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# HC Verma Class XII Physics Chapter 14 – Some Mechanical Properties of Matter

#### Exercise : Solution of Questions on page Number : 297

The ratio of stress to strain will decrease.

Beyond the elastic limit, the body loses its ability to restore completely when subjected to stress. Thus, there occurs more strain for a given stress. At some points, however, the body undergoes strain without the application of stress. So, the ratio of stress to strain decreases.

Let the CSA of the wire be A. The other 12 Mgl is converted into kinetic energy of the mass.

When the mass leaves its initial point on the spring, it acquires a velocity as it moves down. The velocity reaches its maximum at the end point. The spring oscillates. Finally, when the kinetic energy is dissipated into heat, the spring comes to rest.

The elephant has a greater weight than a mouse, but the material that makes their bones is the same. This means that in order to sustain an elephant’s weight, one’s bones need to suffer less stress. Stress = Force/area. A greater cross-sectional area reduces stress on the bones. This is why an elephant’s bones are thicker.

Let my length = L
Let the increase in length = l
Strain Yes, the yield point is much less than the 1% strain because the human body consists of joints and not one uniform solid structure.

The energy of vibration dissipates as heat from the shock absorber.

When a compressed spring dissolves in an acid, the acid molecules leave the sold lattice of the spring faster than the uncompressed spring. This in turn increases the kinetic energy of the solution. As a result, the temperature of the acid also increases. However, this temperature increase will be very small because the mechanical energy content in the spring is lesser than its chemical energy content.

It floats because of the surface tension of water. The surface of water behaves like a stretched membrane. When a blade is placed on the water surface, it’s unable to pierce the stretched membrane of water due to its low weight and remains floating.

However, if the blade is placed below the surface of water, it no longer experiences the surface tension and sinks to the bottom as the density of the blade is greater than that of water.

A liquid wets a surface when the angle of contact of the liquid with the surface is small or zero. Due to its fibrous nature, cloth produces capillary action when in contact with water. This makes clothes have very small contact angles with water. When wax is rubbed over cloth, the water does not wet the cloth because wax has a high contact angle with water.

No, the water will neither rise nor fall in the silver capillary.

According to Jurin’s law, the level of water inside a capillary tube is given by Thus, the water level neither rises nor falls.

No, we cannot conclude the surface tension to be zero solely by the fact that the liquid neither rises nor falls in a capillary.

The height of the liquid inside a capillary tube is given by h=2Tcosθrρg. From the equation, we see that the height (h) of the liquid may also be zero if the contact angle θ between the liquid and the capillary tube is 900 or 2700.

When water is poured in a glass, it reaches the brim and rises further. The edge of the glass lies below the water level. In this case, the force of attraction due to molecules of the glass surface is not perpendicular to the solid. Here, the contact angle can be greater than the standard contact angle for a pair of substances.

As the angle of contact is 0, there is no force between the surface of the tube and the liquid. The diameter of the liquid surface is pulled on both sides by equal and opposite forces of surface tension. This results in no net force remaining on the surface of the liquid. Hence, the liquid stays in equilibrium. No, it does not violate the principle of conservation of energy.

There is a force of attraction between glass and water, which is why the liquid rises in the tube. However, when water and glass are not in contact, there exists a potential energy in the system. When they are brought into contact, this potential energy is first converted into kinetic energy, which lets the liquid rush upwards in the tube, and then into gravitational potential energy. Therefore, energy is not created in the process.

A mosquito thrown into water has its wings wetted. Now, wet wing surfaces tend to stick together because of the surface tension of water. This does not let the mosquito fly.

The forces act tangentially to the bubble surface on both sides of a given line but they have one component normal to the bubble surface. This component balances the force due to excess pressure inside the bubble.

In the figure, let us consider a small length AB on the surface of the spherical bubble. Let the surface forces act tangentially along A and B. On producing the forces backwards, they meet at a point O. By the parallelogram law of forces, we see that the resultant force acts opposite to the normal. This balances the internal forces due to excess pressure. No. The average intermolecular distances do not increase with an increase in the surface area.

A soap bubble’s layer consists of several thousand layers of molecules. An increase in the surface area causes the surface energy to also increase. This in turn allows more and more molecules from the inner liquid layers of the bubble to attain potential energy, enabling them to enter the outer surface of the bubble. Hence, the surface area increases.

#### Exercise : Solution of Questions on page Number : 298

No. For a liquid at rest, no viscous forces exist.

Viscous forces oppose relative motion between the layers of a liquid. These layers do not exist in a liquid that is at rest. Therefore, it is obvious that viscous forces are non-existent in a static liquid.

The motion of any liquid is dependent upon the amount of stress acting on it. The motion of one layer of liquid is resisted by the other due to the property of viscosity. A river bed remains in a static state. Therefore, any immediate layer of liquid in contact with the river bed will also remain static due to the frictional force. However, the next layer of liquid above this static layer will have a greater velocity due to lesser resistance offered by the static layer. Moving upwards, subsequent layers provide lesser and lesser resistance to the movement of the layers above it. Finally, the topmost layer acquires the maximum velocity. Therefore, for a river, the surface waters flow the fastest.

