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Solved Paper 2023
Physics
Class-XII
Time : 3 Hours Max. Marks : 70
General Instructions : Read the Following Instructions very carefully and follow them : (i) This question paper contains 35 questions. All questions are compulsory. (ii) Question paper is divided into FIVE Sections— Section A, B, C, D and E. (iii) In section — A: question number 1 to 18 are Multiple Choice (MCQ) type questions carrying 1 mark each. (iv) In section — B: question number 19 to 25 are Short Answer-1 (SA-1) type questions carrying 2 marks each. (v) In section — C: question number 26 to 30 are Short Answer-2 (SA-2) type questions carrying 3 marks each. (vi) In section — D: question number 31 to 33 are Long Answer (LA) type questions carrying 5 marks each. (vii) In section — E: question number 34 and 35 are case-based questions carrying 4 marks each. (viii) There is no overall choice. However, an internal choice has been provided in 2 questions in Section— B , 2 questions in Sec- tion — C , 3 questions in Section — D and 2 questions in Section— E. (ix) Use of calculators is NOT allowed. c = 3 × 10^8 m/s h = 6.63 × 10–34^ Js e = 1.6 × 10–19^ C mo = 4p × 10–7^ T m A– eo = 8.854 × 10–12^ C^2 N–1^ m– 1 4 πε 0
= 9 × 10^9 N m^2 C–
Mass of electron = 9.1 × 10–31^ kg Mass of Neutron = 1.675 × 10–27^ kg Mass of Proton = 1.673 × 10–27^ kg Avogadro’s number = 6.023 × 10^23 per gram mole Boltzmann constant = 1.38 × 10–23^ JK–
Delhi Set-1 55/5/
SECTION - A
1. An electric dipole of length 2 cm is placed at an angle of 30° with an electric field 2 × 10^5 N/C. If the dipole experiences a torque of 8 × 10–3^ Nm, the magnitude of either charge of the dipole, is
(A) 4 μC (B) 7μC (C) 8 mC (D) 2 mC Ans. Option (A) is correct. Explanation: θ = 30° E = 2 × 10^5 N/C
216 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th
τ = 8 × 10–3^ Nm l = 2 cm = 2 × 10–2^ m τ = pE sinθ [ p = ql ] ∴ τ = ql Esinθ Or, q =
τ lE sinθ
Or, q = 8 10 2 10 2 10 30
3 2 5
×
× × × × °
− − (^) sin ∴ q = 4mC
2. Two long parallel wires kept 2 m apart carry 3A current each, in the same direction. The force per unit length on one wire due to the other is (A) 4.5 × 10–7^ Nm–1, attractive (B) 4.5 × 10–7^ N/m, repulsive (C) 9 × 10–7^ N/m, repulsive (D) 9× 10–7^ N/m, attractive 1 Ans. Option (D) is correct Explanation: i 1 = i 2 = 3A r = 2m F/l = 20 2 2
i i r
μ (^) × π 0 2 10 –7^ – 2 A
μ = × N π
Or, F/l = 2×10–7^ × 4 10 7 3 3 4 2
π × − × × π ∴ F/l = 9 × 10–7^ N Force per unit length = 9.0 × 10–7^ N/m Since, the currents are in same direction so there would be an attractive force between them.
3. Which of the following has its permeability less than that of free space? (A) Copper (B) Aluminium (C) Copper chloride (D) Nickel 1 Ans. Option (A) is correct Explanation: m r < 1 for diamagnetic substance. Copper is diamagnetic substance. 4. A square shaped coil of side 10 cm, having 100 turns is placed perpendicular to a magnetic field which is increasing at 1 T/s. The induced emf in the coil is (A) 0.1 V (B) 0.5 V (C) 0.75 V (D) 1.0 V 1 Ans. Option (D) is correct. Explanation: N = 100 A = 1 10
× m^2
d dt
B (^) = 1.0 Ts–
emf = NA d dt
B
∴ emf = 100 1 10
× × × 1
= 1 V
5. Which one of the following electromagnetic radiation has the least wavelength? (A) Gamma rays (B) Microwaves (C) Visible light (D) X-rays 1 Ans. Option (A) is correct. Explanation: Wavelength of Gamma rays < 10–3^ nm Wavelength of Microwave: 1 mm – 0.1m Wavelength of Visible light: 400nm – 700 nm Wavelength of X-Rays: 10–3^ nm – 1 nm 6. In a Young’s double-slit experiment, the screen is moved away from the plane of the slits. What will be its effect on the following? (i) Angular separation of the fringes. (ii) Fringe-width. (A) Both (i) and (ii) remain constant. (B) (i) remains constant, but (ii) decreases. (C) (i) remains constant, but (ii) increases. (D) Both (i) and (ii) increase. 1 Ans. Option (C) is correct. Explanation: Angular separation is independent of distance of screen from slits. Fringe width is proportional to the distance of screen from slits. 7. The energy of a photon of wavelength λ is (A) hc λ (B) hc / λ (C) λ/ hc (D) λ h / c 1 Ans. Option (B) is correct. Explanation: Energy of photon, E = h n But n = c λ
So, E = hc λ
8. The ratio of the nuclear densities of two nuclei having mass numbers 64 and 125 is (A)^64 125
(B)^4
(C)^5
(D) 1 1
Ans. Option (D) is correct. Explanation: Nuclear density is independent of mass number.
