Kerala Plus Two Physics Question Paper March 2023 with Answers

Reviewing Kerala Syllabus Plus Two Physics Previous Year Question Papers and Answers Pdf March 2023 helps in understanding answer patterns.

Kerala Plus Two Physics Previous Year Question Paper March 2023

Time : 2 1/2 Hours
Maximum : 80 scores

SECTION-A

Answer any 5 questions from 1 to 7. Each carries 1 score. (5 × 1 = 5)

Question 1.
State true of false:
Two field lines never intersect.
Answer:
True

Question 2.
The SI unit of resistance is …………
Answer:
ohm(Ω)

Question 3.
Current loop behaves as a ………… (Magnetic dipole/ Electric dipole)
Answer:
Magnetic dipole

Question 4.
An accelerating charge produces ………….. waves.
(a) electric
(b) magnetic
(c) electromagnetic
(d) none of these
Answer:
(c) electromagnetic

Question 5.
When the speed of light is independent of direction, the secondary waves are
(a) Spherical
(b) Cylindrical
(c) Plane
(d) Rectangular
Answer:
(a) Spherical

Question 6.
X-rays were discovered by in 1895.
(a) Roentgen
(b) J.J. Thompson
(c) William Crookes
(d) Rutherford
Answer:
(a) Roentgen

Question 7.
Atoms of same element differing in mass are called ………..
(a) Isotones
(b) Isobars
(c) Isotopes
(d) Isomers
Answer:
(c) Isotopes

SECTION-B

Answer any 5 questions from 8 to 14. Each carries 2 scores. (5 × 2 = 10)

Question 8.
Define magnetisation. Give its dimension.
Answer:
Manetisation is the net magnetic moment per unit volume.
Dimension of magnetisation is M = \(\frac{m}{V}\)

Question 9.
State laws of electromagnetic induction.
Answer:
Faraday’s law of electromagnetic induction states that the magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.
Mathematically, the induced emf is given by
∈ = \(\frac{d \phi}{d t}\)
Lenz’s law states that the direction emf (or current) is such tfiat it opposes the change in magnetic flux which produces it
Mathematically the Lenz’s law can be written as
∈ = \(\frac{-d \phi}{d t}\)

Question 10.
Obtain the expression for the current flowing through a resistor when an a.c. voltage is applied to it.
Answer:

Consider a circuit containing a resistance ‘R’ connected to an alternating voltage.
Let the applied voltage be
V = V₀ sin ωt ………….(1)
According to Ohm’s law, the current at any instant can be written as
I = \(\frac{V_0 \sin \omega t}{R}\)

Where I₀ = V₀/R is the peak value of current
I = I₀ sin ωt

Question 11.
How Maxwell modified Ampere’s law?
Answer:
Ampere circuital theorem is
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{~d} l}\) = μ₀ I_c
I_C is the conduction current. This equation is not applicable for ac circuits. Maxwell introduced a cor-rection term related to displacment current
I_d =

The modified Maxwell equation is
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{~d} l}\) = μ₀ (i_c + i_d)

\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{~d} l}\) = μ₀ I_c + μ₀ ε_c(\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{~d} l}\))

Question 12.
What is total internal reflection?
Answer:
When light passes from denser medium to rarer medium and angle of incidence is greater than critical angle, light gets reflected back to denser me¬dium itself. The phenomenon is called total internal reflection.

Question 13.
Explain work function.
Answer:
The minimum energy required by an electron to escape from metal surface is called work function of metal.

Question 14.
Differentiate between nuclear fission and nuclear fusion.
Answer:
In nuclear fission, a heavier nuclei splits into lighter ones releasing hugeenprgy. In nuclear fusion, lighter nuclei combine to form heavier nuclei releasing energy.
Fission can be controlled. But fusion cannot be controlled.

