Class : XII (ABCD)
Time : 3 hour Max. Marks : 243
INSTRUCTIONS
1. The question paper contains 28 questions and 16 pages. Q. No. 1-20 carry3 marks each and Q. No. 21 to
28 carry 5 marks each and all of them are compulsory. There is NEGATIVE marking. 1 mark will be
deductedfor each wrong answer.
Please ensure that the Question Paper you have received contains all the QUESTIONS and
Pages. If you found some mistake like missing questions or pages then contact immediately to the
Invigilator.
2. Indicate the correctanswer for each question byfillingappropriate bubble inyour answer sheet.
3. Use onlyHB pencil fordarkeningthe bubble.
4. Use ofCalculator, LogTable, SlideRule and Mobile is not allowed.
5. TheanswerofthequestionsmustbemarkedbyshadingthecirclesagainstthequestionbydarkHBpencilonly.
6. The answer(s) ofthe questions must bemarked byshading thecircles against the questionbydark HB
pencilonly.
For exampleifonly'B'choice is correct then, thecorrect method for fillingthe bubble is
A B C D
thewrongmethodfor fillingthebubbleare
(i) A B C D
(ii) A B C D
(iii) A B C D
The answer of the questions inwrong or anyothermanner will be treated as wrong.
USEFUL DATA
Atomic weights:Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P= 31,Ag = 108, N = 14,
Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.4,
Radius of nucleus =10–14 m; h = 6.626 ×10–34 Js; me = 9.1 ×10–31 kg, R = 109637 cm–1
PART TEST-1
OBJE CTIVE

Select the correct alternative (Only one alternative is correct)
Q.1 AtubeoflengthLisfilledcompletelywith anincompressibleliquidof massMandclosed atboth ends.
Thetubeisthenrotatedinahorizontalplaneabout oneofitsends withauniformangularvelocity.The
force exerted bythe liquid at theother end is:
(A*) ML2/2 (B) ML2 (C) ML2/4 (D) ML22/2
[Sol: Equivalent System   F = ML2/2 ]
Q.2 Ablockofmass15kgisrestingonaroughinclinedplaneasshowninthefigure.
The block is tied up bya horizontal string which has a tension of 50 N. The
friction force between the surfaces of contact is (g = 10m/s2)
(A*) 2
50 N (B) 2
100 N (C) 50 N (D) none of these
[Sol: N =
2
Mg
T 
=
2
150
50 
=
2
200
= 2
100
F =
2
2
T
Mg
 =
2
50
150 
=
2
100
= 2
50
F = fs  2
50 = ]
Q.3 Asimplependulumconsistsofasmallwoodenbobofmassmandlightstringoflengthl.Abulletofmass
m1 isfiredhorizontallytowards thependulum witha speed v1 .Thebullet emergesout ofthebob with a
speed v1/3 and thebob just completes motion along avertical circle. Then v1 is :
(A) gl
m
m
5
1








(B*) gl
m
m
5
2
3
1








(C) gl
m
m
5
3
2 1






(D) gl
m
m





 1
[Sol: Linearmomentumconsevrationinhorizontaldirection
m1V1 = m1 3
1
V
+mV, V = speed of after collision
V = 1
1
3
2
V
m
m
For to just complete vertical circle, V = gl
5
 gl
5 = 1
1
3
2
V
m
m
 V1 = gl
m
m
5
2
3
1
]
Q.4 Particles P andQ of masses 20 gm and 40 gm respectivelyare projected from
the positionsAand B on the ground. The initial velocities of P and Q make
angles of450 and1350 respectivelywiththe horizontal as shownin the figure.
Each particle has an initial speed of 49 m/sec. The separationAB is 245 m.
Bothparticles travel in the same vertical plane and undergo acollision.After the collision Pretraces its
path.Thepositionof Q fromAwhenit hits theground is:
(A) 245m (B)
3
245
m (C*)
2
245
m (D)
2
245
m

