HBSE Class 12th Physics Solved Question Paper 2022

HBSE Class 12th 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. HBSE Class 12 Physics Paper Solution 2022. Haryana Board Class 12th Physics Question Paper 2022 Pdf Download with Answer.

SET-A,B,C,D (Subjective Questions)

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)

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

This is the law of refraction.

Q12. Derive an expression for the force between two long straight parallel conductors carrying current in 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

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.

A circular 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.

SET-A (Objective Questions)

Q1. In a p-type semiconductor the minority charge carriers are :

(A) Electron

(B) Hole

(D) None of these

Ans. (A) Electron

Q2. At absolute zero temperature, a crystal of pure germanium behaves as :

(A) a perfect conductor

(B) a perfect insulator

(D) None of these

Ans. (B) a perfect insulator

Q3. Number and type of nucleons in the nucleus of Helium (2He⁴) will be :

(A) 2 protons

(B) 2 protons and 2 neutrons

(D) 2 neutrons

Ans. (B) 2 protons and 2 neutrons

Q4. In hydrogen atom, the potential energy of electron in an orbit of radius r is given by :

(A) -1/4π∈ × e²/r

(B) 1/4π∈ × e²/2r

(D) -1/4π∈ × e²/2r

Ans. (A) -1/4π∈ × e²/r

Q5. The de-Broglie wavelength associated with a moving particle is :

(A) directly proportional to its mass

(B) inversely proportional to its mass

(D) directly proportional to its momentum

Ans. (B) inversely proportional to its mass

Q6. Electron emission from a metallic surface is possible only, when wavelength of the incident light is :

(A) less than threshold wavelength

(B) twice of the threshold wavelength

(D) no effect of wavelength

Ans. (A) less than threshold wavelength

Q7. Speed of light in air is 3 x 10⁸ m/s. For the glass of refractive 1.5, the speed of light in glass will be :

(A) 1.5 x 10⁸ m/s

(B) 2 x 10⁸ m/s

(D) 2.5 x 10⁸ m/s

Ans. (B) 2 x 10⁸ m/s

v = c/μ = (3 × 10⁸) ÷ 1.5 = 3 × 10⁸ × 1/1.5

= 2 × 10⁸ m/s

Q8. The angle of minimum deviation for a prism is 30° and the angle of prism is 60°. The refractive index of the material of the prism is :

(A) 2

(B) √2

(D) 1/√2

Ans. (B) √2

μ = sin(A+δm)/2 ÷ sinA/2 = sin(60°+30°)/2 ÷ sin60°

= sin45° ÷ sin60° = 1/√2 ÷ 1/2 = √2

Q9. Double-convex lenses are to be manufactured from a glass of refractive index 1.5, with both faces of the same radius of curvature. If the focal length is to be 20 cm, radius of curvature required is :

(A) 10 cm

(B) 15 cm

(D) 30cm

Ans. (C) 20 cm

1/F = (μ – 1)[1/R1 – 1/R2] = (1.5 – 1)[1/R – (-1/R)]

= (0.5)[1/R + 1/R] = (0.5)[2/R] = 1/R

1/F = 1/R

F = R

R = F = 20 cm

Q10. Rainbow is a phenomena due to :

(A) Refraction

(B) Dispersion

(D) All of the above

Ans. (D) All of the above

Q11. Lenz’s law is a consequence of the law of conservation of :

(A) Charge

(B) Momentum

(D) Mass

Ans. (C) Energy

Q12. For a coil having self-inductance 3 mH current flows at a rate of 10³ ampere/sec in it. The emf induced in it is :

(A) 1 Volt

(B) 2 Volt

(D) 4 Volt

Ans. (C) 3 Volt

emf = – L.dI/dt = 3mH × 10³

= 3 × 10-³ × 10³ = 3 volt

Q13. The force (F) acting on a particle of charge q moving with velocity (v) in magnetic field (B) is :

(A) q/v×B

(B) v×B/q

(D) v×q×B

Ans. (C) q(v×B)

