Class 12 Physics Sample Paper Term 1 With Solutions Set F

Please refer to Class 12 Physics Sample Paper Term 1 With Solutions Set F below. These Class 12 Physics Sample Papers will help you to get more understanding of the type of questions expected in the upcoming exams. All sample guess papers for Physics Class 12 have been designed as per the latest examination pattern issued by CBSE. Please practice all Term 1 CBSE Sample Papers for Physics in Standard 12.

Sample Paper Term 1 Class 12 Physics With Solutions Set F

SECTION – A

All questions are compulsory. In case of internal choices, attempt any one of them.

Question 1. What is the forbidden energy gap (in joule) for a germanium crystal ?
Answer. bE_g = 0.7 eV = 0.7 × 1.6 × 10^–19 J = 1.12 × 10^–19 J

Question 2. Plot a graph showing variation of induced e.m.f. with the rate of change of current flowing through a coil.
OR
Predict the polarity of the capacitor in the situation described below.

Answer.

OR
Polarity of plate A will be positive with respect to plate B in the capacitor, as induced current is in clockwise direction

Question 3. An electron is moving along +ve x – axis in the presence of uniform magnetic field along +ve y – axis. What is the direction of the force acting on it?
Answer.  Using Fleming’s right hand rule, the direction of force will be along –ve z-axis.

Question 4. Long distance radio broadcasts use short-wave bands. Why?
OR
Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?
Answer. Electromagnetic waves in frequency range of short-wave band reflect from ionosphere where lower frequency radio waves i.e., medium waves are absorbed. So, short-waves are suitable for long distance radio broadcast.
OR
Atmosphere absorb X-rays, while visible and radiowaves can penetrate through it. Hence optical telescope can work on ground but X-ray astronomical telescopes only work above atmosphere, hence installed on the satellite orbiting around earth.

Question 5. What is the effect on the velocity of the emitted photoelectrons if the wavelength of the incident light is decreased?
Answer. 1/2 mv² = hu – W₀ = hc/λ – W₀
So, if l of incident light is decreased, energy hu of photon increases and hence K.E. and velocity of emitted photoelectron also increases.

Question 6. In a half wave rectifier circuit operating from 50 Hz mains frequency, what would be the fundamental frequency in the ripple ?
Answer. As the output voltage obtained in a half wave rectifier circuit has a single variation in one cycle of ac voltage, hence the fundamental frequency in the ripple of output voltage would be = 50 Hz.

Question 7. What is the purpose of heavy water in nuclear reactors?
OR
Compare the radii of two nuclei with mass numbers 1 and 27 respectively.
Answer. Heavy water is used as a moderator in some reactors to slow down the fast moving neutrons.
OR

Question 8. The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.
(i) What is the kinetic energy of electron in this state ?
(ii) What is the potential energy of electron in this state ?
Answer.
(i) E_K = – E = – (–3.4) = + 3.4 eV
(ii) E_P = 2E = 2 × (–3.4) = – 6.8 eV

Question 9. A potential barrier of 0.3 V exists across a p-n junction. If the depletion region is 1 mm wide, what is the intensity of electric field in this region?
OR
When the voltage drop across a p-n junction diode is increased from 0.65 V to 0.70 V, the change in the diode current is 5 mA. Find the value of the dynamic resistance of the diode.
Answer.

Question 10. Depict the direction of the magnetic field lines due to a circular current carrying loop.
For question numbers 11, 12, 13 and 14, two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false
Answer.
Magnetic field lines due to a circular wire carrying current I :

Question 11. Assertion (A) : If a dielectric is placed in external field then field inside dielectric will be less than applied field. 
Reason (R) : Electric field will induce dipole moment opposite to field direction.

Question 12. Assertion (A) : If a convex lens is kept in water its convergent power decreases.
Reason (R) : Focal length of convex lens in water increases.

Question 13. Assertion (A) : In a cavity within a conductor, the electric field is zero.
Reason (R) : Charges in a conductor reside only at its surface.

Question 14. Assertion (A) : When a charged particle moves in a circular path. It produces electromagnetic wave.
Reason (R) : Charged particle has acceleration.

SECTION – B

Question 15. A Rheostat is used in applications that requires the adjustment of current or varying of resistance in an electric circuit. As shown in figure, a variable rheostat of 2 kW is used to control the potential difference across a 500 W load. Here, the source emf is 50 V and resistance AB is 500 W.

