HBSE 12th Model Paper Physics | Haryana Board 12th Physics Sample Paper
Looking for HBSE 12th Model Paper Physics Pdf? If Yes, here we have published HBSE 12th 2023 Model Question Paper Solution for Physics Subject.
Students of HBSE (12th) Board can download from here www.bseh.org.in Model Paper Physics. Haryana 12th Class Physics Model Paper / Sample Paper will help students to prepare for Board Exam 2023.
Physics
Section – A
1.) An electric dipole consisting of charges +q and –q separated by a distance L is in stable equilibrium in a uniform electric field E → . The electrostatic potential energy of the dipole is:
(a) q LE
(b) zero
(c) – q LE
(d) –2 q LE
Ans –(c) – q LE
2.) For a glass prism, the angle of minimum deviation will be smallest for the light of:
(a) red colour
(b) blue colour
(c) yellow colour
(d) green colour
Ans –(a) red colour
3.) Photons of energies 1 eV and 2 eV are successively incident on a metallic surface of work function 0.5 eV. The ratio of kinetic energy of most energetic photoelectrons in the two cases will be:
(a) 1 : 2
(b) 1 : 1
(c) 1 : 3
(d) 1 : 4
Ans –(c) 1 : 3
4.) A cell of internal resistance r connected across an external resistance R can supply maximum current when
(a) R = r
(b) Rr
(c) R =
(d) R = 0
Ans –(d) R = 0
5.) In a current carrying conductor the ratio of electric field and the current density at a point is called
(a) Resistivity
(b) Conductivity
(c) Resistance
(d) Mobility
Ans –(a) Resistivity
6.) Above curie temperature, a
(a) ferromagnetic material becomes diamagnetic
(b) ferromagnetic material becomes paramagnetic
(c) paramagnetic material becomes ferromagnetic
(d) paramagnetic material becomes diamagnetic
Ans –(b) ferromagnetic material becomes paramagnetic
7.) Displacement current exists only when
(a) electric field is changing
(b) magnetic field is changing
(c) electric field is not changing
(d) magnetic field is not changing
Ans –(a) electric field is changing
8.) An electron is released from rest in a region of uniform electric and magnetic fields acting parallel to each other. The electron will:
(a) move in a straight line
(b) move in a circle
(c) remain stationary
(d) move in helical path
Ans –(a) move in a straight line
9.) When two nuclei (A ≤ 10) fuse together to form a heavier nucleus, the
(a) binding energy per nucleon increases
(b) binding energy per nucleon decreases
(c) binding energy per nucleon does not change
(d) total binding energy decreases
Ans –(a) binding energy per nucleon increases
10.) At equilibrium, in a p-n junction diode the net current is–
(a) due to diffusion of majority charge carrier
(b) due to drift of majority charge carrier
(c) zero as diffusion and drift currents are equal and opposite
(d) zero as no charge carrier cross the junction
Ans –(c) zero as diffusion and drift currents are equal and opposite
11.) In a n-type semiconductor, the donor energy level lies–
(a) at the centre of the energy gap
(b) just below the conduction band
(c) just above the valence band
(d) in the conduction band
Ans –(b) just below the conduction band
12.) The electric flux emerging out form IC charge is
(a) 1/ ∈0
(b) 4π
(c) 4π/ ∈0
(d) ∈0
Ans –(a) 1/ ∈0
13.) In Bohr’s model of hydrogen atom, the total energy of the electron in nth discrete orbit is proportional to
(a) n
(b) 1/n
(c) n2
(d) 1/n2
Ans –(d) 1/n2
14.) In an a.c. circuit the applied voltage and flowing current are E = E0 sinωt and I = I0 sin (ωt + 2 π ) respectively. What is the average power consumed in one cycle in this circuit?
(a) Ev Iv
(b) zero
(c) ∞
(d) –1
Ans – (b) zero
Directions (15-18): Two statements are given one labelled Assertion (A) and other labelled Reason (R). Select the correct answer from codes (a), (b), (c) and (d) given below:
(a) Both A and R are true and R is correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false and R is also false
15.) Assertion (A) : A dentist uses a concave mirror to examine a small cavity.
Reason (R) : A dentist uses a concave mirror so as to form a magnified, virtual image of an object.
Ans –(a) Both A and R are true and R is correct explanation of A.
16.) Assertion (A) : In a series combination of capacitors, charge on each capacitor is same.
