HBSE Class 12 Sample Paper 2027 Physics Marking Scheme

FOR HARYANA CLASS 12 PREPARATION

HBSE Class 12 Sample
Paper 2027 Physics
Marking Scheme
EXAM YEAR TYPE SUBJECT

Haryana Class 12 2027 Answer Key Physics

Notes · Sample Papers · Previous Year Papers · Mock Tests

co m .
. s e
em l a

as a g

BSEH MODEL PAPER (2026-27)
Marking Scheme Physics

m
co
section-A

c om m .
1. (a)
2. (a)

m . se
3. (a)

s e g l a
l a a
4. (d)

ag
5. (b)
6. (c)
7. (b)
8. (a)
9. (a)
10. Diffraction.
11. scaler
→ →
12. The direction of magnetic force is ⊥ to plane containing v and B or is in the direction of

m
co

 

.
V
B .

em



s
h

a
13. λ = .

g l
p
14. Balmer.
15. decreases. a
16. (b)
17. (b)
18. (a)

section-B
19. Total charge present on a body is fixed. Total charge present on a body is intergral number

m
of the smallest unit of charge i.e. electronic charge.

c. o
Q = ± ne

m
c. o m
n- integer

s e
e-1.6 × 10–19C.

s em la
At macroscopic level, total charge present on a body is much larger than the charge on an

la ag
electron. So, grainry nature of electrons destroy. Hence, it is

g
ignored at macroscopic level.

a
Or
qA = 2.5 × 10–7 C
qB = –2.5 × 10–7 C
q = qA + qB = 0
AB = 2 a = 15 + 15
= 30 cm = 0.30 m

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q = q (2a)
= 2.5 × 10–7 × 0.30
= 7.5 × 10–8 C-m along –ve z-axis.
20. According to the Kirchhoff's first rule, the algebraic sum of various currents meeting at a
junction in a closed electrical circuit is always zero. i.e.,
SI = 0
It is based on the law of conservation of charge.
According to Kirchhoff's second rule, the algebraic sum of emfs is equal to algebraic sum
of products of various resistances and the respective currents flowing through them in a
closed electrical circuit i.e.,
SE = SIR
It is based on the law of conservation of energy.
Or
R27 = 100 W
RT = 117 W
a = 1.70 × 10–4/°C
The temperature coefficient of resistance is given by
dR/R0 RT − R27
a= =
dT R27 (T − 27)
117 − 100
1.70 × 10–4 =
100 (T − 27)
1.70 × 10–2 (T – 27) = 17
17
T – 27 =
1.7 × 10 −2
T = 1000 + 27 = 1027 °C.
21. When a very high value of resistance is connected in series with galvanometer then it
works as a voltmeter as shown in figure. 2

V = measured voltage
Ig = full scale deflection current
R = high value of resistance
G = resistance of galvanometer
R and G are in series combination and its total resistance is R + G. The measured voltage
V is given by
V = Ig (R + G)
V
R+G =
Ig
V
R= −G
Ig

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co m .
. s e
em l a

as a g

The effective resistance of converted voltmeter is given by
Reff = R + G
22. Microwaves are electromagnetic waves of frequency range 1 GHz to 300 GHz. 2

m
co
Uses of Microwaves:

om .
(i) Microwave ovens are used for cooking.

.c e m
(ii) Microwaves are used in radar systems for aircraft navigation.

em a s
(iii) Microwaves are used for observing the movement of trains.
23.
a s g l
l a
ag

 2

m
co
Position of the image doesnot change and intersity (brighness) of the image will decrease,

.
if the lower half of the mirror’s reflecting surface is painted black.

em
OR

a s
Dentists use concave mirrors because concave mirror produces a highly magnified, virtual

l
g
and erect image of tooth. Convex mirrors are used as side-view mirrors in car because

a
convex mirrors form a virtual, erect and diminished image, it allows a driver to see a much
wider field of view.
24. When pentavalent impurity is added into a pure semiconductor, we get n-type
semiconductor. On the other hand, if trivalent impurity is added into a pure semiconductor,
we get p-type semiconductor. Electrons are the majority charge carriers in n-type and holes
are the majority carriers in p-type.
25.

m
m c. o
c. o s e m
s em la
gla ag
a 1
As r0 ∝
K .E.
When the K.E. of a-particle is doubles, r0 becomes halved.

