Sample Model Papers For Physics CBSE Board for students online

General instructions:
1. Questions 1-8 carry 1 mark each.
2. Questions 9-18 carry 2 marks each.ss
3. Questions 19-27 carry 3 marks each.
4. Questions 28-30 carry 5 marks each.

Q 1 Is electric flux a scalar or vector quantity? Give its SI units.

Ans. 1 Electric flux is a vector quantity. Its SI unit is Newton meter²/ coulomb.

Q 2 By which wave phenomenon can we distinguish light waves from sound waves?

Ans. 2 Polarization.

Q 3 What is a wave front?

Ans. 3 Wave front is a surface over which the oscillations in a wave have the same phase. The surface is normal to rays in an isotropic medium.

Q 4 What will be the effect on the drift speed of an electron in a metallic conductor if the temperature of the conductor is increased?

Ans. 4 The drift speed will decrease due to the increase in the number of collisions thus decreasing the time constant.

Q 5. A magnetic needle which is free to rotate in a vertical plane, orients itself with the axis vertical at a certain place on the earth. What are the values of horizontal component of earth's magnetic field and angle of dip at this place?

Ans. 5 The horizontal component of earth's magnetic field is 0^° and the angle of dip is 90^°.

Q 6 Write down the names of waves used in RADAR. What is the range of their wavelength?

Ans. 6 Sound waves with microwave frequencies are used in RADAR. The range of the wavelength is approximately 1m to 10^-3m.

Q 7 Give an example of a non-ohmic device, which shows negative resistance.

Ans. 7 Thyristor.

Q 8. Two potentiometers X and Y showing variations of potential difference with length as shown in the figure below. Which one of the potentiometers is more sensitive?

Ans 8. The Y-potentiometer is more sensitive.

Q 9. Define Kirchoff's law for an electric network.

Ans. 9 There are two laws given by Kirchoff,
1. The algebraic sum of the electric currents that meet at any point in a network is zero.
2. In any closed electric circuit the algebraic sum of the products of current and resistance in each part of the network is equal to the algebraic sum of the electromotive forces in the circuit.

Q 10 What is interference of light? Write down the necessary conditions for the interference of two light waves.

Ans. 10 Interference of light is a phenomenon arising when two beams of coherent lights, which have traveled different distances, are superimposed. The beams should be approximately of equal intensity.
The conditions are:
1. The waves should be coherent.
2. The beams should be of nearly same intensity.
3. The beams should not intersect at too large an angle.

Q 11 Write down the principle of reversibility of light. Hence prove,
1m2 = 1 / 2m1

Ans. 11 According to the principle of reversibility, if a ray of light is reversed, it always travels along its original path.
If ₂m₁ is the refractive index from medium 1 to medium 2 then,
₂m₁ = sin i/sin r
Where i and r are the angles of incidence and reflection respectively. From the principle of reversibility, if the ray of light is reversed its original path will be retraced.
₁m_{2 =} sin r/sin i
or, ₁m₂ = 1 / ₂m₁

Q 12 Radius of curvature of each surface of a convex lens of refractive index 1.5 is 50 cm. Calculate its power.

Ans. 12 R = 50 cm = .5 m
f = (½) * R
f = 0.25 m
P = 1/f
P = 1/0.25 = 4 diaptors.

Q 13 Derive a relation between surface temperature of the sun with the solar constant.

Ans. 13 Consider sun to be a black body at temperature T and radius R₀ at the center of a hollow sphere of radius r, r R₀.
Surface area of the hollow sphere = 4pr².
All the energy emitted by the sun falls normally on the inner surface of the hollow sphere. If S is the solar constant, the solar luminosity is given by,
L₀ = 4pr²S
According to Stephen's law, the energy emitted per sec per unit area is given by,
E = s T⁴
Surface area of sun = 4pR0²
Total energy emitted per second by the sun = 4pR₀²E = 4pR₀²s T⁴
As solar luminosity is also equal to the total energy radiated per second by the sun. Therefore,
4pR⁰²sT⁴ = 4pr²S
T = [r²S/R₀²s]^1/4
Knowing the value of S, s, r and R0, T can be determined.