SECTION-C

Answer any 6 questions from 15 to 21. Each carries 3 scores. (6 × 3 = 18)

Question 15.
Explain the basic properties of electric charge.
Answer:
Basic properties of electric charges are:
(a) Unlike charges attract and like charges repel.
(b) Charge is conserved
(c) Electric Charge is Quantized
(d) Additivity of Charges

Question 16.
(a) Derive the expression for the capacitance of a parallel plate capacitor.
Answer:

Expression for capacitance of a capacitor:
Potential difference between two plates
V = Ed
= \(\frac{\sigma}{\varepsilon_0}\) d (E = \(\frac{\sigma}{\varepsilon_0}\))
V = \(\frac{Q}{A \varepsilon_0}\) d ………..(1) (σ = \(\frac{Q}{A}\))

Capacitance C of the parallel plate capacitor,
C = \(\frac{Q}{V}\) …………..(2)
Sub. eq. (1) ineq. (2)
C = \(\frac{Q}{\frac{Q}{A \varepsilon_0} d}\)
C = \(\frac{A \varepsilon_0}{d}\)

(b) What happens to the capacitance if a medium of dielectric constant K is introduced between the plates?
Answer:
Capacitance increases.

Question 17.
(a) State Biot-Savart law.
Answer:
The magnetic field at any point due to an element of current carrying conductor is

1) Directly proportional to the strength of the current (I)
2) Directly proportional to the length of the element(dl)
3) Directly proportional to the sine of the angle (θ) between the element and the line joining the midpoint of the element to the point.
4) Inversely proportional to the square of the distance of the point from the element
I dl sinθ
dB ∝ \(\frac{I dl sinθ}{r^2}\)

dB =\(\frac{\mu_0}{4 \pi}\) \(\frac{I dl sinθ}{r^2}\)

(b) Obtain the expression for the magnetic field on the axis of a circular current loop.
വൃത്താകൃതിയിലുള്ള കറണ്ട് ലൂപ്പിന്റെ അക്ഷത്തിലൂടെ യുള്ള കാന്തിക മണ്ഡല തീവ്രതയുടെ സമവാക്യം രൂപീകരിക്കുക. (2)
Answer:

Consider a circular loop of radius ‘a’ and carrying current “I”. Let P be a point on the axis of the coil, at distance x from A and r from ‘O’. Consider a small length dl at A.
The magnetic field at ‘p’ due to this small element dl,
dB =\(\frac{\mu_0 \text { Idl } \sin 90}{4 \pi x^2}\)

dB = \(\frac{\mu_0 \mathrm{Idl}}{4 \pi \mathrm{x}^2}\) …………(1)

The dB can be resolved into dB cosϕ (along Py) and dB sinϕ (along Px).
Similarly consider a small element at B, which pro-duces a magnetic field ‘dB’ at P. If we resolve this magnetic field we get.
dB sinϕ (along px) and dB cosϕ (along py1)
dB cosϕ components cancel each other, because
they are in opposite direction. So only dB sinϕ components are found at P, so total filed at P is
B = ∫ dB sinϕ
= \(\int \frac{\mu_0 \mathrm{Idl}}{4 \pi \mathrm{x}^2} \sin \phi\)
but from AOP we get, sin ϕ = a / x

Question 18.
Differentiate between paramagnetic, diamagnetic and ferromagnetic substances.
പാരാമാഗ്നറ്റിക്, ഡയാമാഗ്നറ്റിക്, ഫെറോമാഗ്നറ്റിക് വസ്തു ക്കളെ തരം തിരിക്കുന്നതെങ്ങനെ?
Answer:

Diamagnetism: Diamagnetic substances are those which have tendency to move from stronger to the weaker part of the external magnetic field.

Examples : Some diamagnetic materials are bismuth, copper, lead, silicon, nitrogen (at STP), water and sodium chloride.
Paramagnetism : Paramagnetic substances are those which get weakly magnetized in an external magnetic field. They get weakly attracted to a magnet.

Some paramagnetic materials are aluminium, sodium, calcium, oxygen (at STP) and copper chloride.

Ferromagnetism : Ferromagnetic substances are those which gets strongly magnetized in an external magnetic field. They get strongly attracted to a magnet.
Common examples of ferromagnetic substances are Iron, Cobalt Nickel, etc.