9. During the formation of a p - n junction: (A) diffusion current keeps increasing. (B) drift current remains constant. (C) both the diffusion current and drift current remain constant. (D) diffusion current remains almost constant but drift current increases till both currents become equal. **1 Ans. Option (D) is correct.
- The diagram shows four energy level of an electron in Bohr model of hydrogen atom. Identify the transition in which the emitted photon will have the highest energy.**
218 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th
17. Assertion (A): The equivalent resistance between points A and B in the given network is 2R. Reason (R): All the resistors are connected in parallel 1 2 R
A 2 R B 2 R
2 R
2 R Ans. Option (C) is correct. Explanation: The given resistance network represents in the bridge balance condition, so the equivalent resistance between A and B will be 2R, Hence assertion (A) is correct but Reason (R) is false because all resistances are not in parallel.
18. Assertion (A): The deflecting torque acting on a current carrying loop is zero when its plane is perpendicular to the direction of magnetic field. Reason (R): The deflecting torque acting on a loop of magnetic moment m
→ in a magnetic field B
→ is
given by the dot product of m
→ and B
→
. 1 Ans. Option (C) is correct. Explanation: τ = NBIA sinθ θ is the angle between the area vector and the direction of magnetic field. If the plane of loop is perpendicular to the electric field, then the angle between the area vector and the direction electric field is 0^0. Hence, τ = 0. So, the assertion is true. The torque is the cross product of the magnetic moment of the loop and the magnetic field. Hence the reason is false.
SECTION - B
19. Draw a graph showing the variation of potential energy of a pair of nucleons as a function of their separation. Indicate the region in which the nuclear force is (a) attractive and (b) repulsive. 2 Ans. Graph showing variation of potential energy of a pair of nucleons as a function of their seperation: 20. (a) How will the De Broglie wavelength associated with an electron be affected when the (i) velocity of the electron decreases? and (ii) accelerating potential is increased? Justify your answer. 2 OR (b) How would the stopping potential for a given photosensitive surface change if (i) the frequency of the incident radiation were increased? and (ii) the intensity of incident radiation were decreased? Justify your answer. 2
Ans. (a) (i) From De Broglie equation λ = h mv As velocity ( v ) decreases, the wavelength (λ) increases. (ii) λ = h 2 meV As accelerating potential (V) increases, wavelength (λ) decreases. OR (b) (i) From Einstein’s photoelectric equation, hv = φ 0 + KE KE = e V s , where VS is the stopping potential. (i) Stopping potential depends on the frequency of the incident radiation. If frequency increases KE increases (Since, φ 0 remains constant). Hence, stopping potential increases. (ii) There is no intensity term in Einstein’s equation. Hence, stopping potential is independent of intensity of incident radiation.
21. Identify the electromagnetic wave whose wavelengths range is from about (a) 10 –12^ m to about 10–8^ m. (b) 10 –3^ m to about 10–1^ m. Write one use of each. 2 Ans. (a) X-Ray. Used as diagnostic tool in medical science. (b) Microwave. Use in radar system. 22. Depict the orientation of an electric dipole in (a) stable and (b) unstable equilibrium in an external uniform electric field. Write the potential energy of the dipole in each case. 2
Ans. (a) Stable equilibrium: When angle between p
→ and
E
→ is 0^0.
E 2a q
®
® p –q
Potential energy of dipole = – p Ecosθ In this case θ = 0^0 , so Potential energy = – p E
(b) Unstable equilibrium: When angle between p
→ and
E
→ is 180^0.
Potential energy of dipole = – p Ecosθ In this case θ = 180^0 , so Potential energy = p E
23. (a) Write the expression for the Lorentz force on a particle of charge q moving with a velocity v
→ in a
magnetic field B
→
. When is the magnitude of this force maximum? Show that no work is done by this
SOLVED PAPER - 2023 (PHYSICS) [ 219
force on the particle during its motion from a point r 1
→ to point r 2
→
. 2 OR (b) A long straight wire AB carries a current I. A particle (mass m and charge q ) moves with a velocity v
→ , parallel (^) to the wire, at a distance d from it as shown in the figure. Obtain the expression for the force experienced by the particle and mention its directions. 2 B
I
A
d
Particle
v^ ®
Ans. (a) Expression for Lorentz force:
F
→ = q v (^^ B )
→ → × The force is maximum when the angle between v B
→ → and is 90^0.
Here, F
→ is perpendicular to v
→
. So, no work is done by this force on the particle during its motion OR (b) Magnetic field produced by the current carrying from a point r 1
to point r 2
wire, B = 0 2
I
d
μ π The direction of field is ⊗
Force acting on the particle = q v (^^ B )
→ → × Here, θ = 90° So, Force = qvB = μ π
0 I
qv d Its direction is towards right. Repulsive.
