Students must start practicing the questions from CBSE Sample Papers for Class 12 Physics with Solutions Set 1 are designed as per the revised syllabus.
CBSE Sample Papers for Class 12 Physics Set 1 with Solutions
Maximum Marks: 70
Time Allowed: 3 Hours
General Instructions:
- There are 35 questions in all. All questions are compulsory.
- This question paper has five sections: Section A, Section B, Section C, Section D and Section E. All the sections are compulsory.
- Section A contains eighteen MCQ of 1 mark each, Section B contains seven questions of two marks each, Section C contains five questions of three marks each, section D contains three long questions of five marks each and Section E contains two case study based questions of 4 marks each.
- There is no overall choice. However, an internal choice has been provided in section B, C, D and E. You have to attempt only one of the choices in such questions.
- Use of calculators is not allowed.
Section – A
Question 1.
According to Coulomb’s law, which is the correct relation for the following figure?
(a) q1 q2 > 0
(b) q1 q2 < 0
(c) q1 q2 = 0
(d) 1 > q1/ q2 > 0
Answer:
(b) q1q2 < 0
Explanation: According to CouLomb’s low,
\(\overrightarrow{\mathrm{F}}_{12}\) = –\(\vec{F}_{21}\)
There is a force of attraction and the charge must be unlike charge.
and we know that F = \(\frac{q_1 q_2}{4 \pi \varepsilon_0 r^2}\)
For attractive force, q2q2 < 0
q1 = +
q2 = –
q1q2 < 0
Question 2.
The electric potential on the axis of an electric dipole at a distance r from it’s centre is V. Then the potential at a point at the same distance on its equatorial line will be
(a) 2V
(b) -V
(c) \(\frac{V}{2}\)
(d) Zero
Answer:
(d) zero
Explanation
Electric potential at P due to dipole
[As shown in figure]
Let OP be equatorial line.
Let r be the distance of each charge from point P
We know that,
Potential due to charge + q at point P,
V1 = +\(\frac{k q}{r}\)
and
Potential due to charge – q at point P,
V2 = \(\frac{-k q}{r}\)
Hence,
Total potential at point P
V = V1 + V2
⇒ V = \(\frac{k q}{r}\) – \(\frac{k q}{r}\) = 0
Hence, the net potential at any point on the equatorial line of an electric dipole is zero. Hence, option (d) is correct
Related Theory
The electric field at each position is equal to the inverse of the potential gradient there. The negative sign indicates that the electric field is pointing in the direction of decreasing potential.
Question 3.
The temperature (T) dependence of resistivity of materials A and material B is represented by fig (i) and fig (ii) respectively.
Identify material A and material B.
(a) material A is copper and material B is germanium
(b) material A is germanium and material B is copper
(c) material A is nichrome and material B is germanium
(d) material A is copper and material B is nichrome [1]
Answer:
(b) material A is germanium and material B is copper
Explanation: For Semiconductor: (Fig. (i))
Number of free electron is very less. Due to temperature increase, number of free electron increase that electron act as a carrier therefore resistance of the conductor decrease.
For Conductor: (Fig. (ii))
Due to temperature increase number of free electron increase and its amount goes larger then they start to resist motion of each other therefore resistance of the conductor will increase.
So, we can say that material A is germanium and material B is copper.
Question 4.
Two concentric and coplanar circular loops P and Q have their radii in the ratio 2:3. Loop Q carries a current 9 A in the anticlockwise direction. For the magnetic field to be zero at the common centre, loop P must carry:
(a) 3A in clockwise direction
(b) 9A in clockwise direction
(c) 6A in anti-clockwise direction
(d) 6A in the clockwise direction. [1]
Answer:
(d) 6 A in the clockwise direction.
Explanation: We know that:
Given that: \(\frac{R_1}{R_2}\) = \(\frac{2}{3}\)
and l2 = 9A
l1 = ?
Magnetic field at center of circular coil
B = \(\frac{\mu_0 l}{2 r}\)
B ∝ \(\frac{1}{r}\)
According to the que.
BP = BQ
In opp. direction of l2
So, in clockwise direction.
Caution
Students are frequently puzzled by the direction of charged particles in electric and magnetic fields. The charged particle Will travel in either a parallel or anti-parallel direction to the electric and magnetic fields.
Question 5.
A long straight wire of circular cross section of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnitudes of magnetic field at a point distant \(\frac{a}{2}\) above the surface of wire to that at a point distant \(\frac{a}{2}\) below its surface is:
(a) 4 : 1
(b) 1 : 1
(c) 4: 3
(d) 3 : 4 [1]
Answer:
(c) 4 :3
Explanation:
Question 6.
If the magnetizing field on a ferromagnetic material is increased, its permeability:
(a) decreases
(b) increases
(c) remains unchanged
(d) first decreases and then increases [1]
Answer:
(a) decreases
Explanation: The magnetic permeability of a substance is defined as,
µ = \(\frac{B}{H}\)
where, B is the established magnetic field inside the material, H is the applied external magnetic field.
When the external applied magnetic field H is increased, the established magnetic field a for a ferromagnetic material remains the same. Thus the magnetic permeability decreases.
Question 7.