Question 19.
(a) State the principle of a.c. generator.
എ.സി. ജനറേറ്ററിന്റെ തത്വം പ്രസ്താവിക്കുക. (1)
Answer:
Electromagnetic Induction

(b) Obtain the expression for the emf generated by an a.c. generator.
എ.സി. ജനറേറ്റർ ഉത്പാദിപ്പിക്കുന്ന ഇ.എം. എഫിന്റെ സമവാക്യം രൂപീകരിക്കുക. (2)
Answer:

Take the area of coil as A and magnetic field pro¬duced by the magnet as B .Let the coil be rotating about an axis with an angular velocity ω.
Let θ be the angle made by the areal vector with the magnetic field B. The magnetic flux linked with the coil can be written as
ϕ =B. A
ϕ = BA cos θ
ϕ = BA coscot [since θ = ωt]
If there are N turns
ϕ = NBA cos ωt

∴ The induced e.m.f. in the coil,
ε = \(\frac{-\mathrm{d} \phi}{\mathrm{dt}}\)
ε = \(\frac{-\mathrm{d}}{\mathrm{dt}}(\mathrm{NBA} \cos \omega \mathrm{t})\)
ε = NBA cos ωt ω
Let ε = NBA ω,
then ε = ε₀ sinωt

Question 20.
Derive the expression for the refractive index of a prism with the help of a diagram.
ഒരു പ്രിസത്തിന്റെ റിഫ്രാക്ടീവ് ഇൻഡക്സിന്റെ സമവാക്യം ചിത്രത്തിന്റെ സഹായത്തോടെ രൂപീകരിക്കുക.
Answer:

ABC is a section of a prism. AB and AC are the refracting faces, BC is the base of the prism. ∠A is the angle of prism.
A ray PQ incidents on the face AB at an angle i₁. QR is the refracted ray inside.the prism, which makes two angles r₁ and r₂ (inside the prism). RS is the emergent ray at angle i₂.
The angle between the emergent ray and incident ray is the deviation‘d’.
In the quadrilateral AQMR,
∠Q + ∠R = 180°
[since N,M and NM are normal]
ie, ∠A + ∠M = 180° —(1)
In the ∆QMR,
∴ r₁ + r₂ + ∠M = 180° —(2)
Comparing eq (1) and eq (2)
r₁ + r₂ = ∠A —(3)
From the ∆QRT,
(i₁ – r₁) + (i₂ – r₂) = d
[since exterior angle equal sum of the opposite inte-rior angles]
(i₁ + i₂) – (r₁ + r₂) = d
but, r₁ – r₂ = A
.-. (i₁ + i₂) – A = d
(i₁ + i₂) = d + A …….(4)

It is found that for a particular angle of incidence, the deviation is found to be minimum value ‘D’.

At the minimum deviation positio
i₁ = i₂ = i, r₁ = r₂ = r and d = D
Hence eq (3) can be written as,
r + r = A
or r = \(\frac{A}{2}\) ……(5)
Similarly eq (4) can be written as,
i + i = A + D
i = \(\frac{A+D}{2}\) ………(6)
Let n be the refractive index of the prism,
then we can write,
n = \(\frac{sin i}{sin r}\) ……….(7)
Substituting eq (5) and eq (6) in eq (7),
n = \(\frac{\sin \frac{A+D}{2}}{\sin \frac{A}{2}}\)

Question 21.
Explain Rutherford’s, alpha particle scattering experiment.
റൂഥർഫോർഡിന്റെ ആൽഫാ പാർട്ടിക്കിൾ സ്കാറ്ററിംഗ് പരി ക്ഷണം വിവരിക്കുക.
Answer:
Rutherford’s scattering experiment

Experimental arrangement : a particles are incident on a gold foil (very small thickness) through a lead collimator. They are scattered at different angles. The scarred particles are counted by a particle detector.

Observations : Most of the alpha particles are scattered by small angles. Afew alpha particles are scattered at an ang|e greater than 90°.

Conclusions:
1) Major portion of thfe atom is empty space.
2) All the positive! charges of the atom are concentrated in a small portion of the atom.
3) The whole mass of the atom is concentrated in a small portion of the atom.

Rutherford’s model of atom:
1) The massive part of the atom (nucleus) is concentrated at the centre of the atom .
2) The nucleus contains all the positive charges of the atom.
3) The size of the nucleus is the order of 10^-15m.
4) Electrons move around the nucleus in circular orbits.
5) The electrostatic force of attraction (between proton and electron) provides centripetal force.