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section-C
26. (a) Mass of rectants = 2 × 12.000000 = 24.000000 u
Mass of products = 19.992439 + 4.002603
= 23.995042 u
Mass defect (Dm) = 24.000000 u – 23.995042 u
= 0.004958 u
Since Dm > 0, mass is converted to energy, making it an exothermic reaction.
(b) Ratio of densities = 1 : 1.
Or
According to Bohr's postulate, electron can revolve only in certain discrete non-radiating
orbits called stationary orbits for which total angular momentum of the revolving electron
is an integral multiple of h/2p where ‘h’ is Planck’s constant.
nh
mvr =
2p
The energy of electron revolving in a stationary orbit is of two types: kinetic energy which
is due to the velocity and potential energy which is due to the position of electron. Total
energy of electron in its orbit is given by
E = K.E. + P.E. …(i)
Potential energy of electron is given by
kZe 2
P.E. = −
r
According to Bohr’s first postulate,
mv 2 kZe 2 1 kZe 2
= ⇒ mv 2 =
r r2 2 2r
From eqn. (i), we have
kZe 2 kZe 2 1 kZe 2
E= − = − …(ii)
2r r 2 r
n2 h 2
But, r=
4 p mkZe 2
2

Now, substituting value of r in eqn. (ii) we have
−1 4 p2 mkZe 2
E= kZe 2 .
2 n2 h 2
−2 p2 mk 2 Z 2 e 4 −13.6
= 2 2
= eV
n h n2
27. Principle: It is based on the principle of electromagnetic induction.
Let N = no. of turns in the coil
A = area of each turn
→
B = strength of mag field
→ →
f = N( B . A) = NBA cos q
= NBA cos wt

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co m .
. s e
em l a

as a g

df −d
e= (NBA cos wt)
dt dt
= NBAw sin wt

m
co
e will be max. if sin wt = 1


c om m .
\ emax = e0 = NBAw

.


m
e = e0 sin wt
se
a
\

e


a s
Or
g l
l a
ag
Transformer is based on the principle of mutual induction. Step-up transformer converts
low input voltage into high output voltage. Step-down transformer converts high input
voltage into low output voltage.
Factors responsible for energy loss in transformer:
(i) Copper loss.
(ii) Eddy current loss (Any two).
28. The phenomenon of emission of electrons from the metal surface when electromagnetic
radiation of suitable frequency falls on it, is called photoelectric effect. The emitted electrons
are called photoelectrons and the current so produced is called photoelectric current.

m
. co
s em
g l a
a
Stopping Potential:
The minimum negative potential given to anode for which the photoelectric current
becomes zero is called cut-off potential or stopping potential. Its value is different for
different metals and is independent of the intensity of incident light.
29. (i) It is the phenomenon of reflection of light into a denser medium from an interface of

m
this denser medium and a rarer medium.

c. o
Two essential conditions of TIR:

m
c. o m
1. Light should travel from denser to rarer medium.

s e
2. Angle of incidence in denser medium should be greater than critical angle for the

em la
pair of media in contact.

las ag
ag

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1 1
(ii) m= = = 2
Sin C Sin 45°
C C
m= ⇒ v=
v m
3 × 108
\ v= m/s.
2
30.

The threshold voltage (or cut-in voltage) of a diode is the minimum forward voltage
required to overcome the built in potential barrier, allowing significant current to flow
through the diode.
For Si = 0.7 V
For Ge = 0.3 V.
section-D (Case Study)
I
31. (i) (a) .
A
(ii) (c) 0.1 W
(iii) (b) [ML3 T–3 A–2]
Or
(c) CD.

E 
(iv) r =
 1 R

V 
32. (i) (b) 311 V
(ii) (c) Zero.
(iii) (a) 0
(iv)

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co m .
. se
em l a

as ag

Or
XL = w L = 2pn L
22
= 2   100  14  10
3

m
7
= 8.8 W.