Q 14. A sphere S₁ of radius R₁, encloses a total charge q. If there is another concentric sphere S₂ of radius R₂ (R₂ R₁) and there be no additional charges between S₁ and S₂. Find the ratio of the electric flux through S₁ and S₂.

Ans. 14 Flux through S₁ = ± 4pR₁²E (r)
Flux through S1 = ± 4pR₂²E (r)
Ratio = (R₁/R₂)²

Q 15 A voltaic cell has an emf of 2V. Is there a net field inside the cell when,
1) Circuit is open.
2) Circuit is closed.
3) A steady current is drawn from it.

Ans. 15 1. When the circuit is open there is no net electric field inside the circuit. A field of non-electrostatic origin balances the electrostatic field due to the plates.
2. When the circuit is closed current flows in the direction of electrostatic field outside and in the opposite direction of electrostatic field inside the cell. There is a net field inside the cell opposite to the electrostatic field.

Q 16 Calculate the resistivity of material of a wire 10m long, 0.4 mm in diameter and having a resistance of 2W.

Ans. 16 Data provided,
R = 2 W
l = 10m
d = 0.4mm
A = 0.1256 mm²
From the following formula, r is easily obtained.
R = pl/A
r = 2.512 * 10^-8 Wm

Q 17 A charged oil drop falls due to gravity with a speed V. This drop is held stationary by applying electric field in a Millikan's setup and is found to have two excess electrons. Suddenly it is observed that the drop moves up with the same velocity V. Why does this happen? Write down the relevant mathematical expression.

Ans. 17 The drop moves upward due to the attraction of the upper plate as the capacitor plates are charged to a high potential difference. The drop has two excess electrons, which it picks up from the ionized air. The mathematical expression is,
q = [mg/E] t_g[1/t_e + 1/t_g]
where q is the charge, E is the electric field, t_g is the time taken by the droplet during free fall and t_e is the time taken by the droplet when moving in the electric field.

Q 18 In a n-p-n tr
Ansistor, if the emitter and the base have the same doping concentration, how will the collector and base current be affected?

Ans. 18 The n-p-n tr
Ansistor has its emitter-base junction forward biased by connecting the emitter to the negative terminal of V_EB battery and the p-type collector is reverse biased by connecting it to the positive terminal of V_CB battery as shown in the figure below.

The forward biasing of emitter base circuit repels the electrons from the emitter towards the base, setting up emitter current I_E. As the base is very thin and lightly doped, a very few electrons (»1%) from the emitter combine with the holes of the base, giving rise to base current I_B. and the remaining electrons (»99%) are pulled by the collector which has a high potential. The electrons are finally collected by the positive terminal of the battery V_CB giving rise to collector current I_C.

As soon as an electrons from the emitter combines with a hole in the base region, an electron from the negative terminal of battery V_EB enters the emitter and an electron from the base enters the positive terminal of V_EB. This sets a base current I_B. Similarly, corresponding to each electron that goes from collector to the positive terminal of V_EB, an electron enters the emitter from negative terminal of V_EB. Hence,
I_E = I_B + I_C

Q 19 What is a thermocouple? Draw the graph of variation of thermo emf generated as a function of temperature of hot junction. The cold junction is at 0 °C. Also give two applications of thermo couple.

Ans. 19 When two metals are joined together and the two junctions are maintained at two different temperatures, then current flows and the metals are said to constitute a thermocouple.

Applications: In measurement of temperature and in solar batteries.

Q 20 Draw the logic symbol and write the truth table of OR gate. With the help of a circuit diagram describe the realization of OR gate using semiconductor devices.

Ans. 20

	Circuit symbol of a 2-input OR Gate

Q 21 Define the current sensitivity of a moving coil galvanometer. On what factors does it depends? If in a galvanometer current sensitivity is increased by 1.5 times, how will the voltage sensitivity be affected?

