HBSE Class 12th Physics Solved Question Paper 2021

HBSE Class 12th Physics Solved Question Paper 2021

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.


Q12. Draw a Labelled ray diagram showing image formation 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. 


                                        OR 

What is wavefront ? Using Huygen’s principle to verify the Laws of reflection.

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. 

Laws of reflection by Huygen’s principle : 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



Objective Questions 

Q1. Two charges +1µC and +4µC are situated at a distance in air. The ratio of forces acting on them is :  

(A)  1 : 4  

(B)  4 : 1 

(C)  1 : 1 

(D)  1 : 16 

Ans. (C)  1 : 1 

Force AB 1/4π∈ × q1q2/r² 

Force BA 1/4π∈ × q2q1/r²

Force AB : Force BA = 1 : 1 


Q2. Three capacitors of equal capacity C are joined first in series and then in parallel. The ratio of equivalent capacities in both the cases will be :  

(A)  9 : 1  

(B)  6 :1 

(C)  3 : 1  

(D)  1 : 9 

Ans. (D)  1 : 9

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) 

Cs : Cp = C/3 : 3C = 1 : 9 


Q3. The  unit  of  Specific  resistance  is  : 

(A)  ohm-m 

(B)  ohm-m-¹

(C)  ohm-¹ – m-¹

(D)  ohm-¹ – m

Ans. (A)  ohm-m


Q4. The  graph  between  voltage  (v)  and  current  (i)  for  a  conductor  is  a  straight  line which  makes  an  angle  θ  with  y-axis  (representing  i).  The  resistance  of  the Conductor  will  be  : 

(A)  tanθ 

(B)  cotθ 

(C)  sinθ

(D)  cosθ 

Ans. (A)  tanθ 

Ohm’s Law,  V = IR 

R = V/I = P/B = tanθ

 

Q5. An electron enters into a uniform magnetic field perpendicularly to it. The path of the electron would be :  

(A)  Elliptical 

(B)  Circular 

(C)  Parabolic 

(D)  Linear 

Ans. (B)  Circular 


Q6. The force F acting on a particle of charge q moving with velocity v parallel to magnetic field B is :  

(A)  q/(v×B)

(B)  (v×B)/q

(C)  q(v×B) 

(D)  Zero 

Ans. (D)  Zero 


Q7. For  a  coil-having  self  Inductance  2  mH,  current  flows  at  a  rate  of 10³ ampere/sec  in  it.  The  emf  induced  in  it  is  : 

(A)  1 volt 

(B)  2 volt 

(C)  3 volt 

(D)  4 volt 

Ans. (B)  2 volt 

emf = – L.dI/dt = 2mH × 10³ = 2 × 10-³ × 10³ = 2 volt 


Q8. Lenz’s  Law  is  a  Consequence  of  the  Law  of  Conservation  of  : 

(A)  charge   

(B)  momentum 

(C)  energy 

(D)  mass 

Ans. (C)  energy 


Q9. Mirage  is  a  phenomena  due  to  : 

(A)  Reflection 

(B)  Dispersion 

(C)  Scattering 

(D)  Total  Internal  Reflection 

Ans. (D)  Total  Internal  Reflection 


Q10. The two Surfaces of a bi-convex Lens have equal radii of curvature R and its refractive Index n is 1.5. The focal Length of the Lens will be :  

(A)  R/2  

(B)  R  

(C)  -R  

(D)  2R 

Ans. (B)  R  

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 

 

Q11. The angle of minimum deviation for a prism is δm and the angle of prism is A. The refractive index of the material of the prism is : 

(A)  sin(A+δm)/2 

(B)  sin A/2

(C)  (sin A/2) ÷ sin(A+δm)/2

(D)  sin(A+δm)/2 ÷ sinA/2 

Ans. (D)  sin(A+δm)/2 ÷ sinA/2


Q12. Speed of Light in air is 3×10⁸ m/s. For the water of refractive Index 4/3, the speed of light will be :  

(A)  1.5 × 10⁸ m/s  

(B)  2 × 10⁸ m/s 

(C)  1 ×10⁸ m/s 

(D)  2.25 × 10⁸ m/s  

Ans. (D)  2.25 × 10⁸ m/s  

v = c/μ = (3 × 10⁸) ÷ 4/3 = 3 × 10⁸ × 3/4 

             = 9/4 × 10⁸ = 2.25 × 10⁸ m/s 


Q13. Electron  emission  from  a  metallic  surface  is  possible  only,  when  Frequency  of the  incident  Light  is  : 