24. The potential difference applied across a given conductor is doubled. How will this affect (i) the mobility of electrons and (ii) the current density in the conductor? Justify your answers. 2 Ans. (i) Mobility ∝ 1/Potential difference So, if potential difference is doubled, mobility will be halved. (ii) Current density ∝ Potential difference So, if potential difference is doubled, current density will also be double. 25. Two coils C 1 and C 2 are placed close to each other. The magnetic flux φ 2 linked with the coil C 2 varies with the current I 1 flowing in coil C 1 , as shown in the figure. Find
(i) the mutual inductance of the arrangement, and
(ii) the rate of change of current ddt^ I^1 ^
^
that will induce
an emf of 100 V in coil C 2. 2 Ans. (i) Since, N 2 φ 2 = MI 1 From graph, φ 2 = 10Wb corresponding to I 1 = 4A and φ 2 = 10Wb ∴ N 2 × 10 = M × 4 Considering N 2 = 1 M = 10 4
= 2.5H
(ii) Again, N 2 φ 2 = MI 1 Or, d dt
( N 2 φ 2 ) = d dt
( MI 1 )
Or, N d (^2) dt
φ (^2) = M dI dt
1
Considering N 2 = 1 e = M dI dt
1
or, 100 = (^) 2 5. × dI^1 dt
∴ dI dt
(^1) = 40 A/s
SECTION - C
26. (a) A plane wave-front propagating in a medium of refractive index ‘μ 1 ’ is incident on a plane surface making an angle of incidence (i). It enters into a medium of refractive index μ 2 (μ 2 > μ 1 ). Use Huygen’s construction of secondary wavelets to trace the retracted wave-front. Hence, verify Snell’s law of refraction. 3 OR (b) Using Huygen’s construction, show how a plane wave is reflected from a surface. Hence, verify the law of reflection. 3 Ans. (a) A plane wavefront AC is incident on the plane of separation XY of two media of refractive indices m 1 and m 2 (m 2 > m 1 ) making an angle i. This is known as angle of incidence. When the wavefront touches the point A, the point becomes a source of secondary wavelets. Thus, when the whole waveform passes through the XY plane, each point of AF becomes the source of secondary wavelets.
SOLVED PAPER - 2023 (PHYSICS) [ 221
of induced emf is determined by Fleming’s right hand rule. Working of ac generator:
N
B C I A D^ S B 1 R^1
R (^) L
Armature Permanentmagnet
B 2 R (^2)
VI
V
Main components of ac generator are: (i) Armature, (ii) Field magnet, (iii) slip ring (iv) Brush Armature (ABCD) is a copper coil wound on a soft iron core. The armature is rotated by a turbine. The armature is placed in between poles of a strong permanent magnet (NS) known as field magnet. Two ends of armature coil are connected to the slip rings(R 1 and R 2 ). Carbon brushes (B 1 and B 2 ) kept just in firm contact with the rings. External circuit is connected with the brushes. When armature rotates in the magnetic field induced emf is generated which is supplied to the external circuit through the brushes. Expression of instantaneous emf induced: If the armature has N number of turns, then magnetic flux through the coil is
φ = N B A (. )
→ →
= NBA cosθ If ω is the angular velocity, then emf induced
e = - d dt
φ
= NBAωsinωt ∴ e = e 0 sinωt ( where NBAω = e 0 )
29. (a) Briefly describe how the current sensitivity of a moving coil galvanometer can be increased. (b) A galvanometer shows full scale deflection for current Ig. A resistance R 1 is required to convert it into a voltmeter of range (0 – V) and a resistance R 2 to convert it into a voltmeter of range (0 – 2V). Find the resistance of the galvanometer. 3 Ans. (a) Current sensitivity of galvanometer: Current sensitivity of galvanometer = θ/ I = NBA / c So, to increase the current sensitivity: (i) Permanent magnet should be strong so that B is high. (ii) N and A should be high. (iii) Value of c must be low. (b) To convert into a voltmeter of range (0 – V), V = Ig ( Rg + R 1 ) ...(i)
To convert into a voltmeter of range (0 – 2V), 2 V = Ig ( Rg + R 2 ) ...(ii) Dividing equation (i) by (ii) V 2 V
= ( Rg + R 1 )/ (Rg + R 2 )
Or, 1 2
= ( Rg + R 1 )/ ( Rg + R 2 )
∴ Rg = R 2 – 2 R 1
30. (a) (i) Differentiate between ‘distance of closest approach’ and ‘impact parameter’. (ii) Determine the distance of closest approach when an alpha particle of kinetic energy 3. MeV approaches a nucleus of Z = 79, stops and reverses its directions. 3 OR (b) (i) State three postulates of Bohr’s theory of hydrogen atom. (ii) Find the angular momentum of an electron revolving in the second orbit in Bohr’s hydrogen atom. 3 Ans. (a) (i) Difference between “distance of closest approach” and “impact parameter”: Distance of closest approach Impact Parameter Distance of closest approach is the distance of a charged particle from the centre of the nucleus where the total kinetic energy of the charged particle gets converted into potential energy.