An iron cored coil is connected in series with an electric bulb with an AC source as shown in figure. When iron piece is taken out of the coil, the brightness of the bulb will:
(a) decrease
(b) increase
(c) remain unaffected
(d) fluctuate [1]
Answer:
(b) increase
Explanation: As we know that When the current passes through the inductor, magnetic field is produced and it opposes the current flow so when the iron coil is removed, the current flow will be maximum and bulb glows bright.
Related Theory
When iron coil is inserted, there will be induction current effect and bulb glows with less intensity.
Question 8.
Which of the following statement is NOT true about the properties of electromagnetic waves?
(a) These waves do not require any material medium for their propagation
(b) Both electric and magnetic field vectors attain the maxima and minima at the same time
(c) The energy in electromagnetic wave is divided equally between electric and magnetic fields
(d) Both electric and magnetic field vectors are parallel to each other [1]
Answer:
(d) Both electric and magnetic field vectors are parallel to each other
Explanation: We know that Electromagnetic waves do not require any matter to propagate from one place to another as it consists of photons. Therefore option 1 is correct.
In an electromagnetic wave, the electric field and magnetic field very continuously with maxima and minima at the same place and same time. Therefore option 2 is correct.
The energy in an electromagnetic wave is divided equally between electric and magnetic field. Therefore option 3 is correct.
An electromagnetic wave is a perpendicular variation in both the electric field (E) and Magnetic field (B). Therefore option 4 is incorrect.
Related Theory
Electromagnetic waves or EM waves: The waves that are formed as a result of vibrations between an electric field and a magnetic field and they are perpendicular to each other and to the direction of the wave is called an electromagnetic wave. The accelerating charged particle produces an electromagnetic (EM) wave A charged particle oscillating about an equilibrium position is an accelerating charged particle.
Electromagnetic waves do not require any matter to propagate from one place to another as it consists of photons. They can move in a vacuum. Electromagnetic waves move with the velocity of light in free space. Given by c = \(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)
Where c = Velocity of light ε0 = Permittivity of free space. µ0 = Permiability of free space.
Question 9.
A rectangular, a square, a circular and an elliptical loop, all in the (x-y) plane, are moving out of a uniform magnetic field with a constant velocity \(\bar{v}\) = vi. The magnetic field is directed along the negative z-axis direction. The induced emf, during the passage of these loops, out of the field region, will not remain constant for:
(a) any of the four loops
(b) the circular and elliptical loops
(c) the rectangular, circular and elliptical loops
(d) only the elliptical loops [1]
Answer:
(b) the circular and elliptical loops
Explanation: As we know induced emf,
|e| = \(\frac{d \phi}{d t}\) = \(\frac{\mathrm{B} d s}{d t}\)
Now, as the square loop and rectangular loop move out of magnetic field. \(\frac{\mathrm{B} d s}{d t}\) is constant, Therefore |e| is constant. But incase of circular and elliptical loops. \(\frac{d S}{d t}\) changes. Therefore, |e| does not remain constant.
Question 10.
In a Young’s double slit experiment, the path difference at a certain point on the screen between two interfering waves is \(\frac{1}{8}\)th of the wavelength. The ratio of intensity at this point to that at the centre of a bright fringe is close to:
(a) 0.80
(b) 0.74
(c) 0.94
(d) 0.85 [1]
Answer:
(d) 0.85
Explanation: We know that,
Phase difference Δϕ = \(\frac{2 n}{\lambda} \Delta x\)
Δϕ = \(\frac{2 n}{\lambda} \cdot\left(\frac{\lambda}{8}\right)\) [Given, Δx = \(\frac{\lambda}{8}\)]
Δϕ = \(\frac{n}{4}\)
In Young’s double slit experiment
We know that,
Intensity at any point, I = Imax cos2 \(\left(\frac{\Delta \phi}{2}\right)\)
Related Theory
The property of light travelling in a fixed direction is known as rectilinear propagation.
Question 11.
The work function for a metal surface is 4.14 eV. The threshold wavelength for this metal surface is:
(a) 4125 A
(b) 2062.5 A
(c) 3000 A
(d) 6000 A [1]
Answer:
(c) 3000 A
Explanation: Since work function for a metal surface is W = \(\frac{h c}{\lambda_0}\)
where, λ0 is threshold wavelength or cut-off wavelength for a metal surface, here
W = 4.125eV
= 4.125 × 1.6 × 10-19 Joule
so λ0 = \(\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{4.125 \times 1.6 \times 10^{-19}}\) = 3000A
Question 12.
The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10-11 m. The radius of the n = 3 orbit is:
(a) 1.01 × 10-10m
(b) 1.59 × 10-10m
(c) 2.12 × 10-10m
(d) 4.77 × 10-10m [1]
Answer:
(d) 4.77 × 10-10 m
Explanation:
Given that r = 5.3 × 10-11 m
Let r3 be the radius at n = 3.
So, r3 = n2r1 = 4.77 × 10-10 m
Related Theory
The mean radius of an electron s orbit around the nucleus of a hydrogen atom at its ground state, or lowest energy level is defined as the Bohr’s radius.
Question 13.
Which of the following statements about nuclear forces ¡s not true?
(a) The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres.
(b) The nuclear force is much weaker than the Coulomb force.
(c) The force is attractive for distances larger than 0.8 fm and repulsive if they are separated by distances less than 0.8
(d) The nuclear force between neutron-neutron, proton-neutron and proton-proton is approximately the same. [1]
Answer:
(b) The nuclear force is much weaker than the Coulomb force.
