**Haryana Board (HBSE)** Class 12 Physics Question Paper 2023 Answer Key. Haryana Board Class 12th Physics Solved Question Paper 2023 PDF Download. HBSE Board Solved Question Paper Class 12 Physics 2023. HBSE 12th Question Paper Download 2023. HBSE 12th Physics Question Paper Pdf Download 2023. HBSE Class 12 Physics Previous Year Question Paper with Answer. HBSE 12th Physics Solved Question Paper 2023. HBSE 12th Class Physics Solved Question Paper 2023.

**HBSE Class 12 Physics Question Paper 2023 Answer Key (All Sets – A,B,C,D)**

**SET–A**

Q1.(i) At a given place the horizontal and vertical components of earth are equal. The angle of dip at that place will be :

(A) 0°

(B) 45°

(C) 60°

(D) 90°

Ans : (B) 45°

(ii) Two atoms have the same atomic number but different atomic mass. They will be :

(A) Isotopes

(B) Isobaric

(C) Isotones

(D) None of these

Ans : (A) Isotopes

(iii) The stopping potential for photoelectrons depends upon :

(A) Frequency of incident light only

(B) Material of the cathode only

(C) Both the frequency of incident light and the material of the cathode

(D) Intensity of incident light

Ans : (C) Both the frequency of incident light and the material of the cathode

(iv) In intrinsic semiconductor at room temperature, the numbers of electrons and holes are :

(A) Equal

(B) Zero

(C) Unequal

(D) Infinite

Ans : (A) Equal

(v) Microwaves are the electromagnetic waves with frequency in the range of :

(A) Micro hertz

(B) Mega hertz

(C) Giga hertz

(D) Hertz

Ans : (C) Giga hertz

(vi) Is the Junction, diode D is forward or reverse biased in the given diagram ?

Ans : Forward biased

(vii) Write the relation between radius R and mass number A of a nucleus.

Ans : R = R₀A^⅓

(viii) If A = 1, B = 1, find the value of Y in the adjoining logic circuit.

Ans : Y = A.B = 1 (This is AND Gate)

(ix) The mass of an electron is m and has charge e. If this electron is accelerated by a potential difference of V, then write the formula for the de-Broglie wavelength associated with it.

Ans : λ = h ÷ √(2meV) or λ = h/√(2meV)

(x) Are Kirchhoff’s rules applicable to both a. c. and d. c. ?

Ans : Yes

(xi) Tesla is the unit of ……………

Ans : Magnetic flux density

(xii) The magnetic flux passing through a ring is increased from Φ1 to Φ2 at a constant rate in time t. The value of Induced Electromotive force will be …………….

Ans : E = (Φ2 – Φ1)/t

(xiii) Electrostatic force between two point charges in vacuum is F. If the charge are kept at the same distance in water (Dielectric constant K = 80), then the force between them will be …………..

Ans : F’ = F/K = F/80

(xiv) …………… is the SI unit of conductivity.

Ans : Siemens per metre (S/m)

(xv) The velocity of electromagnetic waves in free space can be given by the relation ……………

Ans : c = 1/√(μ₀ε₀), where c is the speed of light in vacuum, μ₀ is the permeability of free space, and ε₀ is the permittivity of free space.

Q2. Define the term electric dipole moment. Is it scalar or vector ?

Ans – Electric dipole moment is the product of the magnitude of charges with the distance of separation between them. The electric dipole moment is a vector quantity (having both magnitude and direction) and has a defined direction that goes from negative charge to positive charge.

Electric dipole moment (p) = q × d

Q3. What is the principle of potentiometer ?

Ans – The basic principle of the potentiometer is that the potential drop across any section of the wire will be directly proportional to the length of the wire, provided the wire is of a uniform cross-sectional area and a uniform current flows through the wire.

Q4. What are magnetic elements at a place ? Define any one.

Ans – Earth behaves as a magnet. The declination (α), the inclination or dip (δ), and the total intensity of earth’s magnetic field (B) are the three magnetic elements of earth’s magnetic field at a place.

Declination can be defined as the angle between the geographic north and magnetic north at a specific location. It is an important parameter for navigation and helps in determining the correct direction while using a compass.

Q5. A 100 Hz (hertz) a. c. is flowing in a 14 mH coil. Find its Reactance.

Ans : Reactance (XL) = 2πfL, where f is the frequency in hertz and L is the inductance in henries.

In this case, f = 100 Hz and L = 14 mH = 0.014 H

Reactance (X) = 2π × 100 × 0.014 = 8.8 Ω (ohms)

Q6. Write two uses each of Microwaves and Gamma rays.

Ans – Microwaves are used for satellite communication and in radar system for aircraft navigation. They are also used in microwave ovens for cooking/warming food.

Gamma rays are used in medical science to kill cancer cells. They are also used in Sterilize medical equipment and Sterilize food (irradiated food).

Q7. Define power of a lens.

Ans – Power of a lens is its ability to converge or diverge the rays of light falling on it. Power of a lens is equal to reciprocal of the focal length of the lens, i.e. P = 1/F. SI unit of power is dioptre (D).

Q8. What are coherent sources of light ?

Ans – Coherent sources of light are those sources which emit a light wave having the same frequency, wavelength and in the same phase, or they have a constant phase difference. A coherent source forms sustained interference patterns when the superimposition of waves occurs, and the positions of maxima and minima are fixed.

Q9. Draw the circuit diagram of Half wave rectifier, showing input and output waves.

Ans –

Q10. State Wheatstone bridge principle and deduce it using Kirchhoff’s rules.

Ans – It works on the principle of null deflection, which means the ratio of their resistances are equal and hence no current flows through the circuit. Under normal conditions, the bridge will be in the unbalanced condition where current flows through the galvanometer. The bridge will be in a balanced condition when no current flows through the galvanometer. One may achieve this condition by adjusting the known resistance and variable resistance.

The Wheatstone bridge principle states that if four resistances P, Q, R, and S are arranged to form a bridge with a cell and key between A and C, and a galvanometer between B and D then the bridge is said to be balanced when the galvanometer shows a zero deflection.

In balanced condition, Ig = 0

so, VB = VD or P/Q = R/S .This is called condition of balance.

Q11. Define resonance in LCR circuit. Find Power factor of series resonance LCR circuit.

Ans – Resonance in an LCR circuit occurs when the inductive reactance (XL) and the capacitive reactance (XC) cancel each other out, resulting in a purely resistive circuit. This happens at a specific frequency called the resonant frequency.

The power factor (pf) of a series resonance LCR circuit is 1.

Q12. Describe briefly with the help of a labelled diagram, principle of step-up Transformer.

Ans – A step-up transformer is a type of transformer that increases the voltage level of an alternating current (AC) power source.

Principle – A transformer works on the principle of mutual induction. Whenever the amount of magnetic flux linked with a coil changes, an emf is induced in the neighbouring coil.

Q13. What is diffraction of light? Explain the intensity distribution graph due to diffraction from a single slit.

Ans – Diffraction of light is a phenomenon that occurs when light waves encounter an obstacle or pass through a narrow opening and then spread out or bend around the edges of that obstacle or opening.

When light passes through a single slit, it undergoes diffraction and creates an intensity distribution pattern on a screen placed behind the slit. This pattern is known as the diffraction pattern or the intensity distribution graph.