Impact parameter is the perpendicular distance between the path of projected charged particles and centre of the nucleus.
r =
2 2 0
Ze πε^ ^ mv
b =
Ze mv
2
0
2
cot( θ/ ) πε ^
(ii) Distance of closest approach = r =
Qq 4 πε 0 E Q = 79e = 79 × 1.6 × 10–19C q = 2e = 2 × 1.6 × 10–19C E = 3.95 MeV = 3.95 × 10^6 × 1.6 × 10-19^ J
∴ r = (9 × 10^9 )
19 19 6 19
− − −
× × × × ×
× × ×
= 576 × 10–16^ m = 5.76 × 10–14^ m OR (b) (i) Postulates of Bohr’s theory: Postulate 1: In an atom electrons are revolving around the nucleus in definite circular orbits. These orbits are called ‘stationary orbits’ and each orbit or shell possesses fixed energy. While revolving in these orbits electrons do not emit any radiation. Postulate 2: Electrons can move only those permissible orbits where the angular momenta of
222 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th
electrons are integral multiples of 2
h π
where h is the
Planck’s constant. Postulate 3: Transition of electrons may occur from one stationary orbit to another. During such transition energy may be emitted or absorbed following the relation Ei – Ef = h n (where Ei ∼ Ef is the difference of energies of the two stable orbits). (ii) Angular momentum of electron revolving in 2nd orbit in Bohr’s Hydrogen atom: Angular momentum = 2
nh π Here, n = 2, ∴ Angular momentum = 2 2
×^ h^ = h π π
SECTION - D
31. (a) (i) Explain how free electrons in a metal at constant temperature attain an average velocity under the action of an electric field. Hence obtain an expression for it. 3 (ii) Consider two conducting wires A and B of the same diameter but made of different materials joined in series across a battery. The number density of electrons in A is 1.5 times that in B. Find the ratio of drift velocity of electrons in wire A to that in wire. B. 2 OR (b) (i) A cell emf of (E) and internal resistance (r) is connected across a variable load resistance (R). Draw plots showing the variation of terminal voltage V with (i) R and (ii) the current (I) in the load. 2 (ii) Three cells, each of emf E but internal resistances 2r, 3r and 6r are connected in parallel across a resistor R. Obtain expressions for (i) current flowing in the circuit, and (ii) the terminal potential difference across the equivalent cell. 3 Ans. (a) (i) In absence of any electric field, the free electrons in metals move haphazardly in all possible directions and hence, develop no net flow of current. When an electric field is applied, a force acts on the electrons and the electrons now tend to move in the direction of the force. When electron collides with lattice, its velocity becomes instantaneously zero and then again it starts moving in a specific direction due to the applied electric field. If the average time between two collisions (relaxation time) is τ, then l = 1 2
a τ^2
Where l = average drift distance a = acceleration = Ee m E = electric field intensity
e = charge of electron m = mass of electron ∴ l = 1 2
Ee m
τ^2
Drift velocity= l τ
= vd = 1 2
Ee m
τ
e m
τ 2
= K, a constant , which depends on the nature
of the material and the temperature. ∴ vd = K × E Thus, free electrons in a metal at constant temperature under the action of an electric field attain a constant average velocity. (b) Since, the wires are joined in series current flowing through then will be same. Let the current in both A and B be I. Diameter being same, there areas of cross section are also same. Let it be A. So, in wire A I = nAeAvdA In wire B, I = nBeAvdB Taking the ratio, 1 = A^ A B B
d d
n v n v
×
Or, 1 = A B
1.5 d d
v v
×
∴ A
B
d d
v
v =^
OR
(b) (i) Terminal voltage vs. load resistance graph: V = E – ir And i = E R + r
∴ V = E − E
r R r
Or, V = E r r R
∴ V =
ER
( r + R )
Terminal voltage vs. current graph: V = E – ir
224 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th
A centre-tapped transformer and two p-n junction diodes are used for a full weave rectifier. Input of the transformer is connected to the ac supply. In secondary there are three terminals – A, B and CT. For positive half cycle, A is positive, B is negative. For the negative half cycle, A is negative, B is positive. CT is always at zero potential. It is always grounded. Anode of one diode (D 1 ) is connected to A and anode of other diode (D 2 ) is connected to B. Cathodes of both the diodes are joined together and ultimately connected to CT through a load resistance (RL). When positive half cycle appears, A is at positive and B is at negative potential. So, diode D 1 is forward biased and hence, it conducts. When negative half cycle appears, B is at positive and A is at negative potential. So, diode D 2 is forward biased and hence, it conducts. The process repeats. For both the half cycles current flowing through the load resistance is unidirectional. Hence, a DC voltage appears across it. Thus, a full wave rectifier works. Input and output wave forms:
AC Input
D 1
output D (^2) output Total output
t
t
t
t Characteristic property of junction diode that makes it suitable for rectification: An ideal p-n junction diode exhibits zero resistance when forward biased and infinite resistance when reversed biased.
