See below CBSE Class 12 Physics Term 1 Sample Paper Set F with solutions. We have provided CBSE Sample Papers for Class 12 Physics as per the latest paper pattern issued by CBSE for the current academic year. All sample papers provided by our Class 12 Physics teachers are with answers. You can see the sample paper given below and use them for more practice for Class 12 Physics examination.
CBSE Sample Paper for Class 12 Physics Term 1 Set F
SECTION – A
Question 1. How does an increase in doping concentration affect the width of depletion layer of a p-n junction diode?
Answer.When there is an increase in doping concentration,the applied potential difference causes an electric field which acts opposite to the potential barrier. This results in reducing the potential barrier and hence the width of depletion layer decreases.
Question 2. A resistance R is connected across a cell of emf e and internal resistance r. A potentiometer now measures the potential difference between the terminals of the cell as V. Write the expression for r in terms of e, V and R.
Define electrical resistivity of a material.
It is defined as the resistance offered by a unit length and unit area of cross-section of the wire.
Question 3. A test charge q is made to move in the electric field of a point charge Q along two different closed paths. First path has sections along and perpendicular to lines of electric field. Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases ?
Answer.Since the electric field is conservative in nature,work done in both the cases is zero.
Question 4. What will be the shape of the interference fringes in Young’s double slit experiment when D (distance between slit and screen) is very large as compared to fringe width?
Question 5. Draw a plot showing the variation of resistivity of a conductor with the increase in temperature.
The V-I characteristic of a silicon diode is shown in figure. What is the resistance of the diode at ID = 15 mA.
Answer.The resistivity of a conductor increases nonlinearly with increase in temperature.
Question 6. If the Earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now ?
Answer.Lower because of absence of green house effect.
Question 7. A step down transformer converts transmission line voltage from 11000 V to 220 V. The primary of the transformer has 6000 turns and efficiency of the transformer is 60%. If the output power is 9 kW, then the value of input power?
Answer.Here, Vp = 11000 V, Vs = 220 V
Np = 6000, η = 60%
Po = 9 kW = 9 × 103 W
Question 8. What are the directions of electric and magnetic field vectors relative to each other and relative to the direction of propagation of electromagnetic waves?
It is an electric field associated with an electromagnetic wave in vacuum is given by E = 40 cos (kz − 6 × 108t)iˆ,where E , z and t are in volt per meter, meter and second respectively. Find the value of wave vector k.
Answer.In an electromagnetic wave E,B and direction of propagation are mutually perpendicular.
Compare the given equation with
E = E0cos(kz – wt)
We get, w = 6 × 108 s–1
Question 9. What is the impedance of a capacitor of capacitance C in an ac circuit using source of frequency n Hz?
Answer.Impedance of a capacitor, Z = XC
Question 10. In the magnetic meridian of a certain place the horizontal component of earth’s magnetic field is 0.25 G and dip angle is 60°. What is the magnetic field of the earth at this location?
What will be the nature of force, due to earth’s horizontal magnetic field?
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.Here, HE = 0.25 G and δ = 60°
Earth’s horizontal magnetic field is uniform. The lines of force due to the field are parallel and straight.
Question 11. Assertion (A): An electric field is preferred in comparison to magnetic field for detecting the electron beam in a television picture tube.
Reason (R): Electric field requires low voltage.
Question 12. Assertion (A) : Work function of copper is greater than that of sodium. But both will have same value of threshold frequency and threshold wavelength.
Reason (R) : The frequency is directly proportional to wavelength.
Question 13. Assertion (A) : Magnetic force is always perpendicular to the magnetic field.
Reason (R) : Electric force is along the direction of electric field.
Question 14. Assertion (A) : Dipole oscillations produces electromagnetic waves.
Reason (R) : Accelerated charge produces electromagnetic waves.
SECTION – B
Question 15. The capacity of a parallel plate air capacitor is C0 =ε0 .A/d When air is replaced fully by an insulating medium of dielectric constant K, its capacity becomes
(i) The capacitance of a capacitor will decrease if we introduce a slab of
(d) none of these.
(ii) When area of parallel plate air capacitor is halved and distance between the plates is doubled, its capacity becomes n times, where n is
(iii) A metal plate of thickness t = d/2 is introduced in between the plates of an air capacitor. The increase in its capacity is
(iv) Very thin metal foil is introduced in between the plates of an air capacitor of capacitance C. The new capacity is
(b) 2 C
(d) 3 C
(v) Potential drop in a dielectric is equal to _______
(a) electric field strength × thickness
(b) electric field strength × area of a cross-section
(c) electric field strength
Question 16. A town has a population of 1 million. The average electric power needed per person is 300 W. A reactor is to be designed to supply power to this town. The efficiency with which thermal power is converted into electric power is aimed at 25%.