Castor oil will come to rest more quickly because it has a greater coefficient of viscosity than water.

Castor oil has a higher viscosity than water. It will therefore, lose kinetic energy and come to rest faster than water.

Correct option: (d) 2000 N

F1 = 500 NLet the required breaking force on the 2 cm wire be F.Breaking stress in 1 cm wire = F1A1 = 500π0.0122. Breaking stress in 2cm wire = F2A2 = F2π0.0222.
The breaking stress is the same for a material.
⇒ 500π0.0122 = F2π0.0222
= > F2 = 2000N

Correct option: (a)

Breaking stress depends upon the intermolecular/ inter-atomic forces of attraction within materials. In other words, it depends upon the material of the wire.

Correct option: (b) 20 kg

As the wire is cut into two equal parts, both have equal cross-sectional areas. Therefore, a weight of 20 kg exerts a force of 20g on both the pieces. Breaking stress depends upon the material of the wire. Since 20g of force is exerted on wires with equal cross-sectional areas, both the wires can sustain a weight of 20 kg.

Correct option: (a) 1/8

Let the Young’s modulus of the wire’s material be Y.Here: Force = F A1 = πr2 L1 = l A2 = πr22 = πr24 L2 = 2l Let the elongation in A be x and that in B be y.Since the Young’s modulus for both the wires is the same :
Y = FA1xl = FA2y2l
⇒ xy = A22A1
⇒ xy = 18

Correct option: (b) 1.0 mm

Let the Young’s modulus of the material of the wire be Y. Force = Weight = W (given)Let C.S.A. = Ax = 1 mm = Elongation in the first caseLength = LY = WAxL = WLAx. Let y be the elongation on one side of the wire when put in a pulley.When put in a pulley, the length of the wire on each side = L2 WAyL2 = Y
⇒ WAyL2 = WLAx
⇒ y = x2
Total elongation in the wire = 2y = 2 x 2 = x = 1mm

Correct option: (a) s​mallest at the top and gradually increases down the rod

As the rod is of uniform mass distribution and stretched by its own weight, the topmost part of the rod experiences maximum stress due to the weight of the entire rod. This stress leads to lateral strain and the rod becomes thinner. Moving down along the length of the rod, the stress decreases because the lower parts bear lesser weight of the rod. With reduced stress, the lateral strain also reduces. Hence, the diameter of the rod gradually increases from top to bottom.

Correct option: (a) ∆ll

C.S.A. = ALength = l Volume of the wire V = Al Assuming no lateral strain when longitudinal strain occurs:Increase in volume :
∆V = A∆l
⇒ ∆V V= A∆lAl = ∆llSo,
∆V V is directly proportional to ∆ll.

Correct option: (c)

Let the Young’s modulus be Y.C.S.A. = AActual length of the wire = LFor tension T1:Y = T1AL-l1L…(1)
For tension T2:Y = T2AL-l2L…(2)
From (1) and (2) : T1AL – l1L = T2AL – l2L
⇒ T1L – l1 = T2L – l2
⇒ L = T2l1 – T1l2T2 – T1

Correct option: (b)

If the velocity of the mass is a maximum at the bottom, then the string experiences tension due to both the weight of the mass and the high centrifugal force. Both these factors weigh the mass downwards. The tension is therefore, maximum at the lowest point, causing the string to most likely break at the bottom.

Correct option: (d)

None of these is the correct option. The decreased gravitational potential energy transforms partly as elastic energy, partly as kinetic energy and also in the form of dissipated heat energy.

The correct option is (c).

The surface of a liquid refers to the layer of molecules that have higher potential energy than the bulk of the liquid. This layer is typically 10 to 15 times the diameter of the molecule. Now, the size of an average molecule is around 1 nm = 10 – 9m, so a diameter of 10 to 15 times would be of order ​10 × 10 – 9 = 10-8 m.

Correct option: (b)
As the ice cube melts completely, the water thus formed will have minimum surface area due to its surface tension. Any state of matter that has a minimum surface area to its volume takes the shape of a sphere. Therefore, as the ice melts, it will take the shape of a sphere.

Correct option: (a)

As the water droplets merge to form a single droplet, the surface area decreases. With this decrease in surface area, the surface energy of the resulting drop also decreases. Therefore, extra energy must be liberated from the drop in accordance with the conservation of energy.

Correct option: (c)
​Dimension of modulus of elasticity: FA∆ll = MLT – 2L2 = ML – 1T-2
Dimension of moment of force: ​FL = MLT – 2[L] = ML2T – 2
Dimension of surface tension: ​FL = MLT – 2L = MT – 2
Dimension of coefficient of viscosity : ​FLAv = MLT – 2LL2LT – 1 = ML – 1T – 1

Correct option: (d)

No. of surfaces of a soap bubble = 2Increase in surface area = 4π(2r)2 – 4π(r)2 = 12πr2
Total increase in surface area = 2 × 12πr2 = 24πr2
Work done = change in surface energy = S × 24πr2 = 24πr2S

Correct option: (a)

Excess pressure inside a bubble is given by: P = 4Tr.
When air is pushed into the bubble, it grows in size. Therefore, its radius increases. An increase in size causes the pressure inside the soap bubble to decrease as pressure is inversely proportional to the radius.

