Kerala Plus Two Physics Board Model Paper 2023 with Answers

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

Kerala Plus Two Physics Board Model Paper 2023

Time : 2 Hours
Total Score: 60

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

Question 1.
Name the physical quantity whose S.I. unit is NC^-1.
Answer:
Electric field intensity (E)

Question 2.
Two capacitors of equal capacitance when connected in series have a capacitance of C₁. When connected in parallel, they have a capacitance of C₂. What is the value of C₁/C₂?
Answer:
\(\frac{1}{4}\)

Question 3.
Lenz’s law is in accordance with law of conservation of …………
Answer:
Energy

Question 4.
Name the physical quantity which can be expressed by the ratio of the amplitude of electric field to the magnetic field in an electromagnetic wave.
Answer:
Velocity

Question 5.
“Critical angle depends on the colour of light”. Whether the statement is true or false?
Answer:
True

Question 6.
A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature of the lens?
Answer:
Converging lens

Question 7.
The binding energy of a nucleus is a measure of its
(a) Charge
(b) Mass
(c) Stability
(d) Momentum
Answer:
(c) Stability

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

Question 8.
What happens to the drift velocity of charge carriers of a wire if its ,
(a) Potential difference is doubled. (1)
Answer:
Drift velocity doubled

(b) Length is increased. (1)
Answer:
Drift velocity decreases

Question 9.
(a) S.l. unit of magnetic moment of a dipole is (1)
(a) Am
(b) Am²
(c) JT
(d) JT^-2
Answer:
(b) Am²

(b) Figure shows a magnetic substance placed in an external magnetic field. Identify the substance. (1)

Answer:
Diamagnetic

Question 10.
(a) Can a transformer be used to step up or step down a DC voltage? Justify your answer. (1)
Answer:
No.
Transformer works on the principle of mutual induction. A change in current in the primary coil produces instantaneous emf in the secondary coil. In the case of dc, there is no change in current and hence transformer can not be used to step up or step down a dc voltage.

(b) Why do we use soft iron core in a transformer? (1)
Answer:
We use soft iron core. In transformer because of low hysteresis loss.

Question 11.
Match the following
(a) Radio waves – Night vision camera
(b) UV rays – Cellular phone
(c) Micro waves – Water purifier
(d) Infrared – Radar
Answer:
(a) Radio waves – Cellular phone
(b) UV rays – Water purifier
(c) Micro waves – Radar
(d) Infrared – Night vision camera

Question 12.
Sketch the wavefront corresponding to light from the following sources:
(i) A point source (1)
Answer:

(ii) A distant source (1)
Answer:

Question 13.
(a) In Rutehrford scattering experiment, most of the a particle go unscattered while some of them are scattered through large angles. What can be concluded from this? (1)
Answer:
The most space inside an atom is empty. The entire mass of atom is concentrated in a small volume called nucleus.

(b) What will be the angle of scattering when impact parameter is 0? (1)
Answer:
180°

Question 14.
Calculate binding energy of ₂₀Ca⁴⁰ from the following data:
mass of proton: 1.007825 a.m.u
mass of neutron: 1.008665 a.m.u
mass of ₂₀Ca⁴⁰ : 39, 962589 a.m.u
Answer:
Binding energy
= ∆mC² – [zm_p + (A – Z)m_n – M]C²
A = 40, Z = 20
m_p = 1.007825 amu
m_n =1.008665 amu
M = 39.962589 amu
Am = Zm_p + (A – Z)m_n – M
= 20 × 1.007825 + 20 × 1.008665 – 39.962589
= 20.1565 + 20.1733 – 39.962589
= 0.367211 amu
= 0.60957 × 10^-22 kg

E_b = AmC² = 0.60957 × 10^-22 × (3 × 10⁸)²
= 5.486 × 10^-11 J

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

Question 15.
An electric dipole of dipole moment p is placed in a uniform electric field E.
(a) Define dipole moment. (1)
Answer:
Electric dipole moment is the product of magnitude of charge and distance of seperation.
\(\overrightarrow{\mathrm{P}}\) = q × \(\overrightarrow{2 \mathrm{l}}\)

(b) Obtain an expression for torque acting on it. (2)
Answer:

Consider an electric dipole of dipole moment P = 2aq kept in a uniform external electric field, inclined at an angle θ to the field direction.
Equal and opposite forces +qE and -qE act on the two charges. Hence the net force on the dipole is zero. But these two equal and opposite forces whose lines of action are different. Hence there will be a torque.
torque = any one force × perpendicular distance (between the line of action of two forces )
τ = qE × 2 a sin θ
Since P = 2aq
τ = P E sin θ
Velocity \(\vec{\tau}\) = \(\vec{P}\) × \(\vec{E}\)

Question 16.
(a) Draw an equipotential surface due to an isolated point charge. (1)
Answer:

(b) If the electric field intensity at a given point is zero, will the electric potential necessarily be zero at that point? Why? (2)
Answer:
No. inside a charged shell electric field, intensity is zero. But the electric potential is equal to potential on the surface.

Question 17.
(a) How will you convert a galvanometer into voltmeter. (1)
Answer:
A galvanometer can be converted in to voltmeter by connecting a large resistance in series to it.

