(agra region) session ending examination 2017-18 …...describe first law of thermodynamics. apply...

KENDRIYA VIDYALAYA SANGATHAN(AGRA REGION)

SESSION ENDING EXAMINATION 2017-18CLASS – XI

SUBJECT - PHYSICS(SOLVED PAPER)

Time: 3 Hrs. M.M.: 70

General Instructions: (i) All questions are compulsory. (ii) There are 26 questions in total. (iii) Question No. 1 to 5 is very short answers and carries one mark each. (iv) Question No. 6 to 10 carry two marks each Question No. 11 to 22 carry 3 marks each Question No. 23 is a value based question which carries 4 marks Question No. 24 to 26 carry 5 marks each. (v) There is no overall choice however an internal choice has been provided in one question of two marks, one question of three

marks and all three questions of five marks each. You have to attempt only one of the choices in such questions. (vi) 15 minutes time has been allotted initially to read this question paper. The students will read the question paper only and

will not write any answer on the answer during this period. (vii) Use of calculators is not permitted however you may use log Tables and trigonometric table if necessary. (viii) You may use the following values of physical constants wherever necessary.

Earth’s radius Re = 6.3781×106 m, c=3 × 108 m/s, h=6.6 × 10-34 Js, e=1.6 × 10-19 C, NA = 6.023 × 1023/mole mn = 1.67 × 10-27 kg, me = 9 × 10-31 kg, k (Boltzmann constant) = 1.38 × 10-23 m2 kg s-2 K-1 and g = 9.8 m/s2

Section A 1. Write the dimensional formula for (a) Planks constant & (b) Surface Tension. 2. Is it possible to have constant rate of change of velocity when velocity changes both in magnitude and direction? 3. What is thermal conductivity of perfect heat conductors? 4. An ideal gas is compressed at a constant temperature; will its internal energy increase or decrease? 5. The absolute temperature of a gas is increased 3 times. What will be the increase in root mean square velocity of

the gas molecule?

Section B

6. A physical quantity, Pa bcd

=3 2

the percentage errors in measurement in a, b, c, d are 1%, 3%, 4% & 2% respectively.

What is the percentage error in measurement of quantity P? 7. Show that impulse of a force is equal to change in linear momentum produced by the force. 8. What are two angles of projection of a projectile projected with velocity 30 m/s, so that the horizontal range is

45 m. Take g = 10 m/s2.OR

If a projectile has a constant initial speed and angle of projection, find the relation between the changes in the horizontal range due to change in acceleration due to gravity.

9. A ball of 0.1 kg makes an elastic collision with a ball of unknown mass that is initially at rest. If the 0.1 kg ball rebounds at one third of its original speed, what is the mass of the other ball?

10. y(t) =(sin ωt - cosωt) represents a Simple harmonic motion,determine the period of SHM.

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2 ] Oswaal CBSE Solved Paper - 2018, PHYSICS, Class-XI

Section C 11. Prove that second law is the real law of motion. 12. A lift of mass 400 kg is hung by a wire. Calculate the tension in the wire when the lift is (a) at rest, (b) moving

upward with a constant velocity of 2.0 m/s,(c) moving upward with an acceleration of 2.0 m/s2 and (d) moving downward with an acceleration 2.0 m/s2.

13. Define Conservative forces, with the help of suitable example show that gravitational forces are conservative in nature.

14. Derive an expression for torque in Polar coordinate system, with the help of appropriate figure. 15. A ring of diameter 0.4 m and of mass 10 kg is rotating about its geometrical axis at a rate of 35 rotations per second.

Find (a) Moment of inertia, (b) angular momentum and (c) rotational kinetic energy of the ring. 16. Derive an expression for acceleration due to gravity at a depth ‘d’ from earth’s surface. Is it increases or decreases. 17. A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25 × 106 m above the surface of

earth. Find the orbital speed and the period of revolution of satellite. Given: Earth’s radius Re = 6.38 × 106 m and g = 9.8 m/s2.

18. Explain the elastic behaviour of wire under an increasing load, with the help of plotting a stress-strain relationship curve.

19. A two wires A and B of length l, radius r and length 2l, radius 2r having same Young’s modulus Y are hung with a weight mg as shown in figure. What is the net elongation in the two wires?

20. Explain the two specific heats of gas? Show that the difference between two specific heats is equal to the Universal gas constant.

OR Describe first law of thermodynamics. Apply this law to describe, (i) isothermal process and (ii) adiabatic process. 21. Write Newton’s formula for the speed of sound in a gas. Why and what correction was applied by Laplace in this

formula. 22. Derive an expression for the time period and frequency of a simple pendulum.

Section D 23. An old man walks 10m due east from his house and then turns to his left at an angle of 60° with east. He then

walks 10m in that direction and falls down on the ground & got injured. His grandson observing him moves straight towards him from the initial position of his grandfather, helped him to stand and take him safely to home.

(i) Should the boy follow the same path followed by the old man? If not, why? (ii) What are the values you suggest for the boy reply?



Oswaal CBSE Solved Paper - 2018, PHYSICS, Class-XI [ 3

Section E 24. State & prove the law of conservation of linear momentum. Write its two applications. The linear momentum of

a body can change in the direction of applied force. Comment.OR

(a) Derive an expression for the maximum speed acquired by vehicle on a banked circular track of radius r with coefficient of friction μ between the wheels and road.

(b) A car wheels are 1.5 m apart has its centre of gravity 0.75 m above the ground. Calculate the maximum speed at which it can go round an un banked curve of radius 20 m.

25. State and prove Bernoulli’s theorem. Write any two limitation of Bernoulli’s theorem.OR

(a) Derive an ascent formula for the rise of liquid in a capillary tube. (b) A liquid of density p and surface tension S rises to a height h in a capillary tube of diameter D. What is the

weight of the liquid in the capillary tube? Angle of contact is 0°. 26. Derive an expression for the total energy of a particle executing SHM. Show graphically the variation of P.E. and

K.E. with time in SHM.OR

(a) What are beats? State the necessary condition for the production of beats. (b) Calculate the speed of sound in a gas in which two waves of wavelengths 1.00 m & 1.01 m produce 4 beats

per second.




Section A 1. Planck’s constant – [ML2T-1] Surface Tension – [ML0T-2] 2. Yes, example – body moving upwards or

downwards where acceleration is constant while magnitude and direction changes.

3. Thermal conductivity is the heat energy transferred in unit time from unit area having a unit difference in temperature over unit length. It is expressed as Wm–1 K–1.