Explanation: As we know that nucleus force is a strong force, they are strongest in magnitude.
They are short range because the distance between the nucleon 0.7 fermi which is very small
They are charged independent. They result for interaction of every nucleon with the nearest limited number of nucleons.
Question 14.
If the reading of the voLtmeter V1 is 40 V, then the reading of voltmeter V2 is:
(a) 30 V
(b)58 V
(c) 29 V
(d) 15 V
Answer:
(a) 30V
Explanation:
Question 15.
The electric potential V as a function of distance X is shown in the figure:
The graph of the magnitude of electric field intensity E as a function of X is:
[1]
Answer:
An equipotential surface is one that has the some electric potential at every point on it.
The electric field is a potential difference derivative.
The negative sign indicates that the direction of E is the inverse of the direction of dv, implying that dv decreases aLong the path of E.
Caution
Students should verify that the electrostatic field varies as 1/r2, whereas the electrostatic potential
varies as 1/r. This requirement must be kept in mind when trying tasks, both conceptual and numerical.
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 and 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.
Question 16.
Assertion (A): The electrical conductivity of a semiconductor increases on doping.
Reason (R): Doping always increases the number of electrons in the semiconductor. [1]
Answer:
(c) A is true but R is false
Explanation: The conductivity of semiconductors is increased by adding an appropriate amount of suitable impurity or doping. Doping can be done with an impurity which is electron rich or electron deficient as compared to the intrinsic semiconductor. silicon or germanium. Such impurities introduce electronic defects in them. When silicon is doped with electron rich impurities the extra electron becomes delocalized.
Related Theory
These delocalized electrons increase the conductivity of doped silicon due to the negatively charged electron, hence silicon doped with electron-rich impurity is called n-type semiconductor which electron-deficit impurities increase the conductivity through positive holes and this type of semiconductors are called/p-type semiconductors.
Each pentovalent impurity atom provides one additional electron, which is referred to as the donor. Because negatively charged electrons carry the majority of the current, semiconductors doped with donor type impurities are referred to be n-type semiconductors..
Question 17.
Assertion (A): In an interference pattern observed in Young’s double slit experiment, if the separation (d) between coherent sources as well as the distance (D) of the screen from the coherent sources both are reduced to 1/3rd, then new fringe width remains the same.
Reason (R): Fringe width is proportional to (d/D). [1]
Answer:
(c) A is true but R is false
Explanation: Fringe width is inversely proportional to spacing between slits. If it doubled then fringe width will be halved.
Beta = λ × \(\frac{D}{d}\) where d is fringe width,
D = separation between screen and slits.
Question 18.
Assertion (A): The photoelectrons produced by a monochromatic light beam incident on a metal surface have a spread in their kinetic energies.
Reason (R): The energy of electrons emitted from inside the metal surface, is lost in collision with the other atoms in the metal. [1]
Answer:
(a) Both A and R are true and R is the correct explanation of A
Explanation: electrons being emitted as photoelectrons have different velocities. Actually all the electrons do not occupy the same level of energy but they occupy continuous band and levels. So, electrons being knocked off from different levels come out with different energies. Work function is the energy required to pull the electron out of metal surface. Naturally d-electrons on the surface will require less energy to be pulled out hence will have lesser work function as compared with those deep inside the metaL So, assertion and reason are correct and reason correctly explains the assertion.
Section – B
Question 19.
Electromagnetic waves with wavelength
(A) λ1 is suitable for radar systems used in aircraft navigation.
(B) λ2 is used to kill germs in water purifiers.
(C) λ3 is used to improve visibility in runways during fog and mist conditions.
Identify and name the part of the electromagnetic spectrum to which these radiations belong. Also arrange these wavelengths in ascending order of their magnitude. [2]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 2m(SA-I) | ✓ Identify and write the name of EM waves λ1, λ2, λ3 (0.5m + 0.5m + 0.5m + 0.5m) ✓ write the ascending order of waves (0.5m) |
Answer:
(c) λ1 Microwave
λ2 — ULtraviolet
λ3 — Infrared
Ascending order – λ1 < λ3 < λ1
Explanation:
From question we can conclude that,
λ1 → Microwave, λ2 → UV, λ3 → Infrared
(2) λ2 < λ3 < λ1
Related Theory
Uses Micro waves-RADAR.
UV – LASIK eye surgery
Infrared-Optical communication.
Question 20.
A uniform magnetic field gets modified as shown in figure when two specimens A and B are placed in it.
(A) Identify the specimen A and B.
(B) How is the magnetic susceptibility of specimen A different from that of specimen B? [2]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 2m(SA-I) | ✓ (A) Identify and write the name of substances A and B (0.5m + 0.5m) ✓ (B) write the magnetic susceptibility (0.5m + 0.5m) |
Answer:
(c) A- Diamagnetic
B – Paramagnetic
The magnetic susceptibility of A is small negative and that of B iš small positive.
Question 21.
What is the nuclear radius of 125Fe, if that of 27 Al is 3.6 fermi?