Ans. 21 The current sensitivity of a moving coil galvanometer is defined as:
1/G = NAB/k
where k is the constant of suspension
B is the magnetic field.
A is the area of the coil {length * breadth}
N is the no. of turns of the coil.
From the relation it is clear that the sensitivity depends on N, B and A.

V_sen = I_sen R (equation I)
V'_sen = I'_sen R' (equation II)
Dividing equation II by I, we have
V'_sen/ V_sen = (I'_sen/ I_sen)(R'/R)
= (1.25 I /I)(1.5 R/R)
V'_sen = 1.875 V_sen

Q 22 The voltage across a cathode ray gun is 500 V. Calculate,
1) The energy gained by the electrons.
2) The speed of the electrons.
3) The momentum of the electrons.

Ans. 22 Energy = ½ mv² = eV = 1.6 * 10^-19 * 500 = 8.0 * 10^-17 eV
v = [2eV/m]^1/2
= [2 * 1.6 * 10^-19 * 500 /9.1 * 10^-31]^1/2
v = 1.3 * 10⁷ m/sec
Momentum = mv =9.1 * 10^-31 * 1.3 * 10⁷ kg m/sec.

Q 23 A series LCR circuit consists of R = 10 W, Xe = 60 W and an inductance coil. The circuit is found to resonate when put across 300 V, 100 Hz supply. Calculate induction of the coil and current in the circuit at resonance.

Ans. 23 Given that as LCR circuit resonates when put across a 300 V, 100 Hz supply.
R = 10 W
XC = 60 W
At resonance,
w² = 1/LC
XC = 60 W = 1/wC
n = 100 Hz, X_C = 60 W
Therefore, w = 2pn = 3.14 * 2 * 100 = 628.2
C = 1/w X_C
= 1/ 628.2 * 60 + 2.653 * 10-⁵
Now, L = 1/w²C
= 1/[ * (3.14)² * (100)2 * 2.653 * 10^-5]
= 9.55 * 10^-2
I = V/R = 300/10 = 30 amps

Q 24 Monochromatic light (l = 589 nm) is incident from air to water surface. What are the values of wavelength, frequency and speed of reflected and refracted light?m = 1.33

Ans. 24 Data given,
l = 589 nm
m = 1.33
m = V_air/V_water = 1.33
V_water = V_air/1.33 = 3 * 10⁸/1.33 = 2.25 * 10⁸ m/sec
l_air = 589 nm = 589 * 10^-9 m
C = n l
n = [3 * 10⁸]/[589 * 10^-9]
lwater = V_water/n
= [3 * 10⁸/1.33][589 * 10^-9/ 3 * 10⁸]
= 4.42 * 10^-7 m
Therefore,
Speed of reflected light = 3 * 10⁸ m/sec
Speed of refracted light = 2.25 * 10⁸ m/sec

Q 25 The half-life of Polonium is 138 days against a decay. What fraction of a sample of a pure Polonium will remain undecayed after 276 days?

Ans. 25 It is given that the half-life period of podium is 138 days against a decay. From the definition of half-life it is obvious that half of the initial Polonium will be decayed in 138 days. Another half of the remaining half will decay in the next 138 days. That me
Ans after 276 days, only one fourth of the initial podium remains undecayed.

Q 26 With a circuit diagram explain the use of a n-p-n tr
Ansistor as an amplifier in common emitter mode. Show the phase relationship between collector and base voltage.

Ans. 26 The figure below shows the use of n-p-n tr
Ansistor as a common emitter amplifier. The emitter is common to both input and output circuits. The emitter is forward biased by battery V_BB and the collector is reversed biased by the battery V_CC. This decreases the resistance R_in of the input circuit and increases the resistance Rout of the output circuit. The low a.c. input signal V_I is superimposed on the forward bias V_BE. A load resistance R_L is connected between the collector and the d.c. Supply and the amplified output is obtained between the collector and the ground.

