= 50 Hz
X_L = L = 2L = 2 x 50 x 0.5 = 50
X_C = 1/c = 1/2c = 1/2 x 50 x 10^-4 = 100/
Z = (R² + (X_L - X_C)²) = (50² + (50 - 100/)²) = 134.7
I_rms =V_rms/Z = 110/134.7 = 0.816 A

Q24) Draw a labelled ray diagram to show the image formation in a reflecting type telescope. Write its two advantages over a refracting type telescope. On what factors does its resolving power depend?  (Marks 3)
Ans24) 

Advantages:
1. There is no chromatic aberration as the objective is a mirror.
2. Image is brighter compared to that in a refracting telescope.
Its resolving power depend on 
(i) Diameter of the objective mirror.
(ii) Wavelength of light with which the object is viewed.

Q25) Define the terms 'solar constant' and 'solar luminosity'. Explain how their knowledge helps us to calculate the surface temperature of the sun. Derive the necessary mathematical expression.  (Marks 3)
Ans25) Solar constant : is defined as the amount of radiant energy received per second by a unit area of a perfectly black body surface field at right angle to the direction of the sun rays at the mean distance of the earth from the sun.
Solar luminosity : is defined as the amount of energy emitted per second by the sun in all direction.
Expression for surface temperature of sun : Consider the sun to be a black body at temperature T and of radius R at the center of a bottom sphere of radius r, where r = 1A.U & r R
According to Stefan's law the energy emitted per second per unit area by the sun is 

E = T4 Where is Stefan's constant. Total energy emitted per second by sun = 4R2E = 4R2T4 = Ls where Ls is solar luminosity. Since solar luminosity is also equal to the total energy radiated per second by the sun 4R2T4  = 4R2S or T = (r2S/R2)1/4

Knowing the value of S, , r & R, the value of T can be determined.

Q26) An object is kept in front of a concave mirror of focal length 15 cm. The image formed is three times the size of the object. Calculate two possible distance of the object from the mirror.  (Marks 3)
Ans26) Given f = -15 cm , m = + 3u
Now m = f/(f - u)
For a real image m = -3
So -3 = -15/(-15 - u)  or -3 = 15/(15 + u)
-45 - 3u = 15
-3u = 60 or u = -20 cm
For virtual image m = 3
3 = -15/(-15 - u)
or 45 + 3u = 15 or u = -10 cm
.^.. The two possible distances of object from the mirror are 20 cm, 10 cm

Q27) A voltmeter V of resistance 400 is used to measure the potential difference across a 100 resistor in the circuit shown here.
(a) What will be the reading on the voltmeter?
(b) Calculate the potential difference across 100 resistor before the voltmeter is connected.  (Marks 3)

Ans27)  

(i) Effective resistance of 400 & 100 'R' is given by
1/R = 1/100 + 1/400 or R = 80  
Total distance of the circuit = R' = 80 + 200 = 280
Current drawn from cell = V/R' = 84/280 = 0.3 A 
P.D across 80 resistance = 0.3 x 80 = 24V
P.D across 100 or 400 = 24V
(ii) When voltmeter is not connected, then the total resistance of the circuit is
100 + 200 = 300
Current drawn from cell = 84/300 = 0.28 A 
P.D across 100 = 100 x 0.28 = 28V

Q28) Derive a mathematical expression for the force per unit length acting on each of the two straight parallel metallic conductors carrying current in the same direction and kept near each other. Why do such current carrying conductors attract each other?
Or
Derive a mathematical expression for the force acting on a current carrying straight conductor kept in a magnetic field. State the rule used to determined the direction of this force.  (Marks 5)
Ans28) Consider AB and CD, two infinite long straight conductor carrying I₁, I₂ in the same direction. They are separated by distance d
field B₁ on CD due to current I₁ in AB is given by