(i) The total resistance of the circuit is
(a) 500 Ω
(b) 375 Ω
(c) 875 Ω
(d) 1500 Ω

(ii) The value of total current flowing through the circuit is
(a) 2.87 A
(b) 0.057 A
(c) 0.87 A
(d) 0.677 A

(iii) The potential difference across the load is
(a) 21.43 V
(b) 32.45 V
(c) 17.62 V
(d) 19.83 V

(iv) If the load is removed, the current across the rheostat is,
(a) 1/4 A
(b) 1/20 A
(c) 1/40 A
(d) 40 A

(v) If the load is removed, what should be the resistance at BC to get 40 V between B and C ?
(a) 500 Ω
(b) 375 Ω
(c) 1600 Ω
(d) 1500 Ω

Question 16. Image of a white object is coloured and blurred because m (hence f ) of lens is different for different colours.This defect is called chromatic aberration. As μ0 > μr, therefore, f_r > f_v. The difference (f_r – f_v) is a measure of longitudinal chromatic aberration of the lens. Focal length for mean colour is f = √fr × fv . Using lens maker formula, for mean colour of light, we have 1/f= (μ–1 ) (1/R₁– 1/R₂) where f is focal length of mean colour and m is the refractive index of mean colour.

(i) Focal length of a equiconvex lens of glass μ = 3/2 in air is 20 cm. The radius of curvature of each surface is
(a) 10 cm
(b) –10 cm
(c) 20 cm
(d) –20 cm

(ii) Focal length of the lens in water would be
(a) 20 cm
(b) 80 cm
(c) –20 cm
(d) –80 cm

(iii) If μv = 1.6, μr = 1.5, R₁ = 20 cm and R₂ = –20 cm, then the chromatic aberration of the lens would be
(a) 3 cm
(b) 3.3 cm
(c) –3 cm
(d) –3.3 cm

(iv) A given convex lens of glass (μ = 3/2) can behave as concave when it is held in a medium of μ equal to
(a) 1
(b) 3/2
(c) 2/3
(d) 7/4

(v) Chromatic aberration of a lens can be corrected by
(a) providing different suitable curvature to its two surfaces
(b) proper polishing of its two surfaces
(c) Suitably combining it with another lens
(d) reducing its aperture.

SECTION – C

Question 17. Interstellar space has an extremely weak magnetic field of the order of 10^–12 T. Can such a weak field be of any significant consequence? Explain.
OR
Consider the plane S formed by the dipole axis and the axis of earth. Let P be point on the magnetic equator and in S. Let Q be the point of intersection of the geographical and magnetic equators. Obtain the declination and dip angles at P and Q.
Answer.
weak field of the order of 10^–12 T can cause the charged particles to move along circular paths of very large radii. Over a small distance, we may not be able to notice the deflection in the path of the charged particles but over large interstellar distance the distance is quite noticeable.
OR
From figure,
For point P : Since point P lies in plane S formed by the dipole axis and the axis of the Earth, declination, δ = 0°.
For point Q : Since point Q lies on the magnetic equator, dip, d = 0°
Declination, D = 11.3°

Question 18. A, B and C are the parallel sided transparent media of refractive index n₁, n₂ and n₃ respectively. They are arranged as shown in the figure. A ray is incident at an angle q on the surface of separation of A and B is as shown in the figure. After the refraction into the medium B, the ray grazes the surface of separation of the media B and C. Find the value of sinq.

Answer.

Question 19. Figure shows three points A, B and C in an uniform electric field. At which of the points the electric potential is maximum?

OR
Is it possible to transfer all the charge from a conductor to another insulated conductor?
Answer.V_A > V_B > V_C, direction of electric field is from higher to lower potential.
OR
‘Yes’. By enclosing uncharged conductor inside charged conductor and then by connecting them with wire.

Question 20. In figure, V0 is the potential barrier across a p-n junction, when no battery is connected across the junction. 

Which of P, Q and R corresponds to forward and reverse bias of junction ?
Answer. Height of potential barrier decreases when p-n junction is forward biased and it increases when junction is reverse biased. So, curve R corresponds to forward biasing and P corresponds to reverse biasing.

Question 21. A charged particle q is moving in the presence of a magnetic field B which is inclined to an angle 30° with the direction of the motion of the particle. Draw the trajectory followed by the particle in the presence of the field and explain how the particle describes this path.
Answer. When a charged particle moving in a uniform magnetic field has two concurrent motions.A linear motion in the direction of  B (along z-axis)
as shown in figure (a) and a circular motion in a plane perpendicular to B (in xy-plane). Hence the resultant path of the charged particle will be a helix, with its axis
along the direction of B, as shown in figure (b).

