SOL-P7

UNIT P7

PROPERTIES OF MATTER

Classroom Discussion Exercise

1. A wire rope of length and mass m is under tensile stress due to its own weight.Consider element of length dx distant x from bottom.Tensile force acting on the element is the weight of the part below = x A g

Its extension = (dx) =

Extension of the whole rope

= =

If there is an additional weight M attached to the end of the wire, then the total force on the element dx is F = (xAg + mg)

dx =

=

=

For the suspension order as shown For Section A: Here M = MB + MC

A =

For Section B: Here M = MC

B =

For Section C: M = 0

C =

= =

2.

Consider an element of the ring (angle , radius R) of length R. Let the cross section area be ‘a’. Mass = R.. a.. ( = density)It is under a tensile force T as shown.

Centripetal force = 2T. = T

(for small , sin = )

But centripetal force = mass

= R. . a . = T

For v = vmax, T = Tmax = a ( = Tensile strength)

a = R. . a . v2max =

vmax =

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dx

x

dx

x

M

A , a, , Y

B 2, a, 2, 2Y

C 3, a, 3, 3Y

TT

R/2

/2

2========================================================================================================

But vmax = nmax 2 R and R =

nmax = = 100 rps.

3. Let ‘x’ be height of ice inside waterThen,22 x 1 + 22 1 0.8 = 23 0.9 4x + 4 0.8 = 7.2 x = 1 cmi.e; whole ice is just inside the liquid Volume of ice when it melts =

= 7.2

ccVolume of water causing level difference = Volume added after melting volume displaced when floating = 7.2 8 = 0.8 ccThis volume of liquid is spread inside a vessel of area 10 cm2

Level difference = = 0.08 cm

4.

Taking moments about point O,

mg. sin = B y sin , where B is the

buoyancy and y is the distance from O to B.

mg. = By (1)

Let us calculate B and y,

The length immersed is = z

B = . . . g, where = density of

rod and = density of liquid ––(2)

( = volume per unit length)

Buoyancy force acts at the middle point of the length immersed, upward

Its moment arm y = - = ( +z)

By (1) and (2) mg = .

g.

z2 = 2 = constant

z =

At = 53, x = 0.4

liquid to be poured to reduce 0.32 m to 0.24 m = 0.08 m The liquid level now will be 0.48 m above the bottom of the vessel.

5.

Ice lifts off when buoyant force (due to water formed by melting of ice) exceeds weight of ice.Let be the density of ice, , that of water, a’ be side of ice cube when lifting off and h the height of water at that time. Then,

h =

=

=

= (1)

and a’2.hg a’3 g (2)

a’3 g



b

a'

h

x

mg

y

2

B

O

3========================================================================================================

a3 a’3 a’b2 – a’3 a’

When it just lifts off the surface, a’ =

6. 6rv = r3 ( )g

r = 5 103 m v = 3.5 102m s1

= 900 kg m3; = 200 kg m3

g = 10 m s2

= = 1.1 N m2 s = 1.1 Pa s

7.

By conservation of mass

Final level H =

Let levels at any instant be H1’ & H2’

A1dH1’ = A2dH2’ dH2’ = (1)

Similarly considering only modulus value, (h:level difference at any instant)

dh = |dH1’| + |dH2’| = dH1’ +

= dH1’ (2)Note: Here both dh & dH1’ are decreasing

quantitiesThe rate of flow from the vessel on the left should be same as rate of flow through the bottom.

A1 = av A1dH1’ = av.dt. From equation (2) substituting for dH1’

= avdt = a dt

T = =

=

=

8. Using Bernoulli’s equation relating the three heads between sections 1 & 2

= + 0.3 (1)

v2 = = 4v1 (2)

Equating pressures at the lower mercury level of manometer,

+ 0.25 = + 0.35 + 0.2 13.5 (3)

Using (2) in (1)

(1) 0.3 = (4)

From, (3) = 2.8

Substituting in (4)

v12 = =

v1 = m s1

Flow rate = cross sectional area velocity

= (0.15)2 = 0.032 m3 s1

9.



H2Area A2

Area A1

area a

dhH1

h

h’Final H

PQ

H1

d

T T

4========================================================================================================

If is length of needle, d its diameter, then upward force due to surface tension

= T 2To keep it afloat, 2T mg

m =

2T . g dmaximum =

10.

The surface of the wetting liquid between the plates is cylindrical in shape with radius of

curvature R = , where d = distance between

plates.

Hence additional pressure p = = acting

on surface area A

Force required = p A = A

= 8 (12.5) 104

= 25 N

11.

Let 2 be the original length and x be the central deflection, and T be the tension in the wire

Mg = 2Tsin T = ≃

=

x <<

=

= [ ≃ ]

= (strain = )

x3

12.

