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serena yoy ete badocemaconehncaie
Time: 3 Hours]
PHYSICS (Max. Marks: 70
Q.1, Select and write the correct answer: [10]
(i) A thin ring has mass 0.25 kg and radius
0.5 m. Its M.l. about an axis passing
through its centre and perpendicular to its
plane is
(a) 0.0625 kgm’ {b) 0.625 kgm?
(c) 6.25 kgm? (a) 62.5 kgm?
Ans. (a) 0.0625 kgm? (1 mark)
[Hint Ig = MR? = 0125 » (0.5)? = 0.0625 kgm")
(ii) Absorption of water by filter paper is |
due to sy
{a) cohesion {b) capillarity |
(c) adhesion (4) elasticity
(b) capillarity )
Ans.
(iii) ‘The molar specific heat of an ideal gas at
constant pressure and constant volume is
C, and Cy respectively. If R is the universal
gas constant and the ratio C, to Cy is 7
then Cy
time: Ey and -Cy=R
©, = Cy
& YCy-Cyp"R_ = Opt) =R
(iv) The second law of thermodynamics deals
with transfer of
{a) work done
(c) momentum
Ans. (d) heat a
() energy
(4) heat
(v) A body of m:
1 kg is performing linear
S.H.M. Its displacement x cm at time t sec
Ans. (c) 30h ti mark)
‘Tn the given circuit, capacitors are parallel
Gy +C, = 10+ 10+ 10 = 30uR}
(vi) One beat means that the intensity of sound
shouldbe
(a) once maximum
(b) once maximum and once minimum
(c) once minimum
(a) twice maximum and twice minimum
‘Ans. (b) once maximum and once minimum
(1 mark
(vil) ‘The effective capacitor between A and B in
the following circuit is __
c i Cy
A aes
Tone | 10H 10 ue
10 3 1
Sup (ec) 30uF (a) +
(a) ur (b) FyuF (6) 30HF (a) Sour
(1 mark)
(viii)
ceuracy of potentiometer can be easily
inereased by _
(a) increasing resistance of wire
(b) decreasing resistance of wire
(c) increasing the length of wire
(a) decreasing the length of wire
‘Ans. (c) increasing the length of wire (1 mark)
(ix) Which of the following statement is correct
for diamagnetic materials?
(a) Susceptibility is negative and low
(b) Susceptibility does not depend on
temperature
() a,
Attempt any EIGHT:
Qs.
[16]
Distinguish between centripetal force and
Centrifugal force. |
[ Centripetal force | Centrifugal force
jilCentripetal force is, (Centrifugal force is
ldirected along the radius|directed along the radius|
Iowards the centre of a away from the centre of
lire. lacircle,
Gilitis a real force. lit is a pseudo force. |
Rieismo?
F = mo?
F be the viscous force.
By Newton’s law of viscosity,
du
Fe al ctl
|
dv
Fe nade
MEICE:
where 1 is constant called coefficient of
viscosity of the liquid
F
(1 mark)
[iiit is considered in
iis conseredin |
etaitane a
igasinerial frame of |)
reference reference
(Bach point - Ya mark)
0.4, State and explain Newton's law of viscosity. |
(2)
Ans. ¢Statement: For streamline flow, the
viscous force acting on any layer is directly
Proportional to the area (A) and the velocity
du
oni
* Explanation:
Let A be the area of layer.
(1 mark)
dv |
GaP the velocity gradient
dv
zl
nei ae Aether anes
dx
.5. Explain, on the basis of kinetic theory, how
2
the pressure of gas changes if its volume is
reduced at constant temperature. (2)
Let P - be the pressure exerted by the gas
V - be the volume of the gas
Ans.
N- be the number of molecule of gas
m- be the mass of each molecule of gas.
:, Total mass of the gas, M = Nm.
