CBSE Sample Paper for Class 11 Physics Solutions - Set B

8/13/2019 CBSE Sample Paper for Class 11 Physics Solutions - Set B

1/15

It is called so because this law holds good irrespective of the nature of interacting bodies at allplaces and at all times. 1

It is the ratio of change in dimension of a body to the original dimension when some deformingforce is applied to it,e.g.,longitudinal strain. 1

The upward force on a body, dipped in a liquid is called buoyant force. 1Time period decreases as effective value of acceleration due to gravity increases (i.e.,g=g+ a) and

(T 1 / ).g 1

Intensity = (Amplitude)2 21

(distance)

Amplitude1

distance

Hence,

1

2

A

A =

2

1

r

r =

25

9 1

% error in = 13 2 1002

a b c d

a b c d

+ + + 1

=1

3 1 2 3 4 12

+ + + 1

= 3 + 6 + 2 + 1 = 12%

Mass given bythe equation a=F

M, i.e.,M =

F

ais called inertial mass. Clearly, higher the mass

means lesser the acceleration. Thus mass of a body resists the acceleration, i.e.,rate of change invelocity coming up due to external force or in other words it is the measure of inertia of a body. 2

Or

2 2| | (| | | |)a b a b+ +

= ||+ +

2 2| | | | 2| |a b a b

cos ||2 2| | | | 2| |a b a b

= ||2| |(1 cos )a b

= || 22| |sin2

a b

1

= a negative quanity

Hence, +| |a b

< | + || |a b

It should be greater in night. In the day time the body is pulled by earth and sun in two oppositedirections. This will result into decrease in weight. 1

During night time the earth and the sum pull the body in the same direction so the weight willincrease in night. 1Monoatomic gas :

Cv=F 3

R R2 2

Cp=F

1 + R2

SAMPLE QUESTION PAPER B

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2/15

=3

1 + R2

=

5R

2

= 21F

+ =2

13

+

=5

3= 167 1

Diatomic gas :For diatomic gasf= 5

Cv= R3

f=

5R

2

Cp= 1 R2

f +

=5

1 R2

+ =7

R2

= 21F

+ =

21

5

+

=20

5= 14

0. Motion of piston in a cylinder is another example of vibrating system.Restoric forcein this case is given by :

F = mg= kx 1where mis mass of the piston andxis the displacement time period of vibration is given by

T =Inertiafactor

2Spring factor

=

M

2 k 1. Let there be aplanet P far away from earth. The planet is observed from two different observation

points A and B on earth called observatories. If distance AB = band angle APB= radian, then

being very smallD

b


3/15

(d) Since the beaker is tumbling and then slips off. So, the distance covered by it in reasonableduration of time is not large. so it is not treated as a point object. When the stone is dropped, it falls freely under the acceleration due to gravityg. With respect tothe earth the acceleration of the stone is g. 1Inside the carriage the stone possesses two accelerations :(i) Horizontal acceleration a due to the motion of carriage. 1(ii) Vertical downward acceleration g due to gravity.The acceleration of the stone w.r.t. the carriage

= 2 2a g+Suppose m1and m2be the masses suspended at the ends of a light inextensible string passing overthe pulley. m1= 8 kg, m2= 12 kgSuppose T = Tension in the string = ?a= common acceleration with which m1moves upward and m2moves downward = ?

Ta

aT

m1

m2m g1

m g2

T

m g2

m2m g1

Tm1

The equationsof motion of m1and m2are given byT m1g=m1 a ....(i)

and m2g T =m2 a ....(ii) Adding equations (i) and (ii), we get

(m2 m1)g= (m1+ m2)a

a= 2 1

1 2

( )m m g

m m

+

From equations (i) and (iii), we get

T =2 1

1 11 2

( )m m gm g m

m m

+

+

T = 2 1 2 2 11 2

( )m g

m m m mm m

+ + +

T = 1 21 2

2m mg

m m+...(iv)

Putting m1= 8 kg and m2= 12 kg andg= 10 ms2in equations (iii) and (iv), we get

a=12 8

108 12

+

= 4 1020

= 2 ms2

(a) v=Adx

dt

= 2

and a=2

A2

d x

dt= 0

As per this equation the particle has no acceleration at all. 1

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4/15

(b) v=Bdx

dt

= 3 2t+ 2 + 0

= 6t+ 2

a=2

B2

d x

dt= 6

Here acceleration is uniform. 1

(c) v=Cdx

dt

= 5 3t2+ 4

= 15t2+ 4

a=dv

dt=

2C

2

d x

dt

= 15 2t+ 30t

Here, acceleration depend upon time so it is not uniform. 16.