Before filling water, taking moments about ‘O’

mg cos = N OQ (i)

When water is filled such that half of the rod is inside water

Buoyant force F = A (ii)

(where A = Area of cross section of rod, = density of water)Taking moment about O,

mg cos = F

[Normal reaction = by data]

From (i) & (ii)

mg cos

+ mgcos



d

A C B

TD

T

x

Mg

O

4

mg

2N

F

O

2

Q

mg

N

5========================================================================================================

mg cos = cos

m =

A =

= 0.5

13. vh = (Toricelli’s theorem) = velocity of

efflux

t = (from Kinematics)

D = vht =

=

14.

Since the beaker is having an acceleration (gsin60°), the (gsin60°) component vanishes (consider pseudo force ma). Hence effective g’ is only gcos60°g’ = g cos 60° = 5 m s2

g’h = =

d = = = 0.6 mm

15. Let the separation be x. The force required to move the plate by ‘dx’ (See also solution of Q.10) is

F =

V = A.x where A is the area of spreadA = V/x

F =

W =

= =

16. For first case

1

2

For second case

1’

2’

17. Buoyant force, F = Volume of solid inside g

= V g

For body of mass m, V =

F =

= , remains same as long as there is

no change in the densities or in acceleration due to gravity

18. The one with larger surface area will take less time,

6rvT = mg FB vT

4r2



a = g sinmg

ma(pseudo)

mg cos

Y1, m1

Y2, m2

M Case 1

Y2, m2

Y1, m1

M Case 2

6========================================================================================================

higher surface area means higher terminal velocity less time taken

19. Time t =

A, a are areas of cross section of vessel and hole respectively.

20. Pressure in smaller bubble is

greater air flows to larger until smaller bubble reduces to a drop. (Pressure in the two will not become equal at any stage)

21. p + = p0 + + gh

v =

v2 = = (From equation of continuity)

p = p0 + gh

= p0 + gh = p0 + h’g

Rise in vertical tube h’ = h

22. Inside the tube, velocity is zero

p = p0 + ρv2

= p0 +

23. p’ = p + =

= p0 + gh

24. = Δp =

v2 = = 400 m2 s2

v = 20 m s1

25. ρwatergh =

h =

= = 0.01 m

= 1 cm

26.

For bubble to form, we have:

p2 p1 =

p1 = p0 + pwgh = p0 + (from Q.25)

p2 = p0 + pair v2

p0 + pair v2 p0

pair v2 =

From Q.25 v = 20 m s1

27. For non-stretchable wire KE = mgh = 100 10 1 = 1000

JKE of stretchable wire = 0.9 1000 = 900 JAt the vertical position:

T = mg +

v2 =

T = 100 10 +

Elastic energy =



h

p0

p1’

p2mesh r

water

v(air)

7========================================================================================================

=

=

28. A1v1 = A2v2

v2 > v1

p1 +

(mercury column is depressed on LHS)

p1 p2 =

= k(v22 v1

2) where k =

h = k(v22 v1

2) = k’(v12 v2

2) where k’ = k

=

= k2v22

(b) is

correcth = k(v2

2 v12) = k’(v1

2 v22), where k’ = k

=

= k1v12

Here k1 =

Correct options (b), (c) and (d)

29. The situation as shown in Fig.1 in the Qn. is impossible as the surface tension forces are equal and opposite, having the weight of liquid unbalanced

In Fig.1, the liquid is entirely within the capillary, so that the angle of contact = 60°. This means that the net surface tension force is zero, so it is impossible as it cannot balance the gravity force.At each surface, the upward force is F1 = F2 = 2aT cos . For maximum height, a2hg = 2aT cos 60° 2

h =

Obviously the meniscus will be as shown but if the column length is less than the maximum h, the curvature will be less, > 60°Correct options (b) and (c)

30.(a) The system moves up with acceleration a. The

effective value of g is now g’ = g + a. The pressure distribution is thus the same as above.But all differences are related by a factor

pA = pB (same level)pC = pD (same level)



T

TT F2

F1

T

8========================================================================================================

pA pC = pB pD

(a) p, r

(b)

The liquid surface is as shown, so lines of constant pressure are now inclined to the horizontal. Now pA = pC for a certain value of acceleration and pA = pE for another (larger acceleration). pA = pC (when free surface of liquid is parallel to AC)or pA = pE (when free surface of liquid is parallel to AE)(b) q, s

(c)

In (c) and (d) the surface is parabolic, but the distribution is symmetric about the axis. Since the points are symmetric about the axis pA = pB

For large enough value of , the above will hold pA = pE

Hence for both (c) and (d), the correct options are p and s

Regular Homework Exercise

31. Y = F/A or / = F/AY = a constant, thus strain will be equal in the two wires.

32. = F/AY or 1/Y so s / a = Ya / Ys = Ya /(20/7)Ya = 7/ 20

33. = = 1 mm

L’ = = = 0.5 mm

34. From Y = F /A or = F /AY = F /r2Y /r2 /r is same for all four

(a)

35. = 52 V = 25

V

36. The maximum load depends on stress, which only depends on area of cross section.

37. Energy density = stress strain

= Y strain strain

= 21010

= 3600 J m3

38. W = F = 100 0.05 = 2.5 J

39. The Young’s modulus for most solids decreases due to thermal excitation of atom in the crystal lattice due to increase in temperature.