Correct option: (c)

The smaller bubble has a greater inner pressure than the bigger bubble. Air moves from a region of high pressure to a region of low pressure. Therefore, air moves from the smaller to the bigger bubble.

Correct option : (c) Here :
Radius of the tube = r Net upward force due to surface tension = Scosθ × 2πr
Upward pressure = Scosθ × 2πrπr2 = 2Scosθr
Net downward pressure due to atmosphere = Po
⇒ Net pressure at A = Po – 2ScosθrSince θ is small, cosθ ≈ 1.
⇒ Net pressure = Po – 2Sr

#### Exercise : Solution of Questions on page Number : 299

Correct option: (d)
Let the excess pressure inside the second bubble be P.
​∴ Excess pressure inside the first bubble = 2P
Let the radius of the second bubble be R.
Let the radius of the first bubble be x.

Excess pressure inside the 2nd soap bubble : P = 4SR…(1)
Excess pressure inside the 1st soap bubble : 2P = 4SxFrom (1),
we get : 24SR = Sx
⇒ x = R2
Volume of the first bubble = 43πx3
Volume of the second bubble = 43πR3
⇒ 43πx3 = n43πR3
⇒ x3 = nR3
⇒ R23 = nR3
⇒ n = 18 = 0.125

Correct option: (c)

The relationship between height h and radius r is given by : h = 2ScosθrρgIf S, θ, ρ and g are considered constant,
we have : h ∝ 1r

This equation has the characteristic of a rectangular hyperbola. Therefore, curve (c) is a rectangular hyperbola.

Correct option: (b) Given:l = 10 cmα = 450Rise in water level after the tube is tilted  = h
⇒ l = hcos450
⇒ h = lcos450 = 1012 = 102 cm

Correct option: (d)

Height of water column in capillary tube is given by : h = 2Tcosθ rρg
A free falling elevator experiences zero gravity.
⇒ h = 2Tcosθ rρ0 = ∞ But, h = 20 cm (given) Therefore, the height of the water column will remain at a maximum of 20 cm

Correct option: (d)

Viscosity is one property of fluids. Fluids include both liquids and gases

Correct option: (a)

The force of viscosity arises from molecular interaction between different layers of fluids that are in motion. Molecular forces are electromagnetic in nature. Therefore, viscosity must also be electromagnetic.

Correct option: (c)
The viscous force acts tangentially between two parallel layers of a liquid. In terms of force on a material, it is analogous to a shearing force.

Correct option: (c)
Air has viscosity. During rainfall, the raindrops acquire acceleration due to gravity. However, the increase in velocity is hindered by the viscous force acting upwards. A gradual balance between the two opposing forces causes the raindrops to attain a terminal velocity, thus, falling with a uniform velocity.

Correct option: (b)

The density of wood is less than that of water.When a piece of wood is immersed deep inside a long column of water and released, it experiences a buoyant force that gives it an upward acceleration. The velocity of wood increases as its motion is accelerated by the buoyant force. However, the viscous drag force acts simultaneously to oppose its upward motion. As a result, the initial acceleration decreases and the wood rises with a decreasing upward acceleration.

Correct option: (d)

In vacuum, no viscous force exists. The sphere therefore, will have constant acceleration because of gravity. An accelerated motion implies that it won’t have uniform velocity throughout its motion. In other words, there will be no terminal velocity.

Correct option: (b)
Initially, when the ball starts moving, its velocity is small. Gradually, the velocity of the ball increases due to acceleration caused by gravity. However, as the velocity increases, the viscous force acting on the ball also increases. This force tends to decelerate the ball. Therefore, after reaching a certain maximum velocity, the ball slows down.

Correct option: (a), (b), (c), (d)

All options are correct.
(a) When a weight is loaded on a wire, the length of the wire increases. The relationship between weight and length is linear.
(b) When a weight is loaded, it produces stress on the wire. The relationship between stress and increase in length is also linear.
(c) When stress is applied, strain develops. Therefore, both are linearly related.
(d) Since the value of Y for the wire is unknown, X may also be the increase in its length. Nevertheless, they still show the same linear relationship.

Correct option: (c) & (d)

(c) The surface molecules acquire air and liquid molecules in their sphere of influence.
(d) The surface molecules have different magnitudes of forces pulling them from the top and the bulk. So, they are affected by a net finite force in one direction.

Correct option: (a), (b), (d)

Height of the liquid in the capillary tube is given by : h = 2Scosθrρgh = HeightS = Surface tensionr = Inner radius of the tubeρ = Density of the liquidg = Acceleration due to gravitya) θ and ρ depend upon the material of the capillary tube and the liquid.b) h is dependent on the length of the tube. If the length is insufficient, then h will be low.d) r is the inner radius of the tube.

Correct option: (a), (b)

The angle of contact between a solid and a liquid depends upon the molecular forces of both the substances. Therefore, it depends upon the material of the solid and the liquid.

Correct option: (c) Let the height of the liquid-filled column be L.
Let the radius be denoted by R.

