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8/13/2019 CBSE Sample Paper for Class 11 Physics Solutions - Set C
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. 1 kWh = 1000 3600 Ws = 36 106J. 1
. The normal force exerted by a liquid on any surface is called thrust. 1
. (a) PV= constant of TVr1= constant. (b) W = nCv(Tf Ti).
4. The strong smelling molecules of a gas move rapidly to intermingle with air molecules through thesurrounding. This process takes little longer time because of collision of gas and air molecules.This keeps the odour persisting for some time. 1
. Condition (i) is not sufficient, because it gives no reference of the direction of acceleration, whereasin S.H.M. the acceleration is always in a direction opposite to that of the displacement. 1
6. It is not necessary that a precise measurement has to be more accurate.Let true value of certain length be 3 cm. This length is measured with a measuring instrument oflimit or resolution of 01 cm and the measured value is obtained as 31 cm. This length is againmeasured with another measuring instrument of resolution 001 cm and the length is measured as
28 cm. 1In this case first measurement has more accuracy because it is closer to the true value but lessprecision because the resolution is only 01 cm whereas in the second are it was 001 cm.
7. This graph can be for a ball dropped vertically from a height d. It hits the ground with somedownward velocity and bowances upto height d/2 where its upward velocity becomes zero. 2
Or
8. As the vectors A B+
and A B
are perpendicular to each other, therefore 1
(A B ).(A B)+
= 0
- + -
A . A A . B B . A B . B = 0
or A2 B2= 0 1
A = B, (QA . B
= B . A )
1
Or
Static frictioncomes into play when the horse applies force to start the motionin the cart, on theother hand, kinetic function comes into play when the cart is moving. 2
9. (a)The inability of a body to change its state of uniform rotation about an axis is called rotational
inertia or M.I. of the body. It plays the same roll in rotatory motion as is being played by the mass in translation motion, i.e.,it is rotational analogue of mass. (b) M.I. of a body about a fixed axis of rotation is defined as the sum of the products of the massesand square of the perpendicular distance of various constituent particles from the axis of rotation.1
0. When a simple harmonic system oscillates with a constant amplitude which does not change withtime, its oscillations are called undamped oscillations. When a simple harmonic system oscillateswith a decreasing amplitude, with time its oscillations are called damped. 2
. Let the measured values be :Mass of block (m) = 393 g
Length of block (l) = 512 cmBreadth of block (b) = 256 cm
Thickness of block (t) = 037 cm
SAMPLE QUESTION PAPER C
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The density of the block is given by
=Mass
Volume=
m
l b t
=393g
512cm256cm037cm= 81037g cm3
Now, uncertainty in m= + 001 guncertainty in l= + 001 cm
uncertainty in b= + 001 cmuncertainty int= + 001 cm 1Maximum relative error in the density value is, therefore given by
=
l b t m
l b t m
+ + +
=001 001 001 01512 256 037 393
+ + +
= 00019 + 00039 + 0027 + 00024
= 00358 1Hence, = 00358 81037 03 g cm3.We cannot, therefore, report the calculated value of (= 81037 gm3) upto the fourth decimalplace. Since = 03 g cm3the value of can be regarded as accurate upto the first decimal placeonly. Hence the value of must be rounded off as 81 g cm3and the result of measurements shouldbe reported as= (81 + 03) g cm3. 1As the ball is moving under the effect of gravity, the direction of acceleration dueto gravity remainvertically downwards.(b) If the ball is at the highest point of its motion, its velocity becomes zero and the acceleration isequal to the acceleration due to gravity = 98 ms2in vertically downward direction.