A B

S = 2m

N

mg

u = 4 ms1 v

f = mgr

(a) (b)

Figure (a) here shows the initial position A and the final position B of the object. Let the speed ofparticle at B be v. The particle does work against force of friction which is a non-conservative force.

Ki Kf= W 1

where W is the work done against the force of friction. Figure (b) shows all the forces acting on theobject,

W =fr+ S = mgS

= (015 01 10 2) J

= 03 J 1

Hence 2 21

01 [4 ]2

v = 03

Or 16 v2= 6

v= 10 ms1= 316 ms1 1

7. Velocity of centre of mass of a system,

cmv

= d r

dt

Total external force F

=

2

2Md r

dt

= CMMd v

dt

....(1) 1

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5/15

Internal forces cancel out in pairs, so velocity of centre of mass is not a affected by internal force.

Puting F

= 0 in equation (i),

= CMd v

dt

= 0, (Qm0) 1

i.e.,cmv

= constant

If resultant external force is zero, the velocity of centre of mass remains same. 1

Or

Expression for torque in cartesian co-ordinates from rotation of a particle in a plane : Physicalmeaning of torque :

Fig.Consider a particle of mass mrotating in a plane XY about the origin O. Let P be the position of the

particle at anyinstant, where OP

= r

and XOP = . Let the rotation occur under the action of a

force F

applied at P, along PA

.

In a small time dt, let the particle at P reach Q, where

OQ

= ( )r d r+

and POQ =dIn vector triangle OPQ,

OP PQ+

= OQ

P Q

= OQ OP

= + -

( )r d r r

P Q = d r Small amount of work done in rotating the particle from P to Q is

dW = F .d r ....(i)

If Fx, Fyare rectangular components of force F and dx, dyare rectangular component of displacement

of dr

, then

F = +

( F F )x yi j

d r

= + ( )i dx j dy

From equation (i), we get

dW = +

( F F )x yi j + ( )i dx j dy

dW = Fx dx+ Fydy

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6/15

Let the co-ordinates of the point P be (x,y) as is clear from fig.x=rcos ....(ii) y=rsin ....(iii)

y

x

y

xO

r

P (x,y)

Differentiating equation (iii) w.r.t.

dx

d= ( cos )

dr

d

= (cos )d

rd

=r( sin ) = rsin

= y

....(iv) dx= y d ....(v)Again, differentiating equation (iv) w.r.t.

dy

d= ( sin )

dr

d

=rcos =x ....(vi)

dy=x d ....(vii) Substituting equation (ii)

dW = Fx(y d) + Fy(x d)=xFy dyFx d

dW = (xFyyFx)d ....(viii)dW =(d) ....(ix)

= (xFyyFx) ....(x) is called torque.

8. Suppose the force of gravitation vanish when the rocket is at distancexfrom the earth.

Sun

M M'

Earth

r

Rocketx

Let. M, Mand mbe the masses of the sun, the earth and the rocket respectively.

The force of attraction between the rocket and the sun =2

GM

( )

m

r xand is directed towards the sun

=1

1 2g

=1

2g

=2

g....(v) 1

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7/15

Wd=mgd= 2g

m . [by using equation (v)]

=1

2mg =

1W

2

=1

2502

= 125 N.

Weight of the body half way down to the centre of earth = 125 N. 1Displacement :Consider a reference particle moving on a circle of reference of radius awithuniform angular velocity . Let the particle start from the point X and trace angular radian intime tas it reaches the point P.

Therefore, =L

or = t.

Let the projection of the particle P on diameter YOYbe at M. Then OM =yis the displacement inS.H.M. at time t.

In OPM, sin =OM

OP=

a

or y=asin = asin t ....(i)

Important notes :(1) If projection of P is taken on diameter XOX, then point Nwill be executingS.H.M. Here, ON =x= displacement in S.H.M. at time t

In ONP, cos =ON

OP

or x=acos = acos t ....(ii) (2) If A is the starting position of the particle of reference such that AOX = 0and AOP = t.