40. Energy / volume =

=

=

41. Energy/volume = m3



A

EB

surface

9========================================================================================================42. Let ‘V’ be the volume of wax

Volume of submerged wax = 9 V / 10Weight of floating body = weight of water displaced V g = 9.V/10 1000 g = 900 kg m3

43. Volume of water produced by ice upon melting is the same as that of the water displaced by it. Thus the level of water is unchanged.

44. Pascal’s law states that pressure is same at all points in a fluid at rest, when gravity is neglected.

45. When ice melts, the water produced has lower density than seawater, so it will increase the level of water.M mass of iceV1 volume displaced by iceV volume of water when ice meltsV1seawater = M = Vwater

seawater > water

V1 < V

46. PA = PB

When water rise by 25 cm on LHS it should sink by 25 cm on RHS: as shown in figure Column of water above A (h(water)) = 50 cm A and B are same level in water, hence at same pressure.hoil 0.8 g = 50 1 g

hoil = = 62.5 cm

level difference, x = 62.5 50 = 12.5 cm

47. Vg = V/2 13.6 g + V/2 0.8 g

= V 7.2 g = 7.2 g cm-3

48. Air has such low density that it can be considered massless, Then the level will remain unchanged as the volume of water added will just balance the volume it displaced before melting.

49. Let V be the volume of iceberg and V’ its volume under liquidV 900 g = V ’ 1200 gV’ = 900V/1200 = 3V/4 = 75% V

50. (50 + 6.3)g 16.3 g = V 1 gV = 40 ccVolume of wax, V = 6.3/0.9 = 7 ccVolume of piece of material= 40 – 7 = 33 ccDensity of material = 50/33 = 1.5

51. p2 p1 = hgh = (p2 p1 ) /g = 0.01 13600 10/1.3 10 = 104.6 m

52. L = 12 cmBuoyancy force is equal to weight of the tube in both cases.Hw . 1 g = Hk .0.8 gHw = 12 4 = 8 cm

Hk = = 10 cm

Length of tube outside liquid = 12 10 = 2 cm

53. Refer solution of Question 46h = 2 13.6 = 27.2 cm

54. V1 = (V –V1)2.5[V volume of stopperV1 volume of inside hollow]

1 =

55. Viscosity is due to inter molecular forces.

56. Viscosity is the property of substances that can flow.

57. F = 6vr F r, v < vt



x

hwaterhoil

A B

25 cmoriginal level

hoil

10========================================================================================================58. Neglecting buoyancy force:

mg = 6r1v1 = 6 v2

v2 = 2v1

59. The viscous force exists between layers, and tries to prevent relative motion.

60. mg = 6rv1 ––(i)2 mg = 6(21/3r) v2 ––(ii)

(ii) (i) 2 = 21/3

v2 = 22/3 v1 = 5 22/3 cm s1

61. The initial velocity is zero, and gravity accelerates it. As speed increases, viscous force increases and when they balance terminal velocity is reached. The curve is actually exponential.

62. Initially only buoyancy force acts which is proportional to volume r3

63. Area = a2

v =

a2v = constant

a2 = constant h a4

n = 464. If h is the height of a hole, velocity of water

v =

t =

R = vt = 2

= 2

= 2H

Let = x

x(1-x) =

x =

Two holes with and

x1 x2 = 2

h1 h2 = H

Difference = H

65. R = 2 (See previous problem)

1.2 = 2

h(2 - h) = 0.62 = 0.36h2 2h + 1 = 1 0.36 = 0.64h = 1 0.8 = 0.2 m or 1.8 mF = ( A v).v = Av2

= A 2gh = 103 5 10-3 2 10 (0.2 or 1.8) = 20 N or 180 N

66. pA = p0

pB = pC(no losses)pD = p0(open to atmosphere)

67. Equation of continuity A1 v1 = A2 v2



11========================================================================================================

68. The pressure is lower over the roof due to higher velocity. Thus force is upward.

69. Volume = Velocity x Area of cross section = v.2rt.

70. The cohesive force in the liquid tends to give it a convex surface, while the solid-liquid force opposes it.

71. Oil rises due to capillary effect (Surface tension)

72. h cos h > 0 if < /2

73. Surface tension of liquids decrease with increase in temperature.

74.