Q14. Using Einstein’s photoelectric equation, explain effect of frequency incident radiation on stopping potential.

Ans – Einstein’s photoelectric equation describes the relationship between the frequency of the incident radiation and the energy of the emitted electrons.

The equation is given by, E = hf – φ, where E is the energy of the emitted electron, h is Planck’s constant, f is the frequency of the incident radiation, and φ is the work function of the material.

According to Einstein’s photoelectric equation, the frequency of the incident radiation has a direct impact on the stopping potential. For frequencies below the threshold frequency, the stopping potential is zero. At the threshold frequency, the stopping potential becomes positive. Above the threshold frequency, the stopping potential becomes negative, and its magnitude increases with increasing frequency.

Q15. Define mass defect and nuclear binding energy. For a nucleus X, write the value of mass defect and nuclear binding energy.

Ans – Mass defect refers to the difference in mass between a nucleus and the sum of the masses of its individual nucleons (protons and neutrons). It arises due to the conversion of some mass into energy according to Einstein’s mass-energy equivalence principle during the formation of the nucleus.

Nuclear binding energy represents the energy required to disassemble a nucleus into its individual nucleons. It is the energy bound within the nucleus due to the strong nuclear force, which holds the protons and neutrons together.

For a nucleus X, the value of the mass defect (Δm) can be obtained by subtracting the mass of the nucleus from the sum of the masses of its individual nucleons : Δm = (mass of nucleons) – (mass of nucleus).

The nuclear binding energy (BE) associated with the nucleus X can be calculated using the equation : BE = Δmc² where c is the speed of light.

Q16. Define the term Half-life period and decay constant of a radioactive element. Write the relation between them.

Ans : Half-life period – The half-life period of an element is defined as the time in which the number of radioactive nuclei decay to half of its initial value.

Decay constant – The decay constant of a radioactive element is defined as the reciprocal of time in which the number of undecayed nuclei of that radioactive element fails to 1/e times of its initial time.

The relation between the half-life period (T½) and decay constant (λ) is given by the equation :

T½ = 0.693/λ

Q17. What is doping? State two differences between N-type and P-type.

Ans – Doping is the process of intentionally adding impurities to a semiconductor material to alter its electrical properties.

Two differences between N-type and P-type doping are :

(i) N-type doping involves adding impurities such as phosphorus or arsenic, which have extra electrons, while P-type doping involves adding impurities such as boron or aluminum, which have fewer electrons.

(ii) In N-type doping, the extra electrons from the impurities become majority charge carriers, while in P-type doping, the holes left by the missing electrons from the impurities become majority charge carriers.

Q18. What is a capacitor? Derive an expression for the capacitance of a parallel plate capacitor in which a dielectric medium of dielectric constant K partially fills the space between the plates. Draw the necessary diagram.

Ans – A capacitor is a two-terminal electrical device that can store energy in the form of an electric charge. It consists of two electrical conductors that are separated by a distance. The space between the conductors may be filled by vacuum or with an insulating material known as a dielectric.

To derive the expression for the capacitance of a parallel plate capacitor with a dielectric medium, we consider the basic equation for capacitance :

C = Q/V, where C is the capacitance, Q is the charge on one of the plates, and V is the potential difference between the plates.

Electric field between plates of the capacitor can be found using Gauss Law.

E = σ/Kεo = Q/(KεoA) , here σ = Q/A

Electric field is defined as the gradient of potential.

E = V/d

V = Ed = Qd/(KεoA)

By definition of capacitance,

C = Q/V = KεoA/d

OR

(a) Define Electric flux. Is it a scalar or a vector quantity.

Ans – Electric flux (ϕ) is a measure of the electric field passing through a given area. It is defined as the dot product of the electric field vector and the area vector. Electric flux is a scalar quantity.

(b) Use Gauss’s law to derive the expression for the electric field (E) due to a straight uniformly charged infinite line of charge density λ c/m.

Ans – According to Gauss theorem, the total electric flux (ϕ) through any closed surface (S) in free space is equal to 1/ε times the total electric charge (q) enclosed by the surface, i.e.

ϕ = ∮ E.dS = q/ε ………(i)

The cylindrical gaussian surface is divided into three parts I, II, III.

∮I E.dS + ∮II E.dS + ∮III E.dS = q/ε

∮I E.dS cos90° + ∮II E.dS cos90° + ∮III E.dS cos0° = q/ε

0 + 0 + E ∫dS = q/ε

E(2πrl) = q/ε = λl/ε

E = λ/2πrε

Q19. Derive an expression for the force between two parallel straight conductors carrying current in the same direction. Hence define one ampere.

Ans –

Consider a small length L of the long straight conductor

B1 = μI1/(2πd)

B2 = μI2/(2πd)

F12 = I2B1L

F21 = I1B2L

F = F12 = F21 = μI1I2L/(2πd)

F/L = μI1I2/(2πd)

One Ampere is defined as the amount of electric current that flows through two parallel conductors of infinite length, which are placed one meter apart in a vacuum, that results in a force of exactly 2 x 10-⁷ N/m.

OR

Explain the principle of a moving coil Galvanometer. How can it be converted into an ammeter? Draw necessary diagram.

Ans – A moving coil galvanometer is an instrument which is used to measure electric currents. It is a sensitive electromagnetic device which can measure low currents even of the order of a few microamperes.

Principle – When a current flows through the coil, a torque acts on it. The magnetic torque tends to rotate the coil. Spring provides a counter torque that balances the magnetic torque resulting in a steady angular deflection.

Convert Galvanometer to Ammeter – A galvanometer is converted into ammeter by connecting a small resistance (called shunt) in parallel with it.

Q20. State Huygens principle and prove the law of reflection on the basis of wave theory.

Ans – According to Huygen’s principle, each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave.

Let, PQ be reflecting surface and a plane wavefront AB is moving through the medium (air) towards the surface PQ to meet at the point B.

Let, c be the velocity of light and t be the time taken by the wave to reach A’ from A. Then, AA’ = ct

Using Huygen’s principle, secondary wavelets starts from B and cover a distance ct in time t and reaches B’. To obtain new wavefront, draw a circle with point B as centre and ct (AA’ = BB’) as radius. Draw a tangent A’B’ from the point A’. Then, A’B’ represents the reflected wavelets travelling at right angle. Therefore, incident wavefront AB, reflected wavefront A’B’ and normal lies in the same plane.

Consider ∆ABA’ and ∆B’BA’

AA’ = BB’ = ct [AA’ = BB’ = radii of same circle]

BA’ = BA’ [common]

∠BAA’ = ∠BB’A’ [each 90°]

∆ABA’ ≅ ∆B’BA’ [by R.H.S]

∠ABA’ = ∠B’A’B [corresponding parts of congruent triangles]

incident angle i = reflected angle r

i.e. ∠i = ∠r

OR

Draw a labelled ray diagram showing the formation of image in a compound microscope. Define its magnifying power and write expression for it.

Ans –

Magnifying Power – Magnifying power of a compound microscope is defined as, the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the image and object are situated at the least distance of distinct vision from the eye.