Q14. A proton enters into a uniform magnetic field perpendicularly to it. The path of the proton would be :

(A) Elliptical

(B) Circular

(D) Linear

Ans. (B) Circular

Q15. The graph between voltage (v) and current (i) for a conductor is a straight line which makes an angle θ with x-axis (representing i). The resistance of the conductor will be :

(A) tanθ

(B) cotθ

(D) cosθ

Ans. (A) tanθ

Ohm’s Law, V = IR

R = V/I = P/B = tanθ

Q16. “Ohm-metre” is unit of :

(A) Resistance

(B) Current density

(D) Conductivity

Ans. (C) Resistivity

Q17. Two charges +1µc and +8µc are situated at a distance in air. The ratio of forces acting on them is :

(A) 1 : 8

(B) 8 : 1

(D) 1 : 16

Ans. (C) 1 : 1

Force AB = 1/4π∈ × q1q2/r²

Force BA = 1/4π∈ × q1q2/r²

Force AB : Force BA = 1 : 1

Q18. Three capacitors of equal capacity C are joined first in parallel and then in series. The ratio of equivalent capacities in both the cases will be :

(A) 9 : 1

(B) 6 : 1

(D) 1 : 9

Ans. (A) 9 : 1

1/Cs = 1/C1 + 1/C2 + 1/C3 = 1/C + 1/C + 1/C

1/Cs = 3/C

Cs = C/3 (series)

Cp = C1 + C2 + C3 = C + C + C

Cp = 3C (parallel)

Cp : Cs = 3C : C/3 = 9 : 1

Q19. The unit of Resistance is …………

Ans. Ohm(Ω)

Q20. …………. is a device can be used to measure potential difference, internal resistance of cell and compare emf’s of two cells.

Ans. Potentiometer

Q21. When a metal rod of length l is placed normal to a uniform magnetic field B and moved with a velocity v perpendicular to the field, the induced emf (called motional emf) across its end is ………….

Ans. E = Blv

Q22. The vertical plane containing the longitudinal circle and the axis of rotation of the Earth is called the ………….. meridian.

Ans. Geographic

Q23. The number of charge carriers can be changed by ‘doping’ of a suitable Impurity in pure semiconductors. Such semiconductors are known as …………… semiconductors.

Ans. Extrinsic

Q24. ………… as a whole, is electrically neutral and therefore contains equal amount of positive and negative charges.

Ans. Atom

Q25. ………….. waves are produced by hot bodies and molecules.

Ans. infrared

Q26. The …………. angle for a ray incident from a denser to rarer medium, is that angle for which the angle of refraction is 90°.

Ans. Critical

Q27. The minimum energy needed by an electron to come out from a metal surface is called the …………. of the metal.

Ans. Work Function

Q28. Is the Junction diode D is forward or reverse biased, in the given diagram ?

Ans. Forward Biased

Q29. Two nuclei have mass numbers in the ratio 1 : 27. What is the ratio of nuclear density ?

Ans. 1 : 1 (nuclear density is same for all nuclei and nuclear density is independent of the mass number)

Q30. If the Young’s apparatus be immersed in water in place of air, then what will be the effect on the Fringe width ?

Ans. Decrease (β’ = β/µ)

Q31. Is the speed of light in glass independent of the colour of light ? (Yes/No)

Ans. No (speed of light in glass is depend on colour of light)

Q32. Two thin lenses of power +5 Dioptre and -3 Dioptre are placed in contact. Find the power of this combination.

Ans. P = P1 + P2 = + 5 – 3 = +2 D

Q33. Under what condition, is terminal voltage of a cell equal to its Electromotive Force ?

Ans. When cell is not connected in circuit, then its terminal voltage becomes equal to emf or when no current flow through circuit (means when cell is open circuit)

V = E – ir

V = E when i = 0

Q34. Write down the formula for the relation of current density (J) and drift velocity (Vd).

Ans. J = neVd

Q35. Consider the circuit shown where APB and AQB are semicircles. What will be the magnetic field at the centre O of the circle ?