From kinetic theory of gases,
(4 mark) 292
Pressure exerted by gas in an enclosed vessel is | Ans. Given.
bayer
Kinetic energy at
constant temperature
= constant a)
Nis number which is also constant.
constant
P bs
Vv
R |
Peo (mark)
Vv
Hence, at constant temperature, if pressure |
of the gas is increased then its volume is
reduced. (4s mark)
A steam engine delivers 4.8 x 10°J of work
per minute and services 1.2 x 10° J of heat
Per minute from its boiler. What is the
Percentage efficiency of the engine? (2)
Given: W = 4.8 x 10°J, Qy = 1.2 * 10°
To find: Efficiency (n) = ?
Q6.
Formula: = (/» mark)
From formula,
Percentage = 0.4 x 100= 40%
The percentage efficiency of the steam engine
is 40%.
A proton is accelerated in a cyclotron
in which the magnetic induction is
0.6 Wb/ m2. Find the cyclotron frequency.
(Given: m, = 1.673 = 10?” kg,
e= 1.6 x 10-9)
Given: B = 0.6 Wb/m?, 0
m, = 1.673 * 10-7" ke,
e= 1.6 « 10-9
To find : Frequency (n) =?
eB
Qxm
27.
(2)
Ans.
(% mark)
From formula,
6 * 10-19 x
2% 3.142 « 1.673 * 1
16
a0 x 108 = 9.13 x 10°
10 51g
Frequency n= 9.13 x 10° Hz
Two sound waves travel at a speed of
330 m/s. If their frequencies are also
identical and are equal to 540 Hz, what will
be the phase difference between the waves
at points 3.5 m from one source and 3 m
from other if the sources are in phase? (2) _
iil
(4 mark)
(0 mark)
Q9.
Ans.
Q.10.
. Given: n,
To find: Phase difference (Ag) =
Formulae:
Qn
ear aes (4 mary)
From formula (i),
somes
~ 540 ~ 18 marr
From formula (i),
ane 3.5 - 3)
AGS Srila Seams) 4 mark)
1
_2xnxl 1
oa Teme
lax
Sasi Te gas (4 mark)
The phase difference between the wave ig
1.64%,
125 small liquid drops, each c 4
charge of 0.5 uC and each of diameter
0.1 m form a bigger drop. Calculate the
potential at the surface of the bigger drop.
2
ec ese case
lectrical Potential (V)
Q
aneyR
As single drop is formed by coalescing, the big
drop will have charge
(mark
Q= ng = 125x05*10%C
Radius = 1251/3 x 0.05 ...(As R= n'/9
= (59)1/9 x 0.05
= 5*0.05
Radius r = 0.25 m (mark)
From formula,
ye 2210 * ee SEL Seay
= 2250 x 10?
= 2.25 x 10°V
Electric potential is 2.25 x 10°V
Calculate the change in angular momentum
of electron when it jumps from 3" orbit to
1* orbit in hydrogen atom.
(Take h = 6.33 x 10-% Js) @)
1 (first orbit), n, = 3 (third orbit),
h = 6.33 x 10-3 Js
To find
Change in angular momentum (L, - L,) = ?
(mark) (4 mark) |
iets
6.33 x 10-34
3.142
2.11 * 10 kgm2/s,
‘The change in angular momentum of electron,
when it jump from 3 orbit to 18 orbit |
hnydrogen atom is 2.11 « 10% keemy2/¢"
git. Draw a labelled circuit dij
2 ym ta
‘agram of meter
bridge to determine the unknows
resistance. 12)
ins. Diagram:
Meter scale
© Labellings:
X - Unknown resistance
R, > Rheostat
R - Resistance box
K - Plug key
G - Galvanometer D ~ Null point
E - Battery J ~ Jocky
Q12. State Faraday’s laws of
induction. (2)
Mus. *Faraday’s first law: Whenever there is a
change of magnetic flux in a closed circuit, |
@n induced emf is produced in the circuit.