**SET–B**

Q1.(i) The electrons ejected from metals due to effect of light are called :

(A) Primary electrons

(B) Secondary electrons

(C) Photo electrons

(D) Thermions

Ans : (C) Photo electrons

(ii) The path difference equivalent to π/4 phase difference is :

(A) λ

(B) λ/2

(C) λ/4

(D) λ/8

Ans : (D) λ/8

(iii) Electric field intensity (E) due to an electric dipole varies with distance (r) of the point from the centre of short dipole as :

(A) E ∝ 1/r

(B) E ∝ 1/r⁴

(C) E ∝ 1/r²

(D) E ∝ 1/r³

Ans : (D) E ∝ 1/r³

(iv) The best instrument for accurate measurement of e. m. f. of a cell is :

(A) Wheatstone bridge

(B) Ammeter

(C) A Potentiometer

(D) Voltmeter

Ans : (C) A Potentiometer

(v) Which of the following electromagnetic waves has smaller wavelengths ?

(A) X-rays

(B) Microwaves

(C) γ-rays

(D) Radiowaves

Ans : (C) γ-rays (Gamma rays)

(vi) What will be the bands of Colour in sequence on carbon resistor, if its resistance is 120 Ω .

Ans : Brown, Red, Brown, Gold

(vii) Can we decrease the range of a given ammeter ?

Ans : No (we can only increase range of Ammeter)

(viii) What is speed of light in vacuum ?

Ans : 3 × 10⁸ m/s

(ix) Why are alkali metal surfaces most suited as photo-sensitive surfaces ?

Ans : The work function alkali metal is very low and also low ionization energy. Even the ordinary visible light can bring about emission of photo electrons from these metal.

(x) An inductor acts as a conductor for d. c., why ?

Ans : Inductor offers no resistance menas zero resistance. So, it act as a conductor for d. c.

(xi) The Boolean expression for NAND gate is ……………

Ans :

(xii) Capacity can be increased by connecting different capacitors in …………..

Ans : Parallel

(xiii) Average power consumed / cycle in an ideal capacitor is ………….

Ans : Zero

(xiv) The relation between angle of incidence i, angle of prism A and angle of minimum deviation for a triangular prism is ………..

Ans : i = (A+Δm)/2 or i = (A+δm)/2

(xv) Isotones are the nuclides which contain …………..

Ans : Same numbers of neutron but different number of protons.

Q2. Draw the circuit diagram to determine unknown resistance using meter bridge. Write the equation for determining the unknown resistance.

Ans –

The unknown resistance can be calculated by the equation :

X = R × l/(100-l)

Q3. What are Eddy currents? Explain.

Ans – Eddy currents are loops of electrical current induced within conductors by a changing magnetic field in the conductor according to Faraday’s law of induction. Eddy currents flow in closed loops within conductors, in planes perpendicular to the magnetic field.

Q4. Write two uses each of Radio waves and X-rays.

Ans – Radio waves are used in standard broadcast radio and television, shortwave radio, navigation and air-traffic control, cellular telephony, and even remote-controlled toys.

X-rays are used to check the bags at the airport, identify the fracture, and help in identifying the infection in the bones, teeth, etc.

Q5. Define focal length of a lens and give its relation with power of the lens.

Ans – Focal length is defined as the distance between the optical centre and the focus of the lens.

Power of the lens is inversely proportional to its focal length. i.e. P = 1/F , where power in diopters (D) and focal length in meters (m).

Q6. What is Nuclear fission? Write one example.

Ans – Nuclear fission is a nuclear reaction in which the nucleus of an atom splits into two or more smaller nuclei, along with the release of a large amount of energy.

For example, When Uranium-235 atom is bombarded with a neutron, it splits into two lighter nuclei Barium and Krypton.

Q7. Write the relation between kinetic energy of α-particle and its distance of closest approach from nucleus in Rutherford’s α-scattering experiment.

Ans : Consider α-particle, mass = m, velocity = v

Kinetic Energy (Eα) = ½mv² ………(i)

If ro = closest distance of α-particle from nucleus

Potential Energy (Ep) = 1/(4πεo) × (2Z×2e)/ro = 2Ze²/(4πεoro) ……….(ii)

At equillibrium, Eα = Ep

½mv² = 2Ze²/(4πεoro)

Eα ∝ 1/ro

This equation shows that the kinetic energy of the α-particle is inversely proportional to the distance of closest approach.

Q8. Discuss how AND gate is realized from the NAND gate ?

Ans – If the output y’ of NAND gate is connected to the input of NOT gate (made from NAND gate by joining two inputs), as shown in Fig. then we get back an AND gate.

which is the Boolean expression for AND gate. Thus the combination work as AND gate. The truth table of the combination is shown in Fig., which is the truth table of AND gate.

Q9. What are N-type semiconductors? How they are formed ?

Ans – N-type semiconductors are materials that have an excess of negatively charged carriers or electrons as the majority charge carriers. These materials have an abundance of free electrons, which are responsible for the conduction of electric current.

N-type semiconductors are formed by introducing impurity atoms into a pure or intrinsic semiconductor material. This process is called doping. The impurity atoms used for doping are phosphorus (P), arsenic (As), or antimony (Sb).

Q10. Calculate the equivalent capacitance between the points A and B in the given diagram :

Ans – Here C1, C2 and C3 are in parallel,

so, C1 + C2 + C3 = 4 + 3 + 2 = 9 µF

Account to diagram, 9µF and C4 are in series,

so, 1/C = 1/9 + 1/C4 = 1/9 + 1/9 = 2/9

C = 9/2 = 4.5 µF

Q11. What is electric flux? State Gauss’s law for electric flux. Write important points regarding this law.

Ans – Electric flux (ϕ) is a measure of the electric field passing through a given area. It is defined as the dot product of the electric field vector and the area vector. Electric flux is a scalar quantity.

Gauss theorem states that the total electric flux (ϕ) through any closed surface (S) in free space is equal to 1/εo times the total electric charge (q) enclosed by the surface. i.e. ϕ = ∮ E.dS = q/εo

This law applies to any closed surface, regardless of its shape or size.

Q12. Explain the use of meter bridge for finding an unknown resistance.

Ans – A metre bridge is an apparatus that is used to measure unknown resistance of a wire or a metallic coil by finding a zero deflection in a galvanometer coil when no electric current is passing through the circuit. It can be also used to compare two unknown resistances.

Q13. State Biot-Savart’s Law and write its Mathematical form.

Ans – Biot-Savart’s Law describes the magnetic field produced by a steady current flowing in a wire. It states that the magnetic field at a point in space is directly proportional to the current in the wire and inversely proportional to the square of the distance from the wire. Mathematically, Biot-Savart’s Law can be expressed as :

dB = μ/4π × IdLsinθ/r²

B = ∫ dB = μ/4π × I(2πr)/r² = μI/2r

Q14. What is diffraction of light? Explain the intensity distribution graph due to diffraction from a single slit.

Ans – Diffraction of light is a phenomenon that occurs when light waves encounter an obstacle or pass through a narrow opening and then spread out or bend around the edges of that obstacle or opening.

When light passes through a single slit, it undergoes diffraction and creates an intensity distribution pattern on a screen placed behind the slit. This pattern is known as the diffraction pattern or the intensity distribution graph.

Q15. Using Einstein’s Photoelectric equation, explain the effect of frequency of incident radiation on stopping potential.