Ans. Zero

SET-B (Objective Questions)

Q1. A positively charged body has in it :

(A) Excess of Neutrons

(B) Excess of Electrons

(D) Deficiency of Protons

Ans. (C) Deficiency of Electrons

Q2. Two plates are at 1 cm a part and potential difference between them is 10 volt. The Intensity of electric field between the plates is :

(A) 10 Newton/Coulomb

(B) 500 Newton/Coulomb

(D) 250 Newton/Coulomb

Ans. (C) 1000 Newton/Coulomb

V = 10 volt, d = 1 cm = 0.01 m

E = V/d = 10/0.01 = 10³ = 1000 N/C

Q3. The relation between current density (j) and drift velocity (Vd) is :

(A) j = neVd

(B) j = ne/Vd

(D) j= 1/neVd

Ans. (A) j = neVd

Q4. The length of a metallic wire of R ohm resistance is stretched two times its initial length. Its new resistance is :

(A) 2R

(B) 4R

(D) 8R

Ans. (B) 4R

Resistance of wire R = ρ × L/A = ρ × L/V × L/L = ρ × L²/V

R ∝ L² (Since V and ρ are constant)

New Resistance, R’ = (2L)² = 4L² = 4R

Q5. A moving charge produces :

(A) Electric field only

(B) Magnetic field only

(D) None of the fields

Ans. (C) Both Electric and Magnetic fields

Q6. Two magnetic lines of forces :

(A) Cut each other at the neutral point

(B) Cut each other near north or south pole

(D) Cut at the middle of magnet

Ans. (C) Never cut each other

Q7. The magnetic flux threading a wire loop placed in a magnetic field depend upon :

(A) Area of the loop

(B) Magnitude of the field

(D) All of the above

Ans. (D) All of the above

Q8. Direction of Induced Current is given by :

(A) Lenz’s Law

(B) Fleming’s Left Hand Rule

(D) Ampere’s Law

Ans. (A) Lenz’s Law

Q9. Two lens of power 5D and -3D are placed in contact. Focal length of the combination will be :

(A) 50 cm

(B) -50 cm

(D) -25cm

Ans. (A) 50 cm

P = P1 + P2 = + 5 – 3 = +2 D

F = 1/P = 1/2 m = 0.5 m = 50 cm

Q10. Which phenomena is used in optical fibre ?

(A) Total Internal Reflection

(B) Dispersion

(D) Interference

Ans. (A) Total Internal Reflection

Q11. Condition of minimum deviation for a Prism is :

(A) Angle of Incidence > Angle of Emergence

(B) Angle of Incidence = Angle of Emergence

(D) None of the above

Ans. (B) Angle of Incidence = Angle of Emergence

Q12. In Young’s double slit experiment, the screen is moved away from the plane of the slits, angular separation of the Fringes :

(A) remain constant

(B) increases

(D) interference pattern disappears

Ans. (B) increases

Q13. de-Broglie wavelength associated with an electron, accelerated through a potential difference of 100 Volt is :

(A) 1.127 Å

(B) 11.27 Å

(D) 1.227 Å

Ans. (D) 1.227 Å

λ = 1.227/√V nm = 0.1227 nm = 1.227 Å

Q14. Awarded the Nobel prize in physics for his contribution to theoretical physics and the photoelectric effect in 1921 was :

(A) Millikan

(B) Einstein

(D) Compton

Ans. (B) Einstein

Q15. All nuclides with same mass number A are called :

(A) isobars

(B) isotones

(D) All of the above

Ans. (A) isobars

Q16. The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV. Kinetic energy of the electron in this state is :

(A) -3.4 eV

(B) 6.8 eV

(D) 3.4 eV

Ans. (D) 3.4 eV

Q17. When a forward biased is applied to a p-n junction, it :

(A) Raises the potential barrier

(B) Reduces the majority carrier current to zero

(D) None of the above

Ans. (C) Lowers the potential barrier

Q18. Energy band of substance shown in the figure, where V is valence band and C is conduction band :

Substance is :

(A) Metal

(B) Semiconductor

(D) None of the above

Ans. (A) Metal

Q19. Electric field is in the direction in which the ………….. decreases steepest.