Total perimeter of the curved part=semi-circumference of upper area= πr Total surface tension force = πRS Total perimeter of the flat part = 2R Total surface tension force = 2RS Ratio of curved surface force to flat surface force = πRS, 2RS = π2

Correct option: (c), (d)

If the liquid level does not rise, it may be assumed that the surface tension is zero or the contact angle is 90°, or both. However, we cannot tell for sure whether the surface tension of the liquid is zero or the contact angle is 0°.

Correct option: (b), (c), (d)

(b) There is no gravitational force acting downwards. However, when the starting velocity is 20 m/s, the viscous force, which is directly proportional to velocity, becomes maximum and tends to accelerate the ball upwards.

When the ball falls under gravity,neglecting the density of air:Mass of the sphere = mRadius = rViscous drag coeff.= ηTerminal velocity is given by:mg = 6πηrvT ⇒ 6πηrvTm = g…(1) Now, at terminal velocity, the acceleration of the ball due to the viscous force is given by: a = 6πηrvTm Comparing equations (1) and (2), we find that:a = g

Thus, we see that the initial acceleration of the ball will be 9.8 ms-2.

(c) The velocity of the ball will decrease with time because of the upward viscous drag. As the force of viscosity is directly proportional to the velocity of the ball, the acceleration due to the viscous force will also decrease.

(d) When all the kinetic energy of the ball is radiated as heat due to the viscous force, the ball comes to rest.

#### Exercise : Solution of Questions on page Number : 300

Given:
Mass of the load (m) = 10 kg
Length of wire (L) = 3 m
Area of cross-section of the wire (A) = 4 mm2 = 4.0 × 10 − 6 m2
Young’s modulus of the metal Y = 2.0 × 1011 N m − 2

(a) Stress = F/A

F = mg
= 10 × 10 = 100 N (g = 10 m/s2)
∴FA = 1004 × 10-6 = 2.5 × 107 N/m2

(b) Strain = ΔLL

Or, Strain = StressY

Strain = 2.5 × 1072 × 1011 = 1.25×10 – 4 N/m2

(c) Let the elongation in the wire be ∆L.
Strain = ΔLL
⇒ ΔL = Strain × L = 1.25 × 10 – 4 × 3 = 3.75 × 10 – 4 m

Given:
Radius of cylinder (r) = 2 cm = 2×10-2 m
Length of cylinder (L) = 2 m
Mass of the load = 100 kg
Young’s modulus of the metal = 2 × 1011 N/m2

(a) Stress(ρ) is given by: FA
Here, F is the force given by mg = 100 × 10 = 1000 N ( Taking g = 10 m/s2)
A is the area of cross-section = πr2 = 4π × 10-4 m2

⇒ Stress ρ = mgA = 100 × 104π × 10-4 = 7.96×105 N/m2

(b) Strain is given by:

Strain = ρY = 7.96×1052 × 1011 = 4 × 10-6

(c) Compression of the cylinder:
ΔL = strain × L
= 4 × 10 − 6 × 2 = 8 × 10 − 6 m

Given:
Elastic limit of steel FA=8×105 N/m2Young’s modulus of steel Y = 2 × 1011 N/m2 Length of steel wire L = 12m = 0.5 m

The elastic limit of steel indicates the maximum pressure that steel can bear.
Let the maximum elongation of steel wire be ∆L.
Y = FA L∆L
⇒ ∆L = FLAY
⇒ ∆L = 8 × 105 × 0.52 × 1011 = 2 × 10-3 m = 2 mm

Hence, the required elongation of steel wire is 2 mm.

Given:
Young’s modulus of steel = 2 × 1011 N m−2
Young’s modulus of copper = 1.3 × 10 11 N m−2
Both wires are of equal length and equal cross-sectional area. Also, equal tension is applied on them.
As per the question:
Lsteel = LCuA steel = ACuFCu = F Steel

Here: Lsteel and LCu denote the lengths of steel and copper wires, respectively.
Asteel and ACu denote the cross-sectional areas of steel and copper wires, respectively.
Fsteel and FCu denote the tension of steel and cooper wires, respectively.

(a) Stress of CuStress of Steel = FCuACuA SteelF Steel = 1

(b) Strain of CuStrain of steel = ∆LSteel LSteel∆LcuLcu = FSteel LSteel Acu YcuASteel YSteel Fcu Lcu Using ∆L L = FAY
⇒ Strain of CuStrain of steel = YcuY Steel = 1.3 × 10112 × 1011
⇒ Strain of CuStrain of steel = 1320
⇒ Strain of steel Strain of Cu = 2013

Hence, the required ratio is 20 : 13.

Given that both wires are of equal length and equal cross-sectional area,
the block applies equal tension on both of them.

∴ Lsteel = LCuA steel = ACuFCu = FSteel
Strain of CuStrain of steel = ∆LSteelLSteel∆LcuLcu = FSteel LSteel Acu YcuASteel YSteel Fcu Lcu Using ∆LL = FAY
⇒ Strain of CuStrain of steel = YcuYSteel = 1.3 × 10112 × 1011
⇒ Strain of steel Strain of Cu = 2013 = 1.54

Hence, the required ratio of the longitudinal strains is 20 : 13.

