# Class 12 Physics Sample Paper Term 1 With Solutions Set G

Please refer to Class 12 Physics Sample Paper Term 1 With Solutions Set G provided below. The Sample Papers for Class 12 Physics have been prepared based on the latest pattern issued by CBSE. Students should practice these guess papers for class 12 Physics to gain more practice and get better marks in examinations. The Term 1 Sample Papers for Physics Standard 12 will help you to understand the type of questions which can be asked in upcoming examinations.

## Term 1 Sample Paper for Class 12 Physics With Solutions Set G

SECTION – A

Question 1. Two circular coils can be arranged in any of three situations as shown in the figure. In which of the following situations, the mutual inductance will be maximum?

OR
Depict a graph which shows variation of induced e.m.f. with the rate of change of current flowing through a given coil.
The mutual inductance will be maximum in the first situation.

Question 2. What do you call the angle between magnetic meridian and geographical meridian?
Answer.Angle between magnetic meridian and geographical meridian is known as angle of declination or magnetic declination.

Question 3. How can we shield an object from the influence of a strong electrostatic field?
OR
Two identical conducting balls A and B have charges –Q and +3Q respectively. They are brought in contact with each other and then separated by a distance d apart. Find the nature of the Coulomb force between them.
Answer.As we found that electric field inside the cavity of conductor is zero, even on charging the conductor, so the object can be shielded from the strong electrostatic
fields in its environment, by covering it with a metallic cover.
OR
Final charge on each ball

As both the balls have same nature of charges, hence nature of the Coulomb force is repulsive.

Question 4. Two identical light waves, propagating in the same direction, have a phase difference d. After they superpose find the relation between the intensity of the resulting wave and the phase difference.

Question 5. The nuclear radius of 2713 Al is 3.6 fermi. Find the nuclear radius of 6429Cu .
OR
Four nuclei of an element undergo fusion to form a heavier nucleus, with release of energy. Which of the two — the parent or the daughter nucleus — would have higher binding energy per nucleon?

OR
In nuclear fusion, daughter nucleus would have higher binding energy per nucleon.

Question 6. State the criteria for the phenomenon of total internal reflection of light to take place.
Answer.Essential conditions for total internal reflection:
(i) Light should travel from a denser medium to a rarer medium.
(ii) Angle of incidence in denser medium should be greater than the critical angle for the pair of media in contact.
(ii) Angle of incidence in denser medium should be greater than the critical angle for the pair of media in contact.

Question 7. Calculate the de-Broglie wavelength of the electrons accelerated through a potential difference of 10 kV.
OR
In an experiment of photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10–15 V s. Calculate the value of Planck’s constant.

Question 8. The graph shown in the figure represents a plot of current versus voltage for a given semiconductor. Identify the region, if any, over which the semiconductor has a negative resistance.

Answer.Region BC of the graph has a negative slope, hence in region BC semiconductor has a negative resistance.

Question 9. State one use of photodiode.
Answer.Photo diodes are used to detect optical signals.

Question 10. Red, blue, green and violet colour lights are one by one made to incident on a photocathode. It is observed that only one colour light produces photoelectrons. Identify that light.
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
The energy of incident light is
E = hu
where h is the Planck’s constant and u is the frequency of incident light.
As uviolet > ublue > ugreen > ured
Eviolet > Eblue > Egreen > Ered
Since the incident energy is maximum for violet
colour, therefore violet light produces photoelectrons.

Question 11. Assertion (A) : Electron has higher mobility than hole in a semiconductor.
Reason (R) : Mass of electron is less than the mass of hole.

B

Question 12. Assertion (A) : X-ray astronomy is possible only from satellites orbiting the earth.
Reason (R) : Efficiency of X-rays telescope is large as compared to any other telescope.

C

Question 13. Assertion (A) : Critical angle of light passing from glass to air is minimum for violet colour.
Reason (R) : The wavelength of blue light is greater than the light of other colours.

C

Question 14. Assertion (A) : For best contrast between maxima and minima in the interference pattern of Young’s double slit experiment, the intensity of light emerging out of the two slits should be equal.
Reason (R) : The intensity of interference pattern is proportional to square of amplitude.

B

SECTION – B

Question 15. Effect of dielectric on capacity : A parallel plate capacitor consists of two parallel metallic plates separated by a small distance. If a dielectric medium of dielectric constant K is filled completely between the plates then,capacitance increases by K times. Now, consider
Two parallel plate capacitors A and B have the same separation d = 8.85 × 10–4 m between the plates. The plate area of A and B are 0.04 m2 and 0.02 m2 respectively. A slab of dielectric constant (relative permittivity K = 9) has dimensions, such that it can exactly fill the space between the plates of capacitor B.

(i) A dielectric can be made a conductor by
(a) compression
(b) heating
(c) doping
(d) freezing

B

(ii) The dielectric slab is placed inside A as shown in figure. A is then charged to a potential difference of 110 V.Calculate the capacitance of A.