Reason (R) : In such a combination, charge can not move only along one route.
Ans –(a) Both A and R are true and R is correct explanation of A.
17.) Assertion (A) : The bending of an insulated wire increases the resistance of wire.
Reason (R) : The drift velocity of electrons in bent wire decreases.
Ans –(d) A is false and R is also false
18.) Assertion (A) : Large angle scattering of alpha particles led to the discovery of atomic nucleus.
Reason (R) : Entire positive charge of atom is concentrated in the central core.
Ans – (a) Both A and R are true and R is correct explanation of A.
SECTION -B
19.) Draw energy band diagrams of n-type and p-type semiconductors at temperature T > 0 K, depicting the donor and acceptor energy levels. Mention the significance of these levels.
Ans –
Significance
n-type semiconductor– small energy gap b/w donor level and conduction band which can be easily covered by thermally excited electrons. ½ p-type semiconductors– small energy gap b/w acceptor level and valence band which can be easily covered by thermally excited electrons.
OR
Discuss the formation of depletion layer in p-n junction diode.
Ans –When p-type semiconductor is joined with n-type semiconductor, e from the n-side diffuse towards p-side and holes from p-side diffuse towards n-sides leaving behind a layer of immobile +ve ions on n-side and immobile –ve ions on p-side leading to formation of depletion layer.
(Note: Award 1 mark, if a student draws a diagram showing depletion layer)
20.) Arrange the following electromagnetic radiation in the ascending order of their frequencies: X-rays, microwaves, gamma rays, radiowaves Write two uses of microwaves.
Ans –Radiowaves < microwaves < X-rays < gamma rays
Uses of microwaves
i.) Microwave oven
ii.) in Rader system
21.) (i) Define the SI unit of power of a lens.
(ii) A plano convex lens made of glass of refractive index 1.5. The radius of curvature of the convex surface is 25 cm. Calculate the focal length of the lens.
Ans –One Dioptre is the power of a lens whose focal length is one metre.
1 f = (μ-1)
1 /R1 – 1/ R2
R₁ = ∞, R₁ = -25 cm, μ = 1.5
1/f = (1.5-1) (1/∞ + 1/25)
f = 50 cm
22.) Derive expression for energy stored in a capacitor.
Ans – as V = q/ C
dW is the W.D. in giving additional charge dq
dW = V dq = q/C dq
W= ∫ Dw = 1/C ∫Q0 q dq = 1/C[q2/2]Q0
= Q2/2C
This work done is stored in the form of energy of capacitor.
E= Q2/2C=1/2CV2=1/2QV |
OR
A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7 C distributed uniformly on its surface. What is the electric field
(a) inside the sphere
Ans –Inside a conductor the electric field is zero because charges resides on the surface of sphere.
(b) just outside the sphere
Ans – E= 1/4π∈0 q/R2
E= 9 X 109 X (1.6X10-7)/(12X10-2)2
= 105 NC-1
23.) Explain the limitations of ohm’s law?
Ans –There are some devices which do not obey ohm’s law and have non-linear I-V characteristics are called non-ohmic devices. 1 e.g., semiconductor devices like p-n junction diode, thermistors etc.
V/ I = R ≠ constant
(for p-n junction diode)
24.) Bar magnet acts as an equivalent solenoid. How?
Ans – Magnetic field lines for a bar magnet and a current carrying solenoid resembles very closely.
Magnetic field on the axial line of a bar magnet is equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field.
B = µ0/4 π 2M/r3
25.) How total internal reflection is used to bend rays by 90° and 180° or to invert image without changing its size in prisms?
Ans – Totally reflecting prisms are used to change the path of a ray by 90° or 180°. These are the right angled glass prisms of refractive index 1.5 and critical angle 41.8°. In totally reflecting glass prisms, angle of incidence is made 45° (> C). Hence light suffers total internet reflection.
SECTION C
26.) Draw Schematic diagram of a reflecting telescope (cassegrain). Write its two advantages over refracting telescope.
Ans –
Advantages: 1. There is no chromatic aberration as the objective is a mirror. 2. Image is bright as compared to refracting type telescope.
OR
In a Young’s double slit experiment using light of wavelength 600 nm, the slit separation is 0.8 mm and the screen is kept 1.6 m from the plane of the slits. Calculate:
(i) the fringe width
Ans – β= λD/d = 600x 10-9x1.6/0.8×10-3
=1.2mm
(ii) the distance of (a) third minimum and (b) fifth maximum from the central maximum.