[Sol: Momentumconservationinhorizontaldirection
1000
40
2
49
1000
20
2
49
1000
40
2
49
1000
20






















V
 V = 0, (Horizontalcomponent ofvelocityofQafter
collision)
X coordinate of point of colllision  mid point ofAB 
2
245
m fromA
A
So Q falls at
2
245
m fromA. ]
Q.5 AuniformrodofmassMislyinginthestateofrestonaroughhorizontalplane.
The rod being rotated in vertical plane very slowly by the help of a variable
force F always perpendicular to the length of the rod such that rod can rotate
about thepoint O without anyangular acceleration. The maximum value of
static frictionforce acting atthe point O is equal to
(A) Mg (B)
2
Mg
(C*)
4
Mg
(D) none
[Sol: f=Fsin
F · l = mg
2
l
cos F =
2
mg
cos
 f =
2
mg
sin·cos =
4
mg
sin2
fmax =
4
mg
when =
4

(C) ]
Q.6 Asoliduniformsphereisrollingwithoutslippingonafrictionlesssurface,
shown infigurewitha translational velocityvm/s.Ifitis toclimb onthe
inclinedsurfacethenv shouldbe:
(A) > gh
7
10
(B*) > gh
2 (C) 2gh (D)
7
10
gh
For Problem 7 to 9
Aboat of mass M sails with velocityv0 iˆ.At t = 0, its engine stops, and at the same time aball of mass
m(m < < M) is thrown from the boat with initial velocity u ĵ . Here x - direction is horizontal and y is
vertically up. The water exerts a friction drag force proportional to the boat's velocity (v), so that
i
Kv
F ˆ



, where K is positive constant. The acceleration of the boat can be written as:
i
M
Kv
a ˆ



or i
M
Kv
dt
dv ˆ



 i
Ce
v M
Kt ˆ
/




WhereC isaconstant emergingduetointegrating. Nowassumethat ballis observedfromthereference
frameofboat(X'Y').Theoriginofthisreferenceframecoincided withthestaticframe(XY)att=0.The
accelerationoftheball inthereferenceframeofboatcanbewritten as j
a
i
a
a y
x
ˆ
'
ˆ
'
' 


andin thestatic
frame it can be written as j
a
i
a
a y
x
ˆ
ˆ 


.
Q.7 The value ofconstant C is
(A) v0/2 (B*) v0 (C) v0(M/m) (D) v0 (m/M)
[Sol: i
Ce
V M
kt
ˆ



, at t = 0, i
V
V ˆ
0


 C = V0 ]
Q.8 Values of ax and ax' respectivelyare:
(A*) ax = 0; ax' =
 
M
Kt
e
M
Kv /
0 
(B) ax = 0; ax' =
 
M
Kt
e
M
Kv /
0 

(C) ax = ax' =
 
M
Kt
e
M
Kv /
0 
 (D) ax = 0; ax' =
 
M
Kt
e
M
Kv /
0 

[Sol: ax = 0; '
x
a

=
dt
V
d

 = – i
M
k
Ce M
kt
ˆ








 '
x
a =
M
kV0
M
kt
e

]
Q.9 Thehorizontalcomponent ofball's velocityas seenfrom theboat will be:
(A) v0e–Kt/M (B) v0 (C) v0eKt/M (D*) v0(1 – e–Kt/M)
[Sol: '
x
a =
dt
dVx '
= M
kt
e
M
kV 
0
 
'
0
'
x
V
x
dV = 

t
M
kt
e
M
kV
0
0
=
M
kV0
×  
M
k

1
t
M
kt
e
0

 Vx' = V0










M
kt
e
1 ]
Q.10 Aconcaveandaconvexlenshavingthefocallengthof20cm&10cmrespectivelyandareputinmutual
contact toformlenscombination.Thecombination isusedtoseeanobject5cm highkeptat20cmfrom
the lens combination.As compared tothe object, the imagewill be
(A*)realmagnifiedandinverted (B) real, smallerand errect
(C)virtual,smaller anderrect (D) none of these
Q.11 Whentheplanesurfaceofplano-convex lensis silvered, thelensbehaves likeaconcavemirroroffocal
length30cm,butwhentheconvexsurfaceissilvered,itbehaves likeaconcavemirroroffocallength10
cm.Therefractiveindex ofmaterial oflensis
(A)
3
4
(B) 2 (C)
2
3
(D*) 1.5