(a) Given:
Breaking stress of wire = 8 × 108 N/m2 Area of cross-section of upper wire (Au) = 0.006 cm2 = 6 × 10 – 7 m Area of cross-section of lower wire (Al) = 0.003 cm2 = 3 × 10 – 7 mm1 = 10 kg, m2 = 20 kg

Tension in lower wire Tl = m1g + w
Here: g is the acceleration due to gravity
∴ Stress in lower wire = TlAl = m1g + wAl
⇒ m1g + wAl = 8 × 108
⇒ w = 8 × 108 × 3 × 10 – 7 – 100
⇒ w = 140 N or 14 kg

Now, tension in upper wire T2 = m1g + m2g + w

∴ Stress in upper wire = TuAu = m2g + m1g + wAu
⇒ m2 g + m1g + wAu = 8 × 108
⇒ w = 180 N or 18 kg

For the same breaking stress, the maximum load that can be put is 140 N or 14 kg. The lower wire will break first if the load is increased.

(b) If m1 = 10 kg and m2 = 36 kg:
Tension in lower wire Tl = m1g + w
Here : g is the acceleration due to gravity
∴ Stress in lower wire:

⇒ TlAl = m1g + wAl = 8 × 105
⇒ w = 140 N

Now, tension in upper wire T2 = m1g + m2g + w

∴ Stress in upper wire:

⇒ TuAu = m2g + m1g + wAu = 8 × 105
⇒ w = 20 N

For the same breaking stress, the maximum load that can be put is 20 N or 2 kg. The upper wire will break first if the load is increased.

Given:
Force (F) applied by two persons on the rope = 100 N
Original length of rope L = 2 m Extension in the rope ∆L = 0.01 m Area of cross-section of the rope A = 2 × 10 – 4 We know that:Young’s modulus Y = FA × L∆L = 1002 × 10 – 4 × 20.01
⇒ Y = 1 × 108 N/m2

Hence, the required Young’s modulus for the rope is 1 × 108 N/m2

Given:
Cross-sectional area of steel rod A = 4 cm2 = 4 × 10−4 m2
Length of steel rod L = 2 m
Compression during night hours ΔL = 0.1 cm = 10 −3 m
Young modulus of steel Y = 1.9 × 1011 N m−2

Let the tension developed at night be F.
Y = FA × L∆L
⇒ F = YA∆LL = 1.9 × 1011 × 4 × 10 – 4 × 10 – 32 = 3.8 × 104 N

∴ Required tension developed in steel rod during night hours = 3.8 × 104 N

Given:
Force (F) = m2g/2
Area of cross-section of the string = A
Young’s modulus = Y
Let a be the acceleration produced in block m2 in the downward direction and T be the tension in the string.
From the free body diagram:
m2g – T= m2a …iT-F = m1a …(ii) From equations (i) and (ii), we get:
a = m2g – Fm1 + m2 Applying F = m2g 2
⇒a = m2g2m1 + m2

Again, T = F + m1a

On applying the values of F and a, we get:

⇒ T = m2g2 + m1m2g2m1 + m2

We know that:

Y = FLA∆L ⇒ Strain = ∆LL = FAY
⇒ Strain = m22 + 2m1m2g2m1 + m2 AY = m2g2m1 + m22AY m1 + m2

∴ Required strain developed in the string = m2g2m1 + m22AY m1 + m2.

Given:
Mass of sphere (m) = 20 kg
Length of metal wire (L) = 4 m
Diameter of wire (d = 2r) = 1 mm
⇒ r = 5 × 10 − 4 m
Young’s modulus of the metal wire = 2.0 × 1011 N m−2
Tension in the wire in equilibrium = T
T= mg
When it is moved at an angle θ and released, let the tension at the lowest point be T’​.
⇒ T’= mg + mv2r
The change in tension is due to the centrifugal force.
∴​ ∆T = T’-T
∆T = mv2r …1

Now, using work energy principle:
12mv2 – 0 = mgr1 – cosθ
⇒ v2 = 2gr1 – cosθ …2

Applying the value of v2 in (1):

∆T = m 2gr1 – cosθr = 2mg1 – cosθ

Now, F = ∆TAlso, F = YA∆LL
⇒ YA∆LL = 2mg1 – cosθ
⇒ cosθ = 1-YA∆LL2mg
⇒ cosθ = 1 – 2 × 1011 × 4 × 3.14 × 52 × 10 – 8 × 2 × 10 – 34 × 2 × 20 × 10
⇒ cosθ = 0.80Or, θ = 36.4°

Hence, the required maximum value of θ is 35.4˚

#### Exercise : Solution of Questions on page Number : 301

Given:
Original length of steel wire (L) = 1 m
Area of cross-section (A) = 4.00 mm2 = 4 × 10 − 2 cm2
Young’s modulus of steel (Y) = 2 × 1011 N/m2
Acceleration due to gravity (g) = 10 m s − 2

Let T be the tension in the string after the load is suspended and θ be the angle made by the string with the vertical, as shown in the figure: cosθ = x x 2 + l2 = xl1 + x2l2 – 1/2

Expanding the above equation using the binomial theorem:
cosθ=xl1-12x2l2 neglecting the higher order termsSince x<<l, x2l2 can be neglected.
⇒ cosθ = xl
Increase in length:
ΔL = (AC + CB) − AB
AC = (l2 + x 2)1/2
ΔL = 2 (l2 + x 2)1/2 − 2l

We know that:
Y = FAL∆L ⇒ 2 × 1012 = T × 1004 × 10- 2 × 2502 + x21/2- 1 00

From the free body diagram:
2Tcosθ = mg2T x 50 = 2.16 × 103 × 980
⇒ 2 × 2 × 1012 × 4 × 10 – 2 × 2502 + x2 12 – 100 x 100 × 50 = 2.16 × 103 × 980

On solving the above equation, we get x = 1.5 cm.
Hence, the required vertical depression is 1.5 cm.