Ans – Einstein’s photoelectric equation describes the relationship between the frequency of the incident radiation and the energy of the emitted electrons.

The equation is given by, E = hf – φ, where E is the energy of the emitted electron, h is Planck’s constant, f is the frequency of the incident radiation, and φ is the work function of the material.

According to Einstein’s photoelectric equation, the frequency of the incident radiation has a direct impact on the stopping potential. For frequencies below the threshold frequency, the stopping potential is zero. At the threshold frequency, the stopping potential becomes positive. Above the threshold frequency, the stopping potential becomes negative, and its magnitude increases with increasing frequency.

Q16. Define the term Half-Life period and decay constant of a radioactive element. Derive the relation between them.

Ans : Half-life period – The half-life period of an element is defined as the time in which the number of radioactive nuclei decay to half of its initial value.

Decay constant – The decay constant of a radioactive element is defined as the reciprocal of time in which the number of undecayed nuclei of that radioactive element fails to 1/e times of its initial time.

The relation between the half-life period (T½) and decay constant (λ) is given by the equation :

T½ = 0.693/λ

Q17. Draw a circuit diagram of full-wave rectifier. Explain its working principle. Draw the input/output wave forms indicating clearly the function of two diodes used.

Ans – A full-wave rectifier is a circuit that converts an alternating current (AC) input signal into a direct current (DC) output signal.

Working principle – The underlying working principle of full wave rectifier is that the p-n junction conducts when it is forward biased and does not conduct when it is reverse biased.

Q18. Explain the principle of a moving coil galvanometer. How can it be converted into a voltmeter ? Explain with the help of diagram.

Ans – A moving coil galvanometer is an instrument which is used to measure electric currents. It is a sensitive electromagnetic device which can measure low currents even of the order of a few microamperes.

Principle – When a current flows through the coil, a torque acts on it. The magnetic torque tends to rotate the coil. Spring provides a counter torque that balances the magnetic torque resulting in a steady angular deflection.

Convert Galvanometer to Voltmeter – A galvanometer is converted into a voltmeter by connecting a high resistance in series with it.

OR

State the principle of Cyclotron. Explain its working with the help of a labelled diagram.

Ans : Cyclotron – Cyclotron is a device used to accelerate charged particles to high energies or speed.

Principle – Cyclotron works on the principle that a charged particle moving normal to a magnetic field experiences magnetic lorentz force due to which the particle moves in a circular path.

Working – When a positive ion of charge q and mass m is emitted from the source, it is accelerated towards the Dee having a negative potential at that instant of time. Due to the normal magnetic field, the ion experiences magnetic lorentz force and moves in a circular path. By the time the ion arrives at the gap between the Dees, the polarity of the Dees gets reversed.

Hence the particles is once again accelerated and moves into the other Dee with a greater velocity along a circle of greater radius. Thus the particle moves in a spiral path of increasing radius and when it comes near the edge, it is taken out with the help of a deflection plate (D.P.). The particle with high energy is now allowed to hit the target T.

When the particle moves along a circle of radius r with a velocity v, the magnetic Lorentz force provides the necessary centripetal force.

Bqv = mv²/r

v = qBr/m

Period of revolution, T = 2πr/v

T = 2πrm/qBr = 2πm/qB

Q19.(a) Define self induction and coefficient of self inductance.

Ans – Self induction refers to the phenomena in which an electromagnetic field is produced in a coil when the current flowing through it changes. It is the property of an electric circuit in which a changing current in the coil causes a voltage to be induced in the same coil.

Coefficient of self inductance (L) is a measure of the ability of a coil to produce an electromagnetic field due to self induction. It is defined as the ratio of induced electromotive force (EMF) to the rate of change of current in the coil. The unit of coefficient of self inductance is henry (H).

(b) What e. m. f. will be induced in a 10 H inductor in which current changes from 10 A to 7 A in 9 x 10-² s ?

Ans – Given,

Inductance (L) = 10 H

Initial current (I1) = 10 A

Final current (I2) = 7 A

Time taken (t) = 9 × 10-² s = 0.09 s

The rate of change of current (di/dt) can be calculated as : di/dt = (I2-I1)/t = (7-10)/0.09 = -33.33

e. m. f = -L × di/dt = -10 × (-33.33)

e. m. f = 333.3 V

OR

Explain the principle and working of a Transformer.

Ans : Transformer – An electrical device that can change the alternate current or voltage is known as a transformer.

Principle – A transformer works on the principle of mutual induction. Mutual induction is the phenomenon by which when the amount of magnetic flux linked with a coil changes, an E.M.F. is induced in the neighboring coil.

Working – The transformer works on the principle of Faraday’s law of electromagnetic induction and mutual induction. There are usually two coils primary coil and secondary coil on the transformer core. The core laminations are joined in the form of strips. The two coils have high mutual inductance. When an alternating current pass through the primary coil it creates a varying magnetic flux. As per faraday’s law of electromagnetic induction, this change in magnetic flux induces an emf (electromotive force) in the secondary coil which is linked to the core having a primary coil. This is mutual induction.

Q20. Using Huygen’s construction, explain reflection of a plane wave by a plane surface.

Ans – According to Huygen’s principle, each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave.

Let, PQ be reflecting surface and a plane wavefront AB is moving through the medium (air) towards the surface PQ to meet at the point B.

Let, c be the velocity of light and t be the time taken by the wave to reach A’ from A. Then, AA’ = ct

Using Huygen’s principle, secondary wavelets starts from B and cover a distance ct in time t and reaches B’. To obtain new wavefront, draw a circle with point B as centre and ct (AA’ = BB’) as radius. Draw a tangent A’B’ from the point A’. Then, A’B’ represents the reflected wavelets travelling at right angle. Therefore, incident wavefront AB, reflected wavefront A’B’ and normal lies in the same plane.

Consider ∆ABA’ and ∆B’BA’

AA’ = BB’ = ct [AA’ = BB’ = radii of same circle]

BA’ = BA’ [common]

∠BAA’ = ∠BB’A’ [each 90°]

∆ABA’ ≅ ∆B’BA’ [by R.H.S]

∠ABA’ = ∠B’A’B [corresponding parts of congruent triangles]

incident angle i = reflected angle r

i.e. ∠i = ∠r

OR

Draw a labelled ray diagram showing the formation of image in a compound microscope. Define its magnifying power and write expression for it.

Ans –

Magnifying Power – Magnifying power of a compound microscope is defined as, the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the image and object are situated at the least distance of distinct vision from the eye.