Ans. Potential

Q20. Conductivity arises from mobile charge carriers. In metals, these mobile charge carriers are …………

Ans. Free Electrons

Q21. The value of magnetic field at any point in the open space inside the toroid is …………..

Ans. B = µi = 0

Q22. SI unit of self-inductance is …………..

Ans. Henry

Q23. One common way to generate …………. is to bombard a metal target by high energy electrons.

Ans. X-rays

Q24. Snell experimentally obtained the laws of ………… of light.

Ans. Refraction

Q25. Energy of photon having frequency v is ………….

Ans. E = hv

Q26. In ……………., lighter nuclei combine to form a larger nucleus.

Ans. Nuclear Fusion

Q27. Two important process occur during the formation of a p-n junction : diffusion and …………..

Ans. Deplation Layer

Q28. What is the effect on relaxation time of electrons in metals, with increase in temperature.

Ans. Decrease

Q29. Two unequal resistance are connected in parallel. What is same for them ?

Ans. Potential Difference (Voltage)

Q30. What is the value of angle of dip at the Earth’s magnetic poles ?

Ans. Angle of dip, δ = 90°

Q31. A point source of light is placed at a distance of 20 cm from a convex lens of focal length 10 cm. Where should image formed on the other side of lens.

Ans. at 2F (real & inverted)

Q32. Monochromatic light is refracted from air into a medium of refractive index n. What is the ratio of the wavelengths of the incident and the refracted waves.

Ans. Wavelength of incident wave : Wavelength of refracted wave = λ1 : λ2 = n : 1

Q33. Two light waves of equal amplitude and wavelength are superimposed. What is the phase difference between the waves so that amplitude of the resultant wave will be maximum.

Ans. Φ = 0

Q34. The radius of the first electron-orbit in hydrogen atom of Bohr model is r. What will be the radius of the second orbit.

Ans. Radius of second orbit = 4r

Q35. In a semiconductor, the concentration of electrons is 8 x 10¹³ cm-³ and that of holes is 5 x 10¹² cm-³. Is it a p-type or n-type semiconductor ?

Ans. N-type (ne > nh)

SET-C (Objective Questions)

Q1. A negatively charged body has in it :

(A) Excess of Neutrons

(B) Excess of Electrons

(D) Excess of Protons

Ans. (B) Excess of Electrons

Q2. The electric potential energy of a system of two positive point charges of 1 µc each placed in air 1 metre a part is :

(A) 1 Joule

(B) 1 electron-volt

(D) Zero

Ans. (C) 9 x 10-³ Joule

q1 = q2 = 1 µc = 1×10-⁶ c = 10-⁶ c

Potential energy = (k × q1 × q2)/r = (9 × 10⁹ × 10-⁶ × 10-⁶)/1 = 9 × 10-³ J

Q3. The relation between current density (j) and electric current (I) is :

(A) j = VdlA

(B) j = A/I

(D) j = IA

Ans. (C) j = I/A

Q4. Energy band of substance shown in the figure where V is valence band and C is conduction band :

Substance is :

(A) Metal

(B) Semiconductor

(D) None of the above

Ans. (C) insulator

Q5. When a reverse biased is applied to a p-n junction, it :

(A) Raises the potential barrier

(B) increases the majority carrier current

(D) None of the above

Ans. (A) Raises the potential barrier

Q6. The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV. Potential energy of the electron in this state is :

(A) -3.4 eV

(B) 6.8 eV

(D) -1.7 eV

Ans. (C) -6.8 eV

E = K.E. + P.E.

-3.4 = 3.4 + P.E.

P.E. = – 3.4 – 3.4 = -6.8 eV

Q7. Nuclides with same neutron number N but different atomic number Z are called :

(A) isobars

(B) isotones

(D) None of the above

Ans. (B) isotones

Q8. In 1905, which scientist proposed a radically new picture of electromagnetic radiation to explain photoelectric effect ?