(i) The release in energy in nuclear fission is consistent with the fact that uranium has
(a) more mass per nucleon than either of the two fragments
(b) has more mass per nucleon as the two fragment
(c) has exactly the same mass per nucleon as the two fragments
(d) less mass per nucleon than either of the two fragments.
(ii) Assuming 200 MeV of thermal energy to come from each fission event on an average, find the number of events on an place every day.
(a) 3.24 × 1024
(c) 3.24 × 1022
(d) 3.24 × 1020
(iii) Assuming the fission to take place largely through U235, at what rate will the amount of U235 decrease?
(a) 1.2643 kg
(b) 1.367 kg
(c) 3.0 kg
(d) 4.56 kg
(iv) If uranium enriched to 3% in U235 will be used, how much uranium is needed per month (30 days)?
(a) 1000 kg
(b) 1264 kg
(c) 3000 kg
(d) 1462 kg
(v) If the number of fissions are 2.375 × 1024, what is the mass of U235 isotope used?
(a) 789.23 g
(b) 879.32 g
(c) 926.65 g
(d) 980.74 g
SECTION – C
Question 17. When a capacitor is connected in series with a series LR circuit, the alternating current flowing in the circuit increases. Explain why ?
Question 18. An a-particle and a proton are accelerated from rest by the same potential. Find the ratio of their de Broglie wavelengths.
The photoelectric threshold wavelength for silver is λ0. Find the energy of the electron ejected from the surface of silver by an incident wavelength λ(λ < λ0).
Question 19. 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.
Answer.Step-up transformer (or transformer) is based on the principle of mutual induction.
An alternating potential (Vp) when applied to the primary coil induced emf in it.
In case of dc voltage flux does not change. Thus, no emf is induced in the circuit.
Question 20. The electric field inside a parallel plate capacitor is E. Find the amount of work done in moving a charge q over a closed rectangular loop a b c d a.
Answer.Electric field inside a parallel plate capacitor = E
Here, electric field is conservative.
Work done by the conservative force in closed loop is zero.
So, required work done = 0.
Question 21. A 80 mF capacitor is charged by a 50 V battery. The capacitor is disconnected from the battery and then connected across another uncharged 320 mF capacitor. Calculate the charge on second capacitor.
Answer. Common potential after connection is
Question 22. When viewing through a compound microscope, our eyes should be positioned not on the eye-piece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. Given reason.
Answer.The eye-piece makes a real image of the objective aperture at a short distance on the outer (eye) side. This is called “exit-pupil” or “eye-ring”. Therefore, the eye must have its pupil on this exit-pupil for best viewing.It is only a few millimeters from the eye-piece.
Reflection and refraction arise through interaction of incident light with atomic constituents of matter which vibrate with the same frequency as that of the incident light. Hence frequency remains unchanged.
Question 23. The following table provides the set of values, of V and I, obtained for a given diode:
Assuming the characteristics to be nearly linear, over this range, calculate the forward and reverse bias resistance of the given diode.
Energy gap in a p – n photodiode is 2.8 eV. Can it detect a wavelength of 600 nm? Justify your answer.
Question 24. Why should the objective of a telescope have large focal length and large aperture? Justify your answer.
Answer. Objective of a telescope is a convex lens of large focal length and a large aperture. It faces the distant object and forms bright image of the distant objects.
The aperture of the objective is taken large so that it can gather sufficient amount of light from the distant objects.
Question 25. Even though an electric field E exerts a force qE on a charged particle yet the electric field of an EM wave does not contribute to the radiation pressure (but transfers energy). Explain.
Answer. Electric force on a charged particle is oscillating because electric field of an electromagnetic wave is an oscillating field. Since direction changes every half cycle, so electric force averaged over an integral number of cycles is zero. That is why electric field is not responsible for radiation pressure.
SECTION – D
Question 26. In the figure shown, calculate the total flux of the electrostatic field through the spheres S1 and S2. The wire AB, shown here, has a linear charge density, l, given by l = kx where x is the distance measured along the wire, from the end A.