Q.12 One face of a prism of refractive angle 300 and ref. index 2 is silvered.At
whatangle(i)mustarayoflightfallontheunsilveredfacesothatafterrefraction
into the prism and reflection atthe silvered surface it retraces its path
(A) 300 (B*) 450 (C) 600 (D) none
[Sol: 1 · sin i = 2 × sin r
= 2 ×
2
1
=
2
1
V
i = 450 (B) ]
Q.13 The distance between an object and screen is d.Aconvex lens of focal length f is placed between the
object andthescreen. Ifm is the transverse magnificationofimagethen
(A*) f =
 2
1 m
md

(B) f =
2
1 m
m
d
 (C) f =
m
md

1
(D) f =
 
m
m 2
1
d
[Sol: m =
x
y
 y = x m
x + y = d
x + xm = d
x =
1

m
d
, y =
1

m
dm
V
1
–
U
1
=
f
1

y
1
–
x

1
=
f
1
solvingf=
 2
1 m
md

(A) ]
Q.14 Ablock of 4 kg is placed on a plank having mass 8 kg.Aforce F = 20 N is
applied on plank.Then find the friction force between 4 kg block and plank.
Here coefficient of friction between 4 kg & 8 kg is  = 0.4 (g = 10m/s2)
(A)
3
10
N (B) 16 N (C*)
3
20
N (D) zero
[Sol:
4
16
max 
upper
a = 4m/s2
 singlemass
 Fmax = 12(4) = 48 N
F < Fmax , so both masses move with same acceleration
 20 = 12 a  a =
3
20
3
5
 N

 a
  f = 4 ·
3
5
=
3
20
N
 a =
3
5
m/s2 ]
Q.15 Thediagram showsathinuniform semicirculardiscofmass Mand radius R.
What is the moment of inertia of disc about an axis in theplane of disc
(showninfigure)
(A)
3
MR2
(B*)
4
MR2
(C) MR2 (D) none
[Sol: MIabout an axis through centre & perpendicular to plane = ½ MR2
By | axis theorem, I + I = ½ MR2  I =
4
1
MR2 ]
Q16. Adiskrollswithoutslidingonahorizontal surface.Iftherelativevelocitybetween
the pair of pointA1B1,A2B2 andA3B3 are V1, V2 and V3 respectivelythen
(A*) V1 = V2 = V3 (B) V1 > V2 > V3
(C) V1 < V2 < V3 (D) Data's areinsufficient to decide
[Sol: Sorelative velocityin everycase isof magnitude2V
 V1 = V2 = V3 = 2V ]
Q17. Aparticle is projected towards afixed wall from pointA(at a distance'x' from the
wall)andcollideselastically. IfR=horizontalrangeinabsenceofwallthentotaltime
offlightwillbeminimumif
(A) x = R/2 (B) x = R/ 2 (C) x = 2 R (D*)independentof x
[Sol: e = 1 At time of collision, vertical component remains same. Horizontal component's magnitude
remainssame,butdirectionreversed.So,timeofflightremainssameasverticalcomponentisunchanged.
Hence, it is independent of x. ]
Q18. Two ends ofa helical uniform springof force constant K andmass M are
projectedwithvelocityVinits natural state(as showninfigure).The
maximumextensioninspringwillbe
(A)V
K
m
(B*) V
K
m
3
(C)
K
m
2
(D) V
K
m
6
[Sol: Let u is velocityat anyx from O. O is the mid point which is at rest.We can take,
u(x) =  
2
L
x
V =
L
x
2
V