Question 22. When an electric field is applied across a semiconductor what happens to electrons and holes ?
Answer. Electrons in conduction band get accelerated and acquire energy by the application of electric field and move from lower energy level to higher energy level. While holes in valence band move from higher energy level to lower energy level.

Question 23. Starting from the expression for the energy W = 1/2LI², stored in a solenoid of self-inductance L to build up the current I, obtain the expression for the magnetic energy in terms of the magnetic field B, area A and length l of the solenoid having n number of turns per unit length. Hence, show that the energy density is given by B²/2m₀
Answer. 

Question 24. Show, by giving a simple example, how e.m. waves carry energy and momentum.
OR
An e.m. wave is travelling in a medium with a velocity v = v i^. Draw a sketch showing the propagation of the e.m. wave, indicating the direction of the oscillating electric and magnetic fields.
Answer. Consider a plane perpendicular to the direction of propagation of the wave. An electric charge, on the plane will be set in motion by the electric and magnetic fields of e.m. wave, incident on this plane.This illustrates that e.m. waves carry energy and momentum.
OR
In figure, the velocity of propagation of e.m. wave is along X-axis v = vi and electric field E along Y-axis and magnetic field  B along Z-axis.

Question 25. For a single slit of width ‘a’, the first minimum of the interference pattern of a monochromatic light of wavelength l occurs at an angle of λ/a. At the same angle of  λ/a, we get a maximum for two narrow slits separated by a distance ‘a’. Explain.
Answer. For a single slit of width “a” the first minima of the interference pattern of a monochromatic light of wavelength l occurs at an angle of (l/a) because the light from centre of the slit differs by a half of a wavelength.Whereas a double slit experiment at the same angle of (l/a) and slits separation “a” produces maxima because one wavelength difference in path length from these two slits is produced.

SECTION – D

Question 26. Define resistivity of a conductor. Plot a graph showing the variation of resistivity with temperature for a metallic conductor. How does one explain such a behaviour, using the mathematical expression of the resistivity of a material.
OR
Derive an expression for the drift speed of electrons in a good conductor in terms of the relaxation time of electrons.
Answer.R = ρ l/A 
If l = 1, A = 1 ⇒ r = R
Thus, resistivity of a material is numerically equal to the resistance of the conductor having unit length and unit cross-sectional area.

The resistivity of a material is found to be dependent on the temperature. Different materials do not exhibit the same dependence on temperature. Over a limited range of temperatures, that is not too large, the resistivity of a metallic conductor is approximately given by,
ρ_T = ρ0 [1 + α(T – T₀)]     …(i)
where ρ_T is the resistivity at a temperature T and ρ₀ is the same at a reference temperature T0. a is called the temperature coefficient of resistivity.
The relation in eqn. (i) implies that a graph of ρ_T plotted against T would be a straight line. At temperatures much lower than 0^°C, the graph, however, deviates considerably from a straight line.
OR
In the absence of an external field, the free electrons in a metal are moving randomly in all directions due to thermal agitation. There is no overall drift and the average velocity is zero.
In the presence of an external electric field E, each electron experience an acceleration

a=eE/m

opposite to the field direction. However, this acceleration is momentary, since the electrons are continuously colliding with vibrating ions and other electrons. After each collision, the electron makes a fresh start and accelerates only to be deflected randomly again. If t is the relaxation time, i.e., the average time between two successive collisions, then the drift velocity of the electrons is given by

Question 27. ₈₆Rn²²² is converted into ₈₄Po²¹⁸ and ₉₃Np²³⁹ is converted into ₉₄Pu²³⁹. Name the particles emitted in each case and write down the corresponding equations.
Answer.