(A)  Less  Then  threshold  Frequency 

(B)  Half  of  the  Threshold  Frequency 

(C)  Greater  than  the  Threshold  Frequency 

(D)  No  effect  of  Frequency 

Ans. (C)  Greater  than  the  Threshold  Frequency 


Q14. The  de-Brogile  wavelength  associated  with  a  moving  particle  is  : 

(A)  Directly  proportional  to  its  mass 

(B)  Directly  proportional  to  its  energy

(C)  Directly  proportional  to  its  momentum 

(D)  inversely  proportional  to  its  momentum

Ans. (D)  inversely  proportional  to  its  momentum


Q15. In hydrogen atom, the Kinetic energy of electron in an orbit of radius r is given by :  

(A)  -1/4π∈ × e²/r 

(B)  1/4π∈ × e²/2r

(C)  1/4π∈ × e²/r

(D)  -1/4π∈ × e²/2r

Ans. (B)  1/4π∈ × e²/2r


Q16. Number  and  type  of  nucleons  in  the  nucleus  of  Deuterium (1H²) will be : 

(A)  2  protons  

(B)  1  proton  and  1  neutron 

(C)  2  neutrons  

(D)  1  proton  and  1  electron

Ans. (B)  1  proton  and  1  neutron 


Q17. Energy  band  of  a  substance  is  shown  in  Figure.  Where  V  is  valence  band  and C  is  conduction  band. 

This  substance  is  : 

(A)  Conductor 

(B)  Semi  conductor 

(C)  Insulator 

(D)  Diamond 

Ans. (B)  Semi  conductor


Q18. In  a    n-type  Semi  Conductor  the  minority  charge  carriers  are  : 

(A)  Electron 

(B) Hole 

(C)  Electron  and  hole 

(D)  None of these 

Ans. (B) Hole 


Q19. Volt  is  unit  of  …………….

Ans. Electric Potential Difference


Q20. Current  through  a  given  area  of  a  conductor  is  the  net  ………………  passing  per unit  time  through  the  area. 

Ans. Charge


Q21. Three  quantities  are  needed  to  specify  the  magnetic  field  of  the  Earth  on  its Surface-the  horizontal  component,  the  magnetic  declination  and  …………… 

Ans. Angle  of  Dip


Q22. In  an  AC  generator,  mechanical  Energy  is  converted  to  …………. 

Ans. Electrical Energy


Q23. X-rays,  γ-rays  and  radiowaves  are  propagating  in  vacuum,  there  ……………  will  be  same  but  frequencies  will  be  different. 

Ans. Velocity


Q24. The  ……………  (Total Internal Reflection)  for  a  ray  incident  from  a  denser  to  rarer  medium,  is that  angle  for  which  the  angle  of  refraction  is 90⁰. 

Ans. Critical Angle


Q25. The  energy  of  photon  of  wavelength  λ  is ……………… 

Ans. E = hv = hc/λ


Q26. In  Rutherford’s  model,  most  of  the  mass  of  the  atom  and  all  its  positive  charge are  concentrated  in  a  tiny  nucleus  and  the ………….  revolved  around  it. 

Ans. Electrons


Q27. The  number  of  electrons  (Ne)  is  equal  to  the  number  of  holes (Nn) i………… semiconductors.   

Ans. intrinsic


Q28. Write  down  the  formula  for  the  relation  of  current  density  (J)  specific conductivity  (σ)  and  the  electric  field  (E). 

Ans. σ = j/E 


Q29. Write  relation  between  Electromotive  force  and  terminal  potential  difference  for a  cell. 

Ans. E = V + ir 


Q30. Consider  the  circuit  shown  where  APB  and  AQB  are  parts  of  square.  What  will be  the  Magnetic  field  at  the  centre  O  of  the  square.  

Ans. Zero 



Q31. Two  thin  Lenses  of  power  +5  dioptre  and  -5  dioptre  are  placed  in  contact.  Find the  power  of  this  Combination. 

Ans. P = P1 + P2 = + 5 – 5 = 0   


Q32. For  which  spectral  colour  of  Light  is  the  speed  Maximum  in  glass  ?  

Ans. Red 

 

Q33. In  Young’s  experiment  what  will  be  the  effect  on  fringe-width  on  using  yellow colour  light  of  sodium  Lamp  in  place  of  violet  colour  light  ? 

Ans. Fringe width will increase 


Q34. Two  nuclei  have  mass  numbers  in  the  ratio  1  :  8.  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)


Q35. Is  the  Junction  diode  D  is  forward  or  reverse  biased,  in  the  given  diagram 

Ans. Reverse Biased 



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