4. As per first law of thermodynamics, conservation of energy is dq = du + pdV, where

dq = thermal energy of gas du = change in internal energy of gas pdV = work done by gas in expanding due to

change in volume dV If T = const., dT = 0 and d(pV) = 0 as pressure and

volume are inversely proportional. Also, dq = pdV, which shows that there is no change in internal energy.

5. V in terms of Boltzmann constant (k)

T =3T

r.m.s =3kTm

Now, Vr.m.s.=3k3T

m= 3 Vr.m.s∴

If the temperature of the gas increases 3 times, then the root-mean-square speed of the gas particles increases by a factor of √3.

Section B

6. P =a b

cd

%error in PP

P= 3

aa

+2b

b+

cc

+d

dP

3 2

∆ ∆ ∆ ∆ ∆

×

∆

12

100

So, PP

100 = 3a

a100+2

bb

100+c

c100

× ×∆

× ×∆

× ×∆

×12

+d

d100

12

×∆

×

Also, % error in P = 3(% error in a) + 2(%% error in b)

+ (% error 12

in c) + (% error in d)12

= 3×1%++2×3%+12

×4%+

= 3%+6%+2%+1%

12

2× %

=12% 7. According to Newton’s 2nd law

F=

dvdt

dPdt

m* =

or F.dt = dP If P1 and P2 are linear moments of the body at time

t and t = 0, so on integrating the expression Fdt = dP within the limits :

Since, I= Fdt

F dt = dPP

P

o

t

1

2

≡≡

or F.(t – 0) = P2 – P1

or, Fd = DP or, I = DP 8. Horizontal Range

R =U sin2

g

sin2

sin2 =R.g30

=45×1030×30

=12

sin2 =12

2

2

θ

θ

θ

θ

= R*g/U2

== sin30

=15 and 90 15 =75

o

o o o o∴ −( )θ

OR

R =U sin2

g; R =

U sin2g

; R =U sin2

g

RR

=gg

2

1

2

12

2

2

1

2

2

1

θ θ θ

∴

Hence, horizontal range is inversely proportional to the acceleration due to gravity.

9. For elastic collision

V =m m U +2m U

m +m

m = 0.1kg, U = V; U = 0; V =V3

V3

=

11 2 1 2 2

1 2

1 1 2 1

−( )

−

∴−00.1 m V+0

0.1+mCross multiplying, we see

V(0.1) V(m )= 0

2

2

2

−( )

− − ..3V 3V(m )0.4V = Vm

Calculate m = 0.2kg.

2

2

2

−− −

∴2

10. y t = sin t cos t

= 212

sin t12

cos t

= 2 sin t cos4

cos

( ) −

−

−

ω ω

ω ω

ωπ

ωωπ

ωπ

t sin4

= 2 sin t4

It is S.H.M. where angular

−

frequency is = 2 / T

Period T =2

ω ππ

ω

Section C 11. (a) First law is contained in the 2nd law F =ma No external force F =0 \ ma =0 Q m ≠0 \ a=0 So there will be no acceleration if no external force

is applied. (b) Now we will prove that Third law is contained

in Second law. Let the body A exerts force F1

��

SOLUTIONS




(action) time ∆t on B and B exerts force F2

��

(reaction) on A. Change in linear momentumA = F × t2

�D and of B = F × t1

�D

Total momentum = 0

F t +F t = 0

F t = F t

F = F

1 2

1 2

1 2

��

� ��

��

∆ ∆

∴ ∆ ∆

∴

−

− Action = Reaction: From above, we conclude that Newton’s first and

third law are contained in Newton’s second law, so second law is the real law of motion.

12. (a) Lift at rest T =W=mg=4000N (b) Moving up with 2m/s, a=0; T = W=mg = 4000N (c) Moving up with a = 2m/s2 T=mg+ma=4000

+ 400 × 2 = 4800N (d) Moving down with a= −2m/s2; T=mg – ma=4000

− 400 × 2 = 3200N 13. Conservative force is a force which has a property

that the total work done while moving an object between the two points is independent of the path taken. A conservative force is dependent only on the position of the object.

F= G

M MR1 2

2

Since work done is independent of the path followed in gravitational field, hence gravitational force is a conservative force.

14. Expression for torque in polar coordinates Suppose the position vector of the particle at point

P(x, y) is r�

and it makes an angle θ with x-axis. Let the angle between the force vector F

��and

position vector r�

be φ . If F��

makes an anglea with the positive direction of x-axis, then

f a - qa a

= Now = F cos and = F sinF Fx y

If x and y are coordinates of the point P where PO= and anglr ee XOP= , thenq

q q

= cos and = sinx r y rWWe have seen that : = yt -xF yFx (1)

�x

y

r

F

O

�

�

F sin� F sin

�

PP (x,y)

dN

�

YF

O

�

�r

��

x

Putting the various values in equation 1, we have,

t q a - ( q)( a)=( cos )(F sin ) sin F cos

= F si

r r

r nn cos cos sin

= Fsin

a q - a q

a - q( )

( )r

= rFsin =

= r Fs

f f a - q\ t

∵( )iinf

The above expression is the expression for torque in polar co-ordinates. Note that torque due to a F��

force depends upon the magnitude of force and displacement ( r

�) of the force from the axis of

rotation (point O). 15. (i) Moment of inertia

I =12

MR

=12

×10×0.2×0.2 = 0.20 Kgm

2

2

(ii) Angular Momentum

L = I = 2 n = 2 ×35 =70L = 0.2×70

ω ω ππ π

π==14×3.142

= 43.99 Kg m s2 -1

(iii) Rotational K.E

=12

I

=12

×0.2×70 ×70 J

= 4836 J

2w

p p

16. Acceleration due to gravity at a depth ‘d’ Consider earth to be a homogeneous sphere of

radius R and mass M with centre at O . Let g be the value of acceleration due to gravity at a point A on the surface of earth,

then g =GMR2 ...(1)

A

d BR

O

(R-d)

If r is uniform density of material of the earth, then

M =43

R3p r

Putting value of M in equation (1) and equating it, we get:

\p r

p r g =G× 4

3R

R=

43

G R

3

2 ...(2)

Let g’ be the acceleration due to gravity at the point B at a depth d below the surface of earth.

The distance of the point B from the centre of the earth is ( R – d). The earth can be supposed to be made of a smaller sphere of radius ( R – d ) and a spherical shell of thickness d.




The body at B is inside the spherical shell of thickness d. The force on body of mass m at B due to spherical shell is zero.

The body at B is outside the surface of smaller sphere of radius (R – d). The force on the body of mass m at B is due to smaller sphere of earth of radius (R - d) is just as if the entire mass M' of the smaller sphere of earth is concentrated at the centre O.

∴−( )

−( )

∴

g'=GM'

R d and M'=

43

R d

2

3π ρ

g'=G×

43

R d

R d =

43

G R d

3

2

π ρπ ρ

−( )−( )

−( ) ...(3)

Dividing (3) by (2), we get

g'g

=

43

G R d

43

GR=

R dR

=RR

dR

or g'= g 1dR

π ρ

π ρ

−( ) −− −

Hence, as depth increases, value of acceleration due to gravity decreases.

17. Orbital speed

V = Rg

R+h= 6.38×10

9.86.38+0.25 10

=6.38×10 ×3.13

2.57×10=7.

66

6

3

( )

88×10 m / s3

Period of revolution

T =2 (R+h)

R

=2×3.1426.38×10

6.63×6.63×6.63×109.8

=

6

18

p3

2

g

00.98×17.07×103.13×10

= 5.34×10 s.9

63

18. Stress-strain curve for an elastic material The behaviour of the wire under increasing load

can be studied by calculating the stress and strain against each load through a graph.

Suppose a metallic wire of uniform cross-section is suspended from a rigid support. When the load on the other end is gradually increased, the length of the wire goes on increasing. If graph between stress and strain is plotted, the shape of the curve will be as shown in Fig.

....

Stress

Elasticlimit

Yieldpoint

Limit ofproportionality

Hooke's lawbeing obeyed

StrainO O1

A BC D

(a) Portion OA: The portion OA of the graph is Stress a straight showing that upto point A, strain produced in the wire is directly proportional to the stress i.e., strain a stress. In this portion, the material of the wire obeys Hooke’s law. The point A is called the limit of proportionality. The proportionality limit is the greatest stress a material can sustain without the departure from a linear stress-strain relation. If the applied force is removed at any point between O and A, the wire regains its original length.

(b) Portion AB: The portion AB of the graph is not a straight line showing that in this region, strain is not proportional to the stress. Note that the slope of the graph is decreased; this means that strain increases more rapidly with stress. Nevertheless, if load is removed at any point between O and B, the wire will return to its original length. The point B is called the elastic limit. The elastic limit is the maximum stress which a body can sustain and still regain its original size and shape once the load has been removed.

(c) Portion BC: If the stress is increased beyond the elastic limit, a point C is reached at which there is marked increase in extension. This point is called yield point. Between B and C, the material becomes plastic i.e. if the wire is unloaded at any point between B and C, the wire does not quite come back to its original length. The extension not recoverable after removing the load is known as permanent set. However, this permanent deformation is not serious enough to be important. In practice, we must keep the stress below the yield point. Here OO’ is the permanent set.

(d) Portion CD: If the stress is increased beyond point C, the wire lengthens rapidly until we reach point D at the top of the curve. The point D is called the utlimate strength or breaking stress. Beyond point D, even a stress smaller than at C may continue to stretch the wire until it breaks.

19. Wire A; length l and radius r Wire B; length 2l and radius 2r Young’s modulus of wires A and B; Y Net elongation

D D D

p p

p p

l = l + l

=mgL

r Y+

mglr Y

=mglr Y

+mg 2l

2r

1 2

1

12

22

2

2

( )( )22 2 2

2

2

Y=

mglr Y

+2mgl4 r Y

=4mgl+

r Y

=32

mglr Y

p p

p

p

2mgl4

OR 20. Apply first law of thermodynamics.




Heat added = increase in internal energy + external work done by the system \ dQ = du+dw Specific heats of gases are expressed as molar

specific heats which are constant pressure Cp and constant volume Cv.

From enthalpy, H = u + Pv Relationship between specific heats for ideal gases

be written as H = u + Pv dh = du + d(RT) Cpdt = Cvdt + Rdt Cp = Cv + R or R = Cp - Cv where R is gas constant

OR First Law of thermodynamics: Whenever heat is

added to a system, it transforms to an equal amount of energy in some other form. i.e

(a) Isothermal process – Temperature remains constant, the internal energy remains constant

\ du = 0 From first law of thermodynamics, dQ = du + dw = 0 + dw \ dQ = dw (b) Adiabatic process – no heat energy enters or

leaves. i.e. dQ = 0 \ dQ = du+dw \ du = – dw 21. Newton’s formula for the velocity of sound

V =Pr

Where P is pressure and r is the density since the motion of sound is an isothermal process.

The calculation showed speed of sound at NTP is 280ms-1 which is much less than the actual speed 332ms-1.

Laplace corrected by saying that it is adiabetic and within the adiabatic process temperature will

increase & decrease So, V =Pγ

r

So,

γ is constant =

CC

P

V = 1.41

22.

lT

mg cos �mg

�

x

mg sin �

T

From the figure shown, mg - Weight of bob mg cos θ - opposite to tension T mg sin θ - towards O.

T=mg cos θ Restoring force F= – mg sin θ

q \ q q

\ - q

-

-

is small sin = =x

F= mg

=mgx

since F= kx

l

l(I)

(II)Comparing equations (I) andd (II),

k =mgl

Substituting value for k in SHM equation for pperiod of oscillating system, which is shown as:

T = 2inter

pttia factor

Spring factor

2 = 2m

mg /= 2

g

Frequency =1T

=1

T= m/kp p pl

l

f22

g

2g

p

p

ll

Time Period: T=

Section D 23. N

W E

S

10m

C

A 60°

House 10m east B

(a) Boy should take the shortest path, ie direct from A to C.

N

W E

S

60°Shorte

st path

A

C

BHouse 10cm east

10m

(b) (i) Caring, by holding and supporting his grandfather from his shoulders.

(ii) Call nearby people for help. (iii) Gives first aid to his grandfather at home. (iv) Taking grandfather to Hospital.

Section E 24. Law of conservation of Linear momentum In an isolated system (i.e., a system with no external

force), mutual forces between pairs of particles in the system can cause changes in linear momentum of individual particles. But as the mutual forces for each pair are equal and opposite, the linear momentum changes cancel in pairs, and the total linear momentum remains unchanged. It explains the law of conservation of momentum that describes that when no external force acts on a system, then complete linear momentum of




system remains conserved. This principle is an important consequence of second and third laws of motion.

Let us consider an isolated system comprising of two bodies A and B, with initial linear momenta PA

��and PB

��. Let them collide for a small time and

separate with final linear momenta P A

��' and P B

��'

respectively. During collision, If FAB

��is force on A exerted by B, and FBA

��is force on

B exerted by A, then according to Newton's second law,

F × t = change in linear momentum of A = P'AB A A

� ��∆ − P (1)

F × t = change in linear momentum of B = PBA B B

� ��∆ ' − P (2)

According to Newton’s second law, F = FAB BA

� �−

\ From eqns. (1) and (2),

P P = P' P or P +P = P +PA A B B A B A B

�� ' ' '− − −( )

It shows that total final linear momentum of an isolated system is equal to its total initial linear momentum. Hence it proves the principle of law of conservation of linear momentum which holds for collision as elastic or inelastic.

Applications: (a) While firing the gun, the gun must be held

tightly to the shoulder. (b) When a man jumps out of the boat the boat is

pushed away, which pushes the man forward. The statement is correct. It is in accordance with

Newton’s second law of motion.OR

(a) Consider a vehicle of weight mg moving on circular level road of radius r with constant velocity v. While taking the round, the tyres of the car tend to leave the road and move away from the centre of curve. In such case, forces of friction f1 and f2 will act inward towards the two tyres. If R1 and R2 are the normal reactions of ground on the tyres, then

f1 = μs R1 and f2 = μs R2 where μs = coefficient of static friction Now total frictional force, f = μs R1 + μs R2 f = μs (R1 + R2) f = μs R where R = reaction of ground on the vehicle. As total frictional force f cannot exceed μs R, hence

f ≤ μs R This frictional force will provide the necessary

centripetal force f = mv2/r As R = mg (weight), so mv2/r ≤ μs R mv2/r ≤ μs mg v2 ≤ μs rg v ≤ √ μs rg Hence the maximum speed, vmax =

µ rgs

If the vehicle is driven at a speed greater than vmax, then it will skid and goes off the road in a circle of radius greater than r as the maximum available friction is inadequate to give required centripetal force.

(b) For rotational equilibrium Given: Distance between car wheels 2x = 1.5 m Height of Centre of gravity g=0.75 m Radius =20 m Apply the law of conservation of energy

m Vr

×h = m gx 2x =1.5m

V =g

2

( )

xxrh

=9.8× 0.75 ×20

0.75=14 m / s

25. (a) Bernoulli theorem. For the streamline flow of an ideal liquid, the total energy (Pressure energy, P. E, K.E) per unit mass remains constant at every cross section throughout the liquid flow

B

P2

P1

h1

V2a2

A

V1a1

P1 - Pressure on liquid at A P2- Pressure at the end B a1, a2- area of cross section at A and B h1, h2 – height above the ground v1, v2 = normal velocity of the liquid flow at section

A and B respectively, r=density of the ideal liquid flowing through the

tube. As the liquid flows from A to B, then, P1 > P2 The mass m of the liquid crossing per second

through any section of tube in accordance with the equation of continuity is

a v = a v = m or a v = a v = m / = V say

As, a > a 1 1 2 2 1 1 2 2

1 2

ρ ρ ρ ( ) v > v2 1∴

Now force on the liquid at section A = P1 a1 and force on the liquid at section B = P2 a2

Work done/second on the liquid by the pressure energy at section A = P1 a1 × v1 = P1V

Work done/second on the liquid against the pressure energy at section B = P2 a2 × v2 = P2V

Net work done/second on the liquid by the pressure energy in moving the liquid from section A and B

P1V – P2V When the mass m of the liquid flows in one second

from A to B, its height increases from h1 to h2 and its velocity increases from v1 to v2.

\ Increases in potential energy/sec of the liquid from A to B = mgh2 – mgh1

Increases in kinetic energy/sec. of the liquid from

A to B =

12

mv12

mv22

12−

According to work energy principle Work done/sec on the liquid by the pressure energy

= increases in P.E./sec + increases in K.E./sec




or P V P V = mgh mgh +12

mv12

mv1 2 2 1 22

12− − −( )

or P V+mgh +12

mv = P V+mgh +12

mv1 1 12

2 2 22

Dividing throughout by m, we get

or

P Vm

+gh +12

v =P Vm

+gh +12

v

P+gh +

12

v =P

+gh +12

v

11 1

2 22 2

2

11 1

2 22 2

2

ρ ρρ∵ ==

mV

Hence,

P+gh+

12

v = constant2

r

Limitations – (1) Viscous drag is not taken into account.

(2) It is assumed that there is no energy loss.OR

(a) Let θ =Angle of contact h =Height to which the liquid rises r =Radius (internal)of the tube p =Density of the liquid T=Surface tension of the liquid.

Weight of liquid

Liquid

R=T R=T

T sin φ T sin φ

T cos φ T cos φ

φ φ

φ φT T

h

A

The surface tension forces cause the liquid to exert a downward directed force T on the walls of the tube. This force T acts along the tangent at the point of contact ‘A’. In accordance with Newton’s third law, the tube exerts an equal and opposite reaction R(= T). The reaction R(= T) can be resolved into two rectangular components viz, T sin θ and T cos θ . The horizontal component T sin θ plays no part and the vertical component T cos θ pulls the liquid upward. The component T cos θ acts along the whole circumference of the meniscus.

\ Total upward force = T cos ×2 rθ π Volume of liquid in the tube above the free surface

of liquid is V = Volume of cylinder of height h and radius

r + Volume of the cylinder of height r and radius r – Volume of hemisphere of radius r

= r h+ r ×r12

43

r

= r h+ r23

r = r h+r3

2 2 3

2 3 3 2

π π π

π π π π

−

−

Weight of liquid column = V g = r h+r3

g2ρ π ρ

The upward force = T cos ×2 rθ π( ) supports the weight of liquid column above the free surface of the liquid.

T r r h

rgcosq p p r× = +

23

2

h

r Trg

+ =3

2 cosqρ

h =

2T cos r g

r3

h =2Tcos

r g;

qr

qr

− or

r is very small.

(b) Height of rise water in the capillary is given as

h =

2Scosr g

qr

Here, r =

D2

, q = 0°, cos 0° = 1

So, h =

2 1

2

SD g

×

× ×r

h =

4SDgr

Now, weight of liquid in the capillary tube w = mass × g = (volume × density) × g = (pr2 × h) × r × g

=

p

rr

D SDg

g2

44

× × ×

w = pDS This is the required result. 26. A particle executing SHM possess two types of

energy: (i) Potential energy (ii) Kinetic energy Potential energy particle of mass m, angular frequency w time t and displacement Y

Y = a sin wt

Y = a sin t

VelocityV =dYdt

a cos t

acceleration a =dvdt

= a s2

q

w w

- w iin t = Y

Restoring force = m×a = m Y = kYWork done dw = Fdy = k

2

2

w -w

- w -- YYdy

Now, Integrating above

w = kY dY =12

kY =12

m a sin t2 2 2\ w w≡ Kinetic energy

K =12

m a cos t

=12

m a 1 sin t

=12

m a Y =12

k

2 2 2

2 2 2

2 2 2

w w

w - w

w -

( )

( ) aa Y

Total energy =12

kY +12

k a Y

2 2

2 2 2

-

\ -

( )

( ) =

12

ka =12

m a2 2 2w




Energy

Total egergy

K E

O

P E

T4

T4

T4

T4

OR (a) Beats : The periodic variations in intensity at a

given point due to the superposition of two sound waves of slightly different frequencies travelling in the same directions are called beats.

The necessary condition is that the human ear can hear about 10 beats per second due to the

persistence of hearing. If it is more than 10 we cannot hear beats.

(b)

f =V

=V

1.00 f =

V=

V1.01

f - f = 4V

1.00-

V1.01

= 4

1.01V -

11

22

1 2

λ λ

11.00V = 4×1.010.01V = 4.04

V =4.040.01

⇒

∴ == 404m / s.




KENDRIYA VIDYALAYA SANGATHAN(JAMMU REGION)

SESSION ENDING EXAM - 2017-18CLASS : XI

SUBJECT : PHYSICS(SOLVED PAPER)

TIME : 3 HOURS MAX. MARKS : 70

General Instructions (a) All questions are compulsory. (b) There are 26 questions in total. Questions 1 to 5 carry one mark each Questions 6 to 10 carry two marks each Questions 11 to 22 carry three marks each Question 23 Carry four marks Questions 24 to 26 carry five marks each. (c) There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three

marks and all three questions of five marks each. You have to attempt only one of the given choices in such questions. (d) Use of calculator is not permitted. (e) You may use the following physical constants wherever necessary. e = 1.6 × 10-19 C c = 3 × 108 m/s h = 6.6 × 10-34 Js μ0 = 4p× 10-7 Tm/A2

kB = 1.38 × 10–23 J/K NA = 6.023 × 1023/mole mn = 1.6 × 10-27 Kg

Section A 1. Name the forces having the longest and shortest range of operation. 2. What is the difference between mN and Nm? 3. Which physical quantities are expressed by the following? (i) Rate of change of angular momentum (ii) Moment of momentum 4. Which of the two forces - deforming or restoring is responsible for elastic behaviour of substance? 5. Which thermodynamic variable is defined by Zeroth law of thermodynamics?

Section B 6. If heat dissipated in a resistance can be determined from the relation: H = I2Rt joule , If the maximum error in the measurement of current, resistance and time are 2% ,1% , and 1%

respectively, What would be the maximum error in the dissipated heat? 7. The velocity of a moving particle is given by v=6+18t + 9t2 (x in meter, t in seconds) what is it’s acceleration at

t=2s 8. Explain why two bodies at different temperatures T1 & T2 if brought in thermal contact do not necessarily settle

to mean temperature (T1+T2)/2?OR

Define stress. A heavy wire is suspended from a roof and no weight is attached to its lower end. Is it under stress? 9. A vessel is filled with a mixture of two different gases. State with reason (i) will the mean K.E. per molecule of

both the gases be equal? (ii) will the root mean square velocities of the molecules be equal? 10. At what displacement (i) the P.E. and (ii) K.E. of a simple harmonic oscillator is maximum?

Section C 11. The Frequency (n) of transverse wave on a string may depend upon (i) length I of string (ii) tension T in the string

and (iii) mass per unit length m of the string. Derive the formula for frequency with the help of dimensions.




12. Plot the position-time graph for an object (a) moving with positive velocity (b) moving with negative velocity, and (c) at rest.

13. The position of an object moving along x-axis is given by x = a + bt2 where a= 8.5m, b= 2.5ms-2 and t is measured in seconds. What is its velocity at t=0s and t= 2.0s. What is the average velocity between t=2.0s and t=4.0s? Give your conclusion.

14. A block of mass 3 kg slides down an incline of angle 30° with acceleration g/4.

Complete the free body diagram and find the coefficient of kinetic friction. 15. The sign of work done by a force on a body is important to understand .State carefully if the following quantities

are positive or negative: (a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) Work done by friction on a body sliding down an inclined plane. (c) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity. 16. Using theorem of parallel axis and perpendicular axis, find moment of inertia of a disc about (i) its diameter (ii) a

tangent in its own plane. 17. A body weighs 63N on the surface of the Earth. What is the gravitational force on it due to the Earth at a height

equal to half the radius of Earth? 18. Define escape velocity. Derive an expression for escape velocity. 19. What is terminal velocity? Derive an expression for it.

OR State Pascal’s law of transmission of fluid pressure. Explain working of Hydraulic Lift with the help of a diagram? 20. Draw block diagram of refrigerator. Explain its coefficient of performance. 21. Derive an expression for the apparent frequency of sound as heard by a stationary observer in a still medium,

when the source is moving towards the observer with a uniform velocity. 22. Show that the average kinetic energy per molecule is directly proportional to the absolute temperature of the gas.

Hence, give kinetic interpretation of temperature.

Section D 23. According to Newton’s law of gravitation ,everybody in this universe attracts every other body with a force,

which is directly proportional to the product of their masses and is inversely proportional to the square of the

distance between their centres,i.e., Fαm mr1 2

2 or F G=

m mr1 2

2 where G is universal gravitational constant =6.67 ×

10-11Nm2kg-2. Read the above passage and answer the following questions: (i) What is the value of G on the surface of moon? (ii) How is the gravitational force between two bodies affected when distance between them is halved? (iii) What values of life do you learn from it.

Section E 24. (i) For an object projected with velocity u at an angle of θ from the horizontal direction, deduce expressions for

(a) Time of flight (b) Horizontal range (c) Maximum height reached by the object. (ii) A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flight in the two

cases, then prove that TTRg1 2

2=

OR (i) Derive an expression for the centripetal acceleration of a body moving with uniform speed v along a circular

path of radius r. (ii) A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the

stone makes 14 revolutions in 25 seconds, what is the magnitude and direction of acceleration of the stone? 25. (i) State and Prove Bernoulli’s theorem. (ii) In streamline flow, water entering a pipe having diameter of 2 cm and the speed of water is 1.0ms-1. Eventually,

the pipe tapers to a diameter of 1 cm. calculate the speed of water where diameter of pipe is 1 cm.




OR (i) What is the phenomenon of Capillarity? (ii) Derive an expression for the rise of the liquid in a capillary tube. What will happen if length of the capillary tube is smaller than the height to which the liquid rise? Explain briefly. 26. What is the effect of (a) Pressure (b) Temperature (c) density (d) humidity (e) wind velocity on the velocity of

sound in gases?OR

A Transverse harmonic wave on a string is describe by Y (x, t) = 3 Sin (36 t + 0.018 x + 4) Where ‘x’ and ‘y’ are in cm and ‘t’ in sec. The positive direction of x is from left to right. (a) What is speed of this wave? (b) What is its amplitude? (c) What is its frequency? (d) What is the initial phase at the origin? (e) What is its wave length?

SOLUTIONSSection A

1. Longest Range of operation = Gravitational Force Shortest Range of operation = Weak Nuclear Force

2. mN – Milli newton, smallest unit of force Nm – Newton meter, unit of work 3. (i) External Torque (ii) Angular momentum 4. Restoring force is responsible for the elastic

behaviour of a substance. 5. Temperature.

Section B 6. = 2H I RT

∆ ∆ ∆ ∆= + +2

H I R TH I R T

In terms of % error

∆ ∆ ∆ ∆ × = × + × + × 100 2 100 100 100

H I R TH I R T

= (2 (2) + 1 + 1)% = 6% 7. Given:-

26 18 9v t t= + +

Since acceleration = +, so 18 18dv

a tdt

Now acceleration at time t=2s will be:

( ) 2

2 sec

18 18 2 54 / st

dvm

dt =

= + =

8. In thermal contact, heat flow from higher temperature to lower temperature till the temperature becomes equal.

Expression:- Let us consider two bodies of masses m1 and m2

having their temp T1 and T2 respectively where (T1 > T2). When they are in thermal contact, then final Temp become T.

Heat given by first body = ( )1 1 1m s T T−

Heat gained by second body = ( )2 2 2m s T T−

Acc. to the Principle of calorimetry Heat given by hot object = Heat gained by cool

object

( ) ( )1 1 1 2 2 2m s T T m s T T− = −

or,

1 1 1 2 2 2

1 1 2 2

m s T m s TT

m s m s+

=+

From the above, expression, it is clear that it is not a mean temp

NOTE:- The final temperature become mean

temperature 1 2

2T T+

only when the specific heat

capacities of two bodies are same.OR

Stress is the restoring force which is applied per unit area to a material.

Stress = Restoring force

Area

Unit - 2/N m Yes the heavy wire is under stress as the weight of

the wire will act as a deforming force. 9. (i) As we know that

Average K.E. of one molecule =32 BK T

It does not contain mass of molecule, so different gases at same temperature have same average kinetic energy per molecule. Hence if we mix two or more gases at same temp. then, there will be no exchange of energy among their molecules.

(ii) Since, vrms is inversely proportional to the square root of mass i.e.

α

1rmsV

Mass

Hence, if the masses of a gases differs then, vrms will also different.




10. (i) P.E. of simple Harmonic Oscillator = 2 212

U m y= ω P.E. will be max. only when y = a. i.e.,

2 21

2U m a= ω

(ii) K.E. of simple Harmonic oscillator

KE =

12

m2 (a2 – y2)

K.E. is max. only when y = 0

i.e.

2 21

K2

m a= ω.

Section C 11. Let the frequency of vibration of string depends

upon – L, m and T n µ La (1) n µ mt (2) n µ Tc (3) Combining them

a t cL M T∝n

n=kLaMtTc (4) Putting Dimensions on both the sides

[ ]1 1 2t ca

T L ML MLT− − − =

0 0 1 2t c a t c cM L T M L T− + − + − =

Equating Dimensions of M, L,T we get,

0t c+ = 0a t c− + = 2 1c− = − On solving we get

1a = − 1

2t

−=

12

c = Putting value of a, b, c in eq (4)

k TL m

=n

12.

Po

siti

on

(a) +ve velocity

Time

Po

siti

on

Time0

+

Po

siti

on

Time

(b) moving with negative velocity

(c) at rest

13. Given:- 2x a bt= +

Instantenous velocity dx

vdt

=

( )2da bt

dt= +

2bt=

0

0t

dxv

dt =

= =

At t = 2 s

dxdt t s

=2

( )( )2 2.5 2 10 / sm= =

Now, At t =2 s x = a + 4b and At t = 4 s x = a + 16b

Average velocity 2 1

2 1

x xt t−

=−

( ) ( )16 44 2

a b a b+ − +=

−

126

2b

b= =

Average velocity = 6(2.5) = 15 m/s 14. R

mg

30˚

f

mg

cos 30

mg

sin30

Let the mass of block be m The forces on the Block is as shown in fig.

(a) mg downward by earth (gravity) (b) R Normal force by incline. (c) f up the plane , (friction) by incline. Taking components mg sinθ – f = ma

sin 30

4mg

mg f∴ ° − =

∵ag

=

4

or 4

mgf =

There is no acceleration perpendicular to inclined plane, so

cosR mg θ=

= ° =

3cos30

2mg mg

As Block is sliding on inclined plane, then coefficient

of friction,

12 33

42

s

f mgR

mg

= = =

µ




15. (a) work done is positive as both, force and displacement are in same direction

(b) work done is negative as direction of friction is opposite to direction of motion

(c) work done is positive, as Body is moving on a rough horizontal plane; frictional force oppose the motion of body. Therefore in order to maintain velocity uniform force must be applied to the body. Since, applied force act in the direction of motion of a body. Hence work is positive.

16. (a) Moment of inertia about any diameter:

A

Y

D

B

Y

C

90

Let AB and CD are two mutually ⊥ diameters in the plane of disc then,

Using Perpendicular Axis theorem

AB CD YYI I I ⊥+ =

Or 212D DI I MR+ =

[∵ M.I. of disc through centre = 212

MR ]

212

2DI MR=

214DI MR=

(b) M.I. of disc about a tangent in its plane

C

A B

E

I T

F

D

Let IT be the moment of inertia along its tangent EBC, then this tangent is || to diameter CD of the disc then,

Applying Parallel Axis theorem

IT = moment of inertia of disc about 2CD MR+

2 214TI MR MR= +

254TI MR=

17. Let g′ be the acceleration due to gravity at a height h above the earth surface

′ =

+

2

1Re

gg

h

Let m be the mass of body then, its weight at a

height h above the earth surface.

′ =

+

2

1Re

mgmg

h

But = eR2

h

2/ 21

mgmg

ReRe

=′ +

\

263 463 28

911+

2

= = × =

N

18. Escape Velocity:- If we go on increasing the initial

velocity, then at a certain velocity the body will go out of the gravitational field of earth and never return back to earth. This velocity is known as Escape velocity.

Expression for Escape velocity The energy required to pass through a body and to

reach to infinity is

eReGM m

(1)

Let Ve the escape velocity, then the K.E. of a body

= 212 emV (2)

Comp (1) and (2)

= 2

e

1R 2

ee

GM mmV

=e

2R

ee

GMV (3)

Also, = 2eR

eGMg

Using this relation in (3)

then e2 ReV g=




But eR og V= (orbital velocity)

= 2e oV V 19. Terminal Velocity:- The magnitude of the viscous force goes on

increasing with increase in the velocity of the ball in liquid and a stage is reached when the opposing viscous force become equal to the effective driving force of the ball. Hence the net force on the ball is zero and the ball attains the constant velocity. This constant velocity is called as Terminal velocity.

EXPRESSION

� � �R ( – )g

343

6 nrv�

Let us consider a small ball, whose radius is r and density is ρ , let density of liquid is σ and coefficient of viscosity η

Effective force acting downward

= ( )V gρ σ−

( )343

r gπ ρ σ= − (1)

Viscous force acting downward 6 v= πηγ (2) Since, ball attains constant velocity, then, there is no

acceleration. Hence net force acting on it is zero ∴ comp (1) and (2)

( )π ρ σ πη− =34

63

r g rv

( )229

−=

r gv

ρ ση

OR Pascal's Law:- Pressure in a fluid in equilibrium is

same everywhere if the effect of gravity is neglected. Working : If the pressure is charged in some part of

a liquid, then this charge is transmitted equally to all parts of a liquid.

Hydraulic lift It consists of a two cylinder’s of cross-section area

A1 and A2 ( )1 2A A<< Cylinder’s are filled with incomposite fluid.

F1

A1 A2

F2

LOAD

Suppose a force F1 is applied on the piston having a smaller area of cross – section A1 then pressure

exerted over the liquid 1

1

= =F

PA

This pressure is transmitted to piston of large cross- section area 2A .

∴ upward force 2 2= ×F P A

12 2

1

FF A

A

=

22 1

1

AF F

A

=

Since, 2 1>>> …A A or F2A1 = F1A2

Hence, heavy load placed on large piston is lifted up.

20. Refrigerator:-

W

Q1

= Q1 - Q2

Q2

T1

Hot Body

( T2 )Cold Body

Sink

Coefficient of performance:- Measure of performance of refrigerator is expressed by coefficient of performance. It is represented by K.

2

1 2

TK

T T=

− 21. When Source is moving towards the observer with

uniform velocity Let S be a source and O be the observer. Let n be the

true freq. of source and v be the velocity of sound, then n waves will be emitted from source in 1sec.




SOURCE STATIONARY

Source Observer

VS V - VS

Then

Wavelength

( ) v

nλ =

Now, Let source is moving with velocity vs towards

observer O then n waves emitting in 1sec will spread in a dis ( )sv v− only then, wavelength will be shortened

sv v

n−′λ =

Hence, freq. of sound appears to be changed. If this

freq. will be n′ then,

( )λ′ = =

′ − /s

v vn

v v n

s

vn n

v v

′ = −

Since,

1

s

vv v

>−

∴ n'< n i.e. Apparent freq. is greater than true freq. When,

sound source appears to observer shrill(of high pitch)

22. Let the mass of 1 mole of gas be M and the mean square speed of its molecule be 2v then

K.E. of 1 mole of gas = 212

MV

21 32

RTM

M

=

32

RT=

There are N molecules in 1 mole of gass

∴ Average K.E. of one molecule =

3

2RT

N

32

RT

N =

But B

Rk

N= =Bottzmann’s constant

Average K.E. of one molecule = 32 Bk T .

Hence, From above expression it is clear that average K.E. is

directly prop to absolute temp.

Section D 23. (i) The value of G at moon will be same as

G = 11 2 26.67 10 Nm kg−× -

(ii) Given:-

F =

Gm mr1 2

2

...(i)

Now,

'

2r

r =

Then, New force of attraction

1 2

2

m mF G

r′ =

′

( )= 1 2

2/2

m mG

r

= 1 224

Gm mr

4F F′ = ...using equation (i)

(iii) Every body in the universe attracts other body with a certain amount of force.

Section E 24. (a) Time of flight:-

v = vx

Let us consider, time taken by the body to reach the max. height be t, then vertical velocity of body at P be zero.

sinyu u u θ= =

∴ = −yv u gt v=0 and uy= u sinθ

θ= −0 sinu gt

sinu

tgθ

= (1)

Time of flight 1 2T t t= + (Here, t1 = t2 = t) =2t

2 sinuT

gθ

= ...using eqn. (1)




(b) Range of projectile

xR u T= ×

( ) 2 sin

cosu

ug

θθ

=

( )22 2sin cos2 sin cos uug g

θ θθ θ= =

2

sin 2u

Rg

θ=

For Max. Horizontal Range

sin 2 1θ = or θ =2 90�

45θ = �

2

max

uR

g=

(c) Height of projectile:- Let h be the vertical height attain by projectile and vertical velocity at P is zero, Apply 3rd kinematic equation of motion.

= −2 2 2v u gh

= = 0yv v and sinyu u u θ= = Putting above

( )θ= −2

0 sin 2u gh

2 2sin2

uh

gθ

=

If θ = °90

∴ 2

max 2u

hg

=

Hence body will go to max height. (ii) Let T1 be the time of flight for a angle of

projection θ Then, T2 be the time of flight for angle of projection of

angle ( )θ−90�

∴

1

2 sinuT

g=

q

(1)

and

( )2

2 sin 90uT

g°

=- q

(2)

Now,

( )1 2

2 sin 902 sin uuT T

g g ° θθ

= -

2 sin 2 cosu ug g

θ θ =

2

2

2

sin 22

2 sin 2

θ=

θ

=

ug

ug g

2

2

2

sin 22

2 sin 2

θ=

θ

=

ug

ug g

1 2

2RT T

g=

∵R

ug

=

2 2sin q

OR (i) Consider a body which is moving in a circular

motion of radius r at constant speed v which is always being accelerated where its acceleration is at right angles to the direction of motion having a magnitude of v2 / r. The direction of acceleration is obtained by considering the symmetry where acceleration will point out of plane of the circle and the body leaves the plane of circle. When the acceleration points in any direction other than left or right, in such case, it will speed up or slows down. For magnitude, let the distance travelled by the body over a small time increment be Δt.

v S

r

v

∆θ

From the above figure, arc length s can be obtained as:

distance travelled by body, distance = rate * time = v Δt

From description of radian, arc = radius × angle in radians = r Δθ

Now, s = vΔt or rΔθ or, rΔθ= vΔθ or, Δθ / Δt=v / r or, =v / r Hence the angular velocity of object is v / r From the figure shown, the tails of the two v vectors

are together with Δθ.

v

∆θ ∆v

Now

sin

Dq2

=

D vv21

.

∆v = 2

2vsin

Dq

Further

DDq

v

= 2v× sin

2×

1DqDq

DDq

v

=

v ×

×sin

D qD q21

2

d vdq

= limD q→ 0

v ×

×sin

DqDq21

2

dvdq

=

v ×

×→

lim cosD q

Dq0 2

1212

=v




dvdt

=

d vd

ddtq

q× = v

vr

×

=

vr

2

(ii) Given: Radius of horizontal circle, r=80cm or 0.80m Time taken by stone to complete 14 revolutions in

25 seconds Now angular speed of revolution of stone:

w =

qt

or θ=2pvt

= 2p vt

Putting the values in the expression, we get:

w=2

227

1425

× ×

w=88/25 rad/s Now we see that: magnitude of acceleration produced in the stone =

magnitude of centripetal acceleration So, rw2

= 0.80 x (88/25)2

= 9.90 m/s2

It is observed that direction of acceleration will be towards the centre of the circle along its radius.

25. (i) Bernoulli’s Theorem When an in-compressive and non-viscous liquid or

gas flows in a stream lined motion from one place to another then, total energy per unit vol. (pressure energy + K.E + P.E) remain constant. i.e.

212

P V gh+ ρ + ρ = constant

Proof:-

A1

A2

V1

V2

X

Y

Work – done on the liquid entering through end X = FORCE × DISTANCE

= P1A1v1 (∵ liquid which enters through the x covers a dis =v1)

Similarly, Work – done per sec by the liquid in leaving through

end y = P2A2v2 Net work - done = P1A1v1 – P2A2v2 But using Principle of continuity

1 1 2 2

mA v A v= =

ρ

Net work-done on liquid = 1 2

m mP P ρ ρ

=

( )1 2

mP P−

ρ (1)

Now,

K.E. of liquid entering at end x is 1sec = 21

12

mv

K.E. of liquid entering at end y in 1 sec = 22

12

mv

Increase in K.E. =

( )2 2

2 1

12

m v v−

Similarly, Decrease in P.E = ( )1 2mg h h− Now, Net increase in energy of liquid

=

( ) ( )2 2

2 1 1 2

12

m v v mg h h− − −

(2)

This increase in energy is due to Net work-done on the liquid

Hence comp (1) and (2)

( ) ( ) ( )2 2

1 2 2 1 1 2

12

mP P m v v mg h h− = − − −

ρ

( ) ( )2 2

1 2 2 1 1 2

12

P P v v g h h− = ρ − − ρ −

2 21 1 1 2 2 2

1 12 2

P v gh P v gh+ ρ + ρ = + ρ + ρ

212

P v gh+ ρ + ρ = constant

(ii) Given:-

2

1 2 2 10d a m−= = × , or, 2

1 1 10r −∴ = × m

2

2 1 1 10d a m−= = × , or

22

110

2r −= ×

m

Sp. of water entry = v1 = 1m / sec v2= ? Using principle of continuity A1v1 = A2v2

2 2

1 1 2r v r v=p p

( ) ( )

222 2

2

11 10 1 10

2v− − × = ×

v2 = 4m / sec

OR (i) The rise phenomena of rise and fall of liquid in a

capillary tube is called as capillarity. (ii) Rise of liquid in capillarity tube Let θ =Angle of contact h =Height to which the liquid rises r =Radius (internal)of the tube p =Density of the liquid T=Surface tension of the liquid.




Weight of liquid

Liquid

R=T R=T

T sin � T sin �

T cos � T cos �

� �

� �T T

h

A

The surface tension forces cause the liquid to exert a downward directed force T on the walls of the tube. This force T acts along the tangent at the point of contact ‘A’. In accordance with Newton’s third law, the tube exerts an equal and opposite reaction R(= T). The reaction R(= T) can be resolved into two rectangular components viz, T sin θ and T cos θ . The horizontal component T sin θ plays no part and the vertical component T cos θ pulls the liquid upward. The component T cos θ acts along the whole circumference of the meniscus.

∴ Total upward force = T cos ×2 rθ π Volume of liquid in the tube above the free surface

of liquid is V = Volume of cylinder of height h and radius

r + Volume of the cylinder of height r and radius r – Volume of hemisphere of radius r

= r h+ r ×r12

43

r

= r h+ r23

r = r h+r3

2 2 3

2 3 3 2

π π π

π π π π

−

−

Weight of liquid column = V g = r h+r3

g2ρ π ρ

The upward force = T cos ×2 rθ π( ) supports the weight of liquid column above the free surface of the liquid.

T r r h

rgcosq p p r× = +

23

2

h

r Trg

+ =3

2 cosqρ

h =

2T cos r g

r3

h =2Tcos

r g;

qr

qr

− or

r is very small.

⇒ if the length of tube is less than h then liquid rises up to the top of the tube and then spread until radius of curvature R increase to R′

Hence, liquid cannot emerges in the form of fountain from upper end.

26. Factor Affecting Speed of Sound in Gas (a) Effect of pressure:-

Speed of sound in gases =

Pv

γ=

ρ

At constant Temp. PV = const.

But

mP

V=

mV

P=\

mP ρ

\ = constant

i.e.

Pρ

= constant

i.e. Pressure has No effect on speed of sound in gas. (b) Effect of Temperature:- Speed of sound is directly proportional to square

root of its absolute temp. i.e.

v T∝

(c) Effect of density:- Speed of sound in gases is inversely proportional to the square root of its density

i.e.

1v

density∝

(d) Effect of Humidity:- As speed of sound is

inversely prop. to the square root of density so sound travel faster in Moist Air than in DRY Air

(e) Effect of wind:- As sound source is carried by air so, its velocity is affected by wind velocity.

When wind flow in direction of sound, resultant velocity = v+w

When wind flow in vopporite direction of sound, then, resultant and velocity = v – w

OR Given: Transverse harmonic wave equation is

given as

( ), 3sin 36 0.018

4Y x t t x = + +

π

Comparing with

( ) 0

2 2, siny x t A t x

Tπ π = + + θ λ

\ A = 3,

2 236, 0.018

Tπ π= =

λ and

0 4

πθ =

(a) Speed of wave = vTλ

=

/ 2/ 2T

λ π=

π

( )1 / 0.0181 / 36

= = 20m / s

(b) Amplitude A = 3cm

(c) Frequency 1 36 18

2 3.14T= = =

πn

= 5.73 s–1

(d) Initial phase of origin = 4π radians

(e) Wave length

20.018 =

pl

= 349 cm = 3.49 m


(agra region) session ending examination 2017-18 …...describe first law of thermodynamics. apply...

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