Y

Y'

X XO

wt

q0

N

P

=XOP = t 0Here () 0is called the initial phase or epoch of S.H.M.If B is the starting position of the particle of reference such that

BOX =0

BOP =t.

X X

Y

Y'

M

ywt

N

q

f0

P

=XOP = t+ 0From equation (i), y=asin (t+ 0)From equation (ii), x=acos (t+ 0)

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8/15

Here (+) 0is called initial phase or epoch of S.H.M.0. (a) The formula for velocity of sound in air is given by

v=P

....(1)

where = constant for a given gas, i.e.,air= density of the gas

From gas equation,PV = RT ....(1)

Or P = RTV

....(2)

From equations (1) and (2), we get

=RT

V

=

RT

M

....(3)

where V = M= molecular weight of air or gas.For a given gas, m= constantR is also constant.When T = constant, then from equation (3), we conclude that vindependent of the pressure of air(gas) if temperature remains constant. 1(b) Effect oftemperature :We know that

PV = RT

Or P =RT

V

Also v=P = V

RT =

RT

M

where M = V = molecular weight of the gas. Since , R and M are constants, sov T .

It means velocity of sound in a gas is directly proportional to the square root of its temperature,hence we conclude that the velocity of sound in air increases with increase in temperature. 1

(c) Effect of humidity :The presence of water vapours in air changes the density of air, thus thevelocity of sound changes with humidity of air.

Let m= density of moist aird= density of dry air

vm= velocity of sound in moist air

vd= velocity of sound in dry air

From relation v= P , we get

vm=P

m

....(i)

and vd=P

d

g

r....(ii)

Dividing equation (i) by equation (ii), we get

m

d

vv

= dm

Also we know that density of water vapours is less than the density of dry air. It means dry air isheavier then water vapours as the molecular mass of water is less than that of N2(28) and O2(32),som< d

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9/15

Or d

m

> 1 ....(iv)

from equations (iii) and (iv), we get

m

d

v

v> 1

vm>vdIt means velocity of sound in air increases with humidity, i.e.,velocity of most air is greater than

the velocity of sound in dry air. That is why sound travels faster on rainy day then a dry day. 1Given : Radius, r= 10 cm = 102cm

Length of tube, l= 15 mDensity of glycerine, = 13 103kg m3

Mass of glycerine flowing per sec,M = 4 103kg s1

Viscosity of glycerine, = 083 Pas.= 083 Nm2s.

Suppose P = pressure difference between to ends of the tube = ?Reynold number, NR= ?

Suppose V = Volume of glycerine flowing per second

=M

=3 1

3 3

4 10 kgs

13 10 kgm

=4

13

= 106 m3s1

Now from Poiseulles formula, we have

V =

4

8

hpr

lor P = 8lV/r4

P =

6

2 4

48 083 15 10

13

3142 (10 )

= 97537 102Pa= 98 102Pa. 1

To check the assumption of laminar flow in the tube, let us find out the value of Reynolds numbergiven by

NR=D cv

....(i)

Where vc= critical velocity

=volumeflowingoutpersec

area of cross- section

= 2/M

r

....(ii)

For Laminar flow, NRlies between 0 to 2000.Now from equations (i) and (ii), we get

NR =2

M 1D

r

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10/15

= 22 Mr

r , (QD = 2r)

=2M

r

=3

2

2 4 10

3142 10 083

= 307 101= 0307 = 031. 1

2. Molar specific heat of a substance is defined as the amount of heat required to raise the temperatureof one gram mole of the substance through a unit degree. By definition one mode of many substanceis a quality of the substance whose mass in gram is numerically equally to the molecular mass M.Thus c= Mc ....(i) 1n no. of molesm mass of substance in grams.

i.e., n=M

m

or m=nM 1

From equation (i), c= ( )m t

= M( )m t

Mc=1

n t

Using equation (i) C =1

n t

1

3. (a) Rahim loves animals and feeds them, dont frighten animals with cveapong. 1(b) (i) The tension developed in the string when the monkey climbs up withan acceleration of6 m/s2is given by T = m(g+ a) = 40 (10 + 6) = 640 N. 1

(ii) The tension developed when the monkey climbs down with an acceleration of 4 m/s2is given byT = m(g a) = 40 (10 4) = 40 6 = 240 N. (iii) When the monkey climbs with uniform speed of 5 m/s acceleration is zero and the tension inthe string is T = mg= 40 10 = 400 N. (iv) As the monkey falls down the rope nearly under gravity, the tension in the string is given by T= m(g a) = m(gg) = 0Since the string can withstand a maximum tension of 600 N, hence the rope will break only in thefirst case (i)

4. (a) Let a constant force F

be applied to a body moving with initial velocity u

,so that its velocity

becomes v along the direction of force when sis its displacement. Using Newtons second law ofmotion we get magnitude of force F = maand from equation of motion, we get v2 u2= 2as, whereais the acceleration of the body. 1Multiplying both sides by m, we get

mv2 mu2= 2mas

i.e.,2 21 1

2 2mv mu =mas

i.e.,2 21 1

2 2mv mu = Fs= W

i.e., K.E.(f) K.E.(i)= W 1where K.E.(f)is final kinetic energy and K.E.(i)is initial kinetic energy.Thus work done on a body by a net force is equal to the change in kinetic energy of the body. 1

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11/15

(b) Suppose, W be the amount of work done in a small time interval t, when Pavbe the averagepower, then

Pav=W

t

....(i)

When P be the instantaneous power, then by def.

P =0

Lt Pavt

= 0

W

Ltt t

P =Wd

dt= (W)

d

dt....(ii)

Now W = F .s ....(iii)

where F = constant force producing a displacementFrom equations (ii) and (iii), we get

P = ( F . S )d

dt

= SFd

dt

= F . .v

Or

(a) In physicswork is not said to be done, if (a) The applied force (F) is zero. A body moving withuniform velocity on a smooth surface has some displacement but no external force so in the casework done is zero. 1

(b) The displacement (s) is zero. A labourer standing with a load on his head does no work.

(c) The angle between force and displacement () is /2 rad or 90. Then cos =cos 90 = 0. Thuswork done is also zero.

In circular motion, instantaneous work done is always zero because of this reasons. 1

(d) The change in kinetic energy (KE) is zero (b)If roads were to go straight up, the slope would have been large, the frictional force (mgcos )would be small. The wheels of the vehicle would slip. Also for going up a large slope, a greaterpower shall berequired. 2

It states that if gravity effect is neglected, the pressure at every point of liquid in equilibrium ofrest is same.

Proof :Consider two points C and D inside the liquid in a container which is in equilibrium of rest.Imagine a right circular cylinder with axis CD of uniform cross-section area A such that points Cand D lie on float faces of the cylinder in figure. 1

The liquid inside the cylinder is in equilibrium under the action of forces exerted by the liquid

outside the cylinder. These forces are acting every where to the surface of the cylinder. Thusforce on the flat faces of the cylinder at C and D will to the forces on the curved surface of the

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12/15

cylinder. Since the liquid is in equilibrium therefore, the sum of forces actingon the curved surface

of the cylinder must be zero. If P1and P2are the pressure at points C and D and F1and F2are the

forces acting on the flat faces of the cylinder due to liquid, then

F1= P1A

and F2= P2A

Since the liquid is in equilibrium, therefore

F1= F2or P1A = P2A

or P1= P

2It means the pressure at C and D are the same.

Experimental Proof: Consider a spherical vessel having four cylindrical tubes A, B, C and D

each fitted with air tight frictionless piston are of cross-section a, a/2, 2aand 3arespectively.

F A

DB3F

F/2

2F

C

Fill the vessel with an incompressible liquid so that no air gap is left inside the vessel and pistonfitted in the various cylindrical tubes. Push the piston A with force F. The pressure developed onthe liquid = F/a= P (say)

It is seen that all other pistons will be pushed outwards. To keep the pistons at their respectiveoriginal positions, the force F/2. 2F and 3F respectively required to be applied on pistons of tubesB, C and D respectively to hold them. Now pressure developed on liquid in tubes B, C and D are F/2, 2F/2a, 3F/3a i.e.,such equal to F/a. This indicates that the pressure applied is transmittedequally to allparts of liquid. This proves Pascal law. 1

Or

(i) Inside a liquid drop :Consider a liquid drop of radius R. The molecules lying on the surfaceof liquid drop, due to surface tension will experiencearesultant force acting inward to thesurface.Let S = Surface tension of liquid drop

P = excess pressure inside the dropDue to excess of pressure, let there be increase in the radius of the drop by a small quanity R, infigure.

