Sample Paper Class 12 Physics Term 1 Set D
Please refer to Sample Paper Class 12 Physics Term 1 Set D with solutions provided below. We have provided CBSE Sample Papers for Class 12 Physics as per the latest paper pattern and examination guidelines for Standard 12 Physics issued by CBSE for the current academic year. The below provided Sample Guess paper will help you to practice and understand what type of questions can be expected in the Class 12 Physics exam.
CBSE Sample Paper Class 12 Physics for Term 1 Set D
SECTION – A
All questions are compulsory. In case of internal choices, attempt any one of them.
Question 1. Is the source of magnetic field an analogue to the source of electric field?
Answer.No. The source of electric field is an electric charge. The source of magnetic field is not a magnetic charge. Infact, moving electric charges produce magnetic field.
Question 2. What is the speed of e.m. wave in a medium of μr = 5 and εr = 20?
A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the direction of electric and magnetic field vectors?
The electric and magnetic field vectors are perpendicular to each other and also perpendicular to the direction of propagation of the electromagnetic wave. If a plane electromagnetic wave is propagating along the z-direction, then the electric field is along x-axis, and magnetic field is along y-axis.
Question 3. A p-n junction photodiode is made of a material with a band gap of 2.0 eV. What is the minimum frequency of the radiation that can be absorbed by the material? [h = 6.6 × 10–34 J s]
Question 4. On which principle is ‘elecromagnetic damping’ based upon?
A bar magnet falls from height ‘h’ through a metal ring. Will its acceleration be equal to ‘g’ ? Give reason for your answer ?
Its acceleration will be less than g because the induced current in the ring opposes the motion of the magnet.
Question 5. An electron is not deflected, while moving through a certain region of space. Can we say that there is no magnetic field in the region?
Answer.No. Magnetic field may exist in the region and electron is moving in the direction of magnetic field, and hence experiences no force due to magnetic field.
Question 6. Based on the band theory of conductors, insulators and semiconductors, the forbidden energy gap is smallest in which one?
Answer.According to band theory the forbidden energy gap in conductors is Eg ≈ 0, in insulators Eg > 3 eV and in semiconductors Eg < 3 eV. Thus, the forbidden energy gap is smallest in conductors.
Question 7. What should be the distance between nucleons for effective nuclear forces?
Heavy stable nuclei have more neutrons than protons. Why?
Answer.Nuclear force is powerfully attractive between nucleons at a distance of about 1 femtometre, i.e., 1 × 10–15 m.
The stability of a nucleus depends on its neutron to proton ratio. More is the number of protons in the nucleus, greater is the electrostatic forces between them. Therefore, more neutrons are needed to provide the strong attractive force necessary to keep the nucleus stable.
Question 8. The given graphs show the variation of photoelectric current (I) with the applied voltage (V) for two different materials and for two different intensities of the incident radiations. Identify the pairs of curves that correspond to different materials but same intensity of incident radiations.
Answer.(1, 3) and (2, 4).
As photoelectric saturation current for them is same,so incident radiations on them are of same intensity, however their stopping potential are different, so they correspond to different materials.
Question 9. In an unbiased p-n junction, holes diffuse from the p-region to n-region. Why ?
In half wave rectification, what is the output frequency if the input frequency is 50 Hz ?
Answer.The hole concentration on p-side is greater than on n-side. So, diffusion take place.
In half wave rectification, only one ripple is obtained per cycle in the output. But the output frequency remain same. So, output frequency = 50 Hz.
Question 10. In hydrogen atom, if the electron is replaced by a particle which is 200 times heavier but has the same charge, how would its radius change ?
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.Radius of nth orbit of H-atom with electron is
So, if electron is replaced with a charged particle of same charge but of mass 200m, then radius of nth orbit of H-atom will become
Question 11. Assertion (A) : A charged particle free to move in an electric field always moves along an electric field line.
Reason (R) : The electric field lines diverge from a negative charge and converge at a positive charge.
Answer.(d) : Only a charged particle initially at rest moves along the electric field line. In case, the charged particle is in motion and its initial velocity is at an angle with the electric field line, the particle may follow a parabolic path.