33. (a) (i) Draw a ray diagram to show the working of a compound microscope. Obtain the expression for the total magnification for the final image to be formed at the near point. 3 (ii) In a compound microscope an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eye-piece has a focal length of 5 cm and the final image is formed at the near point, find the magnifying power of the microscope. 2 OR (b) (i) Draw a ray diagram for the formation of image of an object by an astronomical telescope, in normal adjustment. Obtain the expression for its magnifying power. (ii) The magnifying power of an astronomical telescope in normal adjustment is 2.9 and the
objective and the eyepiece are separated by a distance of 150 cm. Find the focal lengths of the two lenses. 5 Ans. (a) (i) Ray diagram of compound microscope:
In a compound microscope there are two lenses – objective (O) and Eyepiece (E). Object PQ is placed in front of the objective at a distance more than the focal length of the objective. An inverted, magnified, real image P 1 Q 1 is formed in front of eyepiece within the optical centre and the focus of the eyepiece. This acts as the object of the eyepiece. An (erect with respect to P 1 Q 1 , inverted with respect to PQ), magnified, virtual image P 2 Q 2 is formed at a distance D (minimum distance of distinct vision) from the eyepiece. Magnification: For objective: Object distance = u Image distance = v
Magnification = m o =
v u
P Q
PQ
Applying the lens formula, 1 1 v u
f 0
Or, 1 +^
v u =^
v f 0
Or, v u
v f 0
For eyepiece:
Magnification = me = 1 +^
D
fe
Magnification of the combination of objective and eyepiece = m = mO × me
Or, m =
v u
D
fe
× +
Or, m =
v f
D
0 fe
SOLVED PAPER - 2023 (PHYSICS) [ 225
P 1 Q 1 image is formed very close to the eyepiece, hence v can be approximated as the distance between the two lenses i.e., the length of the tube(L).
∴ m =
L
f
D
0 fe
Since, fe << D and f 0 << L , hence the above expression may be approximated as,
m =
L
f
D
0 fe
×
(ii) Given, u = 1.5 cm, f 0 = 1.25 cm, fe = 5 cm, D = 25 cm Here, all alphabets are in their usual meanings Applying lens formula for objective lens, 1 1 v u
f 0
Or, 1 1 v 1 5
∴ v = 7.5 cm
Magnification = m =
v u
D
fe
× +
Or, m = 7 5 1 5
× +
^
∴ m = – 30 OR (b) (i) Astronomical telescope in normal adjustment:
In an astronomical telescope there are two lenses – objective (O) and Eyepiece (E). The two lenses are so placed during focussing that the foci of the lenses meet at a point. Objective is directed towards the object at infinity. Parallel rays coming from the object meet at the focus of the objective and forms an inverted, real image P 1 Q 1 in front of eyepiece. This point is the focus of eyepiece too. This acts as the object of the eyepiece. An (inverted with respect to P 1 Q 1 , erect with respect to original object), highly magnified, real image is formed at infinity. Magnification = m
=
Angle subtended at eye by the final image Angle subtended att eye by the object
Or, m = Angle subtended at eyepiece by the final image Angle subtendded at objective by the object
Or, m = β α
Or, m =
Q EP
Q OP
1 1 1 1
Or, m =
tan tan
Q EP
Q OP
1 1 1 1 [α and β being very small, tanα = α and tanβ = β]
Or, m =
Q P
Q E
Q P
Q O
1 1 1 1 1 1
Or, m =
Q O
Q E
1 1
∴ m = o e
f f
(ii) Since, m = o e
f f
Or, 2.9 = o e
f f ∴ fo = 2.9 fe Also, fo + fe = 150 Or, 2.9 fe + fe = 150
∴ fe = 150 3 9. = 38.46 cm fO = 2.9 fe = 2.9 × 38. = 111.54 cm
SECTION - E
34. A lens is a transparent optical medium bounded by two surfaces; at least one of which should be spherical. Considering image formation by a single spherical surface successively at the two surfaces of a lens, lens maker’s formula is obtained. It is useful to design lenses of desired focal length using surfaces of suitable radii of curvature. This formula helps us obtain a relation between u, v and f for a lens. Lenses form images of objects and they are used in a number of optical devices, for example microscopes and telescopes. 4 (i) An object AB is kept in front of a composite convex lens, as shown in figure. Will the lens produce one image? If not, explain.
SOLVED PAPER - 2023 (PHYSICS) [ 227
Initially the capacitor between A and B is not considered. So, the circuit may be redrawn as
This is like a balanced Wheatstone bridge. So, the capacitor between E and F may not be considered. So, the circuit is further modified as
A B
E
F
C
C
C
C
C
C/
C/
B
A A^ B
C
The capacitance between A and B is C. Now capacitor between A and B (left out initially) is considered.
So, the equivalent capacitance between A and B is 2C.
(ii) When a dielectric is placed between charged plates of a capacitor, the polarization of dielectric produces an electric field opposing the field produced by the charges on the plate. Thus, the resultant electric field between the plates decreases. (iii) Both capacitors will attain a common potential, VC.