When current I_C flows in the output circuit, the potential drop across load resistance is I_CR_L. Hence the output voltage is,
V₀ = V_CE = V_CC - I_CR_L

When the input signal is fed to the base-emitter circuit, the base emitter voltage changes. This changes the emitter current I_E and hence the collector current I_C. the output voltage V₀ varies in accordance with the above relation. These variations in the collector voltage appear as amplified output.

Phase relationship between input and output signals. When an a.c. signal is fed to the input circuit, its positive half cycle increases the forward bias of the circuit which in turn increases the emitter current and hence the collector current. The increase in collector current increases the potential drop across R_L, which makes the output voltage V₀ less positive or more negative. So as the input signal goes through its positive half cycle, the amplified output signal goes through a negative half cycle. Similarly as the input signal goes through its negative half cycle, the amplified output signal goes through its positive half cycle. Hence in a common emitter amplifier, the input and output voltages are 180° out of phase or in opposite phases.

Q 27 Explain the working of a simple microscope and show that its magnification is given by, m = 1 + D/f.

Ans. 27 A microscope consists of a converging lens of short focal length. The object is placed within the focal length of the lens so that an erect, magnified and vertical image of the object is formed. The closest comfortable distance for viewing the object is the near point [D = 25 cm]. The magnification is given by M,
M = image distance/object distance = v/u = v[1/v - 1/f]
In this case v is negative and is equal in magnitude to D.
M = 1 + D/f

Q 28 What are the Bhor's postulates on atomic model. Derive an expression for the total energy of an electron in the n^th orbit.

Ans. 28 Bhor's postulates:
1. Electrons move in circular orbits around the nucleus called the stationary states. It restricts the stationary states to those circular orbits in which the angular momentum is an integral multiple of [h/2p].
2. An additional postulate is that : in violation to electrodynamics, accelerated motion of electronic charge in these orbits is not accompanied by emission of electromagnetic radiation.
3. When a hydrogen atom makes a tr
Ansition from a state with quantum number n_I to the state with quantum number n_f (n_f < n_I), then it does by emitting a photon of energy hn which is the difference in energies of the two states.

The energy of electron in the stationary state is the sum of its kinetic and potential energies.
K.E. = ½ mv²
P.E. = - e²/4pe₀r
The total energy E = ½ m v² - e²/4pe₀r
According to Bhor, a stationary state has its circumference equal to an integral number of De Broglie wavelengths.
So, 2pr = nl = nh/mv
mvr = nh/2p
v = nh/2pmr
Also, mv²/r = e²/4pe₀r
Therefore, v = 1/n[e²/4pe₀r][2p/h]
And r = n²/m [h/2p]2[4pe₀r/ e²]
a is the fine structure constant,
c is the velocity of light,
h is Planck's constant,
r is the radius of the orbit,
e is the elctronic charge.

Q 29. If C₁ = 3 pf and C₂ = 2 pf, calculate the equivalent capacitance of the given network between points A and B.

Ans. 29 We start solving the circuit from the right hand side. All the three capacitors of capacity C₂ 2pf each are in series. This gives the equivalent capacitance,
1/C = 3/C₂ = 3/2 pf. C = 2/3 pf
This is in parallel with the next C2 capacitance, which gives an equivalent capacitance of 8/3 pf.
Now the remaining three capacitors are again in series, which gives an equivalent capacitance of 1/(1/3 + 1/3 + 3/8) or 24/25 pf.

Q 30 A small city having a demand of 880 kW of electric power at 220 V is situated 15 km away from electric plant generating power at 440 V. The resistance of the cable is 15 W. The city gets power through a 4400-220 V step down tr
Ansformer at the substation. Calculate,
1) Power loss in the form of heat.
2) How much power the plant must supply (assuming the power loss in the form of losses is zero).
3) Voltages drop in the line.

Ans. 30 The current in the line can be given by, 880 * 1000/4400 = 200 A
1. The power loss is given by H = I²R = (200)² * 15 = 600 Kw
2. The plant should supply a total power of 880 + 600 = 1480 Kw
3. The voltage drop is given by, 200 * 15 = 3000 V.