HBSE 12th Physics Solved Sample Paper 2023

HBSE 12th Physics Solved Sample Paper 2023 



Q1(i) Angle between equipotential surface and electric field line is :     

      (A) Zero 

      (B) 180°

      (C) 90°

      (D) 45°  


(ii) The unit of permittivity of free space Eo is : 

      (A) C²N-¹M-² 

      (B) N M² C-² 

      (C) C N-² M-¹ 

      (D) C² N-² M-²


(iii) What is the magnitude of force experienced by a stationary charge when placed in a uniform magnetic field ?  

      (A) F = qvB SInQ 

      (B) F = qvB 

      (C) F = Zero 

      (D) None of these   


(iv) What is the number of electrons drifting past a section of a conductor if ‘a’ ampere current is passed through it for ‘b’ second ?  

Charge on one electron, e = 1.6×10-¹⁹C 

n = (Current×Time)/(charge of 1 electron)


n = (ab)/e = (ab)/(1.6×10-¹⁹) = 6.25×10¹⁸×(ab)


(v) The nuclear model of atom was proposed by ………………. 

Ernest Rutherford


(vi) A Proton and an α particle with equal momenta enter normal to a uniform magnetic Field. Their ratio of radii is  

      (A) 2 : 1  

      (B) 1 : 2 

      (C) 1 : 1  

      (D) 1 : 4   

r = mv/qB = p/qB 

r ∝ 1/q 

rp/rα = qα/qp = 2e/e = 2/1 = 2 : 1


(vii) An electric lamp is rated at 110 W and 220 V supply. Find the current flowing through the bulb. 

Vrms = 220V,  P = 110W 

P = V²rms/R 

R = V²rms/P = (220)²/110 = 440 Ω 

Vrms = Irms × R 

Current, Irms = Vrms/R = 220/440 = 1/2 = 0.5 Ampere 


Irms = 0.5 A 


(viii) Total internal reflection takes place when light travels from :  

      (A) Water to glass 

      (B) Glass to diamond 

      (C) Water to air 

      (D) Air to mercury 


(ix) Name the electromagnetic radiation used for viewing objects through haze and fog.  

infrared rays


(x) In the following nuclear reaction :  

6C¹¹ → 5B¹¹ +  B+ + X  

What does X stand for ?   

X is a Neutrino 


(xi) The de-Broglie wavelength associated with a particle of momentum P is given by ……………. 

λ = h/P = h/(mv) 

where h = planck’s constant and P = mv 


(xii) If the intensity of the incident radiation in a photo cell is increased, how does the stopping potential vary ?  

No effect on stopping potential


(xiii) Given total energy of electron in first excited state of hydrogen atom = -3.4ev, what are the values of K.E and P.E in this state ?   

Total energy, E = K.E + P.E

K.E = -(E) = -(-3.4ev) = 3.4 ev 


P.E = -2(K.E) = -2(3.4ev) = -6.8 ev 


(xiv) Net charge on a P- type semi-conductor is ……………… 

Zero because it is Neutral 


(xv) Two thin lenses of power +6D and -2D are in contact, what is the focal length of combination.  

P = P1 + P2 = +6 + (-2) = 6 – 2 = +4 


Focal Length,  F = 1/P = 1/4 = 0.25 m = 25 cm 


Q2. Two coherent sources, whose intensity ratio is 16:1 produce interference fringes. Calculate the ratio of intensity of maxima and minima in the fringe system.          

I1/I2 = a²/b² = 16/1 

a/b = √16/1 = 4/1 

a : b = 4 : 1 

Imax  = (a+b)² = (4+1)² = 5² = 25 

Imin = (a-b)² = (4-1)² = 3² = 9 


Imax : Imin = 25 : 9


Q3. Define the coefficient of mutual induction. Give its units and mention two factors on which mutual inductance between a pair of coil depends.             

The coefficient of mutual induction of two coils is numerically equal to the emf induced in one coil when the rate of change of current through the other coil is unity. 

The S.I unit of coefficient of mutual induction is henry (H). 

The coefficient of mutual induction between a pair of coils depends on the following factors :

(i) Size and shape of the coils, number of turns, and permeability of material on which the coils are wound.

(ii) The proximity of the coils


(iii) Their separation as well as their relative orientation.


Q4. Draw the sketch of variation of resistivity with temperature in the following :         

(i) Conductor 

(ii) Semiconductor   

In conductors, average relaxation time decreases with increase in temperature, resulting in an increase in resistivity. 

In semiconductors, the increase in number density (with increase in temperature) is more than the decrease in relaxation time; the net result is, therefore, a decrease in resistivity. 


Q5. An electric field (E) exerts a torque on an electric dipole moment (P) placed in its. When is the torque  

(i) Zero    

(ii) maximum                

(i) Torque is Zero when angle between P and E is 0°. 


(ii) Torque is Maximum when the angle between P and E is 90°.


Q6. The susceptibility of a magnetic material is 

(i) -4.2 × 10-⁶ 

(ii)  1.9 × 10-⁵ 

What types of material do they represent ?  

(i) Susceptibility is negative, so given material is Diamagnetic.

(ii) Susceptibility is positive, so material is Paramagnetic.  


Q7. Write the truth table for the given circuit.     


A    B    Y
0    0    0
0    1     1
1     0    1
1     1    1



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

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 × (Ze×2e)/ro = 2Ze²/4πεoro  ……….(ii) 

At equillibrium,  

Eα = Ep 

½mv² = 2Ze²/εoro 

ro = 2Ze²/(εo × ½mv²) 


ro ∝ 1/Eα 


Q9. List the factors on which the capacitance of a parallel plate capacitor depend. 

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


Q10. State Gauss’s law in electrostatics. Use it to determine electric field due to infinitely long straight wire which is uniformly charged.             

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 



Q11. Explain the working principle of AC generator with the help of a labelled diagram. 

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. 


Q12. Explain the principle of potentiometer. Draw a labelled circuit diagram to determine internal resistance of the cell.    

We know that a potentiometer is used to determine the internal resistance of a cell. Besides this, we will have a determination of potential difference using a potentiometer.


Principle : The basic working principle of this is based on the fact that the fall of the potential across any portion of the wire is directly proportional to the length of the wire, provided wire has a uniform cross-sectional area and the constant current flowing through it. 

E = V(r/R + 1) 


Q13(i) AC is supplied to a series LCR circuit. Draw the relevant phasor diagram. Obtain the resonance condition.                            


(ii) A series LCR circuit with R=44Ω, C=8µF and L=50H is connected to a variable freq. 220 V ac supply. Calculate angular frequency at resonance condition.  

Angular frequency = ω

ω = 1/√(LC)

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


ω = 50 Hz or 50 rad/s


Q14. A radio can tune into any station in the 7.5 MHz to 21 MHz. What is the corresponding wavelength band? 

Frequency, 7.5 MHz = 7.5 × 10⁶ Hz

Frequency, 21 MHz = 21 × 10⁶ Hz 

Wavelength, λ1 = speed/frequency = c/v1 = (3 × 10⁸) ÷ (7.5 × 10⁶) = 40 m 

Wavelength, λ2 = speed/frequency = c/v2 = (3 × 10⁸) ÷ (21 × 10⁶) = 14.28 m
Wavelength band from 40m to 14.28m


Q15. Plot the graphs showing the variation of photo electric current with anode potential in the following cases :                

(i) Two light beams of same wavelength but different intensity.

(ii) Two light beams of different frequencies but same intensity.  


Q16. Sate the laws of radioactive decay. Define the term decay constant for a radioactive substance. How is related to half life ?  

Laws of radioactive decay : 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.” 

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. 


T½ = (ln 2) = 0.6931/λ 


Q17(i) Show by ray diagrams how a totally reflecting prism of glass can be used to deviate ray of light through 90° and 180°. 

Totally reflecting prism : A prism having on angle of 90° between its two refraction surfaces and the other two angles each equal to 45° is called total reflecting prism because the light incident normally on any of its faces, suffers total internal reflection inside the prims. 

(ii) The refractive index of diamond is 2.42. Calculate the speed of light in diamond. 

Refractive index = speed of light in vaccum/speed of light in diamond 

n = c/v 


Speed of light in diamond, v = c/n = (3×10⁸)/2.42 = 1.239 × 10⁸ m/s 


Q18(i) Explain the principle and working of moving coil Galvanometer with the help of diagram. 

Moving Coil Galvanometer : 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. 

(ii) How can Galvanometer be converted into voltmeter and ammeter ?  

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

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


(i) Explain the principle and working of cyclotron with the help of diagram. Show that time of revolution of ion is independent of its speed or radius of its orbit. 

Cyclotron : Cyclotron is a device used to accelerate charged particles to high energies.

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.

Construction: It consists of a hollow metal cylinder divided into two sections D1 and D2 called Dees, enclosed in an evacuated chamber (Figure). The Dees are kept separated and a source of ions is placed at the centre in the gap between the Dees. They are placed between the pole pieces of a strong electromagnet. The magnetic field acts perpendicular to the plane of the Dees. The Dees are connected to a high frequency oscillator.

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  

T = 2πm/qB = constant 


Thus ‘T’ is independent of ‘v’.

(ii)  State the limitations of this device cyclotron. 



The mass of an electron is very small. Because of this reason, the cyclotron device is incapable of accelerating or speeding up the electron. This device is also incapable of accelerating or speeding up the particles which have no charge (neutral particle).


Q19(i) What is doping? How is it done to make p and n type semiconductors? 

Doping means the introduction of impurities into a semiconductor crystal to the defined modification of conductivity. 


A p-type semiconductor is created when group III elements are doped to a complete semiconductor material. As opposite, an n-type semiconductor is created when group V elements are doped to an intrinsic semiconductor. 

(ii) Draw the energy level diagram for p and n type semiconductor.  


(i) Draw a circuit diagram to study characteristics of transistor (n p n) in common emitter configuration. 


(ii) Draw the sketch of (a) Input characteristics (b) Output characteristics for this configuration.     



Q20(i) What do you mean by polarization? What types of waves can be polarized? Explain polarization by refection and hence deduce Brewster’s law. 

Polarization : When light shines at a certain angle, it is possible for the reflected light to become completely polarized. This phenomenon is called polarization by reflection. It is the plane of polarisation parallel to the plane of the interface.


Transverse (Light) waves can be polarised. 

Polarization by Brewster’s law : Brewster’s angle is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface with no reflection. When unpolarized light is incident at this angle, the light that is reflected from the surface is therefore perfectly polarized. A glass plate or a stack of plates placed in a light beam at Brewster’s angle can thus be used as a polarizer. 


(ii) Calculate refractive index if angle of polarization is 60°.  

Refractive index, μ = tan60° = √3 = 1.73 


(i) What is lens maker formula? What assumptions are made for deriving this formula? Derive lens maker formula using a convex lens.  

Lens maker formula : The relation between the focal length of the lens, the refractive index of its material, and the radii of curvature of its two surfaces is known as the lens maker’s formula. 

1/f = (μ-1)[1/R1 – 1/R2

The following assumptions are taken for the derivation of lens maker formula : 

(i) The object must be point sized. 

(ii) Aperture of the lens which is used should be very small. 


(iii) The thickness of the lens should be less. 

(ii) The radius of curvature of each surface of a convex lens of refractive index 1.5 is 40 cm. Calculate the power. 

μ = 1.5, R1 = 40cm = 0.4 m, R2 = – 40cm = – 0.40 m 

P = 1/f = (μ-1)[1/R1 – 1/R2] = (1.5-1)[1/0.40 + 1/0.40] 

P = (0.5)(5) = 2.5 D 



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