OR
The short wavelength limit for the Lyman series of the hydrogen spectrum is 913.4 A. Calculate the short wavelength limit for the Balmer series of the hydrogen spectrum. [2]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 2m(SA-I) | ✓ Use the relation of radius and mass number (1m) ✓ Claculate the nuclear radius of Fe. (1m) Always write Unit of required value in numerical according to the system. |
Answer:
From the relation R = RoA1/3, where R0 is a constant and A is the mass number of a nucleus
RFe/RAIl) = (AFe/AAl)1/3
= (125/27)1/3
RFe = 5/3 RAl
= 5/3 × 3.6
= 6 fermi
OR
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 2m(SA-I) | ✓ write formula and calculate the wavelength of Lyman series (0.5m + 0.5m) ✓ Use formula and calculate the short wavelength of Balmer series. (0.5m + 0.5m) |
Given short wavelength limit of Lyman series
For the short wavelength limit of Bolmer series n1 = 2, n2 = ∞
Question 22.
A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. will the Lens behave as a converging or a diverging Lens? Justify your answer. [2]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 2m(SA-I) | ✓ write lens maker formula (0.5m) ✓ Find the refractive index (1m) ✓ write your justification (0.5m) |
Answer:
The value of (µ – 1) is negative and ‘f’ will be negative. So it will behave like diverging lens.
Question 23.
The figure shows a piece of pure semiconductor S in series with a variable resistor R and a source of constant voltage V. Should the value of R be increased or decreased to keep the reading of the ammeter constant, when semiconductor S is heated? Justify your answer
The graph of potential barrier versus width of depletion region for an unbiased diode is shown in graph A. In comparison to A, graphs B and C are obtained after biasing the diode in different ways. Identify the type of biasing in B and C and justify your answer.
[2]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 2m(SA-I) | ✓ write the impact of increasing and decreasing the resistance on ammeter |
Answer:
To keep the reading of ammeter constant value of R should be increased as with the increase in temperature of a semiconductor. Its resistance decreases and current tends to increase.
Explanation: So,
σ = q(nµn + pµp)
where µ is mobility,
when T↑ m also↑
⇒ σincreases = Rsemi↓
∴ If Rsemi↓, then the enf. R should increase to keep the current constant.
Caution
Students frequently screw up conductors with semiconductors. Semiconductors have conductivity that fall between conductors and insulators
OR
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 2m(SA-I) | ✓ Write type of Biasing in graph B and C. (0.5m + 0.5m) ✓ Write the reason, how you can identify the biasing. (0.5m + 0.5m) |
B – reverse biased
In the case of reverse biased diode the potential barrier becomes higher as the battery further raises the potential of the n side.
C -forward biased
Due to forward bias connection the potential of P side is raised and hence the height of the potential barrier decreases.
Question 24.
A narrow slit is illuminated by a parallel beam of monochromatic light of wavelength λ equal to 6000 Å and the angular width of the central maximum in the resulting diffraction pattern is measured. When the slit is next illuminated by light of wavelength λ’, the angular width decreases by 30%. Calculate the value of the wavelength λ’. [2]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 2m(SA-I) | ✓ write the formula of angular width (0.5m) ✓ Take a new variable for required wavelength (0.5m) ✓ Find the new angular width (0.5m) ✓ Calculate the required wavelength (0.5m) |
Answer:
Angular width,
2φ = \(\frac{2 \lambda}{d}\)
Given,
λ = 6000 A
In Case of new λ (assumed λ’ here), angular width decreases by 30%
New angular width,
= 0.70 (2φ)
\(\frac{2 \lambda^{\prime}}{d}\) = 0.70 × \(\left(\frac{2 \lambda}{d}\right)\)
λ = 4200A
Question 25.
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.7 × 10-22 c/m2. What is electric fleld intensity E
(A) in the outer region of the first plate, and
(B) between the plates? [2]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 2m(SA-I) | ✓ (A)Find the electric field intensity in the outer region of first plate (0.5m) ✓ (B) Use the correct principle for finding the desired value. (1 m) ✓ Calculate and write the electric field intensity in between the places (0.5m) |
Answer:
Surface charge density of plate
A = +17.7 × 10-22 C/m2
Surface charge density of plate
B = -17.7 × 10-22 c/m2
(A) In the outer region of plate I, electric field intensity E is zero.
(B) Electric field intensity E in between the plates is given by relation,
E = \(\frac{\sigma}{\epsilon_0}\)
Where, ∈0 = Permittivity of free space
= 8.85 × 10-12 N-1C2m-2
∴ E = \(\frac{17.7 \times 10^{-22}}{9.85 \times 10^{-1}}\)
Therefore, electric field between the plates is 2.0 × 10-10 N/C
Section – C [15 marks]
Question 26.
Two long straight parallel conductors carrying currents l1 and l2 are separated by a distance d. If the currents are flowing in the same direction, show how the magnetic field produced by one exerts an attractive force on the other. Obtain the expression for this force and hence define 1 ampere. [3]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 3m(SA-II) | ✓ Draw the diagram (0.5m) ✓ Derive the relation of magnetic force as per given situation (1 m) ✓ write the expression in correct manner (0.5m) ✓ Define 1 ampere according to the derived relation |
Answer:
Diagram
Derivation
The ampere is the value of that steady current which, when maintained in each of the two very long, straight parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would exert on each of these conductors a force equal to 2 × 10-7 newtons per metre of length.
Explanation:
Let the conductors PQ and RS carry currents l1 and l2 in same direction and placed at separation r.
Consider a current-element ‘ab’ of length ΔL of wire RS. The magnetic field produce by current carrying conductor PQ at the location of other wire RS.