PR

R

Then work done by the excess pressure,W = force displacement

= (excess pressure area increase in radius)= P 4R2 R ....(i)

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13/15

Increase in surface area of the drop

= final surface area initial surface area

= 4(R + R)2 4R2

= 4[R2+ 2R (R) + (R)2 R2]=R.R

[Neglecting, (R)2being very very small] Increase in surface energy = increase in surface area surface tension

= 8R (R) S ....(ii)

As the increase surface energy is at the cost of work done by the excess pressure, therefore from(i) and (ii),

P 4R2 R = 8RR S

P =2S

R(ii) Inside a liquid bubble :Consider a soap bubble of radius R the molecules lying on the surfaceof liquid bubble with experience a resultant force acting on water to the surface due to thesurface tension.

Let S = surface tension of the soap solution

P = excess of pressure inside the bubble.

PR

R

Due to it, let there be increase in the radius of the bubble by a small amount R in fig., thenwork done, W = force distance

= (excess pressure area) increase in radius

= P 4R2 R ....(ii)The soap bubble has two free surface one outside the bubble and one inside the bubble when soap

solution and air are is contact.

The effective increase in surface area of the bubble= 2 [final S.A initial S.A.]

= 2 [4(R + R)2 4R2]

= 2 4[R2+ 2R(R) + (R)2 R2]= 8 2R (R), [Neglecting (R)2, being very small]= 16R.R

Increase in surface energy is as the cost of work done by the excess pressure therefore from (iii)and (iv), we get

P 4R2 (R) = 16R (R) S

Or P =4S

R

(iii) Inside a bubble in liquid :Consider an air bubble of radius R. Just inside a liquid of surfacetension S. The air bubble will have only one free surface as shown in figure. It can be shown thatthe pressure inide the air bubble is given by

P =2S

R

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14/15

PR

6. (a)(i) Size of a molecule of a gas is very small as compared to intermolecular distance.

(ii) Behaviour of the molecules of the gas is perfectly elastic.

(iii) Molecules do not exert any force on each other except in collision.(iv) Velocities of the molecules are in all directions.

(v) Molecules are in random motion.

(vi) The duration of collision of instantaneous.

(vii) The molecules move in a straight line between two sucessive collisions.

(viii) Collisions of the molecules with one another and the walls of the container are perfectly

elastic.

(b)Using thegiven relation,

P =

21 ,3

Vwe get

P = -21 M

3 Vv

PV =21 M

3v

Here M is fixed and v2TIf temperature T is fixed then PV = constant.

Or P

1

V

If temperature of a given mass of a gas is kept constant its pressure is inverselyproportional to itsvolume.

Charles law: P = -21 P

3v

= -21 M

3 Vv , we get

V = 21 M3 P

v

v2 T and M is fixedV T.

If pressure of a given mass of gas is kept constant its volume is directly proportional to thetemperature of the gas.

Or

(a) Pressure exerted by on. mole of gas,

P = 21 P3 v =21 M3 Vv

Or PV =21 M

3v

But PV = RT 1

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15/15

21

M3

v = RT

Or21 M

3v =

3RT

2

Or21 (N )

2m v = =

3RT, (M N )

2m

Or 21

M2

v =3 R

T2 N

=3

RT2

1

Total random K.E. for one mole =3

RT2

and average K.E. per molecule =3

KT2

. 1

(b)Root mean square velocity is defined as the square root of the average of the squares of theindividual velocities of the gas molecules i.e.,

vrms=2 2 2 21 2 3 .... nv v v v

n

+ + + +

= 2v

where v1, v2, v3,...., vnare individual velocities

vrms=3P

=3RT

M=

3RT

m

i.e., vrms T . 1

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Documents

CBSE Sample Paper for Class 11 Physics Solutions - Set B