(A) Millikan

(B) Einstein

(D) Compton

Ans. (B) Einstein

Q9. de-Broglie wavelength associated with an electron, accelerated through a potential difference of 100 Volt is :

(A) .1127 nm

(B) 1.127 nm

(D) .1227 nm

Ans. (D) .1227 nm

λ = 1.227/√V = 1.227/√100 = 1.227/10 = .1227 nm

Q10. In Young’s double slit experiment, the screen is moved away from the plane of the slits, the actual separation of the Fringes :

(A) remain constant

(B) increases

(D) interference pattern disappears

Ans. (B) increases

Q11. For yellow light incident on a prism of angle 60°, the angle of minimum deviation is 30°. The angle of incidence in this situation is :

(A) 30°

(B) 45°

(D) 75°

Ans. (B) 45°

incidence angle, i = (A+δm)/2 = (60°+30°)/2= 90°/2 = 45°

Q12. Which phenomena is used in “MIRAGE” ?

(A) interference

(B) Reflection

(D) Total Internal Reflection

Ans. (D) Total Internal Reflection

Q13. Two lens of power 8D and -4D are placed in contact. Focal length of the combination will be :

(A) 50cm

(B) -50cm

(D) -25cm

Ans. (C) 25cm

P = P1 + P2 = + 8 – 4 = +4 D

F = 1/P = 1/4 m = 0.25 m = 25 cm

Q14. The magnetic flux linked with a coil is decreased from 5 Weber to 2 Weber in 1 second. The induced electromotive force in the coil is :

(A) 3 Volt

(B) 30 Volt

(D) .3 Volt

Ans. (A) 3 Volt

Change in magnetic flux Δϕ = 2 – 5 = -3 Weber

Time taken Δt = 1 sec

emf induced in the coil E = − Δϕ/Δt = -(-3/1) = 3 Volt

Q15. The magnetic flux threading a wire loop placed in a magnetic field does not depend upon :

(A) Area of the loop

(B) Magnitude of the field

(D) Shape of the loop

Ans. (D) Shape of the loop

Q16. The tangent to the field line at a given point represents the direction of the net :

(A) Electric Force (F)

(B) Electric Field (E)

(D) Electric Current (I)

Ans. (C) Magnetic Field (B)

Q17. Moving coil galvanometer is converted into ammeter by :

(A) Connecting high resistance in parallel

(B) Connecting low resistance in parallel

(D) Connecting low resistance in series

Ans. (B) Connecting low resistance in parallel

Q18. The length of a metallic wire of R ohm resistance is stretched four times its initial length. Its new resistance is :

(A) 8R

(B) 4R

(D) 16R

Ans. (D) 16R

Resistance of wire R = ρ × L/A = ρ × L/V × L/L = ρ × L²/V

R ∝ L² (Since V and ρ are constant)

New Resistance (R’) = (4L)² = 16L² = 16R

Q19. Electric field lines start from ………….. and end at Negative charges.

Ans. Positive charge

Q20. An important quantity is the mobility µ defined as the magnitude of the ………… per unit electric field.

Ans. Drift Velocity

Q21. The ………….. inside solenoid becomes everywhere parallel to the axis.

Ans. Magnetic Field Lines

Q22. The generation of emf by a solar cell, when light falls on, it is due to the following three basic processes : generation, separation and …………

Ans. Collection

Q23. In …………, heavy nucleus decays into lighter nucleus.

Ans. Nuclear Fission

Q24. Photons are electrically …………..

Ans. Neutral

Q25. The ratio of the sine of the angle of incidence to the sine of angle of refraction is …………..

Ans. Refractive Index

Q26. …………. (physical quantity) is same for X-rays of wavelength 10^-10 m, red light of wavelength 6800 Å and radio waves of wavelength 500 m.

Ans. Speed

Q27. SI unit of coefficient of mutual inductance is …………

Ans. Henry

Q28. In a semiconductor, the concentration of electrons is 5 x 10¹² cm-³ and that of holes is 8 x 10¹³ cm-³. Is it a p-type or n-type semiconductor ?