Given:
Cross-sectional area of copper wire A = 0.01 cm2 = 10 − 6 m2
Applied tension T = 20 N
Young modulus of copper Y = 1.1 × 1011 N/m2
Poisson ratio σ = 0.32

We know that:
Y = FLA∆L

⇒ ∆LL = FAY = 2010 – 6 × 1.1 × 1011 = 18.18 × 10 – 5

Poisson’s ratio,
σ = ∆dd∆LL = 0.32 Where d is the transverse length So, ∆dd = 0.32 × ∆LL
= 0.32 × 18.18 × 10 – 5 = 5.81 × 10 – 5 Again, ∆AA = 2∆rr = 2∆dd
⇒ ∆A = 2∆ddA
⇒ ∆A = 2× 5.8 × 10 – 5 × 0.01 = 1.164 × 10 – 6 cm2

Hence, the required decrease in the cross -sectional area is 1.164 × 10 – 6 cm2.

Given:
Bulk modulus of water (B) = 2.1 × 109 Nm-2
In order to decrease the volume (V) of a water sample by 0.01 %, let the increase in pressure be P.

V × 0.01100 = ∆V
⇒ ∆VV = 10 – 4 From B = PV∆V, we have:
⇒ P = B∆VV = 2.1 × 109 × 10 – 4 = 2.1 × 105 N/m2

Hence, the required increase in pressure is 2.1 × 105 Nm-2.

Given:
Bulk modulus of water B = 2 × 109 N/m2
Depth (d) = 400 m
Density of water at the surface (ρ0) = 1030 kg/m3
We know that:
Density at surface ρ0=mV0Density at depth ρd = mVd ⇒ ρdρ0 = V0Vd …i

Here: ρd = density of water at a depth
m = mass
V0 = volume at the surface
Vd = volume at a depth

Pressure at a depth d=ρ0gd Acceleration due to gravity g = 10 ms2 Volume strain = V0 – VdV0B = Pressure Volume strain ⇒ B = ρ0gdV0 – VdV0 ⇒ 1 – VdV0 = ρ0gdB ⇒ VdV0 = 1 – p0 gdB …ii
Using equations (i) and (ii), we get:

ρdρ0 = 11 – ρ0gdB ⇒ ρd = 11- ρ0ghBρ0
⇒ ρd = 1030 1 – 1030 × 10 × 4002 × 109 ≈1032 kg/m3 Change in density = ρd – ρ0 = 1032 – 1030 = 2 kg/m3

Hence, the required density at a depth of 400 m below the surface is 2 kg/m3.

Given:
Face area of steel plate A = 4 cm2 = 4 × 10 − 4 m2
Thickness of steel plate d = 0.5 cm = 0.5 × 10 − 2 m
Applied force on the upper surface F = 10 N
Rigidity modulus of steel = 8.4 × 1010 N m − 2
Let θ be the angular displacement.
Rigidity modulus m = FAθ
⇒ m = 104 × 10 – 4θ
⇒ θ = 104 × 10 – 4 × 8.4 × 1010 = 0.297 × 10 – 6

∴ Lateral displacement of the upper surface with respect to the lower surface = θ × d
⇒ (0.297) × 10−6 × (0.5) × 10−2
⇒ 1.5 × 10 − 9 m

Hence, the required lateral displacement of the steel plate is 1.5 × 10 − 9 m.

Given:
Length of thread l = 5 cm = 5 × 10 −2 m
Surface tension of water T = 0.76 N/m
We know that:
F = T × l = 0.76 × 5 × 10 − 2
= 3.8 × 10 − 3 N
Therefore, the water surface on one side of the thread pulls it with a force of 3.8 × 10 − 3 N.

Given:
Radius of mercury drop r = 2 mm = 2 × 10 – 3 m Radius of soap bubble r = 4 mm = 4 × 10 – 3 m Radius of air bubble r = 4 mm = 4 × 10 – 3 m Surface tension of mercury TH g = 0.465 N/m Surface tension of soap solution Ts=0.03 N/m Surface tension of water Ta = 0.076 N/m

(a) Excess pressure inside mercury drop:
P = 2THgr = 0.465 × 22 × 10 – 3 = 465 N/m2

(b) Excess pressure inside the soap bubble:
P = 4Ts r = 4 × 0.034 × 10 – 3 = 30 N/m2

(c) Excess pressure inside the air bubble:
P = 2Ta r =2 × 0.0764 × 10 – 3 = 38 N/m2