(c) If the highest point is chosen as the location forx= 0 and t= 0 and vertically downward directionto be the positive direction ofx-axis.Four upward motion, sign of position is negative, sign of velocity negative and the sign ofacceleration positive, i.e.,v< 0, a> 0. For downward motion,sign of position is positive, sign of velocity is positive and the sign ofacceleration isalso positive, i.e.,v< 0, a> 0. (d) Suppose, t= time taken by the ball to reach the highest point.H = height of the highest point from the ground.During vertically upward motion of the ball,Initial velocity, u= 294 ms2
a=g= 98 ms2.Final velocity v= 0, s= H = ?, t= ?Applying the relation, v2 u2= 2as, we get
02 (294)2= 2 98 H
or H =294 294
2 98
= 441 m,
where negative sign indicates that the distance is covered in upward direction.Using relation v= u+ at, we get
0 = 294 + 98 t
t= 29498 = 3 s.
i.e., Time of ascent = 3 s.It is also well known that when the object moves under the effect of gravity alone, the time ofascent is always equal to the time of descent.
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Total time after which the ball returns to the players hand= 2t= 2 3 = 6 s.
3. Step I :Using
H =2 2sin
2u
g
when H = 25 m, u= 40 m/sand g= 98 m/s2
25 =
2 240 sin2 98
i.e., sin2= 2490
40
sin =49040
= 05534
i.e., = 336 1
Step 2 : R =
2 sin2ug
R =240 sin2(336)
98
=240 sin672
98=
240 0921998
= 150514 14. Given, F = 600 N
Suppose m1= 10 kgand m2= 20 kgbe the masses lying on a frictionless horizontal table.
A B
T Tm2m1 F aor
Suppose T betension in the string and a be the acceleration of the system, in the direction of forceapplied.(a)If force is applied on the heavier mass.Then equation of motion of A and B are
m1a= T ....(i)m2a= F T ....(ii)
Dividing equation (ii) by equation (i), we get
2
1
m
m=
F TT
=FT
1
2010
=F
1T
FT = 2 + 1 = 3
T =F3
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=600
3 = 200 N (b)If the force is applied on lighter mass :
A B
m1 m2T' T'
F
Suppose T be the tension in the string in this case, then equations of motion of A and B are,
F T =m1a ....(iii)and T =m2a ....(iv) Equations (iii) and (iv) give
F TT
= 12
m a
m a
F 1T
= 12
m
m=
1020
=12
F
T =
1
1 2+ =
3
2
T = 3 F2
= 2
6003
= 400 N
T =2
6003
= 400 N
According to the principle of conservation of the total kinetic energy for elastic collision,initial kinetic energy = final kinetic energy 1
Here, initial kinetic energy = 2 21 1(200) M(0)2 2
m +
= 21
(200)2
m 1
where mis the mass of the molecule and M is the mass of the container.
and final kinetic energy = 2 21 1
(200) M(0)2 2
m +
= 21
(200)2
m 1
Thus principle of total kinetic energy is obeyed.Using equation for position vector,
r
= 1 1 2 21 2
m r m r
m m
++
Let r
= 0, 1
then 1 1 2 2m r m r+
= 0
i.e., 2 2m r
= 1 1m r
i.e., r2= 1
12
mr
m
1
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Clearly, 2r
and 1r
are opposite to each other. It indicates when centre of mass lies on origin, m1lies on the left side of origin and m2on the right side, i.e.,they lie on a straight line. 1
7. As Keplers third law,2122
T
T=
3132
R
R
Let 1 denotes planet and 2 denotes earth
R13
=
231
222
T
RT 1As the planet is revolving twice as fast as earth,
T1=2T
2
R13=
22
222
T2 (1)T
R13
=31
AU4
R1=1 / 31
AU4
R1= 31
AU4
R1= 062996 A.U. = 063 A.U. 1This result can be stated as the orbital radius of planet is 063 times the orbitalradius of earth.
8. (i) (a)Using F = ky.
The maximum extension, y= Fk
(b)Here the springs has no rigid support so each mass behaves as a support forthe spring againstforce appliedon the other end.
Maximum extension,y=Fk
(ii) (a)Here, time period, T =M
2k
(b) Here the spring is considered of two halves and the time period of the arrangement is depending
upon half spring. When a spring of constant kis cut into two equal halves, the force constant ofeach half becomes 2k.