 Energy of part OA, E0 =
2
1  

2
/
0
2
)
(
L
x
u
dm
E0 =  





2
/
0
2
2
2
1
L
V
L
x
dx

 E0 = 8
3
1
2
2 3
2
2
2
/
0
2
2
2
L
L
V
dx
x
V
L
L





 E0 =
2
2
12
1
12
1
MV
LV 

Energyofwholespringinitially
E1 = 2E0 =
6
1
MV2
Energyconservation
E1 = E2

6
1
MV2 =
2
1
kxm
2
 xm = V
K
m
3
]
Q19. A particle P with a mass 2.0 kg has position vector r = 3.0 m and velocity = 4.0 m/s as shown. It
is accelerated bythe force F= 2.0 N.All three vectors lie ina common plane.Theangular momentum
vector about origin O is
(A*) 12 kgm2
/s out of the plane of the figure
(B) 12 kg m2
/s into the plane of the figure
(C) zero
(D) 24 kgm2
/s out of the plane of the figure
[Sol: p
r
l




 = v
r
m



l

= mrvsin = mvrsin(1800
– 300
)
= l

= mvr sin300
= 12 kgm2
/s
v
r


 points out of the plane,
So, answer is (A) ]
Q20. A small bead of mass m moving with velocity v gets threaded on a
stationarysemicircular ring of mass m and radius R kept on a smooth
horizontal table. The ring can freelyrotate about its centre. The bead

comes to rest relative to the ring. What will be the final angular
velocityof the system?
(A) v/R (B) 2v/R (C*) v/2R (D) 3v/R
[Sol: Pi
= mv  At final position, both the bead and ring are rotating about axis through O and | to the
plane.
Im
= mR2
, Iring
= mR2
 I = Im
+ Iring
= 2mR2
 Lfinal
= I = 2mR2 





R
V '
= 2mRV'
 Pf
=
R
Lfinal
= 2mV'
Pf
= Pi
 2mV' = mV  V' =
2
V
 =
R
V'
=
R
V
2
]
Select the correct alternative(s) (one or more than one alternative may be correct)
Q21. Select the correct alternative(s). The power of a convex lens (R.I.  = 1.5)
(A*) will decrease on immersing in water
(B) will increase on immersing in water
(C) is less for violet rays compared to red rays
(D*) is +10 D if it forms the image of an object placed at 12 cm, at a distance of 60 cm on the
other side.
Q22. The line PQ in the ray diagram represents a thin lens &AB is a principal
axis. Then lens is
(A) converging (B*) diverging
(C*) bi-concave (D*) concavo-convex
Q23. A particle of mass 3 kgis moving under the action of a central force whose potential energyis given
by U(r) = 10r3
Joule here r is in meter. If radius of orbit of particle r = 10 m then
(A*) the energy of particle is 2.5 × 104
Joule
(B) The energy of particle is 0.5 × 104
Joule
(C*)Angular momentum of particle about the centre is 3000 kg–m2
/s
(D)Angular momentum of particle about the centre is 1000 kg–m2
/s
[Sol: U(r) = 10 r3
 F(r) =
dr
r
dU )
(
 = –30r2
r
mV 2
= 30r2
 V =
m
r3
30
 V = 100 m/s
 E = U(r) + ½ mV2
= 10 · (10)3
+ ½ × 3 × (100)2
= 2.5 × 104
J (A)
Angular momentum, L = mvr = 3 · (100)· 10 = 3000 kg m2
/s (C) ]
Q24. A bobbin of mass m and moment of inertia I relative to its own axis is being
pulled alonga horizontal surface bythe light string tightlywrapped as shown

in figure. There is no slipping on the surface throughout the motion
(A*) The angular velocityof bobbin when string is pulled horizontallywith velocity V is=
r
V
3
(B) The angular velocityof bobbin when string is pulled horizontally with velocity V is=
r
V
(C*) If string is pulled by horizontal acceleration a. Then tension in string is T =
9
a