(b) A galvanometer with a coil resistance 12Ω shows full scale deflection for a current of 3 mA. How will you convert the galvanometer into voltmeter of range 0-18 V? (2)
Answer:
R_G = 12 Ω
I_g = 3mA = 3 × 10^-3A
V = 18V
R = \(\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}\) – R_G
R = \(\frac{18}{3 \times 10^{-3}}\) – 12 = 5988Q

Question 18.
(a) State Gauss’s law in magnetism. (1)
Answer:
Net magnetiv flux through ant closed surface is zero.
\(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{ds}}\) = 0

(b) Two substances P and Q have their relative permeability slightly greater and less than one respectively. What do you conclude about P and Q? (2)
Answer:
P → Paramagnetic
Q → Diamagnetic

Question 19.
(a) State Huygens principle. (1)
Answer:
According to Huygen’s principle
1. Every point in a wavefront acts as a source of secondary 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) Using Huygen’s wave theory, verify the law of reflection. (2)
Answer:
Img 9
AB is the incident wavefront and CD is the reflected wavefront, ‘i’ is the angle of incidence and r is the angle of reflection. Let c1 be the velocity of light in the medium. Let PO be the incident ray and OQ be the reflected ray.
The time taken for the ray to travel from P to Q is ,
Img 10
O is an arbitrary point. Hence AO is a variable. But the time to travel for a wave front from AB to CD is a constant. So eq.(2) should be independent of AO. i.e., the term containing AO in eq(2) should be zero.
sin i – sin r = 0
sin i = sin r
i = r

Question 20.
(a) The variation of stopping potential with frequency of sodium is given below. Calculate its work function. (1)
Answer:
ϕ = hυ
υ₀ = 4Hz (check)
∴ ϕ = 6.63 × 10^-34 × 4
= 26.52 × 10^-34

(b) The work function for two metals are in the ratio 1:2. Find the ratio of threshold wavelength of those metals. (2)
Img 2
Answer:
Img 11

Question 21.
Draw the circuit diagram of a full wave rectifier and explain its working. Also, draw the input and output waveform. (3)
Answer:
Img
Full wave rectifier consists of transformer, two diodes and a load resistance R_L. Input a.c signal is applied across the primary of the transformer. Secondary of the transformer is connected to D₁ and D₂ The output is taken across R_L.

Working : During the +ve half cycle of the a.c signal at secondary, the diode D₁ is forward biased and D₂ is reverse biased. So that current flows through D₁ and R_L.

During the negative half cycle of the a.c signal at secondary, the diode D₁ is reverse biased and D₂ is forward biased. So that current flows through D₂ and R_L.

Thus during both the half cycles, the current flows through RL in the same direction. Thus we get a +ve voltage across R_L for +ve and -ve input. This process is called full wave rectification.
Img

Answer any 3 questions from 22 to 25. Each carries 4 scores. (3 × 4 = 12)

Question 22.
(a) In a parallel plate capacitor, how is the capacity of affected when
(i) The distance between the plates is doubled.(1)
Answer:
Halved

(ii) The area of the plates is halved (1)
Answer:
Halved

(b) Figure shows the variation of voltage V across the plates of two capacitors A and B verses increase of charge Q stored on them. Which of the two capacitors has higher capacitance? Given reason for your answer. (2)

Answer:
C = \(\frac{Q}{V}\) = \(\frac{1}{Slope}\)

Question 23.
(a) Kirchoff’s second law is based on laws of conservation of ……… (1)
(i) Charge
(ii) Energy
(iii) Momentum
(iv) Angular momentum
Answer:
(ii) Energy

(b) Use Kirchoff’s rules to analyse the Wheatstore bridge and arrive at Wheatstones condition for balancing the bridge. (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 P, R, Q and S respectively.
Img

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 loop ABDA 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 24.
(a) Define self-inductance. (1)
Answer:
Self inductance is the magnetic flux linked with the coil of unit current passing through it.
ϕ = LI
when I = 1A, ϕ = L

(b) Derive an expression for self-inductance of a coil. (3)
Answer:
Consider a solenoid (air core) of length l, number of turns N and area cross section A. let ‘n’ be the no. of turns per unit length (n = N/l)
The magnetic flux linked with the solenoid,
ϕ = BAN
ϕ = μ₀nlAN (since B = μ₀nl)
but ϕ = LI
∴ LI = μ₀nIAN
L = μ₀nAN
If solenoid contains a core of relative permeability μ_r
the L = μ₀μ_rnAN

Question 25.
(a) Draw a labelled ray diagram of a compound microscope forming an image at the near point of eye. (2)
Answer:
Img

(b) Obtain an expression for the magnification produced by a compound microscope. (2)
Answer:
Working : The object is placed in between F and 2F of objective lens. The objective lens forms real inverted and magnified image (I₁M₁) on the other side of the lens. This image will act as object or eyepiece. Thus an enlarged, virtual and inverted image is formed, (this image can be adjusted to be at the least distance of distinct vision, D)

Magnification : The magnification produced by the compound microscope
m = \(\frac{\text { size of the image }}{\text { size of the object }}\)
ie, m = \(\frac{1}{2}\)
Multiplying and dividing by I₁M₁ we get,
m = \(\frac{\mathrm{I}_2 \mathrm{M}_2}{\mathrm{OB}}\)

but we know, m_e = \(\frac{\mathrm{I}_2 \mathrm{M}_2}{\mathrm{I}_1 \mathrm{~m}_1}\) and m₀ = \(\frac{\mathrm{I}_1 \mathrm{M}_1}{\mathrm{OB}}\)

Where m₀ & m_e are the magnifying power of objective lens and eyepiece lens
m = m₀ × m_e ……….. (1)

Eyepiece acts as a simple microscope.
Therefore m_e = 1 + \(\frac{D}{f_e}\) …………(2)

We know magnification of objective lens
m₀ = \(\frac{V_0}{u_0}\) ……….. (3)
Where v₀ and u₀ are the distance of the image and object from the objective lens.
Substituting (2) and (3) in (1), we get
m = \(\frac{- v_0}{u_0}\) (1 + \(\frac{D}{f_e}\))