**HBSE Class 11th Physics Solved Question Paper 2021**

HBSE Class 11 Physics Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 11 Physics 2021. HBSE 11th Question Paper Download 2021. HBSE Class 11 Physics Paper Solution 2021. Haryana Board Class 11th Physics Question Paper 2021 Pdf Download with Answer.

Objective Type Questions

Q1. Which of the following is not Fundamental Unit ?

(A) Ampere

(B) Second

(C) Mole

(D) Watt

Ans. (D) Watt

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 2% in the measurement of side of a cube. The percentage error in the calculation of its volume of the sphere will be :

(A) 1%

(B) 2%

(C) 3%

(D) 6%

Ans. (D) 6%

Error Formula, 3Δr/r = 3 × 2% = 6%

Q3. The dimensional formula [MLT-¹] represents :

(A) Force

(B) Torque

(C) Power

(D) Momentum

Ans. (D) Momentum

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. The ratio of distance travelled and displacements of particle will be :

(A) π/4

(B) π/2

(C) 3π/4

(D) π

Ans. (B) π/2

Semicircle, 2πr/2 = πr

πr/2r = π/2 or π : 2

Q5. The distance travelled by a particle is directly proportional to the square of the time taken, the acceleration of the particle is :

(A) Increasing

(B) Decreasing

(C) Zero

(D) Constant

Ans. (D) Constant

distance ∝ t²

distance, s = kt²

velocity, v = ds/dt = d(kt²)/dt = 2kt

acceleration, a = dv/dt = d(2kt)/dt = 2k = constant

Q6. 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

Q7. Which of the following is scalar quantity ?

(A) Distance

(B) Displacement

(C) Force

(D) Acceleration

Ans. (A) Distance

Q8. Two objects A and B of masses 2 kg and 10 kg respectively are moving with same velocity. The ratio of their momentum will be :

(A) 1 : 1

(B) 1 : 5

(C) 5 : 1

(D) 1 : 25

Ans. (B) 1 : 5

Momentum = P = mv

P1/P2 = m1v/m2v = 2/10 = 1/5 or 1 : 5

Q9. Which friction is maximum ?

(A) Static Friction

(B) Limiting Friction

(C) Kinetic Friction

(D) None

Ans. (B) Limiting Friction

Q10. Friction force is :

(A) a conservative force

(B) pseudo force

(C) a non-conservative force

(D) centripetal force

Ans. (C) a non-conservative force

Q11. In perfect inelastic collision :

(A) Both Momentum and K. E. are conserved

(B) Both Momentum and Kinetic Energy are not conserved

(C) Momentum is conserved and Kinetic Energy is not conserved

(D) Momentum is not conserved and Kinetic Energy is conserved

Ans. (C) Momentum is conserved and Kinetic Energy is not conserved

Q12. If mass M and radius of solid sphere is R, then moment of Inertia about diameter is :

(A) 2/3 MR²

(B) 2/5 MR²

(C) 3/5 MR²

(D) 7/5 MR²

Ans. (B) 2/5 MR²

Q13. The linear velocity v of a body rotating about an axis with angular velocity ω, then :

(A) v = rα

(B) v = rω

(C) ω = vr

(D) r = vω

Ans. (B) v = rω

Q14. 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 above

Ans. (B) will decrease

Q15. 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

Q16. If a process takes place in a system at a constant volume, the process is known as :

(A) isochoric

(B) isothermal

(C) Adiabatic

(D) isobaric

Ans. (A) isochoric

Q17. 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

Q18. The time period of a particle executing simple harmonic motion is :

(A) T = 2π √Displacement/Acceleration

(B) T = 2π √g/Displacement

(C) T = 2π √Velocity/Displacement

(D) T = 2π √g × Displacement

Ans. (A) T = 2π √Displacement/Acceleration

Q19. ……….. is defined as the mass of a Platinum-Iridium cylinder kept in Paris.