**SET–C**

Q1.(i) The horizontal component of Earth’s magnetic field is zero at :

(A) Magnetic poles

(B) Geographic poles

(C) Every Place

(D) Magnetic equatorial

Ans : (A) Magnetic poles

(ii) The ratio of the intensities of two light waves is 16 : 9. The ratio of maximum and minimum intensities in their interference pattern will be :

(A) 4/3

(B) 49 : 1

(C) 25 : 7

(D) 256 : 81

Ans : (B) 49 : 1

I1/I2 = a²/b² = 16/9

a/b = √16/9 = 4/3

a : b = 4 : 3

Imax = (a+b)² = (4+3)² = 7² = 49

Imin = (a-b)² = (4-3)² = 1² = 1

Imax : Imin = 49 : 1

(iii) The stopping potential for photoelectrons depends upon :

(A) Frequency of incident light only

(B) Material of the cathode only

(C) Both the frequency of incident light and the material of the cathode

(D) Intensity of incident light

Ans : (C) Both the frequency of incident light and the material of the cathode

(iv) In the electromagnetic wave, the phase difference between electric field and magnetic field is :

(A) 0

(B) π/4

(C) π/2

(D) π

Ans : (A) 0

(v) Two electric bulbs of 40 watt each are connected in series. The power consumed by the combination will be :

(A) 20 watt

(B) 60 watt

(C) 80 watt

(D) 100 watt

Ans : (A) 20 watt

P = 1 ÷ [1/P1 + 1/P2] = 1 ÷ [1/40 + 1/40] = 1 ÷ [2/40] = 1 ÷ [1/20] = 20 watts

(vi) A particle of mass m is moving with a velocity v. Write down the formula for the de-Broglie wavelength associated with the particle.

Ans : λ = h/p = h/(mv)

(vii) What are iron loss in a Transformer ?

Ans : Iron loss occurs in the magnetic core of the transformer due to flow of alternating magnetic flux through it. For this reason, the iron loss is also called core loss. We generally use the symbol (Pi) to represent the iron loss. The iron loss consists of hysteresis loss (Ph) and eddy current loss (Pe).

Pi = Ph + Pe

(viii) Is the Junction diode D is forward or reverse biased in the given diagram ?

Ans : Forward biased

(ix) The power of a lens is +5D. Write, the focal length.

Ans : F = 1/P = 1/5 = 0.2 m = 20 cm

(x) Calculate the ratio of the radii of two nuclei of mass numbers 27 and 64 respectively.

Ans : Using, R = R₀A^⅓

R1 : R2 = [R₀(27)^⅓] : [R₀(64)^⅓]

R1 : R2 = 3 : 4

(xi) The electric field due to an electric dipole is ………….. symmetric.

Ans : Cylindrically

(xii) The electromagnetic waves of frequency range from 5 × 10⁵ Hz to 10⁹ Hz are called …………

Ans : Radiowave

(xiii) The conductance of a wire of resistance 2 µΩ is …………..

Ans : C = 1/R = 1/2µΩ = 1/(2×10-⁶) = 5 × 10⁵ = 500000 siemes (S) or mho (Ω-¹)

(xiv) The peak value of 220 V a. c. is …………..

Ans : E₀ = √2 × Erms = 1.414 × 220 = 311 V

(xv) The power factor of an a. c. circuit is given by cosΦ = ………….

Ans : cosΦ = R/Z

Q2. What is Gauss’s law for electric flux ?

Ans – Electric flux (ϕ) is a measure of the electric field passing through a given area. It is defined as the dot product of the electric field vector and the area vector. Electric flux is a scalar quantity.

Gauss theorem states that the total electric flux (ϕ) through any closed surface (S) in free space is equal to 1/εo times the total electric charge (q) enclosed by the surface. i.e. ϕ = ∮ E.dS = q/εo

Q3. Calculate the capacitance of the capacitor C in diagram. The equivalent capacitance of the combination between P and Q is 30 µF.

Ans : In parallel, Cp = C1 + C2 + C3 = 20 + 20 + 20 = 60 µF

In series, 1/Cs = 1/C + 1/60

1/C = 1/Cs – 1/60 = 1/30 – 1/60 = 1/60

C = 60 µF

Q4. Explain Kirchhoff’s Junction rule.

Ans – Kirchhoff’s Junction rule, also known as Kirchhoff’s first law or the law of conservation of charge, states that the total current flowing into a junction in a circuit must be equal to the total current flowing out of the junction. In other words, the algebraic sum of currents at any junction in a circuit is zero.

Q5. Define the term angle of dip and declination.

Ans : Angle of dip – The angle made by the earth’s total magnetic field (B) with the horizontal direction in the magnetic meridian is called angle of dip (δ) at any place.

Angle of declination – The angle between the geographical meridian and the magnetic meridian at a place is called the magnetic declination at that place.

Q6. Write two uses each of microwaves and Gamma rays.

Ans – Microwaves are used for satellite communication and in radar system for aircraft navigation. They are also used in microwave ovens for cooking/warming food.

Gamma rays are used in medical science to kill cancer cells. They are also used in Sterilize medical equipment and Sterilize food (irradiated food).

Q7. Define Total Internal Reflection.

Ans – Total internal reflection is a phenomenon of reflection of ray back to the same medium when passing from denser medium to rarer medium in a such away that angle of incidence greater than its critical angle.

Q8. Write the relation between kinetic energy of α-Particle and its distance of closest approach from nucleus in Rutherford’s α-scattering experiment.

Ans : Consider α-particle, mass = m, velocity = v

Kinetic Energy (Eα) = ½mv² ………(i)

If ro = closest distance of α-particle from nucleus

Potential Energy (Ep) = 1/(4πεo) × (2Z×2e)/ro = 2Ze²/(4πεoro) ……….(ii)

At equillibrium, Eα = Ep

½mv² = 2Ze²/(4πεoro)

Eα ∝ 1/ro

This equation shows that the kinetic energy of the α-particle is inversely proportional to the distance of closest approach.

Q9. Draw the circuit diagram of a full wave rectifier using two p-n Junction diode. Draw input, output wave form also.

Ans – A full wave rectifier is a circuit that converts an alternating current (AC) input signal into a direct current (DC) output signal.

Full wave rectifier –

Input and output wave form –

Q10. Find the expression for electric field intensity E due to uniformly charged spherical shell at a point inside the shell.

Ans – Point Inside the shell :

In this case, we select a gaussian surface concentric with the shell of radius r (r < R).

so, ∮E.ds = E(4πr²)

According to gauss law,

E(4πr²) = (Qend)/εo

Since charge enclosed inside the spherical shell is zero.

so, E = 0

Hence, the electric field due to a uniformly charged spherical shell is zero at all points inside the shell.

Q11. Explain the principle of potentiometer. How is potentiometer superior to that of a voltmeter ?

Ans : Principle – The potential drop across any section of the wire will be directly proportional to the length of the wire, if the wire has uniform cross-sectional area and a uniform current flows through the wire.

The potentiometer does not draw any current while measuring emf whereas the voltmeter draws current to measure emf.

Q12. Define current sensitivity and voltage sensitivity of a moving coil galvanometer. How can a galvanometer be made more sensitive ?

Ans : Current Sensitivity – The current sensitivity of a moving coil galvanometer refers to the deflection of the galvanometer’s pointer in response to a given amount of current flowing through it. It is a measure of the galvanometer’s ability to indicate small changes in current.

Voltage Sensitivity – The voltage sensitivity of a moving coil galvanometer refers to the deflection of the galvanometer’s pointer in response to a given amount of voltage applied across it. It is a measure of the galvanometer’s ability to indicate small changes in voltage.

A galvanometer can be made more sensitive by increasing the number of turns in the coil, reducing the resistance of the coil, increasing the strength of the magnetic field of the galvanometer.

Q13. State Biot-Savart’s Law. Write its Mathematical form.