SECTION-D

Answer any 3 questions from 22 to 25. Each carries 4 scores.
22 മുതൽ 25 വരെയുള്ള ചോദ്യങ്ങളിൽ ഏതെങ്കിലും 3 എണ്ണത്തിന് ഉത്തരമെഴുതുക. 4 സ്കോർ വീതം. (3 × 4 = 12)

Question 22.
(a) What is an electric dipole?
വൈദ്യുത ഡൈപോൾ എന്നാലെന്ത്? (1)
Answer:
Electric dipole : A pair of equal and opposite charges separated by small distance is called electric dipole.

(b) Obtain the expression for the electric field intensity at a point on the axial line of an electric dipole.
വൈദ്യുത ഡൈപോളിന്റെ അക്ഷത്തിലുള്ള ബിന്ദുവിലെ വൈദ്യുത മണ്ഡല തീവ്രതയുടെ സമവാക്യം രൂപീകരി ക്കുക. (3)
Answer:
Electric field at a point on the axial line of an electric dipole : Consider an electric dipole of moment P = 2aq. Let ‘S’ be a point at a distance V from the centre of the dipole.

Electric field at ‘S’ due to point charge at ‘A’
E_A = \(\frac{1}{4 \pi \varepsilon_0}\) \(\frac{q}{(r+a)^2}\)
Directed as shown in figure.
Electric field at ‘S’ due to point charge at ‘B’
E_B = \(\frac{1}{4 \pi \varepsilon_0}\) \(\frac{q}{(r-a)^2}\)
Directed as shown in figure.
Therefore resultant electric field at ‘S’ And its magnitude
E = E_B + -E_A

The direction is along E_B
The field due to an electric dipole is directed from negative charge to positive charge along the axial line.

Question 23.
(a) Derive the expression for the torque on a . rectangular current loop in a uniform magnetic field with the help of a diagram. (2)
Answer:

Consider a rectangular coil PQRS of N turns which is suspended in a magnetic field, so that it can rotate (about yy1). Let ‘l’ be the length (PQ) and ‘b’ be the breadth (QR).
When a current l flows in the coil, each side produces a force. The forces on the QR and PS will not produce torque. But the forces on PQ and RS will produce a Torque.
Which can be written as
τ = Force × ⊥ distance ………..(1)
But, force = BI l ………(2) [since θ = 90°]
And from ∆QTR , we get
⊥ distance (QT) = b sin θ ………..(3)
Substituting the vales of eq (2) and eq (3) in eq(1) we get
τ = BIl b sin θ
= BIA sin θ [since lb = A (area)]
τ = IAB sin θ
τ = mB sin θ [since m = IA]
τ = m x B
If there are N turns in the coil, then

T = NIAB sin θ

(b) A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A. What is the magnetic moment of this coil? (2)
Answer:
r = 10 cm = 10 × 10-2 m
I = 3.2A
N = 100
M = NIA .
m = 100 × 3.2 × 3.14 × (10 × 10^-2)²
= 10.046 Am²

Question 24.
(a) With a neat diagram, derive lens makers formula. (2)
Answer:
Consider a thin lens of refractive index n2 formed by the spherical surfaces ABC and ADC. Let the lens is kept in a medium of refractive index n₁.
Let an object ‘O’ is placed in the medium of refractive index n₁. Hence the incident ray OM is in the medium of refractive index n1 and the refracted ray MN is in the medium of refractive index n₂.

The spherical surface ABC (radius of curvature R₁) forms the image at I₁. Let ‘u’ be the object distance and ‘v₁’ be the image distance.
Then we can write,
This image I₁, will act as the virtual object for the surface ADC and forms the image at v.
Then we can write

(b) The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive index of glass? (2)
Answer:
R₁ = 10 cm
R₂ = -15 cm
f = 12 cm
\(\frac{1}{f}\) = (n-1)[\(\frac{1}{R_1}\) – \(\frac{1}{R_2}\)]
\(\frac{1}{12}\) = (n-1)[\(\frac{1}{10}\) + \(\frac{1}{15}\)]
\(\frac{1}{12}\) = (n-1) [latex]\frac{15+10}{150}[/latex]
\(\frac{1}{12}\) = (n-1) × \(\frac{1}{6}\)
n – 1 = \(\frac{6}{12}\) = \(\frac{1}{2}\)
n = \(\frac{1}{2}\) + 1 = 1.5

Question 25.
(a) Give the classification of materials based on energy band diagram. (3)
Answer:
Classification on the basis of Energy bands
Conductors: Conduction band is partially filled and valence band is partially empty.