Ans. P-type (ne < nh)

Q29. The radius of the first electron-orbit in hydrogen atom of Bohr model is r. What will be the radius of the third orbit.

Ans. Radius of third orbit = 9r

Q30. Two light waves of equal amplitude and wavelength are superimposed. What is the phase difference between the waves so that amplitude of the resultant wave will be minimum.

Ans. Φ = 180°

Q31. Is the speed of light in glass independent of the colour of light? (Yes / No)

Ans. No (because speed of light in glass is depend on colour of light)

Q32. An object is placed at a distance of 2F from a convex lens of focal length F. Where should image of object formed on the other side of lens.

Ans. at 2F (real & inverted)

Q33. What is the value of angle of dip at the Earth’s magnetic equator ?

Ans. Angle of dip, δ = 0°

Q34. Two unequal resistance are connected in series. What is same for them ?

Ans. Current

Q35. What is the effect on relaxation time of electrons in metals, with decrease in temperature ?

Ans. increase

SET-D (Objective Questions)

Q1. On introducing a dielectric material between two positive charges situated in air, the repulsive force between them will be :

(A) increased

(B) decreased

(D) zero

Ans. (B) decreased

Q2. Two plates are at 2 cm a part and potential difference between them is 10 volt. The Intensity of electric field between the plates is :

(A) 5 Newton/Coulomb

(B) 500 Newton/Coulomb

(D) 250 Newton/Coulomb

Ans. (B) 500 Newton/Coulomb

V = 10 volt, d = 2 cm = 0.02 m

E = V/d = 10/0.02= 1000/2 = 500 N/C

Q3. The relation between electric current (I) and drift velocity (Vd) is :

(A) I = neVd

(B) I = neAVd

Ans. (B) I = neAVd

Q4. Prism designed to bend light by 90° or 180° make use of :

(A) dispersion

(B) total internal reflection

(D) diffraction

Ans. (B) total internal reflection

Q5. For small angle A prism (Refractive Index n), angle of Minimum deviation Dm is :

(A) Dm = (n-1)/A

(B) Dm = (n-1)A

(D) Dm = n – 1

Ans. (B) Dm = (n-1)A

Q6. Two lens of power 10D and -5D are placed in contact. Focal length of the combination will be :

(A) 5 cm

(B) -5 cm

(D) – 20 cm

Ans. (C) 20cm

P = P1 + P2 = + 10 – 5 = +5 D

F = 1/P = 1/5 m = 0.20 m = 20 cm

Q7. In Young’s double slit experiment, the monochromatic source is replaced by another monochromatic source of shorter wavelength, then actual separation of the Fringes :

(A) remain constant

(B) increases

(D) None of the above

Ans. (C) decreases

Q8. de-Broglie wavelength associated with an electron, accelerated through a potential difference of V volt is :

(A) 1227/√V Å

(B) 1.227/√V Å

(D) 122.7/√V Å

Ans. (B) 1.227/√V Å

Q9. Which scientist verified Einstein’s photoelectric equation with great precision, for a number of alkali metals over a wide range of radiation frequencies ?

(A) Millikan

(B) Einstein

(D) Compton

Ans. (A) Millikan

Q10. All nuclides with same Atomic number Z are called :

(A) isobars

(B) isotones

(D) None of the above

Ans. (C) isotopes

Q11. The ground state total energy of hydrogen atom is about -13.6 eV. Kinetic energy of the electron in this state is :

(A) -13.6 eV

(B) 13.6 eV

(D) 27.2 eV

Ans. (B) 13.6 eV

Q12. In an unbiased p-n junction, holes diffuse from the p-region to n-region, because :

(A) Free electron in the n-region attract them

(B) They move across the junction by the potential difference

(D) All of the above

Ans. (C) Hole concentration in p-region is more as compared to n-region

Q13. Carbon, silicon and germanium have four valence electrons each. These are characterized by valence band and conduction bands separated by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which of the following statement is true ?

