HBSE Class 12 Physics Half Yearly Question Paper 2022 Answer Key

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HBSE Class 12 Physics Half Yearly Question Paper 2022 Answer Key 

विद्यार्थी का नाम ………. रोल नंबर ……….
अर्धवार्षिक आकलन – सितम्बर 2022
कक्षा – 12
विषय – Physics
समय : 90 मिनट         कुल अंक : 40

निर्देश : 1. सभी प्रश्न अनिवार्य हैं।
2. प्रश्न संख्या 1 से 15 तक प्रत्येक प्रश्न 1 अंक का है। प्रश्न संख्या 16 से 18 तक प्रत्येक प्रश्न 2 अंक, प्रश्न संख्या 19 से 21 प्रत्येक प्रश्न 3 अंक तथा प्रश्न संख्या 22 से 23 तक प्रत्येक प्रश्न 5 अंक का है।
Instructions : 1. All questions are compulsory.
2. Question numbers 1 to 15 carry 1 mark each, question numbers 16 to 18 carry 2 marks each, question number 19 to 21 carry 3 marks each and question number 22 to 23 carry 5 marks each.

(1-10) सही उत्तर पर (✓) का निशान लगाएँ।
Mark (✓) on the correct answer.
1. The SI unit of permittivity of free space is :
(A) C-²Nm-²
(B) C²N-¹m-²
(C) NC²m-²
(D) N-¹C²m²
Answer – (B) C²N-¹m-²

2. With the increase of temperature, resistance of semiconductor :
(A) decreases
(B) increases
(C) may increase or decrease
(D) does not change
Answer – (A) decreases

3. The magnetic susceptibility of paramagnetic substance is :
(A) small and positive
(B) small and negative
(C) large and positive
(D) large and negative
Answer – (A) small and positive

4. The power factor for a purely capacitive circuit is :
(A) 1
(B) √2
(C) 1/√2
(D) 0
Answer – (D) 0

5. Which of the following is same for X-ray and γ-ray :
(A) wavelength
(B) frequency
(C) velocity in vacuum
(D) energy
Answer – (C) velocity in vacuum

6. Mirage is based on :
(A) reflection
(B) refraction
(C) total internal reflection
(D) interference
Answer – (C) total internal reflection

7. Electric field intensity due to short dipole depends on r as :
(A) 1/r³
(B) 1/r²
(C) 1/r
(D) r²
Answer – (A) 1/r³

8. Junction rule is based on law of conservation of :
(A) mass
(B) charge
(C) energy
(D) potential difference
Answer – (B) charge

9. Direction of induced e.m.f. is given by :
(A) Faraday’s first law
(B) Faraday’s second law
(C) Huygen principle
(D) Lenz’s law
Answer – (D) Lenz’s law

10. Among the followings, the wave length is largest for :
(A) X-ray
(B) γ-ray
(C) red light
(D) UV ray
Answer – (C) red light

(11-15) Fill in the blanks.
11. Mathematical form of Gauss law is ………..
Answer : ∮E.ds = q/εo (:. E and ds having vector sign)

12. Phase difference between current and voltage in a.c. circuit containing ‘L’ only is ………….
Answer : π/2 radian

13. Torque is maximum on electric dipole placed in uniform electric field when …………
Answer – dipole should be placed perpendicular to the electric field.

14. Energy stored in capacitor is …………
Answer : ½CV² = ½Q²/C = ½Q/V

15. Value of dip angle at poles is …………
Answer : 90°

(16-18) Write answer of following questions in 20-30 words.
16. How would you connect three capacitors of 6 μF each to get a total capacitance of 2 μF. Also draw a diagram of arrangements.
Answer – Connect all capacitors in series.
1/C = 1/C1 + 1/C2 + 1/C3 = 1/6 + 1/6 + 1/6 = 3/6 = 1/2
C = 2 μF

17. Explain Lenz’s Law.
Answer – Lenz’s law states that the current induced in a circuit due to a change or a motion in a magnetic field is so directed as to oppose the change in flux and to exert a mechanical force opposing the motion.

18. How can you say that our earth behaves like magnet ?
Answer – The crust of the Earth has some permanent magnetization, and the Earth’s core generates its own magnetic field, sustaining the main part of the field we measure at the surface. The deflection of a magnetized needle is due to earth’s magnetic field. So we could say that the Earth behaves like magnet.

(19-21) Write answer of following questions in 30-40 words.
19. A series LCR circuit with R = 44 ohm, C = 4 µF and L = 50 henry is connected to A.C. supply of 220 V. Calculate angular frequency, impedance and current at resonance condition.
Answer – R = 44 ohm, C = 4 µF = 4×10-⁶ F, L = 50
Angular frequency (ω) = 1/√LC = 1/√50×4×10-⁶ = 70.72 rad/s
Impedance (z) = R = 44 Ω
Current (I) = V/R = 220/44 = 5A

20. Calculate electric field intensity due to spherical shell by Gauss theorem at a point outside the shell.
Answer – Let P be the point outside the shell at a distance r from the centre. Since the surface of the sphere is spherically symmetric, the charge is distributed uniformly throughout the surface.

We can say that the electric field at a point outside the shell will remain the same if the entire charge Q is concentrated at the centre of the spherical shell.

21. Write the characteristics of electric field lines.
Answer – Characteristics of electric field lines are following :
• Electric field lines always start from a positive charge and end at a negative charge.
• Electric field lines never cross each other.
• Electric field lines are always perpendicular to the surface of a conductor.
• The direction of the electric field is tangent to the electric field line at any point.
• Electric field lines can be curved or straight, depending on the configuration of the charges producing the field.
• The closer the electric field lines are to each other, the stronger the electric field at that point.

(22-23) Write answer of following questions in 50-60 words.
22. Explain principle and working moving coil Galvanometer.
Answer – 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.

OR

Derive expression for the force per unit length acting on two straight parallel current carrying conductors and hence define 1 ampere of current.
Answer –

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)
One Ampere is defined as the amount of electric current that flows through two parallel conductors of infinite length, which are placed one meter apart in a vacuum, that results in a force of exactly 2×10-⁷ N/m.

23. Draw labeled diagram of microscope and obtain expression for magnifying power when final image is formed at distance of least vision.
Answer –

OR

Draw ray diagram of refracting telescope and derive expression for magnifying power when image is formed at infinity.
Answer – Ray diagram of the image formed at infinity by refracting telescope is shown here which indicates that focal length of objective lens and that of eye piece is fo and fe respectively.
Magnifying power of telescope is defined as the ratio of angle subtended by image on eye (β) to angle subtended by object on eye (α).
Magnifying power of telescope (m) = -β/α
From figure, β = tanβ = A’B’/fe
Also, α = tanα = A’B’/fo
β/α = fo/fe
m = -fo/fe

 

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