Ans – Biot-Savart’s Law describes the magnetic field produced by a steady current flowing in a wire. It states that the magnetic field at a point in space is directly proportional to the current in the wire and inversely proportional to the square of the distance from the wire. Mathematically, Biot-Savart’s Law can be expressed as :

dB = μ/4π × IdLsinθ/r²

B = ∫ dB = μ/4π × I(2πr)/r² = μI/2r

Q14. What are Eddy currents? Write any two applications of Eddy currents.

Ans – Eddy currents are loops of electrical current induced within conductors by a changing magnetic field in the conductor according to Faraday’s law of induction. Eddy currents flow in closed loops within conductors, in planes perpendicular to the magnetic field.

Applications of Eddy currents – They are useful in magnetic brakes in trains, induction furnaces and electromagnetic damping.

Q15. A resistor of 12 ohm, a capacitor of reactance 14 ohm and an inductor of reactance 30 ohm are joined in series and placed across a 230 V, 50 Hz supply.

Calculate :

(i) Impedance

(ii) Current in the circuit

Ans – (i) Impedance = √(R²+(XL-XC)² , where R is the resistance, XL is the inductive reactance and XC is the capacitive reactance.

Plugging in the values, we get:

Impedance = √(12²+(30-14)²) = √(144+256) = √400 = 20 ohm

(ii) Current in the circuit = Voltage/Impedance

Plugging in the values, we get:

Current = 230/20 = 11.5 A

Q16. Derive mathematically the conditions for constructive and destructive interference at an arbitrary point due to coherent sources in terms of phase difference Φ.

Ans – The sources and the waves are said to be coherent if they have the same frequencies, same wavelength, same speed, almost same amplitude and have no phase difference or a constant phase difference.

The conditions for constructive and destructive interference at an arbitrary point due to coherent sources can be derived mathematically in terms of the phase difference Φ as follows :

Constructive Interference: Φ = 2πn, where n is an integer.

Destructive Interference: Φ = (2n+1)π, where n is an integer.

Q17. Using Einstein’s photoelectric equation, explain effect of frequency of incident radiation on stopping potential.

Ans – Einstein’s photoelectric equation describes the relationship between the frequency of the incident radiation and the energy of the emitted electrons.

The equation is given by, E = hf – φ, where E is the energy of the emitted electron, h is Planck’s constant, f is the frequency of the incident radiation, and φ is the work function of the material.

According to Einstein’s photoelectric equation, the frequency of the incident radiation has a direct impact on the stopping potential. For frequencies below the threshold frequency, the stopping potential is zero. At the threshold frequency, the stopping potential becomes positive. Above the threshold frequency, the stopping potential becomes negative, and its magnitude increases with increasing frequency.

Q18. Write Huygen’s principle of secondary wavelets, and use it to explain law of reflection of light.

Ans – According to Huygen’s principle, each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave.

Let, PQ be reflecting surface and a plane wavefront AB is moving through the medium (air) towards the surface PQ to meet at the point B.

Let, c be the velocity of light and t be the time taken by the wave to reach A’ from A. Then, AA’ = ct

Using Huygen’s principle, secondary wavelets starts from B and cover a distance ct in time t and reaches B’. To obtain new wavefront, draw a circle with point B as centre and ct (AA’ = BB’) as radius. Draw a tangent A’B’ from the point A’. Then, A’B’ represents the reflected wavelets travelling at right angle. Therefore, incident wavefront AB, reflected wavefront A’B’ and normal lies in the same plane.

Consider ∆ABA’ and ∆B’BA’

AA’ = BB’ = ct [AA’ = BB’ = radii of same circle]

BA’ = BA’ [common]

∠BAA’ = ∠BB’A’ [each 90°]

∆ABA’ ≅ ∆B’BA’ [by R.H.S]

∠ABA’ = ∠B’A’B [corresponding parts of congruent triangles]

incident angle i = reflected angle r

i.e. ∠i = ∠r

OR

Draw the ray diagram of a compound microscope and obtain an expression for the magnifying power of microscope, when final image is formed at the least distance of distinct vision.

Ans –

Magnifying Power – Magnifying power of a compound microscope is defined as, the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the image and object are situated at the least distance of distinct vision from the eye.

Q19. Define decay constant and half life and derive a relation between them.

Ans : Decay constant – The decay constant of a radioactive element is defined as the reciprocal of time in which the number of undecayed nuclei of that radioactive element fails to 1/e times of its initial time.

Half-life period – The half-life period of an element is defined as the time in which the number of radioactive nuclei decay to half of its initial value.

The relation between the half-life period (T½) and decay constant (λ) is given by the equation :

T½ = 0.693/λ

OR

Explain the laws of radioactive decay. Derive an equation of radioactive decay using these laws.

Ans – It states that “For a particular time, the rate of radioactive decay of an atom is directly proportional to the number of nuclei of the elements present at that time.”

This can be written as, dN/dt ∝ N, where N is the number of nuclei. It further expressed as dN/dt = -λN, where λ is the proportionality constant.

Q20. With the help of necessary circuit diagram, describe briefly how p-n-p transistor in CE configuration amplifies a small sinusoidal input voltage.

Ans –

In a p-n-p transistor in common emitter (CE) configuration, the base is connected to the input signal source, the collector is connected to the output load, and the emitter is grounded. When a small sinusoidal input voltage is applied to the base, it causes a small current to flow through the base-emitter junction. This current causes a larger current to flow from the collector to the emitter through the collector-emitter junction, amplifying the input signal.

The amplification occurs because the input signal causes a change in the amount of current flowing through the transistor, which in turn causes a larger change in the output voltage. The amount of amplification is determined by the gain of the transistor, which is the ratio of the change in output voltage to the change in input voltage.

OR

Draw the circuit diagram for studying the characteristics of an n-p-n transistor in common emitter configuration. Sketch the typical (i) Input and (ii) Output characteristics in this configuration. With the help of these characteristics define input resistance.

Ans –

(i) The Input characteristics : The input characteristics of a transistor in common emitter configuration show the relationship between the base current (IB) and the base-emitter voltage (VBE) while keeping the collector-emitter voltage (VCE) constant. It is represented by a plot of IB on the y-axis and VBE on the x-axis.

(ii) The Output characteristics : The output characteristics show the relationship between the collector current (IC) and the collector-emitter voltage (VCE) while keeping the base current (IB) constant. It is represented by a plot of IC on the y-axis and VCE on the x-axis.

Input resistance (ri): The input resistance is defined as the ratio of the change in base-emitter voltage (ΔVBE) to the change in base current (ΔIB) when the collector-emitter voltage (VCE) is constant. It can be calculated using the formula ri = ΔVBE / ΔIB. The input resistance represents the ease with which the transistor accepts the input signal.

**SET–D**

Q1.(i) At a given place the horizontal and vertical components of earth are equal. The angle of dip at that place will be :

(A) 0°

(B) 45°

(C) 60°

(D) 90°

Ans : (B) 45°

(ii) Two light waves of equal amplitude and wave length are super imposed. The amplitude of the resultant wave will be maximum when the phase difference between them is :

(A) Zero

(B) π/2

(C) π

(D) π/4

Ans : (A) Zero

(iii) If E and B represents electric field and magnetic field vectors of the electromagnetic wave, the direction of propagation of electromagnetic wave is along :

(A) E

(B) B

(C) B × E

(D)

Note : use vector sign (→) on E & B.