Given:
Surface area of mercury drop, A = 1 mm2 = 10 − 6 m2
Radius of mercury drop, r = 4 mm = 4 × 10 − 3 m
Atmospheric pressure, P0 = 1.0 × 105 Pa
Surface tension of mercury, T = 0.465 N/m

(a) Force exerted by air on the surface area:
F = P0A
⇒ F = 1.0 × 105 × 10 − 6 = 0.1 N

(b) Force exerted by mercury below the surface area:
Pressure P’= P0 + 2Tr F = P’A = P0 + 2TrA = 0.1 + 2 × 0.4654 × 10 – 3 × 10 – 6 = 0.1 + 0.00023 = 0.10023 N

(c) Force exerted by mercury surface in contact with it:
P = 2TrF = PA = 2TrA = 2 × 0.4654 × 10 – 3 × 10 – 6 = 0.00023 N

Given:
Surface tension of water T = 7.5 × 10 − 2 N/m
Taking cos θ = 1:
Radius of capillary A (rA) = 0.5 mm = 0.5 × 10 − 3 m

Height of water level in capillary A:
hA = 2T cos θrAρg = 2 × 7.5 × 10 – 20.5 × 10 – 3 × 1000 × 10 = 3 × 10 – 2 m = 3 cm

Radius of capillary B (rB) = 1 mm = 1 × 10−3 m

Height of water level in capillary B:
hB = 2Tcos θrBρg = 2 × 7.5 × 10 – 21 × 10 – 3 × 103 × 10 = 15 × 10 – 3 m = 1.5 cm

Radius of capillary C (rC) = 1.5 mm = 1.5 × 10 − 3 m
Height of water level in capillary C:
hC = 2T cos θrCρg = 2 × 7.5 × 10 – 21.5 × 10 – 3 × 103 × 10 = 151.5 × 10 – 3 m = 1 cm

Let T be the surface tension, r be the inner radius of the capillary tube and ρ be the density of the liquid.

For cos θ = 1, height (h) of the liquid level is given as:

h = 2Tcosθrρg

Now, for mercury:

hHg = 2THg rρHg g …(i)

For water:

hw = 2Twrρwg …(ii)

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

hwhHg = TwTHg × ρHgρw = 0.0750.465 × 13.6 = 2.19

Height of the water level:

hw = 2 × 2.19 = 4.38 cm

Hence, the required rise in the water level in the capillary tube is 4.38 cm.

Given:
Radius of tube r = 1.0 mm
Atmospheric pressure = 76 cm of Hg
Contact angle of mercury with glass θ = 135°
Surface tension of mercury T = 0.465 N/m
Density of mercury = 13600 kg m − 3

Let h be the rise in level in the barometer.

h = 2Tcosθrρg = 2 × 465 × 1/210 – 3 × 13600 × 10 = 0.0048 m = 0.48 cm

∴ Net rise in level in the barometer tube = H − h
= 76 − 0.48
= 75.52 cm

Given:
Radius of capillary tube r = 0.5 mm = 5 × 10 − 4 m
Depth (where pressure is to be found) h = 5.0 cm = 5 × 10 − 2 m
Surface tension of water T = 0.075 N/m
Excess pressure at 5 cm before the surface:
P = ρhg = 1000 × (5 × 10 − 2) × 9.8 = 490 N/m2
Excess pressure at the surface is given by:
P0 = 2Tr = 2 × 0.755 × 10 – 4 = 300 N/m2
Difference in pressure: P0 − P = 490 – 300 = 190 N/m2

Hence, the required difference in pressure is 190 N/m2

Given:
Radius of cylindrical vessel, r = 6.0 cm = 0.06 m
Surface tension of water, T = 0.075 J/m2
Area, A = πr2 = π × (0.06)2
Surface energy = T × A
= (0.075) × (3.14) × (0.06)2
= 8.5 × 10−4 J
Therefore, the surface energy of water kept in a cylindrical vessel is 8.5 × 10−4 J.

Given:
Initial radius of mercury drop R = 2 mm = 2 × 10 − 3 m
Surface tension of mercury T = 0.465 J/m2
Let the radius of a small drop of mercury be r.
As one big drop is split into 8 identical droplets:
volume of initial drop = 8 × (volume of a small drop)
43πR3 = 43πr3 × 8

Taking cube root on both sides of the above equation:

r = R2 = 10 – 3 m
Surface energy = T × surface area

∴ Increase in surface energy = TA’ − TA
= (8 × 4πr2 − 4πR2) T
= 4πT8 × R24 – R2 = 4πTR2
= 4 × (3.14) × (0.465) × (4 × 10 − 6)
= 23.36 × 10 − 6
= 23.4 μJ

Hence, the required increase in the surface energy of the mercury droplets is 23.4 μJ

Given:
Radius of capillary tube r = 1 mm = 10 − 3 m

(a) Let T be the surface tension and ρ be the density of the liquid.

Then, for cos θ = 1, height (h) of liquid level:

h = 2Trρg …(i),
where g is the acceleration due to gravity
⇒ h = 2 × 0.07610-3 × 10 × 100 = 1.52 cm = 1.52 × 10 – 2 m = 1.52 cm

(b) Let the new length of the tube be h’.

h’= 2Tcos θrρgcos θ = h’rρg2T Using equation i, we get : cos θ = h’h = 12 Because h’= h2 ⇒ θ = cos – 112 = 60°
The water surface in the capillary makes an angle of 60∘with the wall.