Ans – (a) X3 = 5/2 λd/d = 5/2×1.2
=3mm
(b) X3 = 5λD/d= 5/2X1.2
=6.0mm
27.) Derive an expression for impedance of an a.c. circuit containing series LCR. Also draw the impedance triangle.
Ans –
Let I = I0 sin ωt
VR = IR represented by OR
VL = I XL represented by OB
VC = I XC represented by OC
(OP) 2 = (OR) 2 + (PR) 2
V2 = VR 2 + (VL – VC) 2
E = V = I √ R2+( XL – XC)2
I= E/ √R2 + (XL –XC)2
Here Z= √R2 + (XL – XC)2= Impedance of the circuit
Resistance offered by the L, C and R to the flow of current.
OR
Calculate the frequency of series resonance circuit. Find out the formula of Q-factor of this circuit.
Ans – At low frequency XL is small but XC is high and at high frequency XL is high but XC is small. But at a frequency when XL = XC called as resonant frequency Impedance of LCR is minimum and current through the circuit is maximum.
XL = XC
ωL = 1/ ωC
ω2 = 1/LC
ω = 1/ √ LC ω-Resonating angular frequency
v =1/ 2π √LC
I0=E0/R (max. at resonance)
Q-factor is the sharpness of curve at resonance
Q-factor = Potn drop across or C/ Potn drop across at resonance
=I0 XL/I0 R= ωL/R AND ω=1/ √ LC
Q-factor = 1/R √L/C |
28.) How energy bands (conduction band and valance band) are formed in solids?
Ans –
The process of splitting of energy levels can be understood by considering the different situations.
(i) When r = d – No modification of energy levels.
(ii) When r = c – Interaction b/w outermost shell electrons of neighbouring Si atoms increases and energy gap b/w 2N – 3s levels and 6N – 3p levels goes on decreasing.
(iii) When b < r < c – Instead of single 3s and 3p level, we get large no. of closely packed level.
(iv) When r = b > a – The energy gap b/w 3s and 3p levels completely disappears and all the 8N levels are continuously distributed. One can say that 4N levels are filled and 4N levels are empty.
(v) When r = a – 4N filled levels get separated from 4N empty levels. 4N filled level form a band called valance band and 4N empty levels form a band called conduction band.
29.) Derive an expression for electric field at a point on the equitorial line of an electric dipole.
Ans – Let us consider two charges –q and +q separated by certain distance 2a form a dipole of moment p = q (2a)
30.) Calculate the current in each branch of the network show in fig.
Ans – Apply KVL on loop EABCE
10 = 10 (I1 + I2 ) + 10 I1 + 5 (I1 – I3 )
2 = 5I1 + 2I2 – I3
Apply KVL on loop ABDA
10 I1 + 5 I3 – 5 I2 = 0
I2 = 2 I1 + I3
Apply KVL on loop BCDB
5(I1 – I3) = 10 (I2 + I3) – 5 I3 = 0
I1 = 2I2 + 4 I3
On solving we get
I1 = 4/17 A, I2=6/17A/ I3=-2/17A
I1+I2 = 10/17A
I1 – I3 =6/17A
I2+I3= 4/17A
SECTION D
(CASE STUDY)
31.) The large scale transmission and distribution of electrical energy over long distance is done with the use of transformer. The voltage output of the generator is stepped-up (so that current is reduced and consequently the FR loss is cut down). It is then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our home.
(i) A transformer works on the principle of:
(a) Converter
(b) Inverter
(c) Mutual Induction
(d) Self induction
Ans – (c) mutual Induction
(ii) Quantity that remains unchanged in a transformer is
(a) voltage
(b) current
(c) frequency
(d) none of these
Ans – (c) frequency
(iii) The best material for the core of a transformer is
(a) stainless steel
(b) mild steel
(c) hard steel
(d) soft iron
Ans – (d) soft iron
(iv) In a transformer, number of turns in the primary is 140 and that in secondary is 280. If current in primary is 4A, then that in the secondary is 1
(a) 4A
(b) 2A
(c) 6A
(d) 10A
Ans – (b) 2A
OR
The core of any transformer is laminated, so as to:
(a) reduce the energy loss due to eddy currents
(b) make it light weight
(c) make it robust and strong
(d) increase the secondary voltage
Ans – (a) reduce the energy loss due to eddy currents
32.) Photoelectric effect thus gave evidence to the strange fact that light in interaction with matter behaved as if it was made of quanta or packets of energy, each of energy hv. Is the light quantum of energy to be associated with a particle? Einestein arrived at the important result that the light quantum can also be associated with momentum (hv/c). This particle was later named photon. Each photon moves with speed of light c. Photons are electrically neutral and are not deflected by electric and magnetic fields.