Ans : (D)

(iv) Which of the following has minimum stopping potential ?

(A) Blue

(B) Yellow

(C) Voilet

(D) Red

Ans : (D) Red

(v) The best instrument for accurate measurement of emf of a cell is :

(A) Wheatstone bridge

(B) Ammeter

(C) A Potentiometer

(D) Voltmeter

Ans : (C) A Potentiometer

(vi) The mass of a charged particle is m and has charge q. If this particle is accelerated by a potential difference of V, then write the formula for the de-Broglie wavelength associated with it.

Ans : λ = h ÷ √(2mqV) or λ = h/√(2mqV)

(vii) Calculate the ratio of the radii of two nuclei of mass numbers 1 and 27 respectively.

Ans : Using, R = R₀A^⅓

R1 : R2 = [R₀(1)^⅓] : [R₀(27)^⅓]

R1 : R2 = 1 : 3

(viii) If A = 1 and B = 0, find the value of Y in the adjoining Logic circuit :

Ans : Y = A + B = 1 + 0 = 1 (This is OR Gate)

(ix) The distance between two charged particles is halved. What will be the effect on the force between them ?

Ans : Four times

F = kq1q2/r²

F’ = kq1q2/(r/2)² = 4 × kq1q2/r² = 4 F

(x) On which factors does the capacitance of a capacitor depend ?

Ans : The capacitance of a capacitor depends on area of each plate, dielectric medium between the plates and distance between the plates.

(xi) The fall of potential per unit length of wire is called …………

Ans : Potential Gradient

(xii) If ω is angular frequency of a. c., then the reactance offered by Inductance L and Capacitance C are XL = ……….. and XC = ………

Ans : XL = ωL and XC = 1/(ωC)

(xiii) An a. c. generator is based on the phenomenon of ………….

Ans : Electromagnetic induction

(xiv) The electromagnetic wave of frequency range from 5 × 10⁵ Hz to 10⁹ Hz are called ………….

Ans : Radiowaves

(xv) For total internal reflection, light must travel …………. to …………

Ans : Denser to Rarer

Q2. Draw circuit diagram of a wheatstone bridge and write the expression for the balance condition.

Ans – In balanced condition, Ig = 0. Expression for balance condition is P/Q = R/S

Q3. What do you understand by Eddy currents ?

Ans – Eddy currents are loops of electrical current induced within conductors by a changing magnetic field in the conductor according to Faraday’s law of induction. Eddy currents flow in closed loops within conductors, in planes perpendicular to the magnetic field.

Q4. Write one use of UV-rays and X-rays.

Ans – UV (Ultraviolet) rays destroy germs and bacteria and hence they are used for sterilizing surgical instruments and for purification of water.

X-rays are used to check the bags at the airport, identify the fracture, and help in identifying the infection in the bones, teeth, etc.

Q5. What do you mean by polarization of light? What is Brewster Law ?

Ans – Polarisation of light is a property shown by transverse waves. The light waves which travel only in a single plane are known as polarized light waves. The process of transforming unpolarized light waves to polarized light waves is called the polarisation of light.

According to Brewster’s law, When an unpolarized light of known wavelength is incident on a transparent substance surface, it experiences maximum plan polarization at the angle of incidence whose tangent is the refractive index of the substance for the wavelength.

Q6. What do you mean by mass defect of nucleus? How mass defect related to binding energy of nucleus ?

Ans – The mass defect of a nucleus is the difference between the mass of an atom’s nucleus and the sum of the masses of its individual protons and neutrons. This difference is due to the energy released when the protons and neutrons come together to form the nucleus. This energy is known as the binding energy of the nucleus.

The binding energy of a nucleus is related to its mass defect through Einstein’s famous equation, E = mc². This equation states that energy (E) and mass (m) are interchangeable, and that the amount of energy released (or absorbed) in a nuclear reaction is directly proportional to the difference in mass (Δm) between the reactants and products. The binding energy of a nucleus is therefore equal to the mass defect multiplied by the speed of light squared (c²).

Q7. What do you mean by Nuclear Fusion? Give one example of it.

Ans – Nuclear fusion is a process in which two or more atomic nuclei combine to form a heavier nucleus, releasing a large amount of energy in the process. This is the process that powers stars and is the basis for hydrogen bombs.

An example of nuclear fusion is the reaction that occurs in the Sun, where hydrogen nuclei (protons) combine to form helium nuclei. This reaction releases a tremendous amount of energy in the form of light and heat, which is what makes the Sun shine and provides energy for life on Earth.

Q8. Draw the circuit diagram of a half-wave rectifier using two p-n Junction diode. Show wave forms of input and output voltages also.

Ans –

Q9. What is a Light Emitting Diode (LED) ? Mention two important advantages of LED’s over conventional lamps.

Ans – A Light Emitting Diode (LED) is a semiconductor device that emits light when an electric current is passed through it.

LEDs are commonly used in lighting applications as they are energy-efficient, long-lasting, and produce less heat than conventional lamps.

Q10. Derive an expression for electric field intensity E near a thin uniformly charged infinite plane sheet.

Ans – Consider an infinite thin plane sheet of positive charge with a uniform charge density σ on both sides of the sheet. Let a point be at a distance a from the sheet at which the elctric field is required.

The gaussian cylinder is of area of cross section A.

Electric flux crossing the gaussian surface,

ϕ = E × (area of cross section of the gaussian cylinder)

Here, electric lines of force are parallel to the curves surface area of the cylinder, the flux due to the electric field of the plane sheet of charge passes only through two circular sections of the cylinder

ϕ = E × 2A

The charge enclosed within the Gaussian surface is given by :

Q = σ × A

Applying Gauss’s law, we have:

Φ = Q/εo , where εo is the permittivity of free space.

Substituting the values of Φ, Q, and A, we get :

E × 2A = σA/εo

E = σ/(2εo)

The direction of electric field for positive charge is in the outward and perpendicular to the plane of infinite sheet.

Q11. What is capacitor? Derive an expression for the capacitance of a parallel plate capacitor in which a dielectric medium of dielectric constant K fills the space detween the plates.

Ans – A capacitor is a two-terminal electrical device that can store energy in the form of an electric charge. It consists of two electrical conductors that are separated by a distance. The space between the conductors may be filled by vacuum or with an insulating material known as a dielectric.

To derive the expression for the capacitance of a parallel plate capacitor with a dielectric medium, we consider the basic equation for capacitance :

C = Q/V, where C is the capacitance, Q is the charge on one of the plates, and V is the potential difference between the plates.

Electric field between plates of the capacitor can be found using Gauss Law.

E = σ/Kεo = Q/(KεoA) , here σ = Q/A

Electric field is defined as the gradient of potential.

E = V/d

V = Ed = Qd/(KεoA)

By definition of capacitance,

C = Q/V = KεoA/d

Q12. State the principle of potentiometer. Draw a circuit diagram used to compare the emf’s of two primary cells. How can the sensitivity of a potentiometer be increased ?

Ans : Principle – The potential drop across any section of the wire will be directly proportional to the length of the wire, if the wire has uniform cross-sectional area and a uniform current flows through the wire.

Compare the emf’s of two primary cells :

Here AJ1 = l1 and AJ2 = l2

so, E1/E2 = l1/l2

Sensitivity of the potentiometer is increased when increasing the length of the potentiometer wire.