(a) 2 × 10–9 F
(b) 3 ×10–6 F
(c) 2.5 × 10–9 F
(d) 3.6 × 10–6 F

A

(iii) Find the energy stored in capacitor A when a potential of 110 V is applied across it.
(a) 2.2 × 10–5 J
(b) 3.2 × 10–5 J
(c) 1.2 × 10–5 J
(d) 4.2 × 110–5 J

C

(iv) If the battery is disconnected and the dielectric slab is removed from A then find the work done by the external agency in removing the slab from A.

(a) 6.3 × 10–4 J
(b) 6.2 × 10–5 J
(c) 4.84 × 10–5 J
(d) 4.84 × 10–4 J

C

(v) The same dielectric slab is now placed inside B, filling it completely. The two capacitors A and B are then connected as shown in figure. Calculate the energy stored in the system.

(a) 1.1 × 10–5 J
(b) 2.6 × 10–5 J
(c) 3.1 ×10–5 J
(d) 2 × 10–5 J

A

16. An emf induced by the motion of the conductor across the magnetic field is called motional electromotive force. It is given as, e = –Bvl. This equation is true as long as the velocity, field and length are mutually perpendicular. The minus sign associated with the Lenz’s law. As shown in figure, a rectangular loop MNOP is pulled out of a magnetic field with a uniform velocity v by applying an external force F. The length MN is equal to l and the total resistance of the loop is R.

(i) Find the current in the loop.

A

(ii) Find the magnetic force on the loop.

B

(iii) Find the external force F needed to maintain constant velocity

A

(iv) The power delivered by the external force is

D

(v) The thermal power developed in the loop is

A

SECTION – C

Question 17. Calculate the energy in fusion reaction

Energy released = final B.E. – initial B.E.
= 7.73 – (2.23 + 2.23) = 3.27 MeV.

Question 18. Explain how electron mobility changes for a good conductor, when (a) the temperature of the conductor is decreased at constant potential difference and (b) applied potential difference is doubled at constant temperature.
OR
Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1 mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistance RA to RB.
Electron mobility in a conductor is given by,

(a) When the temperature of the conductor decreases, the relaxation time, t of the electrons increases, so mobility m increases.
(b) Mobility m is independent of applied potential difference.
OR
For a solid wire of resistance RA,

Question 19. Consider the junction diode as ideal. Find the value of current flowing through AB is

OR
The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~ mA).
What is the reason, then, to operate the photodiode in reverse bias?
Answer.Here, the p-n junction diode is forward biased, hence it offers zero resistance.

OR
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 carrier concentration when there is no illumination.
Remember Dn = Dp 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., Dp/p).
In general, we can state that the fractional change due to the photo-effects 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 20. Consider two conducting spheres of radii R1 and R2 with R1 > R2. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.

Question 21. The intensity at the central maxima in Young’s double slit experimental set-up is I0. Show that the intensity at a point where the path difference is l/3 is I0/4.
OR
Two independent monochromatic sources of light cannot produce a sustained interference pattern. Give reason.

OR
Two independent monochromatic sources cannot produce sustained interference pattern because the phase difference between the light waves from two independent sources keeps on changing continuously.

Question 22. The magnetic force depends on v which depends on the inertial frame of reference. Does then the magnetic force differ from inertial frame to frame? Is it reasonable that the net acceleration has a different value in different frames of reference?

Here, Fm depends on the inertial frame of reference.
Hence magnetic force is frame dependent. Net acceleration arising from this force is however frame independent for inertial frames.

Question 23. A hemisphere is uniformly charged positively. Give the direction of electric field at a point on the diameter, and away from the centre.
Answer.When the point is on the diameter and away from the centre of hemisphere which is charged uniformly and positively, the component of electric field intensity parallel to the diameter cancel out. So the electric field is perpendicular to the diameter.

Question 24. A proton and an a-particle have the same de-Broglie wavelength. Determine the ratio of
(i) their accelerating potentials (ii) their speeds.
Answer.de-Broglie wavelength of a particle of mass m and
charge q accelerating through a potential V is given by

Question 25. In a semiconductor, 2/3 rd of the total current is carried by electrons and remaining 1/3 rd by the holes. If at this temperature, the drift velocity of electrons is 3 times that of holes, what will be the ratio of number density of electrons to that of holes?
If I is the total current, then

SECTION – D

Question 26. Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
We know for diffraction to take place, size of the obstacle/aperture should be of the order of wavelength.Wavelength of sound waves is of the order of few meters that is why sound waves can bend through the aperture in partition wall but wavelength of light waves is of the order of micrometer, hence light waves can not bend through same big aperture. That is why the two students can hear each other but cannot see each other.

Question 27. When the oscillating electric and magnetic fields are along the x-and y-direction respectively, then
(i) Point out the direction of propagation of electromagnetic wave.
(ii) Express the velocity of propagation in terms of the amplitudes of the oscillating electric and magnetic fields.
OR
(a) How do you show that the e.m. wave carries energy and momentum?
(b) Which e.m. waves lie near the high frequency end of visible part of e.m. spectrum? Give its one use. In
what way this component of light has harmful effects on humans?
Answer.(a) The e.m. wave propagates along z-axis.