Question 12. Assertion (A) : When a white light is passed through a lens, violet light is more refracted than red light.
Reason (R) : Focal length for red light is greater than violet.
Answer.(a) : When white light is passed through a lens, violet light is more refracted than red light because wavelength of violet is less than red light and therefore focal length for red light is greater than violet.
Question 13. Assertion (A) : Dipole oscillations produces electromagnetic waves.
Reason (R) : Accelerated charge produces electromagnetic waves.
Answer.(a) : Assertion is true. The reason is also true according to the classical theory of Maxwell. The dipole oscillations gives radiation because the charges are accelerated.
Question 14. Assertion (A) : The force between the plates of a parallel plate capacitor varies linearly with charge on it.
Reason (R) : Electric force is equal to charge per unit area.
Answer.(d) : The potential energy stored between the plates of a capacitor,
Negative sign shows force is attractive.
SECTION – B
Questions 15 and 16 are Case Study based questions and are compulsory. Attempt any 4 sub parts from each question. Each question carries 1 mark.
Question 15. A wheatstone bridge as showing in figure is almost balanced, where P = 10 Ω W, R = 5 Ω W, S = 11 Ω W, E = 48 V and C is grounded.
(i) The potential of point B is
(a) 16 V
(b) 24 V
(c) 32 V
(d) 36 V
(ii) If a galvanometer is connected between B and D, the direction of current through galvanometer is
(a) B to D
(b) D to B
(c) depends on resistance of galvanometer
(d) in the beginning from B to D and later on form D to B.
(iii) For what additional value of the resistance in arm BC in series/parallel would make the bridge balanced?
(a) 22 W
(b) 2 W
(c) 20 W
(d) 44 W
(iv) In wheatstone method, the instrument used as null detector is
(d) none of these
(v) Balance condition can be obtained by
(a) varying the standard arm resistance
(b) varying the ratio arms resistance
(c) keeping the unknown resistance constant
(d) by making use of a null detector in wheatstone method.
Question 16. Optical power (also referred as dioptric power, refractive power, focusing power, or convergence power) is the degree to which a lens, mirror, or other optical system converges or diverges light. It is equal to the reciprocal of the focal length of the device i.e., P = 1/f. The SI unit for optical power is the inverse meter (m−1), which is commonly called the dioptre. Converging lenses have positive optical power, while diverging lenses have negative power. When a lens is immersed in a refractive medium, its optical power and focal length change.
(i) What is the relation between the object distance u, image distance v and power P in a lens?
(a) P = uv/(u − v)
(b) P =u− v/uv
(c) P =uv/ v + v
(d) P = u+ v/uv
(ii) Two thin lenses are kept near each other at a distance of 4 mm. If the focal length of first and second lens are 20 cm and 25 cm respectively, find the power of the combination.
(a) 7.6 D
(b) 8.9 D
(c) –7.6 D
(d) –8.9 D
(iii) A convex lens of focal length 20 cm made of glass is immersed in a liquid of refractive index 2.3. Its power will be
(b) can be positive or negative depending on the depth
(iv) Focal length of a convex lens of refractive index 1.5 is 2 cm. Focal length of lens when immersed in a liquid of refractive index of 1.25 will be
(a) 5 cm
(b) 4 cm
(c) 2.5 cm
(d) 1.2 cm
(v) Find the focal length of a lens of power (–2.0 D). What type of lens is this?
(a) –0.5 m, concave lens
(b) 0.5 m, concave lens
(c) 0.5 m, convex lens
(d) –0.5 m, convex lens
SECTION – C
All questions are compulsory. In case of internal choices, attempt anyone.
Question 17. Two identical loops, one of copper and another of aluminium are rotated with the same speed in a uniform magnetic field acting normal to the plane of loops. State with reason for which of the coils, the induced
(b) current will be more and why?
Answer.(a) Induced EMF in both the loops will be same, as the two loops are of same area A and are rotated with same speed w in same magnetic field B given by
e = BA w sin wt
(b) As I = ε/R
, so in copper loop with less resistance,
induced current will be more.
Question 18. Find the total energy stored in the condenser system shown in the figure.