So, B1 = \(\frac{\mu_0 I_1}{2 \pi r}\) ….(ii)
According to maxwell’s right hand rule or right hand palm rule number 1, the direction of Bi will be perpendicular to the plane of paper and direct downward. Due to this magnetic field, each element of other wire experiences a force. The direction of current element is perpendicular to the magnetic field; therefore the magnetic force on element ab of length ΔL.
ΔF = B1l2 ΔL
sin 90° = \(\frac{\mu_0 I_1}{2 \pi r}\)l2ΔL
∴ The total force on conductor of length ΔL will be
∴ Force acting per unit length of conductor,
f = \(\frac{F}{L}\) = \(\frac{\mu_0 I_1 I_2}{2 \pi r}\) ….. (ii)
According to Fleming’s left hand rule, the direction of magnetic force will be towards PQ
i.e. the force wilt be attractive.
The force between two parallel current carrying conductors is:
f = \(\frac{\mathrm{F}}{\mathrm{L}}\) = \(\frac{\mu_0 I_1 I_2}{2 \pi r}\)N/m
If I1 = I2 = 1A,
r = 1 m,
then f = \(\frac{\mu_0}{2 \pi}\) = 2 × 10-2N/m
Question 27.
The magnetic field through a circular loop of wire, 12 cm in radius and 8.5Ω resistance, changes with time as shown in the figure. The magnetic field is perpendicular to the plane of the loop. Calculate the current induced in the loop and plot a graph showing induced current as a function of time.
[3]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 3m(SA-II) | ✓ Calculate the electric field interm of area (0.5m) ✓ Calculate the current for time 0 to 2 sec. (0.5m) ✓ Calculate the current for time 2 to 4 sec. (0.5m) ✓ Calculate the current for time 4 to 6 sec. (0.5m) ✓ Draw the graph (1m) |
Answer:
Area of the circular loop
= πr2
= 3.14 × (0.12)2 m2
= 4.5 × 10-2 m2
E = \(\frac{d \phi}{d t}\) = – \(\frac{d}{d t}\)(BA)
= -A\(\frac{d B}{d t}\) = -A.\(\frac{\mathrm{B}_2-\mathrm{B}_1}{t_2-t_1}\)
For 0 < t < 2s
E1 = -4.5 × 10-2 × \(\left\{\frac{1-0}{2-0}\right\}\)
= -2.25 × 10-2V
∴ l1 = \(\frac{E_1}{R}\) = \(\frac{-2.25 \times 10^{-2}}{8.5}\)
A = -2.6 × 10-3
A = -2.6 mA
For 2s < t < 4s,
E2 = -4.5 × 10-2 × \(\left\{\frac{1-1}{4-2}\right\}\) = 0
Question 28.
An a.c. source generating a voltage ε = εo sin t is connected to a capacitor of capacitance C. Find the expression for the current I flowing through it. Plot a graph of ε and I versus ωt to show that the current is ahead of the voltage by \(\frac{\pi}{2}\).
OR
An ac voltage V = V0 sin ωt is applied across a pure inductor of inductance L. Find an expression for the current i, flowing in the circuit and show mathematically that the current flowing through it lags behind the applied voltage by a phase angle of \(\frac{\pi}{2}\). Also draw graphs of V and i versus ωt for the circuit. [3]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 3m(SA-II) | ✓ Draw the figure of required derivation. (0.5m) ✓ Deduce the required relation. (1.5 m) ✓ Draw the graph. (1 m) |
Answer:
Deviation
Explanation: The following figure shows an AC source, generating a voltage ε = ε0 sin ωt, connected to a capacitor of capacitance C. The plates of the capacitor get charged due to the applied voltage. As the alternating voltage is reversed in each half cycle.
The capacitor is alternately charged and discharged. If q is the charge on the capacitor, the corresponding potential difference across the plates of the capacitor is, V = \(\frac{q}{C}\)
∴ q = CV, q and V are functions of time, with V,
ε = ε0 sin ωt.
The instantaneous current in the circuit is,
i = \(\frac{d q}{d t}\) = \(\frac{d}{d t}\)(CV)
= C\(\frac{d V}{d t}\)
= C\(\frac{d}{d t}\) (ε0 sin ωt)
= WCε0 cos ωt
Where i0 = \(\frac{\varepsilon_0}{\left(\frac{1}{\omega C}\right)}\) is the peak value of the current.
The above table shows gives the values of e and i for different values of wt and the following figure shows graphs of e and i versus ωt, i leads e by a phase angle of \(\frac{\pi}{2}\) rad.
Graph of e and i versus ωt for a purely capacity AC circuit.
OR
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 3m(SA-II) | ✓ Derive the desired relation. (1.5 m) ✓ Write the relation in terms of phase angle π/2. (0.5 m) ✓ Draw the graph (1 m) |
Derivation
Explanation: AC circuit containing pure inductance: Consider a coil of self-inductance L and negligible ohmic resistance. An alternating potential difference is applied across its ends. The magnitude and direction of AC changes periodically, due to which there is a continual change in magnetic flux linked with the coil. Therefore according to Faraday’s law, an induced emf is produced in the coil, which opposes the applied voltage. As a result the current in the circuit, is reduced. That is inductance acts like a resistance in ac circuit. The instantaneous value of alternating voltage
V = V0 sinωt. …..(i)
If i is the instantaneous current in the circuit and \(\frac{d i}{d t}\) the rate of change of current in the dt circuit at that instant, then instantaneous induced emf,
ε = -L\(\frac{d i}{d t}\)
According to Kirchhoff’s loop rule
Integrating with respect to time ‘t’:
This is required expression for current
or i = i0 sinε(ωt – \(\frac{\pi}{2}\)) …. (iii)
where, i0 = \(\frac{V_0}{\omega L}\) …. (iv)
is the peak value of alternating current.