Or

Conduction band and valence band are overlaped so that E_g = 0 ev

Due to overlapping, electrons are partially filled in conduction band. These partially filled elec¬trons are responsible for current conduction.
Insulators

Conduction band is empty. Valence band may fully or partially filled. There is a wide energy gap between valence band and conduction band (E_g > 3ev).

Semiconductors : Conduction band may be empty or lightly filled. Valence band is fully filled. The energy gap is very small (< 3ev)

(b) Differentiate between intrinsic and extrinsic semiconductors. (1)
Answer:
A semiconductor in its pure form is called intrinsic semiconductor. The number of of free electrons is equal to number of holes. Dopped semi conductors (n type or p type) are called extrinsic semiconductors.

SECTION-E

Answer any 3 questions from 26 to 29. Each carries 5 scores. (3 × 5 = 15)

Question 26.
(a) Give the relation between electric field and potential. (1)
Answer:
E = \(\frac{-dv}{dx}\)

(b) Derive the expression for the potential due to an electric dipole. (2)
Answer:
Potential due to an electric dipole

Consider dipole of length ‘2a’. Let P be a point at distance r₁ from +q and r₂ from -q . Let r be the distance of P from the centre ‘O’ of the dipole. Let 0 be angle between dipole and line OP.

The potential due to +q , V₊ = \(\frac{1+q}{4 \pi \varepsilon_0 \quad r_1}\)

The potential due to -q , V_– = \(\frac{1-q}{4 \pi \varepsilon_0 \quad r_1}\)

Therefore total potential,
V = \(\frac{1+q}{4 \pi \varepsilon_0 \quad r_1}\) + \(\frac{1-q}{4 \pi \varepsilon_0 \quad r_1}\)
= \(\frac{1}{2}\) (\(\frac{1}{r_1}\) – \(\frac{1}{r_2}\))
= \(\frac{1}{2}\) (\(\frac{r_2-r_1}{r_1 r_2}\)) ……..(1)

From ∆ABC , we get (r₂ – r₁) = 2a cosθ
we can also take r₂= r₁ = r (since ‘2a’. is very small) Substituting these values in equation (1),we get
V = \(\frac{\mathrm{q}}{4 \pi \varepsilon_0}\) (\(\frac{2 a \cos \theta}{r^2}\))
V = \(\frac{1, \mathrm{P} \cos \theta}{4 \pi \varepsilon_0 \quad \mathrm{r}^2}\) (Since P = q2a)
But \(\overrightarrow{\mathrm{P}}\) . \(\hat{\mathbf{r}}\) = P cosθ, where \(\hat{\mathbf{r}}\) = \(\frac{\vec{r}}{|r|}\)
V = \(\frac{1 \overrightarrow{\mathrm{P}} \cdot \hat{r}}{4 \pi \varepsilon_0 \mathrm{r}^3}\)

(c) Calculate the potential at a point due to a charge of 4×10^-7 located 9 cm away. (2)
Answer:
q = 4 × 10^-7 C
r = 9 cm = 9 × 10^-2 m
V = 9 × 10⁹ × \(\frac{4 \times 10^{-7}}{9 \times 10^{-2}}\) = 4 × 10⁴ V

Question 27.
(a) State Kirchhoff’s law. (2)
Answer:
First law (Junction rule): The total current en-tering the junction is equal to the total current leaving the junction.
Second law (loop rule) : In any closed circuit the algebraic sum of the product of the current and resistance in each branch of the circuit is equal to the net emf in,that branch.
OR
Total emf in a closed circuit is equal to sum of voltage drops.

(b) Obtain the balancing condition of Wheatstone’s bridge with the help of a diagram. (3)
Answer:
Four resistances P,Q,R and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I₁, I₂,I₃ and I₄ be the four currents passing through RR,Q and S respectively.