Four point charges –Q, –q, 2q and 2Q are placed, one at each corner of the square. If the potential at the centre of the square is zero then find the relation between Q and q.
Question 19. Name the electromagnetic radiation to which waves of wavelength in the range of 10–2 m belong. Give one use of this part of electromagnetic spectrum.
How are electromagnetic waves produced?
Answer. Radio wave: It is used for communication purposes like radio and television broadcast.
An oscillating or accelerated charge is supposed to be source of an electromagnetic wave. An oscillating charge produces an oscillating electric field in space which further produces an oscillating magnetic field which in turn is a source of electric field. These oscillating electric and magnetic field, hence, keep on regenerating each other and an electromagnetic wave is produced.
Question 20. Consider the circuit shown here where APB and AQB are semi-circles. What will be the magnetic field at the centre C of the circular loop?
Answer.Zero, because magnetic field of each half cancels each other. As it has same magnitude but opposite direction
Question 21. Draw the output waveform across the resistor.
The diode acts as half wave rectifier, it offers low resistance when forward biased and high resistance when reverse biased.
Question 22. In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?
A proton has spin and magnetic moment just like an electron. Why then its effect is neglected in magnetism of materials?
Answer.At geomagnetic north or south pole, angle of dip is 90°, where horizontal component of earth’s magnetic field BH is zero. A compass needle can only turn in horizontal plane, so it can point in any direction as BH = 0, which governs its direction.
Question 23. Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass
sheet when immersed in a liquid?
Answer.When the refractive index of the biconvex lens is equal to the refractive index of the liquid in which lens is immersed then the biconvex lens behaves as a plane glass sheet. In this case, 1/f =0 or f→ ∞.
Question 24. Three photo diodes D1, D2 and D3 are made of semiconductors having band gaps of 2.5 eV,2 eV and 3 eV, respectively. Which ones will be able to detect light of wavelength 6000 Å?
Answer.We know that, energy of incident photon
For the incident radiation to be detected by the photodiode, energy of incident radiation photon should be greater than the band gap. This is true only for D2. Therefore, only D2 will detect this radiation.
Question 25. A parallel beam of light of 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Calculate the width of the slit.
Answer.Position of first minimum in diffraction pattern
SECTION – D
All questions are compulsory. In case of internal choices, attempt any one.
Question 26. Sketch the graphs, showing the variation of stopping potential Vs with frequency u of the incident radiations for two photosensitive materials A and B having threshold frequencies u0 and u0′ respectively such that u0 > u0′.
(a) Which of the two metals A or B has higher work function?
(b) What information do you get from the slope of the graphs?
(c) What does the value of the intercept of graph A on the potential axis represent?
Answer.The variation of stopping potential (Vs) with frequency (u) of the incident radiations for two photosensitive materials is shown in figure.
(a) Metal A has higher work function, as
Question 27. A cell of emf e and internal resistance r is connected across a variable external resistance R. Plot graphs to show variation of
(i) E with R
(ii) Terminal p.d. of the cell (V) with R.
(a) In a meter bridge as shown in figure, the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connection between resistors in a wheatstone or meter bridge made of thick copper strips?
(b) Determine the balance point of the bridge above if X and Y are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?
Answer.(i) As the emf e of cell is independent of resistance R of the circuit, so it does not changes with increase in R.
The connections are made of thick copper strips so as to provide negligible resistance by connecting wires.
(b) If X and Y are interchanged the balance point will be at 60.5 cm from A.
(c) In balanced condition of the bridge, the cell and the galvanometer can be exchanged, the galvanometer will still show zero deflection.
Question 28. Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (m–) of mass about 207 me orbits around a proton].
Answer.In Bohr’s model, the radius of nth orbit
In the given muonic hydrogen atom, a negatively charged muon (μ–) of mass 207me revolve around a proton.
Therefore, radius of electron and muon can be written as
Question 29. We are given the following atomic masses:
92U238 = 238.05079 u, 90Th234 = 234.04363 u, 91Pa237 = 237.05121 u, 1H1 = 1.00783 u, 2He4 = 4.00260 u
(a) Calculate the energy released during a-decay of 92U238.