Question 29.
Radiation of frequency 1015 Hz is incident on three photosensitive surfaces A, B and C. Following observations are recorded:
Surface A: no photoemission occurs
Surface B: photoemission occurs but the photoelectrons have zero kinetic energy.
Surface C photo emission occurs and photoelectrons have kinetic energy.
Using Einstein’s photo-electric equation, explain the three observations.
OR
The graph shows the variation of photocurrent for a photosensitive metal
(A) What does X and A on the horizontal axis represent?
(B) Draw this graph for three different values of frequencies of incident radiation υ1, υ2 and υ3 (υ3 > υ2 > υ1) for the same intensity.
(C) Draw this graph for three different values of intensities of incident radiation I1, I2 and I3 (I3 > I2 > I1) having the same frequency. [3]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 3m(SA-II) | ✓ write the conclusion for surface A. (1m) ✓ Write the conclusion for surface B. (1m) ✓ write the conclusion for surface C. (1m) |
Answer:
From the observations made (parts A and B) on the basis of Einstein’s photoelectric equation, we can draw following conclusions:
- For surface A, the threshold frequency is more than 1015 HZ, hence no photoemission is possible.
- For surface B the threshold frequency is equal to the frequency of given radiation. Thus, photo-emission takes place but kinetic energy of photoelectrons is zero.
- For surface C, the threshold frequency is less than 1015 Hz. So photoemission occurs and photoelectrons have some kinetic energy.
OR
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 3m(SA-II) | ✓ (A) Identify X and A from the graph (0.5m + 0.5m) ✓ (B) Draw the graph for the frequencies (1m) ✓ (C) Draw the graph for intensities (1 m) |
(A) A – cut off or stopping potential
X – anode potential
Question 30.
The ground state energy of hydrogen atom is – 13.6 eV. The photon emitted during the transition of electron from n = 3 to n = 1 state, is incident on a photosensitive material of unknown work function. The photoelectrons are emitted from the material with the maximum kinetic energy of 9eV. Calculate the threshold wavelength of the material used. [3]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 3m(SA-II) | ✓ Calculate the energy of photon as per given data (1m) ✓ write the photoelectric equation and calculate the work function (0.5m + 0.5m) ✓ write the formula and calculate threshold wavelength (0.5m + 0.5m) |
Answer:
For a transition from n = 3 to n = 1 state, the energy of the emitted photon,
hv = E2 – E1 = 13.6\(\left[\frac{1}{1^2}-\frac{1}{3^2}\right]\) eV
= 12.1 eV.
From Einstein’s photoelectric equation,
hv = Kmax + W0
∴ W0 = hv – Kmax = 12.1 – 9 = 3.1 eV
Threshold wavelength,
λth = \(\frac{h c}{W_0}\)
= \(\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{3.1 \times 1.6 \times 10^{-19}}\)
= 4 × 10-7m
Section – D
Question 31.
(A) Draw equipotential surfaces for
(i) an electric dipole and
(ii) two identical positive charges placed near each other.
(B) In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10-3m2 and the separation between the plates is 3 mm.
(i) Calculate the capacitance of the capacitor.
(ii) If the capacitor is connected to 100V supply, what would be the the charge on each plate?
(iii) How would charge on the plate be affected if a 3 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected ?
OR
(A) Three charges -q, Q and -q are placed at equal distances on a straight line. If the potential energy of the system of these charges is zero, then what is the ratio Q : q?
(B) (i) Obtain the expression for the electric field intensity due to a uniformly charged spherical shell of radius R at a point distant r from the centre of the shell outside it.
(ii) Draw a graph showing the variation of electric field intensity E with r, for r > R and r < R. [5]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 5m(LA) | ✓ (A) Draw the figure of equipotential surfaces. (1m + 1m) ✓ (B) (i) Calculate the capacitance (1m) (ii) Calculate the charge on each plate. (1m) ✓ Find the value of affected charge (1m) |
Answer:
Here, A = 6 × 10-3 m2, d = 3 mm = 3 × 10-3m
(B)
(i) Capacitance, C = ∈o\(\frac{\mathrm{A}}{d}\)
= (8.85 × 10-12 × 6 × 10-3/3 × 10-3)
= 17.7 × 10-12 F
(ii) Charge, Q = CV = 17.7 × 10-12 × 100
= 17.7 × 10-10C
(iii) New charge Q = KQ = 6 × 17.7 × 10-10
= 1.062 × 10-8C
OR
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 5m(LA) | ✓ Draw the diagram (0.5m) ✓ (A) write correct formula and find the ratio (0.5m + 1m) ✓ (B)(i) Draw conceptual figure of desired relation. (0.5m)Write the formula used to deduce the relation. (0.5m) Write the principle used and derive relation. (0.5m)Write the derived equation. (0.5m) (ii) Draw the graph (1m) |
(A) Diagram
(B) Electric field due to a uniformly charged thin spherical shell:
(i) When point P lies outside the spherical shell: Suppose that we have calculate field at the point P at a distance r (r > R) from its centre. Draw Gaussian surface through point P so as to enclose the charged spherical shell Gaussian surface is a spherical surface of radius r and centre O.