(1 mark)
* Faraday’s second law: The magnitude of
in
duced emf produced in the circuit is directly
Pr
‘portional to the rate of change of magnetic |
lux linked with the circuit
electromagnetic
(1 mark) |
|
7"
e- ir
But, Instantaneous emf
= Instantaneous v;
falue of potential drop.
™ equations (i) and
(i), we have,
(ii) (4 mark)
R
we have,
We know, peak current jy = ©
Substituting in equation (ii
i= jsinot (iv)
from equation (i) and (v) there is no phase
difference between alternating current and
emfi.e. there is zero phase difference between
current and emf,
= 6) 2in at
Fig.)
Q.14. Define: (a) Threshold frequency
(b) Work function. (2)
(8) Threshold frequency: The minimum
frequency of incident radiation required to
start a photoemission in any photosensitive
material is known as threshold frequency,
(4 mark)
Ans
(1 mark)
(b) Work function: The minimum amount of
energy required to be provided to an electron
to pull it out of the metal from the surface is
called the work function of the metal. (1. mark) no eee
EE
204
> Attempt any EIGHT:
Q.15. Derive an expression for excess pressure
inside a liquid drop.
Ans. Due to surface tension free liquid drops
and bubbles are spherical in shape, if effect
and air resistance are negligible.
A bubble or drop does not collapse because
the resultant of the external pressure and the
maller than the
ure inside a bubble or drop i.e. Py
ac Iron core
pores (stationery)
Q.23.
Former alk Permanent Magnet a
Lower Suspension (i)
Ans.
North > aan
|
emanent 7” soying Cail HOR Coe |
‘ig. Moving Coil Galvanometer (M.C.G.) (1 mark
© Construction: A rectangular coil of thin
insulated copper of N-turns wound over a
non-magnetic frame. It is suspended between
the concave poles of strong horse shoe
magnet. The lower end of the coil is connected
to a light spring. The current enters the coil
through the fiber and leave through the
spring. |
297
small mirror attached to the suspension wire
used to measure the deflection of the coll by
lamp and scale arrangement. The upper end
Of the wire is connected to a rotating screw
head so that plane of the coil can be adjusted
in any desired position. A soft iron cylinder is
fixed inside the coil such that the coil rotate
freely between the poles of the magnet and
the concave pole pieces. Soft iron cylinder
Produces strong radial magnetic field. The
Soft iron cylinder increases the strength of
the magnetic field Pa
A magnetic needle placed in uniform
magnetic field has magnetic moment of
2 10-3 Am? and moment of inertia of
7.2 * 10-7 kg m?. It performs 10 complete
oscillations in 6 sec. What is the magnitude
of the magnetic field? (3)
Given: M= 2% 10-? Am?, I= 7.2 * 10-7 kg m?,
ecemg
Te) 7 ye
{As there are 10 oscillations per 6 sec.)
To find: Magnetic field (B)
For la: T = 2: y i
rmutat = 2xy|
Squaring and rearranging, we get
Al
MT?
oy
10-7 «(3
axio*(3]
4 * 9,872 7.2 * 10-7 25
2*10?%9
B= 3.948 x 10° Wb/m?, (
A light bulb is rated 100 W for 220 V AC
supply of 50 Hz. Calculate:
resistance of the bulb.
the rms current through the bulb.
Given: P,, = 100 W, e;_, = 220 V,
To find: (i) Resistance (R)
iarky
B
(3)
0 Hz,
2
(i) rms current (i
Formulae: (i) P,, = ¢,
2
ms)
Cra
@h52 5
From formula (i),
Py = ¢,
100 =
ley mark)
220 ing
100 _ 10
(220) 23)
= 0.4545 A
(1 mark) Q.24.
Ans,
Q.25.
From formula (ii),
545
In nuclear reactor, neutrons travel with
energies of 5 x 10-2! J, Find their speed
and wavelength.
(Take m, = 1.67 x 10-27 kg)
Given: E = 5x 10-2" J, m,
(3)
.67 * 10-27 kg.
To find: (i) Speed of neutron (v) =?