(A) (Eg)Si < (Eg)Ge < (Eg)C

(B) (Eg)C < (Eg)Ge > (Eg)Si

(D) (Eg)C = (Eg)Si = (Eg)Ge

Ans. (C) (Eg)C > (Eg)Si > (Eg)Ge

Q14. The length of a metallic wire of R ohm resistance is stretched n times its initial length. Its new resistance is :

(A) nR

(B) 2nR

(D) n⁴R

Ans. (C) n²R

Resistance of wire R = ρ × L/A = ρ × L/V × L/L = ρ × L²/V

R ∝ L² (Since V and ρ are constant)

New Resistance (R’) = (nL)² = n²L² = n²R

Q15. Moving coil galvanometer is converted into voltmeter by:

(A) Connecting high resistance in parallel

(B) Connecting low resistance in parallel

(D) Connecting low resistance in series

Ans. (C) Connecting high resistance in series

Q16. The magnetic field lines of a magnet form :

(A) Continuous curve

(B) Continuous closed loops

(D) None of the above

Ans. (B) Continuous closed loops

Q17. The magnetic flux linked with a coil is decreased from 1 Weber to 0.1 Weber in 1 second. The induced electromotive force in the coil is :

(A) 9 Volt

(B) 90 Volt

(D) .09 Volt

Ans. (C) .9 Volt

Change in magnetic flux Δϕ = 0.1 – 1 = -0.9 Weber

Time taken, Δt = 1 sec

emf induced in the coil E = − Δϕ/Δt = -(-0.9/1) = .9 Volt

Q18. On what factor does the coefficient of mutual inductance of two coils not depend ?

(A) Filled with medium inside solenoid

(B) Separation between coils

(D) Their resistances

Ans. (D) Their resistances

Q19. An ……………. surface is a surface with a constant value of potential at all points on the surface.

Ans. Equipotential

Q20. The electrolyte through which current flows has a finite resistance, called the ………….

Ans. internal Resistance

Q21. According to …………. rule, your extended thumb pointing in the direction of the current. Your fingers will curl around in the direction of the magnetic field.

Ans. Right Hand Thumb

Q22. The induced emf can be increased by …………… the number of turns (N) of a closed coil.

Ans. increasing

Q23. A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. The frequency of the electromagnetic waves produced by the oscillator is ………….

Ans. 10⁹ Hz

Q24. Advance sunrise and delayed sunset is due to ……………

Ans. Atmospheric Refraction

Q25. Photons are not deflected by Electric and …………. fields.

Ans. Magnetic

Q26. In Rutherford’s nuclear model of the atom, the entire positive charge and most of the atom are concentrated in the …………….

Ans. Nucleus

Q27. In an n-type silicon, ………….. are minority carriers.

Ans. Holes

Q28. What is the effect on resistivity in metals, with increase in temperature ?

Ans. Resistivity increase

Q29. Current density is scalar quantity or vector quantity.

Ans. Vector Quantity

Q30. The horizontal and vertical components of the earth’s magnetic field at a place are equal. What is angle of dip at the place.

Ans. Angle of dip, δ = 45°

Q31. A point source of light is placed at a distance of 40 cm from a convex lens of focal length 20 cm. Where should image formed on the other side of lens.

Ans. at 2F (real & inverted)

Q32. In intrinsic semiconductors, what is the ratio of the number of free electrons (ne) to the number of holes (nh).

Ans. 1 : 1 (ne = nh)

Q33. The radius of the first electron-orbit in hydrogen atom of Bohr model is r. What will be the radius of the fourth orbit ?

Ans. 16r

Q34. In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size of central diffraction band ?

Ans. Half

Q35. Monochromatic light is refracted from air into glass of refractive index 3/2. What is the ratio of the wavelengths of the incident and the refracted waves ?

Ans. Wavelength of incident wave : Wavelength of refracted wave = λ1 : λ2 = n : 1 = 3/2 : 1 = 3 : 2

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