(a) VC = Total charge Total capacitance
= Q C C
Q
+ C
(b) Final charge on capacitor A = QA = CVC
= CQ
3 C
= Q
So, charge lost by capacitor A = Q – Q^ Q 3
OR
By placing the dielectric slab, it behaves like two capacitors in series C 1 = K^^1 0 A^ K^^0 A 3
ε 6 ε d / d
C 2 = K^^2 0 A^^ K^^0 A
ε ε d / d
C eq^ =^
C 1 C 2
Or,
C eq^ =^
d d 6
K ε 0 A 3 K ε 0 A
Or, 1
C eq^ =^
d K ε A
∴ Ceq = 6 5
K ε 0 A d
Delhi Set-2 55/5/
Note: Except these, all other questions are from Delhi Set-
SECTION - A
1. A charge Q is placed at the centre of a cube. The electric flux through one if its face is
(A)
Q
ε 0^ (B)^
Q
6 ε 0
(C)
Q
8 ε 0 (D)^
Q
3 ε 0^1
Ans. Option (B) is correct. Explanation: According to Gauss’Law total electrical flux =
Q
ε 0
So, electric flux through each face =
Q
6 ε 0
5. Choose the correct option related to wavelengths ( λ ) of different parts of electromagnetic spectrum. (A) λ x -rays < λmicro waves < λradio waves < λvisible (B) λvisible > λ x -rays > λradio waves > λmicro waves (C) λradio waves > λmicro waves > λvisible > λ x -rays (D) λ (^) visible < λmicro waves < λradio waves < λ x -rays 1 Ans. Option (C) is correct Explanation: Wavelength of radio wave > 0.1 m Wavelength of microwave: 1 mm – 0.1m Wavelength of visible light: 400nm – 700nm Wavelength of X-rays: 10–3^ nm – 1 nm
228 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th
11. Figure shows a plot of stopping potential (V 0 ) versus 1 λ
, is the wavelength of the radiation
causing photoelectric emission from a surface. The slope of the line is equal to
B
V (^0) A
1 l
(A) φ 0 (B) h e
(C) hc e
(D) h c e
2 2^1
Ans. Option (C) is correct. Explanation: From Einstein’s equation, e V 0 = hn – hn 0
or, V 0 = h e
ν (^) – h e
ν 0 h c
∵ (^) ν =
or, V 0 =
hc e
hc λ λ e
0 So, the slope of V 0 vs. 1/λ graph is hc / e.
13. An ideal inductor is connected across an AC source of voltage. The current in the circuit. (A) is ahead of the voltage in phase by p (B) lags voltage in phase by p (C) is ahead of voltage in phase by p/ (D) lags voltage in phase by p/2. 1 Ans. Option (D) is correct. Explanation: For a pure inductor AC circuit, if V = V 0 sinωt the current i = i 0 sin (ω t – 90^0 ) Note: In question number 16, 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 Assertion (A) and Reason (R) are true and (R) is the correct explanation of (A). (B) Both Assertion (A) and Reason (R) are true and (R) is NOT the correct explanation of (A). (C) Assertion (A) is true and Reason (R) is false. (D) Assertion (A) is false and Reason (R) is also false. 16. Assertion (A): In insulators, the forbidden gap is very large. Reason (R): The valence electrons in an atom of an insulator are very tightly bound to the nucleus. Ans. Option (A) is correct. Explanation: In insulators forbidden band gap is more than 3eV which is large enough. So, the assertion is true.
Since, the electrons are very tightly bound to nucleus they cannot easily jump to the conduction band. This explains why the forbidden band gap is so large. So, the reason is also true and it explains the assertion.
SECTION - B
21. Identify the electromagnetic radiation and write its wavelength range which is used to kill germs in water purifier. Name the two sources of these radiations. 2 Ans. The electromagnetic radiation which is used to kill germs in water purifier: Ultraviolet rays. Wavelength range of ultraviolet radiation: 1 nm – 400 nm Sources of ultraviolet radiation: Sunlight, electric arcs. 22. An electric dipole of dipole moment ( p
→ ) is kept in a uniform electric field E
→
. Show graphically the variation of (^) torque acting on the dipole ( τ ) with its orientation ( θ ) in the field. Find the orientation in which torque is (i) zero and (ii) maximum. 2 Ans. Graphical variation of torque (τ) with the orientation of electric dipole (θ) in the electric field:
(i) At θ = 0°, torque is zero. (ii) At θ = 90°, torque is maximum.
24. A potential difference (V) is applied across a conductor of length ‘L‘ and cross-sectional area ‘A’. How will the drift velocity of electrons and the current density be affected if another identical conductor of the same material were connected in series with the first conductor? Justify your answers. 2 Ans. Since, R = P L A
I =
V V VA
R L L
A
= (^) ρ = ρ
I = (^) neAVd VA L
ρ when another conductor is joint in series Effective Resistance = R R^2 L A
Effective current = (^2 )
V V VA
R L L
A
= (^) ρ = ρ
230 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th
l Atom exhibits line spectrum. Rutherford’s model failed to explain this. According to Rutherford, if electron continuously emits energy then there should be continuous band spectrum. But that is not the reality. (ii) For Balmer series 1 λ
= R^
^
n For 2nd^ line, n = 4
λ
= R^
^
Or,
R
∴ R = 16
3 × 4861
For 1st^ line, n = 3
λ '
= R^
^
Or, 1 λ '
R
Or, 1/λ’ = 5 16 36 3 4861
×
× ×
Or, λ’ = 36 3 4861 5 16
× ×
×
∴ λ’ = 6562.35Å OR (b) (i) From Einstein’s photoelectric equation. h n = φ 0 + KE max So, KE energy of photoelectron depends on the frequency of incident radiation and the work function of the substance. There is no intensity term in the equation. Hence, the KE is independent of the intensity of the radiation. (ii) From Einstein’s photoelectric equation, eV s = h n – φ 0 and h n = E 2 – E 1
= −^ −^ − ^
= 10.2 eV Putting the Einstein’s equation, 5eV = 10.2eV – φ 0 ∴ φ 0 = 5.2 eV
Delhi Set-3 55/5/
Note: Except these, all other questions are from Delhi Set-1 & 2
SECTION - A
2. Two horizontal thin long parallel wires, separated by a distance r carry current I each in the opposite directions. The net magnetic field at a point midway between them, will be (A) zero
(B)
μ π
0 2
I
r
^
^ vertically downward
(C)
2 μ 0 I r
^
^
vertically upward
(D)
μ π
0 I
r
^
^
vertically downward 1
Ans. Option (D) is correct. Explanation: Magnetic field due to both the wires is 0 4 2
I
r
μ π^
I
r
μ π
So, total field is 2 0 0 2
I I
r r
× μ^ =μ π π The direction will be vertically downward.