(b) The speed of em-waves in vacuum determined by the electric (E0) and magnetic fields (B0) is, c = E0/B0
OR
(a) Electromagnetic waves or photons transport energy and momentum. When an electromagnetic wave interacts with a small particle, it can exchange energy and momentum with the particle. The force exerted on the particle is equal to the momentum transferred per unit time. Optical tweezers use this force to provide a non-invasive technique for manipulating microscopic-sized particles with light.
(b) Ultraviolet rays lie near the high-frequency end of visible part of e.m. spectrum. These rays are used to preserve food stuff. The harmful effect from exposure to ultraviolet (UV) radiation can be life threatening, and include premature aging of the skin, suppression of the immune systems, damage to the eyes and skin cancer.

Question 28. Distinguish between Biot Savart’s law and Ampere’s circuital law.

Question 29. Two inductors of self-inductances L1 and L2 are connected in parallel. The inductors are so far apart that their mutual inductance is negligible. Derive the equivalent inductance of the combination.
OR
Suppose the loop with a small cut as shown in figure is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that field decreases from its initial value of 0.3 T at the rate of 0.02 T s–1. If the cut is joined and the loop has a resistance of 1.6 W, how much power is dissipated by the loop as heat? What is the source of this power?

Answer.Inductances in parallel : For the parallel combination, the total current I divides up through the two coils as I = I1 + I2.

Question 30. (a) The earth’s magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?
(b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why?
(a) Yes, earth’s field undergoes a change with time.
For example, daily changes, annual changes, secular changes with period of the order of 960 years and irregular changes like magnetic storms. Time scale for appreciable change is roughly a few hundred years.
(b) The earth’s core does contain iron but in the molten form only. This is not ferromagnetic and hence it cannot be treated as a source of earth’s magnetism.

SECTION – E

Question 31. (a) Write three characteristic features to distinguish between the interference fringes in Young’s double slit experiment and the diffraction pattern obtained due to a narrow single slit.
(b) 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 a distance of 2.5 mm away from the centre. Find the width of the slit.
OR
A plane wavefront propagating in a medium of refractive index ‘μ1’ is incident on a plane surface making the angle of incidence i as shown in the figure. It enters into a medium of refraction of refractive index ‘μ2’ (μ2 > μ1). Use Huygens’ construction of secondary wavelets to trace the propagation of the refracted wavefront. Hence verify Snell’s law of refraction.

Answer.(a) Difference between interference and diffraction:
experiment to observe diffraction pattern

Let v1 and v2 represents the speed of light in medium 1 and medium 2 respectively. We assume a plane wavefront AB propagating in the direction A′A incident on the interface at an angle i. Let t be the time taken by the wavefront to travel the distance BC.
BC = v1t [… distance = speed × time]
In order to determine the shape of the refracted wavefront, we draw a sphere of radius v2t from the point A in the second medium (the speed of the wave in second medium is v2).
Let CE represents a tangent plane drawn from the point C. Then
AE = v2t

Question 32. Derive an expression for the potential energy of an electric dipole in a uniform electric field. Explain conditions for stable and unstable equilibrium.
OR
A small sphere of radius r1 and charge q1 is enclosed by a spherical shell of radius r2 and charge q2. Show that if q1 is positive, charge will necessary flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2 on the shell is.
Answer.Since net force on electric dipole in uniform electric field is zero, so no work is done in moving the electric dipole in uniform electric field, however some work is done in rotating the dipole against the torque acting on it. So, small work done in rotating the dipole by an angle dq in uniform electric field E is
dW = 𝜏 dθ = pE sinθ dθ
Hence, net work done in rotating the dipole from angle qi to qf in uniform electric field is

If initially, the dipole is placed at an angle qi = 90° to the direction of electric field, and is then rotated to the
angle θf = θ, then net work done is
W = pE [cos90° – cosθ]
or W = – pE cosθ
This gives the work done in rotating the dipole through an angle q in uniform electric field, which gets stored in it in the form of potential energy i.e.,
U = – pE cosθ
This gives potential energy stored in electric dipole of moment p when placed in uniform electric field at an angle q with its direction.
(i) When q = 0°, then Umin = –pE
So, potential energy of an electric dipole is minimum,when it is placed with its dipole moment p parallel to the direction of electric field E and so it is called its most stable equilibrium position.
(ii) When q = 180°, then Umax = + pE
So, potential energy of an electric dipole is maximum,when it is placed with its dipole moment p anti parallel to the direction of electric field E and so it is called its most unstable equilibrium position.
OR
The potential on inner small sphere is

So, charge q1 given to sphere A will flow on the shell B, no matter what the charge on the shell B is.

Question 33. Using postulates of Bohr’s theory of hydrogen atom, show that
(a) the radii of orbits increase as n2, and
(b) the total energy of the electron increase as 1/n2, where n is the principal quantum number of the atom.
OR
(a) Write Rydberg’s formula for wavelengths of the spectral lines of hydrogen spectrum. Mention to which series in the emission spectrum of hydrogen, Ha line belongs.
(b) Using Rydberg formula, calculate the longest wavelength belonging to Lyman and Balmer series of hydrogen spectrum. In which region these transitions lie?