(i) Wavelength of neutron (A) = ?
h
Formulae: (i) © =
nee
ae
mv? (ii)
[2x5xio7r
\ 671077
10-10
Vier x 10-7
v= 2.45 * 103 m/s
h
™,0
0.2447 x 10% m/s
6.63 = 10-
x10
mark)
“275 2.45 * 10°
6.63 x 10-34
167 x 2.45
1.620 x 10~
162A (
x 10-10
Ae
Determine the binding energy per neucleon
of the Americium isotope 744m, given the
mass of 734Am to be 244.06428 u. (3)
Given: Z - 95, N(p) = 95, N(n) = 224-95 = 149
M = 244.06428 u, 1 u = 931.5 MeV/C?
m, = 1.00728 u, m, = 1.00866 u
Ez
To find: B.E. per nucleon “2 = ?
‘The binding energy per nucleon,
E, _ (Zm,+Nm,~-M)C? -
eee (mar
95 «1.00728 + 147 » 100866 -244.06428 |
q 244
(mark)
5.6916 + 150.2904 - 244.06428
(pee 7a }-sonsuev
(4 mark)
Ei (0 mark)
FE = 7.3209 MeV/nucleon
ee
'S Papers
9.26. Explain working of a tran
UTTAM’s XII Physic:
lution
sistor.
amplifier in detail with the neqeet, ® a
elrcuit diagram. “IP of proper
Ans, ¢ Diagrams a om
# Working: 2 mary
The circuit of an amplifier using
transistor in common emitter configura
as shown in the above figure. "t9nis,
When the input voltage Vi. is not app,
oy pling KVL to the out, PD,
write a
Vee Vee + IcR,
Similarly, for input loop, we have
Ven Vor * IR
When input AC signal is applied, V,. is no,
Thus, the potential across the input loop wy
be,
Von + Vin = Vor * InRp + Aly(R, + 7)
(i
‘The AC signal applied adds the current stan,
to the original current flowing through a2
circuit. Therefore the additional potential i,
the input loop will be across resistor Ry,
ie. AlgRy and across the input dynamic
resistance of the transistor (= Al,r,)
From equation(i)
Vin = dlp (Ry +1)
Veer WATar (i
(As Ry is very small, neglecting Ry)
(4 mark
|
The changes in the base current 1, cause
changes potential across the load resistance
because Vo¢ is constant. We can write,
Vee = Veg + 1cR, = 0
| Vee = ~1eR, |
The change in output voltage AVog is the
| changes in the collector current Io. This
|
|
‘output voltage V, hence we can write,
Yo
= AVee = PrcR,Aly 7-5
aon PODS!
any THREE :
7 Mie? Ln expression for angle of nq, 27
7
ot gen avehicle ae ‘A.curved banked
nei friction.
pss
radius of curvature of yo
Om. If angle of banking ig 27. oe
60 imum speed with which hicte
sm turn along this curve. (g= 9.3 m/s) (4)
ression for Angle of Banking: 1,
folowing figure shows the vertical sect
jicle on curved road of radius «
es angle ‘6’ with the horizontal,
a
rhe
1 banked at
(Diagram: mark)
Consider the vehicle to be a point and ignorin
ig
friction and non-conservative forces like aie
resistance.
‘There are two forces acting on vehicle
{) Weight (mg) vertically downwards,
{i) Normal reaction (N) perpendicular to
surface of road.
N resolved into two components:
{j Nsin® - horizontal component being
resultant force, must be the necessary
centripetal force.
(i) Ncos@ - vertical component balances
weight (mg)
mov?
Nsin@ =
=
Neos@ = mg (6 mark)
Dividing equation (i) by (
mv?
Nsin@ E r
Ncos@ mg
v
tang = ~_
ze ,
e = tan (1 mark)
1g
v = \rgtano (4 mark)
v = (60x98 x tana
= 60 9B X 0.5095 (1s mark)
ion of
section]
Q28,
Ans,
299
= 588 %0.