Time period, T = M22k
p 1
Or
Node (N) is a point where the amplitude of oscillation is zero (and pressure is maximum).Antinode (A) is a point where the amplitude of oscillation is maximum (and pressure is minimum).These nodes and antinodes does not coincide with pressure nodes and antinodes. 1In fact, N coincides with pressure antinode and A coincides with pressure node, as is clear from thedifinitions. 1The separation between a node and nearest antinode is /4.As the phase difference between two points separated by A is 2radian, therefore/phase differencebetween two points separated by 1/10 is 2/10 = /5 radian. 1
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It states, the change in pressure is produced in any part of an enclosed fluid, the same is transmittedequally to all points of liquid in all directions. 1
Or
Imaging the effect of gravity, pressure in a fluid at rest is the same at all points.
Or
Equal pressure is exerted by a liquid in all directions.
Applications :This principle is used to manufacture hydraulic lift. It consist of two cylinders, oneof larger areaof cross-section A and the other smaller a. Force is applied to smaller piston toproduce a pressure.
P = F/ a As per Pascals law same pressure is transmitted to larger pistion, then W = P A.
A
W F
Piston
Clearly large area A is producing more lifting force W.Hydraulic brakes are also based upon Pascals law where transmitted to brake drum.The large force then operates the brake shoe. 1(a) Stream line flow of a liquid is that flow in which every particle of the liquid follows exactly thepath of its preceding particle and has he same velocity in magnitude and direction as that of itspreceding particle while crossing through that point.The stream line flow is accompanied by stream lines. 1A stream line is the actual path followed by the procession of particles in a steady flow, which maybe straight orcurved such that tangent to it at any point indicates the direction of flow of liquid atthe point.(b) Important properties of stream lines :
(i) In a streamline flow, no two stream lines can cross each other. If they do so, the particles of theliquid at the point of intersection will have two different directions for their flow, which will destroythe steady nature of the liquid flow.(ii) The greater is the crowding of stream lines at a place, the greater is the velocity of liquidparticles at that place and vica-versa. 1Joule observed that amount of mechanical work W when disappeared leappers in the form of heat,i.e.,
W Hor W = JH
or J = WH
= 418 J cal1 1
He made two equal weights to fall through some heights. This makes the drum D rotate and thuspaddle P chowns water inside the calorimeter.
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T T
D
h
T
mm
This makes the water heated up. The rise in temperature is noted with thermometer T. Theheight of fall of weights helps in working out the mechanical work W. The temperature helps incalculating the heat energy H. It was observed that,
W H. 12. (a)Kelvins statement : It is not possible to have continuous work by cooling a body to a
temperature.Lower than that of coldest of its surroundings. 1Celsius statement : It is not possible to transfer heat from a body at lower temperature toanother at higher temperature without the help of some external energy. 1(b)It is defined as the ratio of the amount of heat removed in a cycle from therefrigerator to the
work done bysome external agency to help the removal of this heat, i.e.,
= 2QW
= 21 2
QQ Q
Also, = 21 2
TT T
For higher efficiency of refrigerator should be higher. 13. (a) Sabita is sincere and hard working and having scientific temper. 2
(b) [k= k1k2/(k1+ k2)]; T = 1 2 1 22 ( ) / m k k k k + . 24. It states that if two vectors can be represented completely (i.e.,both in magnitude and direction) by
the two adjacent sides of a parallelogram drawn from a point then their resultant is representedcompletely byits diagonal drawn from the same point.
Proof :Let P
and Q
be the two vectors represented completely by the adjacent sides OA and OB
of the parallelogram OACB s.t.,
OA = P .OB
= Q
or OA = P, OB = Q
B C
RQ
90
P
ODA
= angle between them AOBIf R be their resultant, then it will be represented completely by the diagonal OC through point O
s.t., OC = R .