Q13. Derive an expression for the force between two parallel straight conductors carrying currents in the same direction. Hence define one ampere.

Ans –

Consider a small length L of the long straight conductor

B1 = μI1/(2πd)

B2 = μI2/(2πd)

F12 = I2B1L

F21 = I1B2L

F = F12 = F21 = μI1I2L/(2πd)

F/L = μI1I2/(2πd)

One Ampere is defined as the amount of electric current that flows through two parallel conductors of infinite length, which are placed one meter apart in a vacuum, that results in a force of exactly 2 x 10-⁷ N/m.

Q14. What is interference of light waves? Write formula of fringe width.

Ans – Interference of light waves is the phenomenon that occurs when two or more waves of light interact with each other, resulting in a pattern of bright and dark fringes known as interference fringes. This occurs due to the constructive and destructive interference of the waves, which can either reinforce or cancel each other out.

The expression of fringe width in interference is β = λD/d

where β is the fringe width, λ is the wavelength of light, D is the distance between the light source and the screen, and d is the distance between the two slits or sources of light.

Q15. Explain effect of potential on photoelectric current.

Ans – The potential difference applied across the material can affect the photoelectric current in two ways :

(i) If the potential difference is too low, it may not be able to overcome the work function of the material, which is the minimum amount of energy required to remove an electron from the material. In this case, no electrons will be emitted and there will be no photoelectric current.

(ii) If the potential difference is high enough to overcome the work function, it can accelerate the emitted electrons and increase the photoelectric current. This is because the potential difference provides an electric field that can pull the electrons towards the collector electrode, increasing their speed and therefore the current.

Therefore, the potential difference plays a crucial role in determining whether or not electrons are emitted and how fast they move, ultimately affecting the photoelectric current.

Q16. What do you mean by half life and decay constant of radioactive substance and write the relation between them ?

Ans : Half life period – The half-life period of an element is defined as the time in which the number of radioactive nuclei decay to half of its initial value.

Decay constant – The decay constant of a radioactive element is defined as the reciprocal of time in which the number of undecayed nuclei of that radioactive element fails to 1/e times of its initial time.

The relation between the half-life period (T½) and decay constant (λ) is given by the equation :

T½ = 0.693/λ

Q17. Write the names of two important processes formation of p-n Junction. Define the depletion region and potential barrier in it.

Ans – Diffusion and Drift are the two processes that take place in the formation of a p-n junction.

Depletion Region – It is the space charge region on either side of the junction, which got depleted of free charges.

Potential barrier – The potential difference developed across the junction and opposes the diffusion of charge and brings equilibrium condition is known as the potential barrier.

Q18. Draw the ray diagram of a compound microscope and obtain an expression for the magnifying power of the microscope, when final image is formed at the least distance of distinct vision.

Ans –

OR

In Young’s Double slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes.

Ans – To deduce the conditions for obtaining constructive and destructive interference fringes, we need to consider the path difference (Δd) between the two interfering waves :

(i) For constructive interference : Constructive interference occurs when the path difference (Δd) is an integral multiple of the wavelength (λ) of the light. Mathematically, this can be expressed as Δd = mλ, where m is an integer (0, 1, 2, 3…).

This condition results in the two waves arriving at the same phase at a specific point on the screen, leading to constructive interference and resulting in a bright fringe.

(ii) For destructive interference : Destructive interference occurs when the path difference (Δd) is an odd multiple of half-wavelength (λ/2). Mathematically, this can be expressed as Δd = (2m+1)(λ/2), where m is an integer (0, 1, 2, 3…).

This condition results in the two waves arriving at opposite phases at a specific point on the screen, leading to destructive interference and resulting in a dark fringe.

Q19. Explain the principle of a moving coil galvanometer. How can it be converted into a voltmeter? Draw necessary diagram.

Ans – A moving coil galvanometer is an instrument which is used to measure electric currents. It is a sensitive electromagnetic device which can measure low currents even of the order of a few microamperes.

Principle – When a current flows through the coil, a torque acts on it. The magnetic torque tends to rotate the coil. Spring provides a counter torque that balances the magnetic torque resulting in a steady angular deflection.

Convert Galvanometer to Voltmeter – A galvanometer is converted into a voltmeter by connecting a high resistance in series with it.

OR

State the principle of cyclotron. Explain its working with the help of diagram.

Ans : Cyclotron – Cyclotron is a device used to accelerate charged particles to high energies or speed.

Principle – Cyclotron works on the principle that a charged particle moving normal to a magnetic field experiences magnetic lorentz force due to which the particle moves in a circular path.

Working – When a positive ion of charge q and mass m is emitted from the source, it is accelerated towards the Dee having a negative potential at that instant of time. Due to the normal magnetic field, the ion experiences magnetic lorentz force and moves in a circular path. By the time the ion arrives at the gap between the Dees, the polarity of the Dees gets reversed.

Hence the particles is once again accelerated and moves into the other Dee with a greater velocity along a circle of greater radius. Thus the particle moves in a spiral path of increasing radius and when it comes near the edge, it is taken out with the help of a deflection plate (D.P.). The particle with high energy is now allowed to hit the target T.

When the particle moves along a circle of radius r with a velocity v, the magnetic Lorentz force provides the necessary centripetal force.

Bqv = mv²/r

v = qBr/m

Period of revolution, T = 2πr/v

T = 2πrm/qBr = 2πm/qB

Q20.(a) Define self induction and coefficient of self inductance.

Ans – Self induction refers to the phenomena in which an electromagnetic field is produced in a coil when the current flowing through it changes. It is the property of an electric circuit in which a changing current in the coil causes a voltage to be induced in the same coil.

Coefficient of self inductance (L) is a measure of the ability of a coil to produce an electromagnetic field due to self induction. It is defined as the ratio of induced electromotive force (EMF) to the rate of change of current in the coil. The unit of coefficient of self inductance is henry (H).

(b) What e. m. f. will be induced in a 10 H inductor in which current changes from 10 A to 7 A in 9 x 10-² s ?

Ans – Given,

Inductance (L) = 10 H

Initial current (I1) = 10 A

Final current (I2) = 7 A

Time taken (t) = 9 × 10-² s = 0.09 s

The rate of change of current (di/dt) can be calculated as : di/dt = (I2-I1)/t = (7-10)/0.09 = -33.33

e. m. f = -L × di/dt = -10 × (-33.33)

e. m. f = 333.3 V

OR

Explain the principle and working of a Transformer.

Ans : Transformer – An electrical device that can change the alternate current or voltage is known as a transformer.

Principle – A transformer works on the principle of mutual induction. Mutual induction is the phenomenon by which when the amount of magnetic flux linked with a coil changes, an E.M.F. is induced in the neighboring coil.

Working – The transformer works on the principle of Faraday’s law of electromagnetic induction and mutual induction. There are usually two coils primary coil and secondary coil on the transformer core. The core laminations are joined in the form of strips. The two coils have high mutual inductance. When an alternating current pass through the primary coil it creates a varying magnetic flux. As per faraday’s law of electromagnetic induction, this change in magnetic flux induces an emf (electromotive force) in the secondary coil which is linked to the core having a primary coil. This is mutual induction.