om . co
.c e m
33. The ratiom s
section-E

s e g l a
a
of electric force in free space (F ) to electric force in medium (F) is called dielectric

l a
0

g
constant.
F

a Consider two parallel plates each of area A and separated by small distance d. The plates of
0
K=
F

capacitor are oppositely charged. The electric field across parallel plates without dielectric
slab is given by
q
E0 = ...(i)
∈0 A
The electric potential difference across parallel plates capacitor without dielectric slab is
given by

m
V0 = E0d ...(ii)
Substituting the value of E0 in eqn. (ii), we have
. c o
∈ eAm
q

s
V = d

q la∈ A
0

g= d
0

Va
0

0

∈0 A  q 
C0 = ...(iii)  C0 = 
d  V0 
When a dielectric slab of dielectric constant k, thickness t and have same area as plates
of capacitor is placed between parallel plate capacitor then induced electric field Ep is
produced in opposite direction of E0 and the dielectric constant is given by
E0 E
= 0 ...(iv)

m
k =

c. o
E0 − Ep E

m
c. o
+ + + +

e m
E0

em a s
E = E0 – Ep Ep

l
d t

las g
a with dielectric slab is
g
– – – –

a
The electric potential difference across the parallel plate capacitor
given by
V = E (d – t) + Et 0
 E 
V = E0 d − t + t ...(v)
 E0 

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Substituting the value of E0 and k in the eqn. (v), we have
q t q ∈0 A
V =  d−t+  ⇒ =
∈0 A  k V t
d−t+
k
∈0 A  q
C =  C = 
 t t   V
d 1 − + 
 d kd 
C0  ∈0 A 
∴
C =  C0 = 
 t t   d 
d 1 − + 
 d kd 
Thus the capacitance of parallel plate capacitor with dielectric slab increases.
Or
1
(i) It states that the surface integral of electric field over closed surface is equal to
∈0
times the charge enclosed in closed surface.
  q

E . ds = ∈ . 0

WAB
(ii) VB – VA =
q0
If points A and B lie on equipotential surface, then VB = VA
WAB = 0
Hence no work is done in moving the test charge from one point of equipotential
surface to the other.
(iii) Suppose an electric dipole consists of two opposite charges (i.e., q and – q) which are
separated by small gap i.e., 2a. Let P be a point on axial line at distance r from the
centre of dipole O as shown in figure.

BP = r – a
AP = r + a
The electric field E1 at point P due to positive charge is given by
q
E1 =
4 p ∈0 (BP)2
q
= along +ve X axis
4 p ∈0 (r − a)2
The electric field E2 at point P due to negative charge is given by
q
E2 =
4 p ∈0 ( AP)2
q
= along –ve X axis
4 p ∈0 (r + a)2

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co m .
. s e
em l a

as ag

The electric field intensity at point P on axial line of dipole is the resultant of E1 and E2.
∴ E = E1 – E2 along +ve X axis
q q
−

m
=

co
4 p ∈0 (r − a)2 4 p ∈0 (r + a)2

c om m .
q  1 1  q  ( r + a) 2 − ( r − a) 2 

. e
=  −  =  

m s
4 p ∈0  (r − a)2 (r + a)2  4 p ∈0  ( r 2 − a 2 )2 

s e g l a
a a
q  r 2 + a 2 + 2 ar − r 2 − a 2 + 2 ar 

l
=

g
 
( r 2 − a 2 )2

a
4 p ∈0  
q  4 ar  2 (2 aq) r
=  2 2 2
=
4 p ∈0  (r − a )  4 p ∈0 (r 2 − a 2 )2
2 pr
=
4 p ∈0 (r 2 − a 2 )2
34. (i) Let us consider a conductor of length l, carrying current I kept in uniform magnetic
→
field B .

m
. co
s em
l a
ag per unit volume, then total no. of electrons in the
If n is the No. of free electrons
conductor is
N = n Al
Force experienced by each electron,
→  
F = – ne Al
 Vd
B 


→ →  
Total force F = N f = – ne Al
 Vd
B  ...(i)



m
m c. o
We know that I = n eA Vd

c. o Put in (i) m
→ →

e
I l = – ne Al Vd

em la s
las →  

ag
g
F = I ( l
B)

a
→
F = I lB Sin q
If q = 90°
Sin 90° = 1
\ F = Ilb = maximum.

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(ii)
Diamagnetic substance Paramagnetic substance
1. They are feebly repelled by magnets 1. They are feebly attracted by magnets
2. Susceptibility is small and negative 2. Susceptibility is small and positive.
(Write any two differences).
OR
Consider a rectangular coil of length l breadth b.
SP = QR = l; PQ = RS = b
q angle made by plane of coil with mag. field.
→
F1 = IbB sin q on RS acting ⊥ to both acting upward given by Right-hand Thumb
Rule
→
F2 = IbB sin q on PQ acting downward given by Right-hand Thumb Rule.
→ →
F1 & F2 are equal and opposite hence cancel each other.
→
F3 = I lB sin 90° = IlB acting on QR inwards by Fleming Left-hand Rule.
→
F4 = I lB sin 90° = IlB acting on SP outwards by Fleming Left-hand Rule.