 2
4
r
I
m
(D) If string is pulled by horizontal acceleration a. Then tension in string is T =
2
2
2
a
m
r
I







[Sol: V0
+ r = V; V0
= 2r  3r + V  =
r
V
3
a0
+ r = a ; a0
= 2r  3r = a   =
r
a
3
 a0
=
3
2a
T – f = Ma0
 T – f = M 





3
2a
...........(1)
Also, f(2r) + T(r) = I
f +
2
T
=
r
I
2

.............(2)
(1) + (2) 
2
3T
= 





 2
4
r
I
m
a
 T =
9
a






 2
4
r
I
m
Ans: (B) & (C) ]
Q25. Ahollowuniformsphericalballisgiveninitialpushupaninclinedofinclination
angle . The balls rolls purely. The coefficient of static friction between the
ball and incline is . During its upwards journey
(A*) friction acts up along the incline
(B) friction acts down along the incline
(C*)  >
5
2
tan
(D)  >
7
2
tan 
[Sol: acm
= R
mgsin – f = macm
..............(1)
f R =
3
2
mR2

 f =
3
2
mR =
3
2
macm

 (1)  mgsin –
3
2
macm
= macm
 acm
=
5
3
gsin
f =
3
2
macm
=
5
2
mgsin, up the incline
N = mgcos,  =
N
f
=
5
2
tan ]
Q26. Uniform squareplateis connected totwo identical vertical idealsprings (as
showninfigure)&systemisinequalibrium.Suddenlytherightspringis
removed,thenthenet accelerationofpointAimmediatelyafterremoving
rightspringis
(A)
2
3
g (B)
2
g
(C*)
8
5
g (D)
4
3
g
[Sol: mg –
2
mg
= ma
 a =
2
g
2
mg
×
2
l
= 
6
2
ml
 =
l
g
2
3
Net conservation of point A is
4
3
cos
2
·
·
2
2
2
2 

 l
a
l
a 






 =
2
8
5
g = g
8
5
(C) ]
Q27. A smooth track in the form of a quarter circle of radius 6 m lies in the
vertical plane.Aparticle moves from P1 to P2 under the action of forces
2
1 F
,
F


and 3
F

. Force 1
F

is always toward P2 and is always 20 N in
magnitude. Force 2
F

always acts horizontally and is always 30 N in
magnitude.Force 3
F

alwaysactstangentiallytothetrackandisofmagnitude
15 N. Select the correct alternative(s)
(A) work done by 1
F

is 120 J (B*) work done by 2
F

is 180 J
(C*) work done by 3
F

is 45  (D*) 1
F

is conservativein nature
[Sol: 2
wF = F2  
2
OP = 30 · (6) = 180 J
3
wF = 
2
1
3
P
path
the
Along
P
dx
F
= 
4
6
2
0
3

dx
F = 15 · (3) = 45J

1
wF = 
2
1
11
P
P
dx
F
1
wF =  






2
/
0
2
4
cos
20




Rd
1
wF = 20 (–2) R ·















2
/
0
2
4
sin



1
wF = (–40) · 6 






4
sin
0

1
wF = 2
120 J
Work done byF1 is independent of path. So it is conservative in nature ]
Q.28 Auniform rodAB& length Lis releasedfrom rest inthe position shown.
It swings down to vertical positionand strikes a second and identical rod
CD is restingonfrictionlesssurface.Ifimpactis prefectlyelasticthen
velocityofrodCDimmediatelyafterimpactis
(A)
4
1
2gL (B*) 3gL
2
1
(C) 4gL
3
1
(D) 2gL
[Sol:
2
0
2
ML
3
1
2
1







= Mg
2
L
L
g
3
0 

def. of e
0L = L + V
AMC
I0 = –I + MVL
Solving V = gL
3
2
1
]