Magnitude of R
:Draw CD to OA produced DAC = AOB
=
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8/13
Now in right angled triangle ODC,OC2= OD2+ DC2
= (OA + AD)2+ DC2
= OA2+ AD2+ 2OA.AD + DC2
= OA2+ (AD2+ DC2) + 2OA.OD ....(i)Also in right angled triangle ADC,
AC2= AD2+ AC2 ....(ii)
Also,ADAC
= cos
Or AD = AC cos ....(iii)
andDCAC
= sin
Or DC = AC sin ....(iv)From eqns.(i), (ii), (iii), we get
OC2= OA2+ AC2+ 2OA.AC cos
Or OC = 2 2OA + AC +2OA.AC cos ....(v)
As OC = R, OA = P, AC = OB = Q ....(vi)From equations (v) and (vi), we get
R = 2 2P + Q +2PQ cos ....(vii)
Equation (iii) gives the magnitude of R
. 1
Direction of R :Let the angle made by R with P
.In right angled triangle ODC,
tan =
DC
OD =
DC
OA+ AD
=ACsin
OA+ AC cos
, [by using (iii) and (iv)]
tan =Q sin
P+Qcos
....(viii)
Special cases :(a) When two vectors are acting in same direction, then= 0
R = 2(P+Q)
= R + Q
and tan =Q.0P+ Q = 0
Or = 0Thus, the magnitude of the resultant vector is equal to the sum of the magnitudes of the twovectors acting in the same direction and their resultant acts in the direction of P and Q. 1(b)When two vectors act in the opposite, then
= 180 cos = 1 and sin = 0
R = 2 2P + Q + 2PQ (1)
= 2 2P + Q 2PQ
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9/13
= 2(P Q) or - 2(Q P)
= (P Q) or (Q P)
and tan =Q0
P+Q(1)= 0
= tan 0 or tan 180 = 0 or 180.Thus, the magnitude of the resultant of two vectors acting in the opposite direction to the difference
of the magnitude of two vectors and it acts in the direction of bigger vector. 1
(c)If = 90, i.e.,if P
Q
, then cos 90 = 0
and sin 90 = 1
R =2 2
1
P + Q
and tan =0
P
1
4. Letsv
and
rv
be the velocities of swimmer and river respectively.
Let v
= resultant velocity of vsand vr.
A
(i)Let the swimmer begins to swim at an angle with the line OA where OA is to the flow ofriver.If t= time taken to cross the river, then,
t=coss
l
v
where l= breadth of riverFor tto be minimum, cos should be maximum,i.e., cos = 1.This is possible, if = 0.Thus, we conclude that the swimmer should swim in a direction to the direction of flow of river.
t
v
vr
vs
O
A
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10/13
(ii) v= 2 2s rv v+
(iii) tan = r
s
v
v =Xl
or X = r
s
vlv
(iv) t=s
l
v
(b) It is defined as a vector having zero magnitude and acting in the arbitrary direction. It is
denoted by O
.
Properties of null vector :
(i)The addition or subtraction of zero vector from a given vector is again the same vector.
i.e., A O+ = A
A O = A
(ii)The multiplication of zero vector by a non-zero real number is again the zero vector.
i.e., On
= O
(iii)If
1An =
2Bn , w h e r e n 1 a n d n 2 a r e n o n - z e r o r e a l n u m b e r s , t h e n t h e r e l a t i o n w i l l h o l d g o o d .
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11/13
As the body is rigid, angular acceleration of all the particles of the body is the same. However,their linear accelerations of the particle from the axis. Ifa1, a2, a3,..., anare the respective linearaccelerations of the particle, then
a1=r1, a2= r2, a3= r3, .... 1Force on particle of mass m,is
f1=m1a1= m1r1Moment of this force about the axis of rotation
f1 r1= (m1r1) r1= m1r12 1Similarly, moment of forces on other particles about the axis of rotation are m2r22, m3r32, ....,mnrn2.Torque acting on the body,
= m1r12+ m2r22+ m3r32,...., mnrn2= (m1r12+ m2r22+ m3r32+ mnrn2)
So, = 21
i n
i ii
m r=
=
= I 1where m1r12= moment of inertia of this body about the given axis of rotation.