= (2995
= 17.31 m/s 1
State and explain Stefan’s law of radiation.
Pnergy is emitted from a hole in an electric
{trace at the rate of 20 W, when the
"mperature of the hole is 727°C. What
is the area of the hole? (c = 5.7 « 10-* J/
Wm? K+) 4)
*Stefan's law: The radiant energy emitted
Per unit time per unit area by a perfectly black
body is directly proportional to the fourth
Power of its absolute temperature. (!2 mark
let R-be the quantity of radiant energy
emitted per unit time per unit area by
the blackbody
T- be the absolute temperature.
Then, R= Tt
R= oT ti) mark)
where,s is constant called as Stefan’s constant
and 6 = 5,7 « 10-8 J m-!s~2K-4 or J Wm~? K~*
Dimension of o = [L°M!'T er).
‘Thus, the power radiated by a perfectly black
body depends only on its temperature and
not on colour, materials, nature of surface,
etc
By definition of emissive power,
ero ee
Ae
Equation (i) becomes
Q
a
For body which is not blackbody, the energy
radiated per unit area per unit time is still
Proportional to the fourth power of absolute
temperature.
R = eoAT,#
Problem:
AQ
dt
T= 727°C = 727 +273
9=5.7* 10-8 J m's-2 K+
To find: Area of the hole (A) = ?
dQ
Sd (4 mark)
Consider given hole to be a perfect blackbody.
e=1
From formula,
20= 5.7 x 10-8 x Ax (1000)
(iil)
mark)
Given:
=20W
1000 K
= eat,
(% mark) 300 i
20
A= 37x 108x107
20 a
= 577 10"
= 0.35 x 10° m?
= 3.5% 10-4 m?
+ The area of the hole is 3.5 x 10-* m?. (1 mark)
Q.29. Obtain the differential equation of linear
simple harmonic motion.
At what distance from the mean position is
the speed of a particle performing SHM half
its maximum speed. (Given: Path length of
‘SHM = 10 cm) (4)
Ans. Consider a particle of mass _m performing
linear SHM with its mean position ‘O'
between two extreme position A and B as
shown in figure
Mean
extreme OSS extreme
6
aoe
Pig. A particle performing linear SHM
Let P-be the any position of particle at
distance x from point O.
F- be the force acting on particle towards
mean position.
By definition of linear SHM
F = -kx iy
where k is force constant. Negative sign
indicates force and displacement are in
opposite direction.
By Newton’s 2" law
F = ma
dx
dt?
ax
“aq is the acceleration.
Comparing equations (i) and (i),
mark)
=m sil) (4 mark)
where a
ax
cade
rex
(4 mark)
=o
Substituting =
ax
fq tex = 0 (ii) (4 mark)
Equation (iii) is the differential equation of
linear SHM.
_UTTAM’s XII Physics Papers Solution
Problem: Given: Path length = 10 cm
Path length _ 10
«. Amplitude (A) = ——3—=— => = 5 em
v
To find: Distance (x) = ?
Formulae: (i) Vine = A
(i) v= oY =
From formula (i),
ota? ‘
mark)
(5 mark)
x = 0.8665 =
4.33 cm
.. The distance at which speed of particle is half
Q.30.
Ans.
of its maximum value is 4.33 cm.
Derive the conditions for occurrence of
dark and bright fringes on screen in Young's
double slit interference experiment. Also
determine fringe width. @)
Let S, and S, be the two sources of light
separated by distance d. A screen is placed at
a distance D from the two sources of light.
Let A be the wavelength of light.
Draw 8,8’, and 8,8’, perpendicular to screen,
Let O be the midpoint of S’,S’,
o's, = O'S, 4
Let ‘P’ be the point at distance y from point ‘0’
on screen. To find whether the point P to be
a bright or dark is depending upon the path
difference. The path difference is S,P - S,P.
(Diagram-1 mark} 1n 8,S/,P, .
(PP = (S84)? + (spe
2 = p?+(y- 4)?
