**Haryana (HBSE)** Class 12 Physics Question Paper 2022 Answer Key. HBSE Class 12 Physics Solved Question Paper 2022. HBSE Class 12 Physics Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 12 Physics 2022. HBSE 12th Question Paper Download 2022 Answer Key. HBSE Class 12 Physics Paper Solution 2022. HBSE Class 12th Physics Question Paper 2022 PDF Download with Answer Key.

**HBSE Class 12 Physics Question Paper 2022 Answer Key**

Q1. A tank is filled with water to a height of 15 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 10 cm. What is refractive index of water ?

Ans. Refractive index (μ) = Real Depth/Apparent Depth

μ = 15/10 = 1.5

Q2. Write any two characteristics of Electromagnetic Waves.

Ans. Electromagnetic waves are transverse in nature as they propagate by varying the electric and magnetic fields such that the two fields are perpendicular to each other.

Electromagnetic wave propagation it does not require any material medium to travel.

Q3. Obtain the resonant frequency ω of a series LCR circuit with L = 5.0 H, C = 80 μF and R= 40 Ω.

Ans. Resonant frequency (ω) = 1/√(LC)

ω = 1/√(5×80×10-⁶) = 1/(20×10-³) = 1000/20

ω = 50 Hz or 50 rad/s

Q4. Define capacitance of a Capacitor.

Ans. The capacity of a capacitor to store the charge is called its Capacitance. The SI unit of capacitance is farad(F).

Q5. What is photoelectric effect ? Draw a graph between the frequency of incident light and stopping potential in photoelectric effect.

Ans. The photoelectric effect refers to the emission, or ejection of electrons from the surface of a metal in response to incident light. This takes place because of the energy of incident photons of light have energy more than the work potential of the metal surface, ejecting electrons with positive kinetic energy.

Graph of frequency of incident light vs stopping potential

Q6. Write Bohr’s postulates for hydrogen atom.

Ans. First Postulate– Electron revolves round the nucleus in discrete circular orbits called stationary orbits without emission of radiant energy. These orbits are called stable orbits or non-radiating orbits.

Second Postulate– Electrons revolve around the nucleus only in orbits in which their angular momentum is an integral multiple of h/2π.

Third Postulate– When an electron makes a transition from one of its non-radiating orbits to another of lower energy, a photon is emitted having energy equal to the energy difference between the two states. The frequency of the emitted photon is then given by, v = (Ei-Ef)/h.

Q7. State Wheatstone bridge principle for electrical circuits giving necessary circuit diagram.

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.

Q8. State and prove Gauss’s Law in Electrostatics.

Ans. According to Gauss’s theorem the net-outward normal electric flux through any closed surface of any shape is equivalent to 1/∈ times the total amount of charge contained within that surface.

Proof of Gauss’s Theorem–

Let’s say the charge is equal to q.

Let’s make a Gaussian sphere with radius = r

Now imagine surface A or area ds has a ds vector

At ds, the flux is:

dΦ = E(vector) ds(vector) cos θ

But , θ = 0

Hence , Total flux Φ = E × 4πr²

Here, E = 1/4π∈ × q/r²

Hence, Φ = 1/4π∈ × q/r² × 4πr²

Φ = q/∈

As per the Gauss law, the total flux associated with a sealed surface equals 1/∈ times the charge encompassed by the closed surface.

Q9. Draw labelled diagram of A.C. generator and write its principle.

Ans. Principle– In an electric generator, mechanical energy is used to rotate a conductor in a magnetic field to produce electricity. It is working on the principle of electromagnetic induction.

Working– When the axle attached to the two rings is rotated such that the arm AB moves up (and the arm CD moves down) in the magnetic field produced by the permanent magnet. Let us say the coil ABCD is rotated clockwise in the arrangement. By applying Fleming’s right hand rule, the induced currents are set up in these arms along the directions AB and CD. Thus an induced current flows in the direction ABCD. If there are larger number of turns in the coil, the current generated in each turn adds up to give a large current through the coil. This means that the current in the external circuit flows from B2 to B1. After half a rotation, arm CD starts moving up and AB moving down. As a result, the directions of the induced currents in both the arms changes, giving rise to the net induced current in the direction DCBA. The current in the external circuit now flows from B1 to B2. Thus after every half rotation the polarity of the current in the respective arms changes. There are two brushes and in the electric generator, one brush is at all times in contact with the arm moving up in the field, while the other is in contact with the arm moving down. Because of these Brushes unidirectional current is produced.

Function of brushes is to transfer the current from coil to load connected in the circuit of the electric generator.

Q10. Draw a circuit diagram of a Full Wave Rectifier using a P-N junction diode. Show waveforms of input and output voltages.

Ans. Rectifier is a circuit which convert ac to unidirectional pulsating output. In other words its convert ac to dc.

Full Wave Rectifier–

Waveforms of input and output voltages–

Q11. Draw a labelled ray diagram showing image formation in an astronomical telescope. Derive expression for its magnifying power.

Ans. Astronomical Telescope– Astronomical Telescope is used to observe objects which are very far from us. Telescopes produce magnified images of distant objects. It produces virtual and inverted image and is used to see heavenly bodies like sun, stars, planets etc. so the inverted image does not affect the observation.

Principle– It is based on the principle that when rays of light are made to incident on an objective from a distant object, the objective forms the real and inverted image at its focal plane. The eye lens is so adjusted that the final image is formed at least distance of distinct vision.

Construction– The refracting type astronomical telescope consists of two convex lenses one of which is called the objective and the other eye piece. The objective is a convex lens of large focal length and large aperture. It is generally a combination of two lenses in contact so as to reduce spherical and chromatic aberrations. The eye piece is also a convex lens but of short focal length and small aperture. The objective is mounted at one end of a brass tube and the eye piece at the other end in a smaller tube which can slide inside the bigger tube carrying the objective.

Magnifying Power– The magnifying power of a refracting type astronomical telescope is defined as the ratio of angle subtended by the final image at eye to the angle subtended by the object at eye.

M= fo/fe (1 + fe/D)

OR

What is Wavefront ? Using Huygen’s principle to verify the Laws of Refraction.

Ans. A wavefront is defined as the continuous locus of all the particles which are vibrating in the same phase. The perpendicular line drawn at any point on the wavefront represents the direction of propagation of the wave at that point. A wavefront is an imaginary surface over which an optical wave has a constant phase. The shape of a wavefront is usually determined by the geometry of the source.

In ∆AA’B,

sin i = A’B/AB …….(i)

In ∆AB’B,

sin r = AB’/AB ……..(ii)

Dividing eqn.(i) by eqn.(ii),

sin i/sin r = A’B/AB ÷ AB’/AB = A’B/AB’ = ct/vt

sin i/sin r = c/v = μ