If = 1, = 1 1 or I =
= I
. 1
Or
Kinetic energy of rotation :K.E. of rotation of a body is the energy possessed by the body on account of its rotation about agiven axis. If figure, we have shown a rigid body rotating inxyplane about z-axis with a uniform angularvelocity .Let the body consists of particles of masses m1, m2, m3,...., mn at distances r1, r2, r3,...., rn
respectively. Let the linear velocity of different particles are v1, v2, v3,....,vn. v1= r1v2= r2v3= r3,.... 1
K.E. of particle mass m1is
21 1
12
m v = 211
( )2
m r
= 2 21 112
m r
Similarly, K.E. of other particles of the body are :
2 2 2 22 2 3 3
1 1, ,
2 2m r m rw w .....
m1 r1
r2
r3
M
Pk
M
M
Axis of rotation
K.E. of rotation of the body = 2 2 2 2 2 21 1 2 2 3 31 1 12 2 2
m r m r m r + + ....
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12/13
= 2 2 2 21 1 2 2 3 31
( ....)2
m r m r m r+ + +
= 2 2
1
1 12 2
i n
i ii
m r=
=
....(1) 1
= 21
I2
i.e., K.E. of rotation = 21
I2
,
where I = 2
1
i n
i i
i
m r=
= . 1
Terminal velocity :It is maximum constant velocity acquired by the body while falling freely ina viscous medium.When a smallspherical body falls freely through a viscous medium, three forces act on it.
(i) Weight of the body acting vertically downwards. (ii) Upward thrust due to buoyancy equal to weight of liquid displaced. (iii) Viscous drag acting in the direction opposite to the motion of body. According to Stokes law.F v, i.e.,the opposing viscous drag goes on increasing with the increasing velocity of the body. As the body falls through a medium, its velocity goes on increasing due to gravity. Therefore, theopposing viscous drag which acts upwards also goes on increasing. A stage reaches when the trueweight of the body is just equal to the sum of the upward thrust due to buoyancy and the upwardviscous drag.At this stage, there is no net force to accelerate the body. Hence it starts falling with
a constant velocity, which is called terminal velocity.Let be the density of the material of the spherical body of radius r and 0be the density of themedium.True weight of the body,
W = volume density g= 343
r g
Upward thrust due to buoyancy,FT= weight of the medium displaced
FT= volume of the medium displaced density g=3
043
r g 1
If vis the terminal velocity of the body, then according to Stokes law.upward viscous drag FV= 6rv
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13/13
When body attains terminal velocity, thenFT+ FV= W
3 043
r g + 6rv = 343
r g
Or 6rv = 3 04
( )3
r g
Or v=2
02 ( )9
r g 1
It depends upon the terminal velocity varies directly as the square of the radius of the body andinversely as the coefficient of viscosity of the medium. It also depends upon densities of the bodyand the medium. 1
Or
Surface energy is defined as the amount of the work done against the force of surface tension, informing the liquid surface of a given area at a constant temperature. 1To obtain a expression for surface energy, take a rectangular frame ABCD having a wire PQ whichcan slide along the sides AB and CD. Dip the frame in soap solution and form a soap film BCQP onthe rectangular frame. There will be two free surfaces of film where air and soap are in contact.
B
C
A
D
Q Q1
P P1
S F
x
Let S = surface tension of the soap solution.l= length of the wire PQ.
Since there are two free surfaces of the film and surface tension acts on both of them, hence totalinward force on the wire PQ is
F = S 2l To increase the area of the soap film we have to pull the sliding wire PQ outwards with a force F.Let the film be stretched by displacing wire PQ through a small distancexto the position P1Q1.The increase in area of film PQQ1P1
=a= 2 (lx)
Work done in stretching film is,E = force applied distance moved
= (S 2l) x= S (2lx)= S a,
where 2lx= a= increase in area of the film in both sides. lx= increase in area of film on both sides.If temperature of the film remains constant in this process, this work done is stored in the film asits surface energy. 1
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