Ans. Kilogram

Q20. The angles of projection for which a projectile covers the same horizontal range, are 30° and ………..

Ans. 90°-30° = 60°

Q21. By using Ball Bearings, sliding friction can be converted into ………… friction.

Ans. Rolling

Q22. The ……….. of a body is defined as its capacity of doing work.

Ans. Energy

Q23. In the universe each particle of matter attracts every other particle. This universal attractive force is called ………….

Ans. Gravitation

Q24. 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

Q25. …………. of work is equivalent to 1 calorie of heat.

Ans. 4.18 Joule

Q26. The number of degrees of freedom of a molecule of a monoatomic gas is ……….

Ans. Degree of freedom = 3N – R = 3(1) – 0 = 3

Q27. 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 …………..

Ans. Forced Vibrations

Q28. If a vector A is multiplied by a scalar m, what will be the resultant vector ?

Ans. mA (:. A means →Vector A)

Q29. What is relation between coefficient of static friction and the angle of friction ?

Ans. μ = tanθ

Q30. What is unit of work in C. G. S. system ?

Ans. Erg

Q31. Write down the formula for work done in rotatory motion.

Ans. w = iθ

Q32. What is time period of Geo-stationary Satellite ?

Ans. 24 hours

Q33. Write relation between coefficient of linear expansion (α) and coefficient of volume expansion (r) of a solid material.

Ans. α = r/3 or r = 3α

Q34. When 100 Joule of heat is given to a gaseous system, then internal energy increases by 30 Joule. Find the work done by the system.

Ans. 70 Joule

Q35. The equation of a simple harmonic motion is y = 5 sin 200t. Give value of amplitude of oscillation.

Ans. A = 5

Subjective Questions

Q2. If a physical quantity is x = a³b²/c and the percentage errors in the measurement of a, b and c are 1%, 2% and 3% respectively. Find the maximum percentage error in the measurement of physical quantity x.

Ans. ∆x/x = 3(∆a/a) + 2(∆b/b) + ∆c/c

= 3(1) + 2(2) + 3 = 3 + 4 + 3 = 10%

Q3. 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

Q4. Write Pascal’s Law.

Ans. This law states that pressure in a fluid in equilibrium is the same everywhere, if the effect of gravity is neglected.

Q5. 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.

Q6. Draw acceleration-time graph of a particle executing simple harmonic motion.

Ans.

Q7. A projectile is thrown at an angle θ from the horizontal with velocity u under the gravitational field of Earth. Find expression for time of flight.

Ans.

Suppose, the body takes time t to reach the highest point P of its path.

In equation (v − u + at)

0 = usinθ − gt

t = usinθ/g

Time of flight T, 2T = 2usinθ/g

Q8. State the laws of Limiting Friction.

Ans.

Laws of Limiting Friction :

(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.

Q9. Obtain expression for the centre of mass of a system consisting of two particles.

Ans. rcm = (m1r1+m2r2)/(m1+m2)

Q10. Explain, Second Law of Thermodynamics.

Ans. There are two equivalent statements of the second law of thermodynamics.

Kelvin-Planck statement : It is impossible to convert all the heat extracted from a hot body into work.

Clasius statement : It is impossible to transfer heat from a cold body to a hot body without expenditure of work by an external energy source.

Q11. Obtain the formula for the escape velocity of a body from the Earth.

Ans.

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

Q12 State and prove Bernoulli’s theorem.

Ans. 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

OR

State and prove Newton’s Law of Cooling.

Ans. According to Newton’s law of cooling, when a hot body is cooled in air, the rate of loss of heat from the body is proportional to the temperature difference between the body and its surroundings.

Q13. Obtain an expression for the time-period of a simple pendulum.

Ans. Simple pendulum–

Let

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

OR

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 sin2πn1t

y2 = a sinw2t = a sin2πn2t

By principle of super position

y = y1 + y2

y = a sin2πn1t + a sin2πn2t

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-n2

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-n2

Hence difference of frequency of sound sources = n1-n2