(b) Calculate the kinetic energy of emitted a-particle.
(c) Show that 92U238 cannot spontaneously emit a proton.
As the Q value is negative, the process cannot proceed spontaneously.
Question 30. A long solenoid S has n turns per metre, with diameter a. At the centre of this coil, we place a smaller coil of N turns and diameter b (where b < a). If the current in the solenoid increases linearly with time, what is the induced emf appearing in the smaller coil. Plot graph showing nature of variation in emf, if current varies as (mt2 + C).
A series LCR circuit is made by taking R = 100 W, L = 2/p H, C = 100/p mF. The series combination is connected across an a.c. source of 220 V, 50 Hz. Calculate
(a) the impedance of the circuit,
(b) the peak value of the current flowing in the circuit.
SECTION – E
All questions are compulsory. In case of internal choices, attempt any one.
Question 31. (a) Draw a ray diagram to show the formation of the image of an object placed between the optical centre and focus of a convex lens. Deduce the relationship between the object distance, image distance and focal length under the conditions stated.
(b) A diverging lens of focal length f is cut into two identical parts, each forming a plano concave lens.What is the focal length of each part ?
(a) A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
(b) Answer the following questions :
(i) When a low-flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
(ii) The principle of linear superposition of wave displacements is basic for understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?
Answer.(a) Ray diagram for the formation of image of an object AB placed between the optical centre C and focus F of convex lens is shown below. The image A′B′ is virtual, erect and magnified.
(b) (i) The low flying aircraft reflects the TV signals.Due to superposition between the direct signal received by the antenna and the reflected signals from aircraft. We sometimes notice slight shaking of the picture on the TV screen.
(ii) Superposition principle states how to explain the formation of resultant wave by combination of two or more waves. Let y1 and y2 represent instantaneous displacement of two superimposing waves, then instantaneous displacement for resultant waves is given by
y = y1 + y2
Question 32. (a) Derive the relationship between the peak and the rms value of current in ac circuit.
(b) The electric current in an AC circuit is given by I = I0 sin wt. What is the time taken by the current to change from its maximum value to the rms value ?
(a) Draw a schematic arrangement for winding of primary and secondary coil in a transformer when the two coils are wound on top of each other.
(b) State the underlying principle of a transformer and obtain the expression for the ratio of secondary to primary voltage in terms of the number of secondary and primary windings and primary and secondary currents.
(c) Write the main assumption involved in deriving the above relations.
(d) Write any two reasons due to which energy losses may occur in actual transformers.
(a) The instantaneous value of ac passing through a resistance R is given by
I = I0 sin wt
The alternating current changes continuously with time. Suppose that the current through the resistance is constant for an infinitesimally small time dt.
The small amount of heat produced in the resistance R in time dt is given by
dH = I2 R dt = (I0 sin wt)2 R dt = I02 R sin2 wt dt
The amount of heat produced in the resistance in time T/2 is
(c) The following three assumptions are involved
(i) The primary resistance and current are small.
(ii) The same flux links both with the primary and secondary windings as flux leakage from the core is negligibly small.
(iii) The terminals of the secondary are open or the current taken from it is small.
(d) 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 magnetizationand 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.
Question 33. (a) Use Gauss’ law to derive the expression for the electric field (E→) due to a straight uniformly charged infinite line of charge density l C m–1.
(b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
(c) Find the work done in bringing a charge q from perpendicular distance r1 to r2 (r2 > r1).
A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed vx. The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL2/(2mvx2).
Answer.(a) Electric field intensity due to line charge or infinite long uniformly charged wire at point P at distance r from it is obtained as :
Assume a cylindrical gaussian surface S with charged wire on its axis and point P on its surface, then ne electric flux through surface S is
Let the point at which the charged particle enters the electric field, be origin O(0, 0), then after travelling a horizontal displacement L, it gets deflected by displacement y in vertical direction as it comes out of electric field.
So, co-ordinates of its initial position are x1 = 0and y1=0
and final position on coming out of electric field are x2 = L and y2 = y
Components of its acceleration are ax = 0