Answer.Total flux through the sphere S1
Φ1 = Q/ε0
Charge on the line enclosed by the sphere S2
Question 27. (a) State the law that gives the polarity of the induced emf.
(b) A 15 mF capacitor is connected to 220 V, 50 Hz source. Find the capacitive reactance and the rms current.
The current flowing through an inductor of self inductance L is continuously increasing. Plot a graph showing the variation of
(i) Magnetic flux versus the current
(ii) Induced emf versus dI/dt
(iii) Magnetic potential energy stored versus the current.
Answer.(a) Lenz’s law gives the polarity of induced emf. The induced emf (or current) always opposes the cause that produces it.
Question 28. The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~ μA)
What is the reason, then, to operate the photodiode in reverse bias ?
Answer.Consider the case of an n-type semiconductor.
The majority carrier density (n) is considerably larger than the minority hole density p (i.e., n >> p).
On illumination, let the excess electrons and holes generated be Dn and Dp, respectively
n′ = n + Δn
p′ = p + Δp
Here n′ and p′ are the electron and hole concentrations at any particular illumination and n and p are carriers concentration when there is no illumination. Remember Δn = Δp and n > > p. Hence, the fractional change in the majority carriers (i.e., Dn/n) would be much less than that in the minority carriers (i.e., Δp/p). In general, we can state that the fractional change due to the photoeffect on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes are preferably used in the reverse bias condition for measuring light intensity.
Question 29. (a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band? Explain.
(b) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot in seen at the centre of the obstacle. Explain why.
The intensity at the central maxima (O) in Young’s double slit experiment is I0. If the distance OP equals one-third of the fringe width of the pattern, show that the intensity at point P would be I0/4
Answer.(a) Intensity distribution of light in diffraction at a signal slit is shown in figure.
Question 30. A cell of emf E and internal resistance r is connected to two external resistances R1 and R2 and a perfect ammeter. The current in the circuit is measured in four different situations :
(i) without any external resistance in the circuit
(ii) with resistance R1 only
(iii) with R1 and R2in series combination
(iv) with R1 and R2 in parallel combination
The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in that order.
Identify the currents corresponding to the four cases mentioned above.
Answer.The current relating to corresponding situations are as follows :
SECTION – E
Question 31. Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field.
(a) State Biot-Savart law, and write its mathematical expression Use this law to derive an expression for the magnetic field due to a circular coil carrying current at a point along its axis.
(b) A long straight wire AB carries a current I. A proton P travels with a speed v, parallel to the wire, at a distance d from it in a direction opposite to the current as shown in the figure. What is the force experienced by the proton and what is its direction?
Answer.When a rectangular loop PQRS of sides ‘a’ and ‘b’ carrying current I is placed in uniform magnetic field B,such that area vector A
an angle q with direction of magnetic field, then forces on the arms QR and SP of loop are equal, opposite and collinear, thereby perfectly cancel each other, whereas
forces on the arms PQ and RS of loop are equal and opposite but not collinear, so they give rise to torque on the loop.
Question 32. Write the basic assumptions used in the derivation of lens maker’s formula and hence derive this expression.
(a) Draw a ray diagram for final image formed at least distance of distinct vision (D) by a compound microscope and write expression for its magnifying power.
(b) An angular magnification (magnifying power) of 30 is desined for a compound microscope using as objective of focal length 1.25 cm and eyepiece of focal length 5 cm. How will you set up the compound microscope?
Answer.Basic assumptions in derivation of lens maker’s formula:
(i) Aperture of lens should be small
(ii) Lenses should be thin
(iii) Object should be point sized and placed on principal axis.
Suppose we have a thin lens of material of refractive index n2, placed in a medium of refractive index n1, let O be a point object placed on principle axis then for refraction at surface ABC we get image at I1.
Question 33. (a) Using de Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second postulate of quantization of energy levels in a hydrogen atom.
(b) The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?
Using Bohr’s postulates, derive the expression for the frequency of radiation emitted when electron in hydrogen atom undergoes transition from higher energy state (quantum number ni) to the lower state, (nf).
When electron in hydrogen atom jumps from energy state ni = 4 to nf = 3, 2, 1. Identify the spectral series to which the emission lines belong.
Answer. (a) According to de Broglie, a stationary orbit is that which contains an integral number of de Broglie waves associated with the revolving electron.
For an electron revolving in nth circular orbit of radius rn,
Total distance covered = Circumference of the orbit = 2πrn
For the permissible orbit, 2πrn = nλ
According to de-Broglie,