75. Work done = 2 4R2. T = 8 (8 102)2 32 103 = 5.1 103 J

76. =

0.15 m

77. Volume of 8 small droplets each of radius r = volume of big drop of radius Ri.e., 8 4/3r3 = 4/3R3 or r = R/2Work done = surface tension increase in surface area

= T[8 4r2 4R2] = T[32(R/2)2 4R2] = 4R2T

78. n =

When droplets coalesce, the reduction in area

A = n4r2 4R2 =

=

E = TA = 32 103 4 104 [n1/3 1] = 4 103

= 4 103

= 100 n = 106

79. A component of surface tension should be upwards to balance the weight of the liquid inside the tube concave.

80. p. A = 2r. T

Aliter:

Excess pressure p =

But p = hg

hg =

h = = 30

cm

Assignment Exercise

81. All bodies have same magnitude of acceleration ‘a’.Let T1 and T2 be the tension in W1 and W2.For the body A, T1 = ma (1)for B, T2 T1 = ma (2)for C, mg T2 = ma (3)

(1) + (2) + (3) a = ;

where g = 10 m s2

T2 = T1 + ma = 2ma = 2a = N ( m = 1 kg)

S = Stress in BC =

= 20 106 N m2



L

12========================================================================================================

Strain in BC =

82. Tension in wire T =

= 13.07 N

= breaking stress = S r =

r = = 1.44 10-2 cm

83. Assume that strain is same for both steel and concrete.

YC =

F = (1)

= (2)

(1) + (2) F =

84. 1 = (1 ) = (1)

2 = (2 ) = (2)

(2) (1)

= 7 cm

Aliter

= = 7 cm

85.

L = (Refer soln. of Q.1)

= 0.2 m

86.

Volume of rod =

Upthrust on rod = Vrod water g = 2 .5 g = 2.5g N

Net torque at hinge = 0 (for equilibrium)mg (6 cos ) + 1.5g (3 cos ) = 2.5g (3 cos)

m =

87.

m = mass; A = area of cross section; Y = Young’s modulus; L = original length

(1)

When revolved force on the wire is T.

T cos = mg



mgmg

r

T

T cos = mg

rmvsinT

2

4 m

TB = 2.5g N

1.5g N

6 m

A

B

F = ‘mg’ N

13========================================================================================================

Extension

88.

167

107

89. Volume of cylinderV = 200 cm3 = 200 106 m3

= 0.8 g cm3 = 800 kg m3

upthrust = Vg = 200 106 800 10 = 1.6 NLet T1 and T2 be the tensions on the wire initially and after the cylinder is immersed. Obviously upthrust F = T1 T2; L1 =

Change in extension = L1 L2

=

= 1.27 104 cmAs it is an upthrust the length of the wire decreases.

90. Volume of cylinder =

Volume of liquid displaced = 62.5 cm3

h = height through which liquid rises in vessel

= =

= 0.3125 103 m Additional pressure at bottom of vessel = gh

= 800 kg m3 10 m s2 0.3125 103 m

= 2.5 Pa.

91. V1 = volume of the cone in top liquidV2 = volume of the cone in bottom liquidV Total volume of the coneV1 = k3;V = kh3; V2 = V V1 = k(h3 3)s(V1 + V2) = V1 + 2V2

sh3 = .3 + 2(h3 3) =

92.

The surface will be normal to g’ represented by vector diagram shown. p = hg + ax

= 0.8 103 10 + 103 3 2 = 14000 N m2

= 14 kN m2

93. Velocity of body on reaching the surface of water =

Force acting on the bodyF = mg Vg

= mg

= mg

= mg

Value of acceleration within water

= constant

[ = density of body, = density of medium] v2 = u2 + 2aS for motion inside water



P

A B

'gmg’ mg

ma(pseudo)

14========================================================================================================94. External radius = R

Internal radius = rDensity of material = Density of water = Weight of water displaced = weight of floating body

R3 = R3 r3 r3 = R3 ( )

= 10

R r = 10 9 = 1 cm

95.

Radius of the vessel = RP is a point on the concave surface of the liquid at a distance r from the centre where the elevation from the centre is s. The surface will be normal to the effective g’ as shown.Slope of the tangent at P

=

ds =

=

= = 7.1 103 m = 7.1

mmAliter:

p + = constant

At the centre, v = 0 and at the extreme circumference v = r

p = hg = (r)2 (where h is

the additional height)Substituting values h =

= 7.1 103 m

96. g’ = g + a = 13 m s2

p = g’h = 850 13 0.18 = 2.0 kPa

97. The bubble rising up with uniform velocity, net force = zero

=

=

= 6.67 104 Pa s

98. Volume of a drop =

m = mass of a drop =

Radius of the large drop R = n1/3. r = 10 0.02 = 0.2 mm

v =

=

5 m s1

99. +

(let density of water = )



OA = R = radius of cylinder

O

2rP

g g’2r

(centripetal)

surface

s

15========================================================================================================

100. Velocity of efflux v =

Time of fall t =

distance x =

x =

This is maximum when ; h <

(1.9H2h) + h (2) = 0

1.9 H = 4h h = < h =

101.

The column of liquid above the level PQ having length 2x has centre of gravity at x above P. When the pushing force is removed, CG of the

mass of the column (m = 2x) falls from x to

above P. Loss in PE = (2x)g = x2g.