→ →

F3 and F4 are equal but not acting along same line and form a couple which moves the
coil in magnetic field.
Torque = force × arm of couple
t = I lB (RT)
= I lB b cos q
t = I AB cos q

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co m .
. s e
em l a

as a g

m
om . co
.c e m
e m l a s
l a s ag
g
a A = lB = area of coils
where
Let a is the angle b/w normal to the coil and magnetic field
then a + q = 90°
\ q = 90 – u
If n– no. of turns then
\ t = n I AB cos (90 – a)
= n I AB sin a

m
t = MB sin a M-mag dipole moment


  
t = M
B.
. c o
m
etwo waves be represented as
s
35. (i) Two sources must be Coherent sources of light for sustained interference of light.

a
l a sin wt and y = a sin (wt + f)
(ii) Let the electric field vectors of the

g
a andaa = amplitudes of two waves
y = 1 1 2 2
where 1 2
f = phase difference between the waves
The resultant wave at point P is given by
Y = y1 + y2 = a1 sin wt + a2 sin (wt + f)
Y = a1 sin wt + a2 sin wt cos f + a2 cost wt sin f
Y = (a1 + a2 cos f) sin wt + a2 sin f cos wt...(i)
Substitute a1 + a2 cos f = A cos q...(ii)

m
and a2 sin f = A sin q...(iii)

m c. o
Equation (i) becomes

c. o m
Y = A cos q sin wt + A sin q cos wt

s e
em
Y = A (sin wt cos q + cos wt sin q)

s la
Y = A sin (wt + q)...(iv)

ag


la
Now squaring and adding equations (ii) and (iii), we have

ag

(A sin q)2 + (A cos q)2 = (a2 sin f)2 + (a1 + a2 cos f)2
A2 = a22 sin2 f + a12 + a22 cos2 f + 2a1a2 cos f
A2 = a22 (sin2 f + cos2 f) + a12 + 2a1a2 cos f
A2 = a12 + a22 + 2a1a2 cos f...(v)
The resultant intensity I is directly proportional to the square of the amplitude of the
resultant wave, therefore
I ∝ A2
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I = KA2
Thus, the resultant intensity at a point P depends on the phase difference f.
Condition for constructive interference: The intensity will be maximum if
cos f = 1 i.e., f = 0, 2p, 4p
or f = 2np where n = 0, 1, 2, 3, ...
Condition for destructive interference: The intensity will be minimum if
cos f = –1 i.e., f = p, 3p, ...
or f = (2n – 1)p where n = 0, 1, 2, 3, ...
Or
Consider a point object O lying on the principal axis of surface. A ray of light OP passes
straight and another ray OA refracted along AI. Both the rays meet at I to form real image
of object O.

Let ∠AOM = a, ∠AIM = b and ∠ACM = g
In D IAC
g = r+b
 (External angle of triangle is equal to interior angles)
r = g–b
In D OAC
i = a+g
By snell's law,
m2 sin i
=
m1 sin r
As angles are small, therefore
m2 i
=
m1 r
m1 i = m2 r
Putting values of ’i’ and ’r’ from above equation, we have
m1 (a + g) = m2 (g – b)
As angles a, b and g are also small. Therefore,
AM AM AM
a= , g= , b=
MO MC MI
 AM AM   AM AM 
So, m1  +  = m2  +
 MO MC   MC MI 
 1 1   1 1 
m1  + = m2  −
 MO MC   MC MI 

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co m .
. s e
em l a

as a g

As aperture of surface is small.
Therefore, MO = PO, MI = PI, MC = PC
m1 m m m
+ 1 = 2 − 2

m
co
PO PC PC PI

om .
Using PO = – u, PI = v and PC = R

.c e m
m1 m 2 m − m1

s
we have, + = 2 .

e m a
−u v R

a s g l
l a
ag

m
. co
s em
g l a
a

m
m c. o
c. o s e m
s em la
gla ag
a

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