Let \(\vec{E}\) be the electric field at point P, then the electric flux through area element of area \(\overrightarrow{d s}\) is given by
dϕ = \(\overrightarrow{\mathrm{E}} . \overrightarrow{d s}\)
Since \(\overline{d s}\) is also along normal to the surface
dϕ = E dS
∴Total electric flux through the Gaussian surface is given by
ϕ = ϕ Eds = E ϕ ds
Now, ϕds = 4πr2 … (i)
= Ex4 πr2
Since the charge enclosed by the Gaussian surface is q, according to the Gauss’s theorem,
ϕ = \(\frac{q}{\epsilon_0}\) …. (ii)
From equation (i) and (ii) we obtain
E × 4 πr2 = \(\frac{q}{\epsilon_0}\)
E = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{r^2}\) (for r > R)
(ii) E(N/C)
Explanation:
Question 32.
(A) Explain the term drift velocity of electrons in a conductor. Hence obtain the expression for the current through a
conductor ¡ri terms of drift velocity.
(B) Two cells of emfš E1 and E2 and internal resistances and respectively are connected in parallel as shown in the figure. Deduce the expression for the
(i) equivalent emf of the combination
(ii) equivalent internal resistance of the combination
(iii) potential difference between the points A and B.
OR
(A) State the two Kirchhoff’s rules used in the analysis of electric circuits and explain them.
(B) Derive the equation of the balanced state in a Wheatstone bridge using Kirchhoff’s Laws. [5]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 5m(LA) | ✓ (A) Write the definition of drift velocity. (0.5m) ✓ Derive the expression of current in terms of drift velocity. (1.5m) (B) (i) Derive the expression for emf (1m) (ii) Derive the expression for internal resistance. (1m) (iii) Derive the expression of potential difference. (1m) |
Answer:
(A) Drift velocity: It is the average velocity acquired by the free electrons superimposed over the random motion in the direction opposite to electric field and along the length of the metallic conductor.
Derivation I = ne A Vd
(B) Here, I = I1 + I2 …(i)
Let V = Potential difference between A and B.
For cell ε1
Comparing the above equation with the equivalent circuit of emf ‘εeq‘ and internal resistance ‘req’ then.
V = eq — $req
t = \(\frac{\mathrm{L}}{\mathrm{V}_d}\)
Current, I = \(\frac{q}{t}\)
= \(\frac{n \mathrm{ALe}}{\mathrm{L}}\) × Vd
= nAeVd
Related Theory
Drift velocity is inversely proportional to the length of the conductor.
OR
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 5m(LA) | ✓ (A) write both kirchoff’s rules. ✓(B) Draw and Labeling figure of wheatstone bridge in balanced condition. (1m + 0.5m) ✓ Dedurce the relation of balanced state of whearszone bridge. (1.5m) |
(A) Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction
Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero.
(B) Derivation
Explanation: (B)
Wheatstone’s bridge is a device used to determine unknown resistance. It consists of 4 arms of resistances P, Q, R and S, a galvanometer of resistance ‘G’ and a battery with plug key. Let ‘I’ be the main current and splits into branch current I1 and I2 respectively. Applying KVL to the mesh ABDA
I1P + IgG – I2R = 0 …(1)
Parallelly to the mesh BCDB
(I1 – Ig)Q – (I2 + Ig)S – IgG = 0 …(2)
The wheatstone bridge is said to be balanced when no current flowing through the galvanometer,
This is the balanced condition of wheatstone bridge.
Caution
When there is rio current flowing in the galvonometer arm, students are frequently confused. The potential difference between points B and D is zero when the Wheatstone bridge is balanced.
Question 33.
(A) Draw the graph showing intensity distribution of fringes with phase angle due to diffraction through a single slit. What is the width of the central maximum in comparison to that of a secondary maximum?
(B) A ray PQ is incident normally on the face AB of a triangular prism of refracting angle 600 as shown in figure. The prism is made of a transparent material of refractive index \(\frac{2}{\sqrt{3}}\). Trace the path of the ray as it passes through the prism. Calculate the angle of emergence and the angle of deviation.
OR
(A) Write two points of difference between an interference pattern and a diffraction pattern.
(B) (i) A ray of light incident on face AB of an equilateral glass prism, shows minimum deviation of 30°. Calculate the speed of light through the prism.
(ii) Find the angle of incidence at face AB so that the emergent ray grazes along the face AC. [5]
Answer:
|
Total Marks |
Breakdown |
| 5m(LA) | ✓ (A) Draw the graph. (1m) Write the value of width. (1m) ✓ (B) Draw the complete figure. (1m) Calculate the angle of emergence. (1m) Calculate the angle of deviation. (1m) |
Answer:
width of central maximum is twice that of any secondary maximum
So the ray PM after refraction from the face AC grazes along A
∴ ∠e = 90°
From ∠i + ∠e = ∠A + ∠δ
or 0° + 90° = 60° + ∠δ
∴ δ = 90° – 60° = 30°
OR
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 5m(LA) | ✓ (A) write any two differences (1m + 1m) ✓ (B) Draw the complete ray diagram (1 m) ✓ (i) Find the speed of light (1 m) ✓ (ii) Find the angle of incidence (1 m) |
(A) (i) The interference pattern has a number of equally spaced bright and dark bands. The diffraction pattern
has a central, bright maximum which is twice as wide as the other maxima. The intensity falls as we go to successive maxima away from the centre, on either side.