This imparts KE for the total length of mass m = (1 + 2 + )PE = KE

x2g = (1 + 2 + )v2;where x =

v = (1 2)

102.v22 = v1

2 + 2gh1 = (4)2 + (2 10 1) = 36v2 = 6 m s1

a1v1 = a2v2 (continuity equation)

a2 =

103.p pressure inside the nozzle

p + (Take g = 10 m s2) p < p0

Hence the water rises on the vertical limb. For water to rise upto h:

p + w.gh = p0 v =

=

= 20 m s1

104.Q = Av =

103 m3 s1.

105.

v = Time to fall is t

= S + (H h) t =

Range R = vt =

= 2

Rmax when : = 0

= 0 h(1) + [S + (H h)] = 0

h = Given S = H; h = H

Hence range is maximum from hole No.5

106.



S

H

Stand

R

h v

a(area)

H

A(area)

1

2

x

xP Q

16========================================================================================================

H = height of water at any instantA = area of the cylindera = area of the hole

H is decreasing hence is ve

A

T =

=

T other factors being constant

When Hm becomes 2Hm, then T becomes

107.

Hole Iarea = adepth = h1

velocity of efflux v1

momentum imparted per second = (av1)v1

= av12 = a 2gh1 = 2agh1

Force of reaction = 2agh1

Similarly force of reaction on the opposite face = 2agh2

Resultant reaction = 2agh1 ~ 2agh2

= 2ag(h1 ~ h2) = 2agh = 2 2 106 1000 10

0.5 = 0.02 N

108. = 1.5 mm

= 1.5 103 mT = 0.075 N m1

=

pw = = 105 50

= 99950 Pa = 9.995 104 Pa

109.p =

= (r1 = 0.14

mm)p = gh

h =

h = 0.05 m of water = 5 cm of water.

110.

W = work done = increase in surface energy = (n 4r2 – 4R2) T

= (n 4 n2/3R2 4R2)T = 4R2(n1/3 1)T W = 4 (103)2 [(106)1/3 1] 0.07 = 87 J

111.T = v = velocity and

inclination of the string with vertically downward direction; at the bottom T becomes maximum since v is maximum and cos is maximum.

112.Not necessary. In unstable equilibrium too, line of buoyant force will be through CM.

113.Loss of internal energy = gain in surface energy.

114.The pressure exerted by the water at the top = 0The pressure exerted by the water at

point O = gh Average pressure =

Net force =

115.Pressure at point ‘P’ = g(hy)Force exerted on the slice of width ‘dy’ at height y from bottom:=g(hy)(dy)



h1

h2I

II

17========================================================================================================

Torque dy = g(hy)y(dy)

Total torque = g

= g

=

116.F.x =

x = =

117.When floating ice melts there is no contribution to the original level. So from case , the volume of embedded body is 10 cc, as the body should have sunk fully into water Volume of external body same in both cases V = 10 cc.

Case () Mass of embedded body : - when body floats the total weight of the embedded body is supported by Buoyant force, equal to the weight of the displaced water.

Ms = 50 1 = 50 g s =

In second case : Ms = 5 g

s =

118.Since from both the depths, the ball moves up the same height, it is coming up with terminal velocity: v0

2 = 2gh = 2 10 0.45 v0 = 3 m s1

r = 4.5 104 m = 0.45 mmOption (a) is not possible because if the initial velocity is the terminal velocity, there is no change in velocity (increase or decrease) Option (b) is not possible because viscous force is a dissipative force

119.

Consider the first case when L = La < h;Let p = atmospheric pressurepa = p hg + Lg

= p (h L) g pa < p So it is concave In second case, because of same argument, Since Lb > h, pb > p, so it is convex. The maximum column the surface tension can support is h0, below the surface level. Hence Lmax = h + h0. Hence when L > (h + h0) the meniscus breaks and water flows down (siphon)

120. v =

(Toricelli’s theorem) Since A is a point within the tube where water flows with velocity v.

pA +

But = gh3

pA = p0 + g(h1 h3)pB = pA gh1 = p0 gh3

pC = pB gh2 = p0 gh3 gh2 = p0 g(h2 + h3)



h0

h

p

pa

pb

pc

La

Lb

Lc

18========================================================================================================

Additional Practice Exercise

121.For horizontal condition, should be same for each rod. R = stress area = Y strain area

RA = YA . 2 104 N

Similarly RBr , RSt and RA + RBr + RSt = 68000 N

2 10-4 1010

= 68000 N

= = 4 104 m

Substituting, RA = 16,000 NRBr = 20,000 NRSt = 32,000 N

122.T = mg +

r = 4.1L = 4.22 (4 + 0.1 2) = 0.02 m

Strain = = 5 103

Stress = Y.strain = 1.324 1011 5 103

= 6.6 108 PaTension = Stress . Area = 6.6 108 106

= 660 N

v2 = =

=

v = 8.2 m s1

123.Let mass of block be m.Buoyant force = BThen spring balance A’ s reading7.5g = mg B (1)Spring balance B’s reading7.5g = 5g + B (2) B = 2.5g = 25 N

(i) (1) m = = 10 kg

(ii) If A is detached now, B will read mass of vessel + liquid + m = 5 + 10 = 15 kg.