Question 28. The work function for the following metals is given :
Na : 2.75 eV; K: 2.30 eV; Mo : 4.17 eV; Ni : 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell?
What happens if the laser is brought nearer and placed 50 cm away?
OR
Why should gases be insulators at ordinary pressures and start conducting at very low pressure?
Answer. The distance between laser source and receiver does not affect the energy of each photon incident, hence does not affect the energy of emitted photoelectrons.
But the reduction in distance will increase the intensity of incident light and hence number of photons. This will increase the photoelectric current.
Given that, wavelength of incident radiation is
λ = 3300 Å
So, energy of incident radiation is

Now, work function of Mo : 4.17 eV, Ni : 5.15 eV is more than energy of incident photon, hence these two metals will not give photoelectric emission.
OR
At ordinary pressures a few positive ions and electrons produced by the ionisation of the gas molecules by energetic rays (like X-rays, γ-rays, cosmic rays etc., coming from outer space and entering the earth’s atmosphere) are not able to reach their respective electrodes, even at high voltages, due to their frequent collisions with gas molecules and recombinations. That is why the gases at ordinary pressures are insulators.
At low pressures, the density of the gas decreases, the mean free path of the gas molecules become large.
Now under the effect of external high voltage, the ions acquire sufficient energy before they collide with molecules causing further ionisation. Due to it, the number of ions in the gas increases and it becomes a conductor.

Question 29. Using the relevant Bohr’s postulates, derive the expression for the speed of the electron in the nth orbit.
Answer.Speed of the electron in the nth orbit : The centripetal force required for revolution is provided by the electrostatic force of attraction between the electron and the nucleus.

Question 30. Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in figure.

Answer. Consider a rectangular strip of small width dx of the square loop at a distance x from the wire as shown in figure.
Magnetic field due to current carrying wire at a distance x from the wire is

SECTION – E

Question 31. Figure shows an experiment setup similar to Young’s double slit experiment to observe interference of light.

Here SS₂ – SS₁ = l/4
Write the condition of (i) constructive, (ii) destructive interference at any point P in terms of path difference, Δ = S₂P – S₁P
Does the central fringe observed in the above setup lie above or below O? Give reason in support of your answer. Yellow light of wavelength 6000Å produces fringes of width 0.8 mm in Young’s double slit experiment. What will be the fringe width if the light source is replaced by another monochromatic source of wavelength 7500Å and separation between the slits is doubled?
OR
(a) What are coherent sources of light? State two conditions for two light sources to be coherent.
(b) Derive a mathematical expression for the width of interference fringes obtained in Young’s double slit experiment with the help of a suitable diagram.
Answer.

OR
(a) Coherent sources are those which have exactly the same frequency and are in the same phase or have a constant difference in phase.
Conditions : (i) The sources should be monochromatic and originating from common single source.
(ii) The amplitudes of the waves should be equal.
(b) Expression for fringe width : Let S₁ and S₂ be two coherent sources separated by a distance d. Let the distance of the screen from the coherent sources be D. Let M be the foot of the perpendicular drawn from O, the midpoint of S₁ and S₂ on the screen. Obviously point M is equidistant from S₁ and S₂. Therefore the path difference between the two waves at point M is zero. Thus the point M has the maximum intensity.
Consider a point P on the screen at a distance y from M. Draw S1N perpendicular from S₁ on S₂ P.

Question 32. (a) An alternating voltage V = V_m sin wt applied to a series LCR circuit drives a current given by i =i_m sin (wt + Φ). Deduce an expression for the average power dissipated over a cycle.
(b) Determine the current and quality factor at resonance for a series LCR circuit with L = 1.00 mH, C = 1.00 nF and R =100 W connected to an ac source having peak voltage of 100 V

 OR
(a) Draw the diagram of a device which is used to decrease high ac voltage into a low ac voltage and state its working principle. Write four sources of energy loss in this device.
(b) A small town with a demand of 1200 kW of electric power at 220 V is situated 20 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 W per km. The town gets the power from the line through a 4000-220 V step-down transformer at a substation in the town. Estimate the line power loss in the form of heat.
Answer.

Principle : When the current flowing through the primary coil changes, an emf is induced in the secondary coil due to the change in magnetic flux
linked with it i.e., it works on the principle of mutual induction.
There are number of energy losses in a transformer.
(i) Copper losses due to Joule’s heating produced across the resistances of primary and secondary coils.It can be reduced by using copper wires.
(ii) Hysteresis losses due to repeated magnetization and demagnetization of the core of transformer. It is minimized by using soft iron core, as area of hysteresis loop for soft iron is small and hence energy loss also becomes small.
(iii) Iron losses due to eddy currents produced in soft iron core. It is minimized by using laminated iron core.
(iv) Flux losses due to flux leakage or incomplete flux linkage and can be minimised by proper coupling of primary and secondary coils.

Question 33. (a) Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law.
(b) Three charges − √2 μC,  √2 μC and −  √2 μC are arranged along a straight line as shown in the figure.
Calculate the total field intensity due to all three charges at the point P. 

Answer.