Working
The voltage across R.
When key is closed, current flows in different branches as shown in figure. Under this situation
The voltage across P, V_AB = I₁P
The voltage across Q, V_BC = I₃Q ………(1)
The voltage across R, V_AD = I₂R
The voltage across S, V_DC = I₄S
The value of R is adjusted to get zero deflection in galvanometer. Under this condition,
I₁ = I₃ and I₂ = I₄ ………(2)
Using Kirchoffs second law in loopABDA and BCDB, we get
V_AB = V_AD …………(3)
and V_BC = V_DC ……….. (4)
Substituting the values from eq(1) in to (3) and (4), we get
I₁P = I₂R …………..(5)
and I₃Q = I₄S ………….(6)
Dividing Eq(5)byEq(6)
\(\frac{I_1 P}{I_3 Q}\) = \(\frac{\mathrm{I}_2 \mathrm{R}}{\mathrm{I}_4 \mathrm{~S}}\)
\(\frac{P}{Q}\) = \(\frac{R}{S}\) [since I₁ = I₃ and I₂ = I₄]
This is called Wheatstone condition.

Question 28.
(a) State the principle of a transformer. (1)
Answer:
Mutual Induction

(b) Explain the working of a transformer. (2)
Answer:

A transformer consists of two insulated coils wound over a core. The coil to which energy is given is called primary and that from which energy is taken is called secondary.

Working and mathematical expression : Let V₁, N₁ be the voltage and number of turns in the primary. Similarly let V₂, N₂ be voltage and number of turns in the secondary.

When AC is passed, a change in magnetic flux is produced in the primary. This magnetic flux passes through secondary coil.

If ϕ₁ and ϕ₂ are the magnetic flux of primary and secondary, we can write ϕ₁∝N₁ and ϕ₂∝N₂.
Dividing ϕ₁ and ϕ₂
\(\frac{\phi_1}{\phi_2}\) = \(\frac{N_1}{N_2}\)
[since ϕ is proportional to number of turns]
or ϕ = \(\frac{N_1}{N_2}\) ϕ₂
Taking differentiation on both sides we get
\(\frac{d \phi_1}{d t}\) = \(\frac{N_1}{N_2} \frac{d \phi_2}{d t}\); V₁ = \(\frac{N_1}{N_2}\) V₂
[since V₂ = \(\frac{-d \phi_2}{d t}\) and V₁ = \(\frac{-d \phi_1}{d t}\)]

i.e., \(\frac{V_1}{V_2}\) = \(\frac{N_1}{N_2}\)

(c) Differentiate between step up transformer and step down transformer. (2)
Answer:
Step up Transformer : If the output voltage is qreater than input voltaqe, the transformer, is called step up transformer. In a step up transformer N₂ > N₁ and V₂ > V₁.
Step down transformer: If the output voltage is less than the input voltage, then the transformer is called step down transformer. In a step down transformer N₂ < N₁ and V₂ < V₁

Question 29.
(a) Sate Huygens principle. (2)
Answer:
Huygen’s principle

1. Every point in a wavefront acts as a source of seconda ry wavelets.
2. The secondary wavelets travel with the same velocity as the original value.
3. The envelope of all these secondary wavelets gives a new wavefront.

(b) Explain the refraction of plane wave using Huygens principle. (3)
Answer:
AB is the incident wavefront and c, is the velocity of the wavefront in the first medium. CD is the refracted wavefront and c2 is the velocity of the wavefront in the second medium. AC is a plane separating the two media.

The time taken for the ray to travel from P to R is
t = \(\frac{PO}{C_1}\) + \(\frac{OR}{C_2}\)

O is an arbitrary point. Hence AO is a variable. But the time to travel a wavefront from AB to CD is constant. In order to satisfy this condition, the term containing AO in eq.(2) should be zero.

AO (\(\frac{\sin i}{C_1}\) – \(\frac{\sin r}{C_2}\)) = 0
\(\frac{\sin i}{C_1}\) = \(\frac{\sin r}{C_2}\)
where 1 n2 is the refractive index of the second medium w.r.t. the first. This is the law of refraction.