(SPV 2
(PP =
D+ yas 2
4
in AS,S',P,
(S:P)? = (S2S',)? + (Spyz
(2! 2
ay
PP = DP+{y+ 2
@
ase
e 4
(SPP = D?+ y+
subtracting equation (j
(S,P)? - (SP?
eee
Dit tude Soa
(S,P-S,PIS,P-S,P)~ 2yq
(ii)
) from (ii),
S,P-S,P = _2yd
SP +S
2ud
SP+s,h
Now D is very large as compared to y and a
. Path difference = a
fringe width: The distance between any two
frnctS*ive fringes or any two successive dark
ee in an interference pattern is called
Ting width or band width.
Fringe width (W) = ys, — yy mark)
We know the equation of ni” bright band
y= Daa
a
Equation for (n+ 1)! bright band
Din+ yr
ye, = POS
+ Fringe width(W) = y,,,-y,
_ Din+ija__Dnar
a a
~ PRinei—n)
ws (5 mark)
d
Similarly, we can determine distance between
two successive dark fringes is same,
ie. we DA
291. Show that the work done in pulling a loop
ive, D >> yand D >> d. through the magnetic field appears as heat
S,P=S,P=D energy in the loop. (4)
meio Qua Ans. Consider a loop ABCD moving with constant
eee DES velocity in a uniform magnetic field B as
a shown in the figure.
Path difference = 49 |... (ii) ;
mark
Case-I: The point P will be bri
ight if path
difference = ni.
wenn
D
— Dna
d
Equation for position of n'® bright band is
Deu tv) mark
na
Case-II: The point P will be dark if path
a
difference = (2n— 1)
Equation for position of n'* dark band is
= D@n- 1a
2d
E ol)
(45 mark)
A current iis induced in the loop in clockwise
direction. Let F,, F, and F, be the forces
acting on side AD, AA’ and DD’ respectively.
‘The dashed line shows limit of magnetic field.
| Loop moving
out with
velocity v
ep Cc
@
© Magnetic field B into plane of the paper
Fig. (a)
(mark)
To pull the loop at constant velocity towards
right, it is required to apply external force F on.
loop so as to over come the magnetic force of
equal magnitude but opposite in direction.
‘The rate of work done on loop is
Work done
Time
Power a
See ee ee ee eee ees
Force * displacment
epaTiras
Force * velocity |
Fo
Let us find the expression for P in terms of
B, resistance, area and width, When the
loop is moved to the right, the distance x
decreases ie, area of loop inside the field
decreases, causing magnetic flux decreases
and it reduces current in the loop
The magnitude of magnetic flux through loop
(14 mark)
® BA
= Bix where A= ix
By Faraday’s law, the induced emf is
eee eo) :
at
acy cal
arr x}
dx
eames Eo ri
7 ia)
e- -Bl {hsv = =8
(=v) ave
e- Bl (i (mark)
v is negative, as time increases distance
x decreases,
The magnitude of induced current is,
lel _ Bw |
a R li)
The force acting on AA’ and DD’ ice. F, and
F, are equal in magnitude and opposite in
direction. Therefore they cancel each other.
|
Pee
UTTAM’s XII Physics Papers Sony
cThe force F; is directed opposite tor,
aren
‘the magnitude of F; is
F, = iBsind
Fr, = iB
(As FLELB, © 0-90" and sing
[rp = [Fle iB ial
putting the value of {in equation (ij, "4
Blo
| -( 2 |e
iri -[FR
Bev
a i
Fie i
‘the rate of mechanical work, i.e. power jg
P= Fo
Bry
ae ie mar
‘The rate of production of heat energy int,
loop is 5
P= ?R
Putting i =
B72 v?
ar Sa aren
Comparing equations (iv) and (v) we find tha,
the rate of doing mechanical work is exact,
samme as the rate of production of heat energ,
in the loop. Thus the work done in loop appean
as heat energy in the loop. (
ae
7