(i) Which one among the following shows particle nature of light?
(a) Photoelectric effect
(b) Interference
(c) Refraction
(d) Polarization
Ans – (a) Photoelectric effect
(ii) Which of the following statement about photon is incorrect?
(a) photons exert no pressure
(b) Momentum of photon is
(c) Energy of photon is hv
(d) Photons are electrically neutral
Ans – (a) Photons exert no pressure
(iii) The rest mass of photon is
(a) hv/c
(b) hv/c2
(c) hv/ λ
(d) zero
Ans – (d) zero
(iv) In a photon-particle collision (such as photon-electron collision), which of the following may not be conserved?
(a) Total energy
(b) No. of photons
(c) Total momentum
(d) both (a) and (b)
Ans – (b) No. of photons
OR
Calculate number of photons in 6.62 J of radiation energy of frequency 10″ Hz.
Given h = 6.62 x 10-34 Js.
Ans E = n hv or n = E/ hv
n = 6.62/ = (6.62x 10-34)x 1012
=1022
SECTION E
33.) Draw a Schematic arrangement of the Geiger-Marsden experiment. Write down the observations of scattering of alpha-particle. What is distance of closest approach and write down its formula.
Ans –
Observations:
1.) Most of the α-particles pass without deflection.
2.) Large no. of α-particles are scattered by very small angle.
3.) Small no. of α-particles are scattered by large angle.
4.) A very few no. of α-particles (1 in 8000) retraced their path.
Distance of closest approach: It is the minimum distance of α-particle from nucleus at which K.E. converts into P.E
OR
(i) With examples differentiate between nuclear fission and nuclear fusion?
Ans – (A > 200) into two or more lighter nuclei.
Nuclear fussion: It is the phenomenon of fusing two or more lighter nuclei to form a single heavy nucleus
2H1 + 2H1 → 4He2 + 24 Mev
(ii) How De Broglie explains Bohr’s second postulate of Quantistaion.
Ans – Acc. to de Broglie, a stationary orbit is that which contains an integral number of de Broglie waves associated with revolving electron
Total distance covered 2 π r
which is the quantum condition proposed by Bohr in second postulate
34.) Explain principle, construction and working of moving coil galvanometer by drawing its diagram.
Ans –
Principle: When a current carrying coil placed in magnetic field, it experiences a torque.
Construction: It consists of a rectangular coil PQRS of large no. of turns of insulated copper wire wound over a non-magnetic material frame. A soft iron cylindrical core is placed such that coil can rotate without touching it. Coil is suspended b/w two cylindrical magnets by a phosphor bronze wire. Upper end of the coil is connected to movable torsion head and lower end is connected to hair spring
Working: Function of cylindrical core and magnet is to provide radial magnetic field.
τ = n I AB
If k is the restoring torque per unit twist and θ be the twist in the wire.
In equilibrium
OR
Derive expression for torque on a rectangular current loop in a uniform magnetic field.
Ans –
35.) (i) Derive mirror formula for concave mirror when image formed is real?
Ans –
(ii) An object is placed at 10 cm in front of a concave mirror of radius of curvature 15 cm. find the position, nature and magnification of the image.
Ans : u = – 10 cm
f = R/2=15/2 cm
1/ f = 1/u + 1/v
-2/ 15 = 1/10 + 1/v
1 /v = -2/15 +1/10
v = – 30 cm
Image is real and inverted
m = -v/u=-(-30)/(-10)=-3
Image is magnified 3 times
Or
What is refraction of light? Prove that µ2/- µ+ µ1/v = µ1– µ2/R when refraction occurs form denser to rarer medium at a convex spherical refracting surface.
Ans – Refraction of light is the phenomenon of change in path of light when it goes from one medium to another.
Refraction from denser to rarer at convex spherical refracting surface.
Apply Snell’s law
Physics Marking Scheme 1 | |
Physics Model Paper 2 | |
Physics Part 1 |
Click Here |
Physics Part 2 |