Given:
Radius of tube r = 1 mm = 10 − 3 m
Contact angle of mercury with glass θ = 135°
Surface tension of mercury T = 0.465 N/m
Let ρ be the density of mercury.

(a) Depression (h) of mercury level is expressed as follows:

h=2Tcosθrρg …(i)

⇒ h = 2 × 0.465 × cos 135°10 – 3 × 13600 × 9.8 = 0.0053 m = 5.3 mm

(b) If the length dipped inside is half the result obtained above:
New depression h’= h2
Let the new contact angle of mercury with glass be θ’.

∴ h’= 2Tcosθ’rρg …(ii)

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

h’h = cosθ’cosθ ⇒ cosθ’= cosθ2
⇒ θ = 112°

Given:
Surface tension of water T = 0.075 N/m
Separation between the glass plates d = 1 mm = 10 − 3 m
Density of water ρ = 103 kg/m3

Applying law of conservation of energy:

T (2L) = [1 × (10 − 3) × h] ρg

⇒ h = 2 × 0.07510 – 3 × 103 × 10 = 0.015 m = 1.5 cm

Therefore, the rise of water in the space between the plates is 1.5 cm.

Given:
Edge of the ice cube (a) = 1.0 cm
The water that is formed due to the melting of ice acquires a spherical surface.
In the absence of gravity, let the radius of the spherical surface be r.
Volume of ice cube = volume of spherical surface of water

⇒ a3 = 43πr3
⇒ r = 3a34π1/3
Surface area of spherical water surface = 4πr2
= 4π3a34π2/3 = 36π1/3 cm2

Given:
Surface tension of soap solution T = 0.030 N m −1

⇒ 2πr = 6.28 cm
⇒ r = 6.282 × 3.14 = 1 cm

The excess pressure inside the loop is expressed as follows:
∆P = 4Tr

T’=∆P×area of loop
⇒ T’= 4Tr × πr2
⇒ T’= 4πTr = 4 × 0.030 × 3.14 × 10 – 2 N = 3.8 × 10 – 3 N

Given:
Radius of metallic sphere r = 1 mm = 10 − 3 m
Speed of the sphere v = 1 0 − 2 m/s
Coefficient of viscosity η = 8 poise = 0.8 decapoise
Mass m = 50 mg = 50 × 10 − 3 kg
Density of glycerin σ = 1260 kg/m3

(a)
Viscous force exerted by glycerine on the sphere F = 6πηrv
⇒ F = 6 × (3.14) × (0.8) × 10 − 3 × (10 − 2)
= 1.50 × 10 4 N

(b)
Let V be the volume of the sphere.
Hydrostatic force exerted by glycerin on the sphere F’= Vσg
⇒ F’= 43πr2σg
= 4 3× 3.14 × 10 – 6 × 1260 × 1 = 5.275 × 10 – 5 N

(c)
Let the terminal velocity of the sphere be v’.
The forces acting on the drops are
(i) The weight mg acting downwards
(ii) The force of buoyance, i.e., 43πr3σg acting upwards
(iii) The force of viscosity, i.e., 6πηrv’ acting upwards
From the free body diagram: 6πηrv’ + 43πr3σg = mg
⇒ v = mg – 43πr2σg6πηr = 50×10 – 3 – 43 × 3.14 × 10 – 6 × 1260 × 106× 3.14 × 0.8 × 10-3 = 500 – 43 × 3.14 × 10 – 3 × 1260 × 106 × 3.14 × 0.8 = 2.3 cm/s

Given:
Radius of the drops r = 0.02 cm = 2 × 10 − 4 m
Viscosity of air η = 1.8 × 10 − 4 poise = 1.8 × 1 0 − 5 decapoise
Acceleration due to gravity g = 9.9 m/s2
Density of water ρ = 1000 kg/m3
Let v be the terminal velocity of a drop.
The forces acting on the drops are
(i) The weight mg acting downwards
(ii) The force of buoyance, i.e., 43 πr3 ρg acting upwards
(iii) The force of viscosity, i.e., 6πηrv acting upwards

Because the density of air is very small, the force of buoyance can be neglected.
From the free body diagram: 6πηrv = mg6πηrv = 43πr3ρg
v = 2r2ρg9η

= 2 × 0.02 × 10 – 22 × 1000 × 9.99 × 1.8 × 10 – 5 = 5 m/s.

Hence, the required vertical speed of the falling raindrops is 5 m/s.

#### Exercise : Solution of Questions on page Number : 302

Given:
Speed of water, v = 6 cm/s = 6 × 10 − 2 m/s
Radius of tube, r = 1 cm = 10 − 2 m
Diameter of tube, D = 2 × 10 − 2 m
Coefficient of viscosity, η = 0.01 poise

Let the Reynolds number be R and the density of water be ρ.

⇒ R = vpDη = 6 × 10 – 3 × 103 × 2 × 10 – 210 – 2 = 120

Here, the Reynolds number is less than 2000.Therefore, it is a steady flow.

error: