HBSE Class 11th Physics Solved Question Paper 2022

HBSE Class 11th Physics Solved Question Paper 2022 

HBSE Class 11 Physics Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 11 Physics 2022. HBSE 11th Question Paper Download 2022. HBSE Class 11 Physics Paper Solution 2022. Haryana Board Class 11th Physics Question Paper 2022 Pdf Download with Answer.

Objective Questions 

Q1(1). Which of the following is Fundamental Unit ? 

(A) Watt

(B) Newton

(C) Ampere

(D) Joule 

Ans. (C) Ampere

There are 7 Fundamental Unit – 

Length – meter (m)

Time – second (s)

Amount of substance – mole (mole) 

Electric current – ampere (A)

Temperature – kelvin (K)

Luminous intensity – candela (cd)

Mass – kilogram (kg)

Q2. There is an error of 3% in the measurement of side of a cube. The percentage error in the calculation of its volume of the sphere will be:

(A) 9%

(B) 6%

(C) 3%

(D) 1% 

Ans. (A) 9%

Error Formula,  3Δr/r = 3 × 3% = 9%

Q3. The dimensional formula [MLT-²] represent :

(A) Force 

(B) Torque

(C) Power 

(D) Momentum

Ans. (A) Force

Force = ma = M¹L¹T-² 

Torque = Fs = M¹L²T-² 

Power = Fv = M¹L²T-³ 

Momentum = mv = M¹L¹T-¹ 

Q4. A particle completes semicircular path of radius r, displacement travelled by particle will be :

(A) r/4 

(B) r/2

(C) 2r

(D) 4r

Ans. (C) 2r

Q5. An object is moving with a constant acceleration. Its velocity time graph is :

Ans. Option A

Q6. Which of the following is scalar quantity ?

(A) Displacement

(B) Velocity 

(C) Work

(D) Force 

Ans. (C) Work

Scalar quantity has only magnitude but vector quantity has magnitude and direction. 

Q7. Two objects A and B of masses 5 kg and 10 kg respectively are moving with same velocity. The ratio of their momentum will be :

(A) 1 : 1

(B) 1 : 5

(C) 10 : 5

(D) 1 : 2 

Ans. (D) 1 : 2 

Momentum = P = mv 

P1/P= m1v/m2v = 5/10 = 1/2  or  1 : 2 

Q8. Which Friction is minimum ?

(A) Static Friction

(B) Limiting Friction

(C) Kinetic Friction

(D) None 

Ans. (D) None 

Rolling friction is minimum & limiting friction is maximum. 

Q9. Gravitational force is :

(A) a conservative force

(B) Pseudo force 

(C) a non-conservative force

(D) None of the above.

Ans. (A) a conservative force

Q10. The linear acceleration(a) of a body rotating about an axis with anguler acceleration α, then : 

(A) α = rω

(B) α = ar

(C) a = rα 

(D) r = aα

Ans. (C) a = rα 

Q11. The potential energy of a particle in SHM at a distance x from the equilibrium position is :

(A) ½mw²x²

(B) ½mw²a² 

(C) ½mw²(a²-x²)

(D) Zero

Ans. (A) ½mw²x²

Q12. If a process takes place in a system at a constant pressure, the process is known as :

(A) isochoric

(B) isothermal

(C) Adiabatic

(D) isobaric

Ans. (D) isobaric

Q13. A body of mass m is moving on a circle of radius r with uniform speed v. The centripetal force on body is :

(A) mv²/r 

(B) mvr 

(C) mv/r 

(D) mv/r² 

Ans. (A) mv²/r

Q14. In perfect inelastic collision : 

(A) both momentum and K.E. are conserved 

(B) both momentum and K.E. are not conserved

(C) momentum is conserved and K.E. is not conserved

(D) momentum is not conserved and K.E. is conserved

Ans. (C) momentum is conserved and K.E. is not conserved

Q15. If mass M and radius of solid sphere is R, then moment of inertia about diameter is :

(A) ⅔MR²

(B) ⅖MR² 

(C) ⅗MR² 

(D) 7/5MR² 

Ans. (B) ⅖MR² 

Q16. The mean kinetic energy of a molecule of any gas at absolute temperature T is proportional to :

(A) 1/T

(B) √T

(C) T 

(D) T²

Ans. (C) T 

Q17. The value of Young’s modulus of elasticity for a perfectly rigid body is : 

(A) Zero

(B) infinite

(C) 1

(D) 100

Ans. (B) infinite

Q18. A body enters in the gravitational field of the earth from external atmosphere. The potential energy of the earth-body system :

(A) will increase

(B) will decrease

(C) will remain unchanged

(D) None of the avove

Ans. (B) will decrease

Q19. ……….. is 1/273.16 part of thermodynamic temperature of the triple point of water.

Ans. Kelvin

Q20. The angle of projection for which a projectile cover the maximum horizontal range is ……..

Ans. 45°

Q21. The rate of doing work by a machine is called …………. of the machine. 

Ans. Power

Q22. The value of …………. is determined to be 6.67 x 10-¹¹ Newton-metre²/kg². 

Ans. Gravitational Force (G) 

Q23. By using ball bearings, sliding friction can be converted into ……….. friction. 

Ans. Rolling

Q24. The number of degrees of freedom of a molecule of a diatomic gas is …………

Ans. 5

Q25. When a body being acted by an external periodic force, vibrates with the frequency of the force, then the vibrations of the body are called ………… vibrations. 

Ans. Force

Q26. 4.18 Joule of work is equivalent to …………. calories of heat. 

Ans. One

Q27. The angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the point of contact is called ……….. for that pair of solid and liquid. 

Ans. Angle of contact 

Q28. The equation of a simple harmonic motion is y = 10sin100πt. Give value of amplitude of oscillation. 

Ans. Amplitude (A) = 10 

Q29. When 200 Joule of heat is given to a gaseous system, then internal energy increases by 60 Joule. Find the work done by the system. 

Ans. Work done = 200 – 60 = 140 Joule 

Q30. Write relation between coefficient of linear expansion (α) and coefficient of surface expansion (β) of a solid material. 

Ans. β = 2α 

Q31. If a vector C is multiplied by a scalar n, what will be the resultant vector. 

Ans. Result Vector = Cn 

Q32. What is relation between coefficient of static friction and the angle of heating friction ? 

Ans. μ = tanθ

Q33. What is time-period of Geo-stationary satellite ? 

Ans. 24 hours 

Q34. Write down the formula for work done in rotatory motion. 

Ans. ω = iθ

Q35. How many erg are there in 1 Joule ? 

Ans. 1 Joule = 10⁷ erg 

Subjective Questions 

Q2. Check the equation v = u+ at by the method of dimensions. 

Where v = final velocity, u= Initial velocity, a = acceleration, t= time. 


LHS : v = L¹T-¹ 

RHS : u + at = L¹T-¹ + L¹T-²(T) = L¹T-¹ + L¹T-¹ = L¹T-¹ 

As per addition rule in dimensions, only quantities having similar dimensions can be added. 

So, LHS = RHS 

Q3. Draw velocity-time graph of a particle executing simple harmonic motion. 


Q4. Write Stoke’s Law. 

Ans. Stoke’s Law states that the force that retards a sphere moving through a viscous fluid is directly proportional to the velocity and the radius of the sphere, and the viscosity of the fluid. 

In Stokes’s law, the drag force F acting upward in resistance to the fall is equal to 6πrηv, in which r is the radius of the sphere, η is the viscosity of the liquid, and v is the velocity of fall.

Q5. The applied force F = 3i + 4j – 5k on a particle produces displacement S = 5i + 4j + 3k. Calculate work done by the force. 

Ans. W = F.S = (3i + 4j -5k).(5i + 4j + 3k) 

= 3×5 + 4×4 – 5×3 = 15 + 16 – 15 = 16 Joule 


Q6. Write down the postulates of kinetic theory of gases. 

Ans. Postulates of kinetic theory of gases :  

(i) A gas consist of a very large number of atoms.  

(ii) The molecules of a gas are in a state of incessant random motion.  

(iii) The size of the gas molecule in very small as compared to the distance between them. 

(iv) The molecules do not exert any force of attraction or repulsion on each other. 

(v) The collisions of the molecules with themselves and with the walls of the vessel are perfectly elastic. 

Q7. Obtain the formula for the escape velocity of a body from the Earth. 


Let earth be a perfect sphere of mass M, radius R with centre at O.  

Gravitational force of attraction on the body at P is F = GMm/x² 

Work done is taking the body against gravitational attraction from P to Q is   

dW = Fdx 

dW = GMm/x² dx 

Total work done in taking body from x = R to           x = α   

W = ∫GMm/x² dx  [:. Limit R to α] 

     = GMm[x-²+¹]/[-2+1]  [:. Limit R to α] 

     = -GMm[1/x]  [:. Limit R to α] 

     = -GMm[1/α-1/R] 

     = GMm/R 

The work done is at the cost of K.E. given to the body at the surface of the earth. 

½mVe² = GMm/R 

Ve = √2Gm/R 

Q8. Write difference between isothermal and adiabatic process. 

Ans. Isothermal Process– Transfer of heat occurs.The pressure is more at a given volume. The temperature remains constant.  Heat can be added or released to the system just to keep the same temperature.  The transformation is slow.

Adiabatic Process– No transfer of heat occurs. The pressure is less at a given volume. The temperature changes due to internal system variations. There is no addition of heat, nor is heat released because maintaining constant temperature doesn’t matter here. The transformation is fast. 

Q9. If two vectors P and Q represent two adjacent sides of parallelogram, then prove that, magnitude of resultant R = √P²+Q²+2PQcosθ. 


Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude and direction by two adjacent sides OA and OD of a parallelogram OABD as shown in figure.

Let θ be the angle between P and Q and R be the resultant vector. Then, according to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q.

So, we have                  

R = P + Q

Now, expand A to C and draw BC perpendicular to OC.

From triangle OCB, 

OB² = OC² + BC² 

OB² = (OA+AC)² + BC²  ………(i) 

In triangle ABC, 

cosθ = AC/AB

or AC = AB cosθ

or AC = OD cosθ 

or AC = Q cosθ    [AB = OD = Q] 

Also  sinθ = BC/AB 

or BC = AB sinθ 

or BC = OD sinθ 

or BC = Q sinθ    [AB = OD = Q] 

Magnitude of resultant :

Substituting value of AC and BC in ( i ), we get

R² = (P+Qcosθ)² + (Qsinθ)² 

R² = P² + 2PQcosθ + Q²cos²θ + Q²sin²θ 

R² = P² + Q² + 2PQcosθ      [sin²θ+cos²θ = 1] 

R = √P²+Q²+2PQcosθ 

Q10. Obtain expression for the centre of mass of a system consisting of two particles. 


rcm = (m1r1+m2r2)/(m1+m2)

Q11. State the laws of limiting friction. 

Ans. (i) The direction of limiting friction force is always opposite the direction of motion.

(ii) It always acts tangential to the two surfaces.

(iii) It is dependent on the material and the nature of the surfaces in contact.

(iv) It is independent of the shape and area.

Q12. Prove that the number of beats heard per second is equal to the difference in frequencies of two sound sources. 

Ans. Let us consider two harmonic waves of same amplitude but slightly different frequencies n1 and n2 moving with same velocity in the same direction. 

y1 = a sinw1t = a sin2πn1

y2 = a sinw2t = a sin2πn2

By principle of super position  

y = y1 + y2 

y = a sin2πn1t + a sin2πn2

y = 2a cos[π(n1-n2)t] sin[π(n1+n2)t] 

y = A sin[π(n1+n2)t] 

Where A = 2a cos[π(n1-n2)t] Amplitude

For Maximum Amplitude

cosπ(n1-n2)t = ±1 

π(n1-n2)t = kπ 

t = k/(n1-n2

Substituting k = 0, 1, 2, 3, …….

t = 0, 1/(n1-n2), 2/(n1-n2), ……. 

Time internal between two maxima = 1/(n1-n2)   

Frequency of max. = n1-n

For minimum amplitude 

cos[π(n1-n2)t = 0 = coskπ/2

π(n1-n2)t = kπ/2   where k = 1, 3, 5 

Substituting k 

t = 1/2(n1-n2), 3/2(n1-n2), ……

Time internal b/w two minima = 1/(n1-n2

Frequency of minima = n1-n

Hence difference of frequency of sound sources  = n1-n


Obtain an expression for the time-period of a simple pendulum. 

Ans. Simple Pendulum : 


l = length of simple pendulum 

m = mass of bob  

At any instant, thread makes an angle θ with vertical.  

Restoring force set up  

F = − mg sin θ 

F = -mg(x/l) {sinθ ≈ θ ≈ x/l} 

F = -ma = -mg(x/l) 

   aα − x  

Motion of pendulum is simple harmonic 

T = 2π √displacement(x)/acceleration (a) 

T = 2π √l/g 

Q13. Derive expression for the formula of rise of water in a capillary tube. 


Lets say capillary of radius r is placed in a beaker containing a liquid of density & and water rises in the capillary the height h, 

PB – PC = 2T/R …….(i) 

PB = PO ……(ii) 

PA = PO ……(iii) 

PC + ρgh = PA …..(iv) 

From (1),(2)&(3),(4) press are gradient 

P– 2T/R = P ρgh 

2T/R = ρgh 

h = 2T/Rρg = 2Tcosθ/ρgr 


State and prove Bernoulli’s theorem. 


Bernolli’s theorem– The theorem states that for the stream line flow of a liquid, the total energy per unit mass remains constant at every cross-section throughout the liquid flow.  

Proof : The energies possessed by a flowing liquid are mutually convertible. When one type of energy increases, the other type of energy decreases and vice-versa. Now, we will derive the Bernoulli’s theorem using the work-energy theorem.

Consider the flow of liquid. Let at any time, the liquid lies between two areas of flowing liquid A1 and A2. In time interval ∆t, the liquid displaces from A1 by ∆x1 = υ1∆t and displaces from A2 by ∆ x2 = υ2 ∆t. Here υ1 and υ2 are the velocities of the liquid at A1 and A2. The work done on the liquid is P1A1∆x1 by the force and P2A2∆x2 against the force respectively. 

Net work done,

So, more the cross-sectional area, lesser is the velocity and vice-versa. 

So, from Bernoulli’s theorem

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