3. Which of the following cannot modify an external magnetic filed as shown in the figure?
(A) Nickel (B) Silicon (C) Sodium Chloride (D) Copper 1 Ans. Option (A) is correct. Explanation: Nickel is a ferromagnetic substance. Ferromagnetic substance does not modify external magnetic field as shown.
7. E, c and v represent the energy, velocity and frequency of a photon. Which of the following represents its wavelength? (A) h c
ν 2 (B)^ h n
(C) hc E
(D) h c
ν (^) 1
Ans. Option (C) is correct. Explanation: E = h n Or, E = hc λ
∴ λ = hc E
9. The energy required by an electron to jump the forbidden band in silicon at room temperature is about. (A) 0.01 eV (B) 0.05 eV (C) 0.7 eV (D) 1.1 eV 1 Ans. Option (D) is correct.
SOLVED PAPER - 2023 (PHYSICS) [ 231
13. What is the ratio of inductive and capacitive reactance in an ac circuit? (A) ω^2 LC (B) LC^2 (C) LC ω^2
(D) ω^2 L 1
Ans. Option (A) is correct. Explanation: Inductive reactant XL = ω L Capacitive reactant X C = 1 ω C
∴
X
X
L C^ =^ ω
2 LC
14. In an interference experiment, third bright fringe is obtained at a point on the screen with a light of 700 nm. What should be the wavelength of the light source in order to obtain the fifth bright fringe at the same point? (A) 420 nm (B) 750 nm (C) 630 nm (D) 500 nm 1 Ans. Option (A) is correct.
Explanation: Since, yn = n d
λ D
For 3rd^ bright fringe,
y 3 = 3 ×^700 D
d
= 2100 D
d For 5th^ bright fringe, y 5 = 5 λ ' D d Since y 5 = y 3 5 λ ' D d
= 2100 D
d
∴ λ’ = 2100 5
= 420 nm
Note: In question number 17, 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 Assertion (A) and Reason (R) are true and (R) is the correct explanation of (A). (B) Both Assertion (A) and Reason (R) are true and (R) is NOT the correct explanation of (A). (C) Assertion (A) is true and Reason (R) is false. (D) Assertion (A) is false and Reason (R) is also false.
17. Assertion (A): The given figure does not show a balanced Wheatstone bridge. 2 W
I
P
R S
Q
3 W
3 W 2 W
G I
Reason (R): For a balanced bridge small current should flow though the galvanometer. 1 Ans. Option (A) is correct. Explanation: Balance condition of Wheatstone bridge
is P
Q
= R
S
In the given circuit, P
Q
≠ R
S
So, the assertion is true. Since, the given bridge is not in balanced condition, so current will flow in the galvanometer.
SECTION - B
19. Plot a graph showing the variation of photo electric current, as a function of anode potential for two light beams having the same frequency but different intensities I 1 and I 2 (I 1 > I 2 ). Mention its important features. 2 Ans. Variation of photoelectric current as a function of anode potential:
I >I 1 2 I 1 I 2
O (^) Anode Potential
Stopping potential
Photo Electric
Current
Important features: (i) As intensity increases, saturation current increases. (ii) Stopping potential is independent of the intensity of light.
21. How are electromagnetic waves produced? Write their two characteristics. 2 Ans. Accelerated charged particles produce electromagnetic waves. Characteristics of electromagnetic waves: (i) Electromagnetic waves are transverse in nature. (ii) All electromagnetic waves travel through vacuum with a speed 3 x 10^8 m/s. (iii) In electromagnetic waves the oscillations of E and B
→ → are perpendicular to each other and also perpendicular to the direction of propagation of the wave.
22. Three points charges Q, q and – q are kept at the vertices of an equilateral triangle of side L as shown in figure. What is
(i) the electrostatic potential energy of the arrangement? and (ii) the potential at point D? 2
SOLVED PAPER - 2023 (PHYSICS) [ 233
Or, A 1 vd 1 ρ l A 1
= A d l (^2 2) A 2 ν ρ
(since R = ρ l A
∴ vd1 = vd 2
SECTION - C
27. An alternating current I = 14 sin (100 p t) A passes through a series combination of a resistor of 30 Ω and an inductor of
5 π
^
^
H. Taking 2 = 1 4. ,
calculate the (i) rms value of the voltage drops across the resistor and the inductor, and (ii) power factor of the circuit. 3 Ans. (i) Vrms drop across resistor, R = I rms × R =
2 × 30
= 294V
Vrms drop across inductor, L = Irms × ω L = (14/√2) x (100p) x (2/5p) = 392V (ii) Power factor, R/Z R Z =^
R
R^2 + X L^2
R
Z =
30
30 100 2 5
2
2
30. (a) Calculate the binding energy of an alpha particle in MeV. Given mass of a proton = 1.007825 u mass of a neutron = 1.008665 u mas of He nucleus = 4.002800 u 1u = 931 MeV/c^2 OR (b) A heavy nucleus P of mass number 240 and binding energy 7.6 MeV per nucleon splits into two nuclei Q and R of mass number 110 and 130 and binding energy per nucleon 8.5 MeV and 8.4 MeV respectively. Calculate the energy released in the fission. 3 Ans. (a) Mass defect = Mass of protons + mass of neutrons
- Mass of Helium nucleus Or, Mass defect = (2 × 1.007825 + 2 × 1.008665 – 4.002800)u ∴ Mass defect = 0.03018 u Binding energy = 0.03018 × 931 = 28.09785 MeV OR (b) According to the question (^240) P 110 Q + 130 R + Q Q = 110 × 8.5 + 130 × 8. - 240 × 7. Or, Q = 935 + 1092 – 1824 ∴ Q = 202 MeV
Outside Delhi Set-1 55/2/
SECTION - A
1. The magnitude of the electric field due to a point charge object at a distance of 4.0 m is 9 N C . From the
same charged object the electric field of magnitude, 16 N C
will be at a distance of
(A) 1 m (B) 2 m (C) 3 m (D) 6 m 1 Ans. Option (C) is correct. Explanation: In 1st^ case, E = kq r^2 Or, 9 = kq 42
[ r = 4.0 m]
∴ kq = 9 × 16 In 2nd^ case,
E ’ =
kq ( ') r 2
Or, 16 = 9 ×^126 ( ') r Or, ( r ’)^2 = 9 ∴ r ’ = 3 m
2. A point P lies at a distance x from the mid point of an electric dipole on its axis. The electric potential at point P is proportional to
(A)^12 x
(B)^13
x
(C)^14 x
(D) 1 2^1
x /^
Ans. Option (A) is correct. Explanation: Expression for electric potential due to
dipole in given as V =
4 πε 0 2
p x
So, V ∝ (^) x^12
234 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th
3. A current of 0.8 A flows in a conductor of 40 Ω for 1 minute. The heat produced in the conductor will be (A) 1445 J (B) 1536 J (C) 1569 J (D) 1640 J 1 Ans. Option (B) is correct.
Explanation: Given, R = 40Ω and I = 0.8A t = 1 minute = 60s Heat produced = H = I^2 Rt Or, H = (0.8)^2 × 40× 60 ∴ H = 1536 Joule
4. A cell of emf E is connected across an external resistance R. When current ‘ I’ is drawn from the cell, the potential difference across the electrodes of the cell drops to V. The internal resistance ‘ r ’ of the cell is
(A)
E V
E
^ − R
^
^
(B)
E V
R
^ −
^
(C) (E^ V)R
I
− (D) (E^ V)
V
^ − R
^
^
Ans. Option (D) is correct. Explanation:
Since, I =
E
R + r So, E = I r + V ∴ r = E^ V I
Or, r =
E V R
V
[ V = IR]
5. Beams of electrons and protons move parallel to each other in the same direction. They (A) attract each other. (B) repel each other. (C) neither attract nor repel. (D) force of attraction or repulsion depends upon speed of beams. 1 Ans. Option (A) is correct. Explanation: Since, electron and proton has opposite charge and moving in the same direction, So there must be an attractive force between them. 6. A long straight wire of radius ‘ a ’ carries a steady current I. The current is uniformly distributed across its area of cross-section. The ratio of magnitude of magnetic field B a^ B
→ → 1 at^2 and^2 at distance 2 a is
(A)^1
(B) 1
(C) 2 (D) 4 1
Ans. Option (B) is correct. Explanation: Magnetic field inside B 1 = μ π
μ π
μ π
0 2
0 2
0 2
ir a
i a a
i a
= (^ /^ )=
(Here, r = a /2)
Magnetic field outside = B 2 =
μ π
μ π
μ π
0 0 0 2 2 2 4
i r
i a
i a
(Here, r = 2 a )
So, The required ratio
B
B
1 2
7. E and B
→ → represent the electric and the magnetic field of an electro- magnetic wave respectively. The direction of propagation of the wave is along
(A) (^) B
→ (B) (^) E
→
(C) E B
→ → × (D) B E
→ → × (^) 1 Ans. Option (C) is correct.
8. A ray of monochromatic light propagating in air, is incident on the surface of water. Which of the following will be the same for the reflected and refracted rays? (A) Energy carried (B) Speed (C) Frequency (D) Wavelength 1 Ans. Option (C) is correct. Explanation: Frequency of light does not depend on medium through which light travels. 9. A beam of light travels from air into a medium. Its speed and wavelength in the medium are 1.5 × 10^8 ms–1^ and 230 nm respectively. The wavelength of light in air will be (A) 230 nm (B) 345 nm (C) 460 nm (D) 690 nm 1 Ans. Option (C) is correct. Explanation: In the medium c = n λ or, 1.5 × 10^8 = n × 230 × 10–
or, n = 1 5^10 230 10
8 9
. ×
× −
∴ n =
. (^) × 10 17 Hz
In air, Frequency remains unchanged. So,
n = 1 5 230
. (^) × 10 17 Hz
v ’ = n λ’