(ii) We calculate the interference pattern by superposing two waves originating from the two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.
(ii) At face AC, let the angle of incidence be r2. For grazing ray,
e = 90°
Let angle of refraction face AB be r1
Now
r1 + r2 = A
∴ r1 = A – r2
= 60° – 45° = 15°
Let angle of incidence at this face be i
µ = \(\frac{\sin i}{\sin r_1}\)
⇒ \(\sqrt{2}\) = \(\frac{\sin i}{\sin 15^{\circ}}\)
∴ i = sin-1(\(\sqrt{2}\).sin 15°)
= 21.5°
Section – E
Question 34.
Case Study :
Read the following paragraph and answer the questions.
A number of optical devices and instruments have been designed and developed such as periscope, binoculars, microscopes and telescopes utilising the reflecting and refracting properties of mirrors, lenses and prisms. Most of them are in common use. Our knowledge about the formation of images by the mirrors and lenses is the basic requirement for understanding the working of these devices.
(A) Why the image formed at infinity is often considered most suitable for viewing. Explain [1]
(B) In modern microscopes multicomponent lenses are used for both the objective and the eyepiece. Why? [1]
(C) Write two points of difference between a compound microscope and an astronomical telescope
OR
Write two distinct advantages of a reflecting type telescope over a refracting type telescope. [2]
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 4m(CBQ) | ✓ (A) Write the reason of image formation at infinity. (1 m) ✓ (B) Write the reason of use of multicomponent lenses. (1 m) ✓ (C) Write any two differences. (1m + 1m) |
Answer:
(A) When the image is formed at infinity, we can see it with minimum strain in the ciliary muscles of the eye.
(B) The multi-component lenses are used for both objective and the eyepiece to improve image quality by minimising various optical aberrations in lenses.
(C)
(i) The compound microscope is used to observe minute nearby objects whereas the telescope is used to observe distant objects.
(ii) In compound microscope the focal length of the objective is lesser than that of the eyepiece whereas in telescope the focal length of the objective is larger than that of the eyepiece.
OR
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 2m(CBQ) | (C) Write any two advantages. (1m + 1m) |
(i) The image formed by reflecting type telescope is brighter than that formed by refracting telescope.
(ii) The image formed by the reflecting type telescope is more magnified than that formed by the refracting type telescope.
Question 35.
Case study: Light emitting diode.
Read the following paragraph and answer the questions:
LED is a heavily doped P-N junction which under forward bias emits spontaneous radiation. When it is forward biased, due to recombination of holes and electrons at the junction, energy is released in the form of photons. In the case of Si and Ge diode, the energy released in recombination lies in the infrared region. LEDs that can emit red, yellow, orange, green and blue light are commercially available. The semiconductor used for fabrication of visible LEDs must at least have a band gap of 1.8 eV. The compound semiconductor Gallium Arsenide-Phosphide is used for making LEDs of different colours.
LEDs of different kinds
(A) Why are LEDs made of compound semi-conductor and not of elemental semi-conductors? [1]
(B) What should be the order of bandgap of an LED, if it is required to emit light in the visible range? [1]
(C) A student connects the blue coloured LED as shown in the figure. The LED did not glow when switch S is closed. Explain why ?
OR
Draw V-I characteristic of a p-n junction diode in:
(i) forward bias and
(ii) reverse bias [2]
| Total Marks |
Breakdown |
| 4m(CBQ) | ✓ (A) Write the reason to use compound semiconductor instead of element semiconductor in LED. (1m) ✓ (B) Write the range of bandgap of LED to emit light in visible range. (1m) ✓ (C) Write the reason for LED not glowing. (2m) |
Answer:
(A) LEDs are made up of compound semiconductors and not by the elemental conductor because the band gap in the elemental conductor has a value that can detect the light of a wavelength which lies in the infrared (IR) region.
(B) 1.8 eV to 3 eV
(C) LED is reversed biased that is why it is not glowing.
Explanation:
(B) The range of energy values that electrons in a solid material (such as an insulator or semiconductor) are forbidden to have, as measured by the energy difference between the valence band and the conduction band A band gap of 1.67eV is connected with a wavelength of 743nm. The light is a bright red. Direct semiconductors, which make up the bulk of LEDs, are employed in their fabrication because no change in momentum is required for an electron in the conduction band to recombine with a hole in the valence band.
OR
| Total Marks | Breakdown (As per CBSE Marking Scheme) |
| 2m(CBQ) | (C) Draw the V-I graph: (i) forward bias, (1 m) (ii) reverse bias. (1 m) |
V-I Characteristic curves of pn junction diode in forward biasing and reverse biasing.
(C) As we know, reverse bias reverse current via a junction diode is caused by minority carriers, which are insufficient at normal temperatures. So a finite change in voltage results in a very tiny (negligible) change in current, making it look time invariant.