(iii) specific gravity =

= = 4

124.Initial:

Consider level at D. Equating pressures on either side of tube.13 SA = 5 SB + SC

13 4 = 5 SB + 1 2 SB = 10

Final:

Total length of liquid B is 9 cm. After the re-distribution column of liquid B will be as shown (assuming length on LHS for B as x)Equating pressures at level E, 11SA = ( 9 – 2x) SB + 3SC

44 = (9 – 2x).10 + 6 9 2x = 3.8 x = 2.6 cm. (9 x) = 6.4 cm.

125.(i) Since density of air << that of oil, terminal

velocity v =



5 cm

2 cm

13cm

B

1 cmC

D

A

x cmB

92x 11cm

C

E

A

(9x) cm

3 cm

19========================================================================================================

=

= 105 m = 10 m

(ii) v’ = v = m s1.

= 0.625 mm s1

126.Lift = A (p1 p2) = A . (v22 v1

2)

= 400 . . (1.25) (1252 752 )

= 25 105 NNet force = Lift mg = 15 105 N

127.

Q1 = 0.6Q

Q2 = 0.4Q

= v2 > v1

Q2 = 0.4 Q 82v2 = 0.4 30 2 v

v2 > v. v2 = 4 m s1

v1 = v2 = 4 = m s1

Q = = 0.05 m3 s1

128.

Upper meniscus is always concave and pressure caused by the curvature is always directed

upwards, = p = where r is radius of the

capillary tube.

p1 = = 300 N m2

Hydrostatic pressure due to liquid column p2 = gh (downwards) = 200 N m2, 400 N m2, 300 N m2

For case i, ii, iii respectively.(i) p1 > p2 (ii) p1<p2 (iii) p1 = p2

hence lower meniscus is (i) concave (ii) convex (iii) flat and pressure caused by lower meniscus is directed(i) downward (ii) upward (iii) nil

and p3 = takes values

(i) p1 = 100 R = 1.5 mm (concave)(ii) p2 = 100 R = +1.5 mm (convex)(ii) p3 = 0 R = (flat)

129.p1 = p0 ; p2 = p0

hg = p1 p2 = 2T

= 2 (0.075). 104 = 750 N m2

h = = 7.5 cm

Aliter

h =

h1 h2 = = 7.5 cm

130.



d

v1,Q1

v,Q

v2,Q2

3 cm2 cm 4 cm

(i) (ii) (iii)

A’h

8.4 cm

x =1.6 cmO

B

10 cm

20========================================================================================================

T = 80 dyne cm1 = 0.08 N m1

Capillary rise ‘h’ =

=

= 0.016 m = 1.6 cm OB = (10 + h x) = (10 + 1.6 1.6) = 10 cmOA ‘ = 20 cm (data)

cos = = 0.5

= cos1 0.5 = 60

131.Due to self weight, L =

=

=

132.Strain energy stored half of loss of P.E

Ratio of massescannot be determined without ratio of AYou will get the same result by

stress strain volume, after integration.

133.L =

L1+L2 =

134.L =

L’ = L1 + L2 =

= ,

is minimum if A1 = A2 =

L’ = 4L

135.Slope =

Slope maximum for maximum

Hence maximum slope for wire a and slope reduces in the order a, b, c, d

[ values are: 4, 1.5, 0.67, 0.25]

136.Strain =

Stress = strain Y =

F =

At bottom position:

F =

v2 = 2g (1cos) = 2 10 10.05 (1cos) = 200(1cos) 1, cos 0, 90

137.Elastic energy = strain volume

= = 4 105

vol.Volume = =



21========================================================================================================

=

= 250 108 m3

Elastic PE = 4 105 250 108 = JThis much energy will be reduced in the final energyTotal energy with non-stretchable string =

= 10.5 10 10.05= 1055.25 J

E = = kv2

E/E =

138.Energy density = =

139.Ratio of energy per unit volume =

Where e is the strain

[Data from previous problem]

140.Work = .x = . Force

elongation

141. 8.1107,

= 8.1 107

L = = 3 km

142.

Stress = Y = 2 1011 4 103

143.

mg = 2T sin ------(1)Small sin

cos 1 =

From the figure:

tan = 0.08

=

=

T = YA

= 2 1011 (2 103)2



mg

25 cm25 cm

m

cm8.25225 22

P Q

R

T sin T sin T

2 cm

22========================================================================================================

= 8 103 NFrom (1):

m = 130 kg

Refer also to solution of Q.11

144. ,

Let mass M be compressed by V in volume;

’ =

’ = 0

= 1% (data)

B =

= 2 109 N m2

145.Let perimeter p

Then. ph. = Ahg

h = ; For h to be minimum

p should be maximum (for given A) Equilateral triangle

146.At all points on the base pressure is same = p0 + (h1 + h2) g

This is also equal to p0 + + h2g

147.