B1 = oI1/2d Force on CD due to field B1 = B1I2l  F2 = oII2l/2d According to Fleming's left hand rule F2 is perpendicular to both direction of the current I2 and B1 and is acting as shown in the figure. Again field B2 on AB due to current I2 in CD is B2 = oI2/2d  Force F1 on AB due to B2 is  F2 = B2I1l or F2 = oI2I1l/2d

It is directed inwards in accordance with Fleming's left hand rule.
Hence wires AB and CD attract each other with a force F = F₁ = F₂ = _oI₁I₂l/2d
Force per unit length = _oI₁I₂/2d
Whenever a current carrying conductor is placed with a magnetic field, moving electrons experience a force. Since the electrons are not free to move outside the body of the conductor, they transmit, their force to the conductor through collisions with the atoms of the conductor. This results in a force on conductor.

Let us consider a wire of length L and area cross-section A placed in magnetic field B acting perpendicular to the plane of the paper and directed inwards. If n is the number of free electrons per unit volume and V_d is the drift velocity, then
Force on each electron m = e(_d x )
In magnitude F_m = e V_d B sin 90^o = e V_d B
Volume of wire = Al
Number of electrons in the wire N = n Al
Total force on the wire F = N F_m
Or F = (n Al) e V_d B = (neV_dA)l B 
But current I = neV_dA
.^..F = I/B
In vector form   = I ( x )
Direction of the force can be determined by using right hand rule.

Q29) Draw a labelled diagram of Thomson's experimental set-up to determine e/m of electrons. Explain by deriving the necessary mathematical expression how of electron can be determined by this method.  (Marks 5)
Ans29)  Construction : it consist of a discharge tube. A P.D. of a few thousand volts is maintained between C and A. Small hole in the anode produces a fine beam of cathode rays which strike the screen coated with ZnS. Two plates are given negative & positive potential resp. They produce electric field that is perpendicular to the path of electrons. A uniform magnetic field is applied perpendicular to both electric field and the flow of electrons is directed into the paper.

Theory : If no electric or magnetic field is applied, the electron beam meets the screen at point M. Apply both the electric and magnetic fields and adjust the strength of electric field and magnetic field B so that the electron beam meets the screen S at its undeflected position M. In this situations the forces on the electron due to electric field magnetic field balance each other.

Thomson's method for the determination of e/m of an electron

Let m and e be mass and charge of electron and v be the velocity of the electron when it comes out of the hole of the anode.
Force on the electron due to electric field = Ee
Force on the electron due to magnetic field = Bev
For the undeflected position of the spot on the screen S, the two forces must be equal and opposite, therefore
Ee = Bev or v = E/B  - (i)
As the electron beam is accelerated from cathode to anode, its potential energy at the cathode appears as gain in its kinetic energy at the anode. If V is the potential difference between cathode and anode, then potential energy of electron at cathode = charge x potential difference = eV
Gain in Kinetic Energy of electron at anode
 = 1/2 mv²
So, we have eV = 1/2 mv²
or e/m = 1v²/2V = (1/2V)(E/B)²
= E²/2VB²  - (ii)

Q30) Define the terms 'potential harrier' and 'depletion region' for a p-n junction. Explain with the help of a circuit diagram, the use of p-n diode as a full wave rectifier. Draw the input and output wave-forms.  (Marks 5)
Ans30)  Potential Barrier : The potential difference created across the junction due to diffusion of free electrons & holes across the junction is called potential barrier.
Depletion region : It is the region around the junction which is devoid of free charge carriers Diode as full wave reflector.

During the positive half of the input A.C. the upper p-n junction diode is forward biased as shown in Fig (a) and the lower p-n junction diode is reverse biased. The forward current flows on account of majority carriers of upper p-n junction diode in the direction shown.
During the negative half cycle of input A.C. the upper p-n junction diode is reverse biased, and the lower p-n junction diode is forward biased, Fig (b) The forward current flows on account of majority carriers of lower p-n junction diode. We observe that during both the halves, current through R flows in the same direction.