**HBSE Class 12 Physics Q****uestion Paper 2021 Answer Key **

HBSE Class 12 Physics Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 12 Physics 2021. HBSE 12th Question Paper Download 2021. HBSE Class 12 Physics Paper Solution 2021. Haryana Board Class 12th Physics Question Paper 2021 Pdf Download with Answer.

Subjective Questions

Q1. State Coulomb’s Law in Electrostatics.

Ans. The force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

F = k × q1 × q2 / r²

where k is coulomb’s constant = 9×10⁹ Nm²C-²

Q2. Obtain the resonant frequency ω of a series LCR circuit with L = 2.0 H, C = 32 µF and R = 10 Ω.

Ans. Resonant frequency (ω) = 1/√(LC)

ω = 1/√(2×32×10-⁶) = 1/(8×10-³)

ω = 125 Hz or 125 rad/s

Q3. What are electromagnetic waves ?

Ans. Electromagnetic waves are also known as EM waves. Electromagnetic radiations are composed of electromagnetic waves that are produced when an electric field comes in contact with the magnetic field. It can also be said that electromagnetic waves are the composition of oscillating electric and magnetic fields. Electromagnetic wave is a transverse wave. They are non-mechanical wave and do not require any medium for propagation.

Q4. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle Lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive Index of water ?

Ans. Refractive Index (μ) = Real Depth ÷ Apparent Depth

μ = 12.5/9.4 = 1.33

Q5. Write Einstein’s Photoelectric Equation. Draw a graph between the stopping potential and Frequency of the Incident light.

Ans. Einstein’s photoelectric equation,

Kmax = hc/λ – φ = hν – φ

Graph of stopping potential vs frequency of incident light

Q6. Obtain an expression for the Intensity of Electric field near a uniformly charged straight wire of Infinite Length, with the help of Gauss’s theorem in Electrostatics.

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

Q7. State Kirchoff’s Laws for electrical circuits giving necessary circuit diagram.

Ans. Kirchhoff’s current law states that the total current flowing into a node or junction in an electric circuit must be equal to the total current flowing out. It is also known as the junction law. Kirchhoff’s Voltage Law states that the algebraic sum of all the voltages in a given circuit will be equal to zero.

Q8. What do you mean by eddy currents ? How it can be produced. Write the name of applications of eddy currents.

Ans. Eddy currents are the currents induced in a metallic plate when it is kept in a time varying magnetic field. Magnetic flux linked with the plate changes and so the induced current is set up. Eddy currents are sometimes so strong, that metallic plate become red hot.

When a metallic cylinder (or rotor) is placed in a rotating magnetic field , eddy currents are produced in it. According to Lenz’s law, these currents tends to reduce to relative motion between the cylinder and the field.

Strong eddy currents produce so much heat that the metal melts. We use this in metal extraction from the ore. The eddy currents used to rotate the rotor.

Q9. Draw the energy Level diagram Fer hydrogen atom and show transitions corresponding to Lines of Lyman and Balmer Series.

Ans. When an electron makes a transition from one orbit to the other in an atom, it gains or loses energy due to which it emits wavelengths corresponding to the energy differences. These wavelengths are observed as lines called spectral lines. Different wavelengths are emitted for different transitions.

Q10. Draw a circuit diagram of a HALF WAVE RECTIFIER using a p-n junction diode. Show waveforms of input and output voltages.

Ans. The circuit diagram for a p-n junction diode as a HALF WAVE RECTIFIER is shown :

Working : During the positive half cycle of the input a.c., the p-n junction is forward biased i.e., the forward current flows from p to n. In the forward biasing, the diode provides a very low resistance and allows the current to flow. Thus, we get output across load.

During the negative half cycle of the input a.c., the p-n junction is reversed biased. In the reverse biasing, the diode provides a high resistance and hence a very small amount of current will flow through the diode which is of negligible amount. Thus no output is obtained across the load. During the next half cycle, output is again obtained as the junction diode gets forward biased. Thus, a half wave rectifier gives discontinuous and pulsating d.c. output across the load resistance.

The waveform for input and output voltage is shown :

Q11. Derive an expression for the force between two Long straight parallel conductors carrying current in opposite direction. Hence define one Ampere.

Ans. Placed two long thin straight conductors parallel to each in vacuum carrying current.

Force per unit length between two long straight parallel conductors. Suppose two thin straight conductors (or wires) PQ and RS are placed parallel to each other in vacuum (or air) carrying current I1 and I2 respectively. It has been observed experimentally that when the currents in the wire in the same direction, they experience an attractive force and when they carry current in opposite directions, they experience a repulsive force. Let the conductors PQ and RS carry currents I1 and I2 in the same direction and placed at separation r.

B1 = (μ I1)/2πr

ΔF = B1 I1 ΔL sin90°

B1 = (μ I1 I2 ΔL)/2πr

F = (μ I1 I2)/2πr × ∑ΔL = (μ I1 I2)/2πr × L

F/L = (μ I1 I2)/2πr N/m

Consider a current-element ‘ab’ of length ΔL of wire RS. The magnetic field produced by current carrying conductor PQ at the location of other wire RS.

According to Maxwell’s right hand rule or right hand Palm rule no. 1, the direction of B1 will be perpendicular to the plane of paper and directed downward. Due to this magnetic field, each element of another wire experiences a force. The direction of the current element is perpendicular to the magnetic field; therefore the magnetic force on element AB of length ΔL.

Definition of ampere : Fundamental unit of current ampere has been defined assuming the force between the two current carrying wires as standard. The force between two parallel current carrying conductor of separation r is 2×10-⁷ N/m.

Thus 1 ampere is the current which when flowing in each of parallel conductors placed at separation 1 m in vacuum exert a force of 2×10-⁷ N/m on 1 m length of either wire.

OR

A circuit Loop of radius R carries a current I. Obtain an expression for the magnetic field at a point on its axis at a distance X from its centre.

Ans. We have a ring of radius R carrying current I as shown in the following figure.

We will take an element on ring of length dl, r is the position vector of point P from the element dl. In the front view, we can see that horizontal component will cancel each other. Only vertical will add up and will give net magnetic field.

Using Biot-savart law :

Magnetic field at point P due to element dl =

dB = (μI/4π) × (dl/r²)

Vertical component = (dB)sinα

Net magnetic field = BNet = ∫(dB) sinα

BNet = ∫(μI/4π) × (dl/r²) sinα = μI/4π ∫R/r × 1/r² dl

BNet = μIR/4πr³ ∫dl

BNet = μIR/4πr³ × 2πR = μIR²/2r³

Along axis | r | = r = √R²+x²

BNet = μIR²/4πr³ ÷ 2(√R²+x²)³

Magnetic field at a point on its axis at a distance x from its centre.