Let area of liquid surface = AVolume of stone = V V = (Ah1)

Then h1 =

h2 =

148.If , 1, 2 be the densities, f1, f2 be the respective fractions immersed, thenVg = V11g + V2 2gV = V1 + V2

=

(Clearly is more than one of them, less than the other)If it floats alone in say 1, immersed by xV of its total volume V, then

x =

(2 < < 1 2 < 1

f1 =

f2 =

x =

149.Clearly < (otherwise it will sink ).

Let total length be and length immersed ’

mg acts with moment arm sin about hinge.

Volume of immersed part = (mass of immersed

part) =

B = acts with moment arm sin

about hinge

mg . sin

x =



h0

h1 h2

B

mg

23========================================================================================================

150.Clearly < (otherwise it will sink).

Total length , length immersed ’.

mg acts with moment arm sin about hinge.

B = m acts with moment arm

sin about hinge.

mg sin

f2 2f + = 0

f = 1 , f = 1

151.Weight of the body W = g VWeight of liquid displacedW’ = ¢gVApparent weight = True weight – up thrust = g V ¢g V = g V (1 ¢/ ) = W (1¢/)

152.If x is the fraction of the volume V immersed then xVwater = Vice

x =

Fraction of ice exposed = 1 x

=

Height of the hole above water surface = 12 0.1

= 1.2 mThis is the length of the rope.

153

The surface will take a contour such that the resultant force on a particle on the surface will be normal to the surface. The typical particle on the surface will have forces as shown

tan =

Angle by which the surface tilts is given by

= tan1 = tan1 = 30

If x is the maximum level above the original surface

x = m

154.Initial contraction of spring is x0

Vg = kx0

Additional contraction of spring = x = 0.1 mResultant downward force = Resultant upward force.F + weight of body = up thrust by water + up thrust by springF + V g = Vg + k x + kx0

F = Vg + k x = 103 103 10 + 100 0.1 = 20 N

155.Up thrust reduces acceleration but viscosity makes acceleration = 0.

156.(1) is obvious, terminal velocity is reached(2) terminal velocity yet to be reached(3) is less obvious. This pertains to body of less density (buoyancy more than mg, hence retardation) with initial velocity, velocity reduces viscous force reduces retardation reduces

157.F = 6rv

158.a1v1 = a2v2

Pressure ratio is not relevant.

159.aA vA = aBvB. Since aA > aB, vA < vB.



B

mg

a

g

ma(pseudo)

mg

surface

30x

1

24========================================================================================================

pA + vA2 + gh = pB + vB

2

(pA pB) = (vB2 vA

2) gh.

RHS can be 0, hence cannot be

concluded.

160.Refer to solution of Q.106

t0 = t2 =

161. = 1 (1)

g’ = g , due to buoyancy.

Time to empty =

60 = = 0.6 (2)

From equation (1) & (2)

1

specific gravity =

162.p = hg hpA 3 0.88 = 2.64 and pB 2 1 = 2pA > pB

163.If x is the length of the liquid, x is also the compression of the spring.

p =

p =

v =

Q = a.v = A

a.

t A3/2 n = 1.5

164.

Two surfaces.

2TRd = 2Fsin = 2F.

F = 2TR

165.

R =

166.Angle of contact 90 Resultant of adhesive and cohesive forces is tangential to solid surface.

167.

p = T

V = A. 2r r = .

Forced required = p. A = .

168.



F Fd R

2T.Rd

R

r

r

4 cm

K

Open to atmosphere

A

liquid

x

25========================================================================================================

(Condition for normal capillary rise). T. 2r = hgr2

For a water column, held in a vertical capillary: (T. 2r) 2 = hg r2 h = 4 cm

169. ; original surface area A =

4r2

Final surface area of 27 drops = 27 = 3

A A = 3 A A = 2 A Surface energy increased = 2 A.s. = 2 4 (1 102)2 s = 8 s 104 J

170.

Since the conditions for capillary rise are identical in both tubes, the water columns should have equal heights unless is less than the expected rise ‘h’ , 1 < h; 1 = 28 mm, 2 = 21 = 56 mm

Water column 2 = 56 mm

T

N m1

171.F1 + F2 = F

Strain = same since is same.

172.Mechanical energy is not conserved. Half of Loss of gravitational potential energy is stored as strain energy. Other half goes as heat (i.e, thermal / internal energy)

173.

p0A = mgcos = Agcos p0 = cos g

= hg.m = A >hA

174.

175.No gravity no weight and no buoyancy force At terminal velocity viscous force should be zero 6rvT = 0 vT = 0 at terminal conditions.The sphere will finally be at rest, not moving with constant velocity.

176.

177.Surface tension is constant.Angle of contact differs for different materials.



mg

N1’N1

N1 = mg + N’1

mg

N2N’2

N2 = mg N’2

h

mgp0A

This should also be < 76 cm

1

2

26========================================================================================================

178.

If FCo = then resultants vertically

downward

179.Isothermal : pV n ( n : no. of mole.)

n

p0r3 + 4Tr2 nFor increasing r, n to be increased more amount of air required but pressure inside decreases.

180.Surface tension for a liquid is constant. When

adhesive force = (cohesive force), angle of

contact = 90° so that excess pressure = 0(Refer also Q.178)

181.Well known formula :

Applying the above.

1 = =

.

[Total elongation is 2 (for bottom half)

=

Which can also derived as

182

mass/unit length consider a small element dx at a distance x from the pivot end. Then the centripetal force this exerts at the end isdF = (.dx)2x. Hence total force at the pivot is

F = = =

183.Extension at the end of length x due to a small

element dx is :

Total extension

=

=

To achieve same extension

2 =

=

184.Apply law of floatation,Upward thrust = total wt.

1 = 3 0.1 +

0.7 = n = 700

185.’a water = 3 a wood +

’= (0.3 + 0.699) = 0.999

186.’a water = 3a wood +

’= (0.3 + 0.350) = 0.650



FCo

Fad

Resultant

Surface normal to resultant

= m2

A, m1

m

, A, no mass

=

, A, m

=

2m

A,2

,2m

dx

x

27========================================================================================================187.Lowest pressure will be at B.

TE at surface : p0 + .22 + 0

TE at B : pB + 122 + g.1

Equating , pB = p0 (122 22) g

= 105 70 103 104 = 2 104 Pa

188.Between surface and C,

h = 4.8 m total length = 1 + 1 + 2 + 1+ 4.8

= 9.8 m

189.Using av = constant,Comparing surface and A,0.1 2 = r2 12

r = = 0.07 m = 7 cm

190.

For wire 1: Y =

For wire 2:

T2 = 4T1 mg = 5T1

T1 =

For rotational equilibrium CM should be in between 1 and 2 and at distance ratio 4 : 1For rotational equilibrium, consider torque about B

5 T1.x = T1 d x = = 0.2 d

Consider torque about A

= T2 y + T1 (y + d)

mg = 5 T1; T2 = 4 T1; = 4d (data)5 T1 . 2d = 4 T1 y + T1 (y + d) y = 1.8 d

191.

Let the density of liquid A : A

Then the density of solid : 0.8 A

RD of A wrt solid = = 1.25

Let the density of liquid B : B, B < A

(If B > A, B will be below. A up, and A’s height is at least 0.8 side of cube. So equal fractions cannot be immersed in the two liquids). B < A

solid will sink further fraction exposed is now 0.1 equal fractions immersed are 0.45 each 0.8A = 0.45B + 0.45 A

B = A

RD of heavier liquid wrt lighter liquid.

=

192.VS = 10 cc; density of solid

=

It will not float in water10 cc of kerosene = 80 g 72 g = 8 g density of kerosene= 0.8 g/cc specific gravity of kerosene = 0.8Loss of weight in water = VSwater

= (10 cc) (1 g/ cc) = 10 g Weight in water = 70 g



0.8

0.10.450.45

B

Ay

B C D

mg = 5 T1

T2 = 4 T1 T1

d

(1)(2)

x d x

2d

4T1 T1

0.8d

2d5T1

0.2d

2 1

28========================================================================================================193.Viscous force F = 6rv

= slope = 6r

r1 > r2 (a)At terminal velocity,b = 6r1a1 = 6r2a2

r1a1 = r2a2

At terminal velocity,

v = r2( )

1 =

r1 > r2 2 > 1

(c) is incorrect

b = m1g = m2g

2 > 1 m1 > m2

194.Av = aV

v2 = (1) If A >> a, v = 0.

acceleration =

(1)

acceleration(magnitude) = ;

If A >> a, acceleration = 0

195.In streamlined motion, the velocity of a particle arriving at a given point is constant

196.If A & a represent areas at A & B, then in both

cases :vAA = vBa vB = vA

pipe 1:

pA +

vB2 vA

2 = (pA pB)

(vB vA) (vB + vA) = (pA pB)

v(vB + vA) = p

v = p (1)

pipe 2:p’A +

(v’B v’A)(v’B + v’A) = (p 2gh)

v (p 2gh) (2)

=

> 2

Also, p 2gh > 0 p > 2gh

197 r2 = a2r = anFor normal capillary rise of column h2r.T = r2hg

anT = ahg h =

But if rise is insufficient lengthH = ( d)

198.



Ela

stic

(pro

porti

onal

ity)

Yield point

UTS

Hooke’s law valid

Elongation

Load

Maximum stress

Plastic region

regi

on

Breaking stress

29========================================================================================================

199.

Force exerted by solid on liquid= Force exerted by liquid on solid= Buoyancy B = mg NWeight of displaced liquid = B

200.

tan =

> 45° : concave = 45° : flat < 45° : convex

i.e, FA > , concave = , flat

< , convex

FA > 0 ; liquid wets solid



Buoyancy

N

mg

Resultant 45°

FA

FC

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SOL-P7