
Chapter 14 – Some Mechanical Properties of Matter
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Question 1:
The ratio stress/strain remain constant for small deformation of a metal wire. When the deformation is made larger, will this ratio increase or decrease?
Answer:
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.
Question 2:
When a block a mass M is suspended by a long wire of length L, the elastic potential potential energy stored in the wire is
12× stress × strain × volume. Show that it is equal to
12Mgl, where l is the extension. The loss in gravitational potential energy of the mass earth system is Mgl. Where does the remaining
12Mgl energy go?
Answer:
Let the CSA of the wire be A.
Stress = Force Area=MgAStrain = lLVolume = ALWe need to calculate the elastic potential energy stored in the wire which is given to be equal to 12×Stress×Strain×Volume.Elastic potential energy = 12×Stress×Strain×Volume=12×MgA× lL× AL=12MglThe other
12Mgl 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.
Question 3:
When the skeleton of an elephant and the skeleton of a mouse are prepared in the same size, the bones of the elephant are shown thicker than those of the mouse. Explain why the bones of an elephant are thicker than proportionate. The bones are expected to withstand the stress due to the weight of the animal.
Answer:
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 crosssectional area reduces stress on the bones. This is why an elephant’s bones are thicker.
Question 4:
The yield point of a typical solid is about 1%. Suppose you are lying horizontally and two persons are pulling your hands and two persons are pulling your legs along your own length. How much will be the increase in your length if the strain is 1%? Do you think your yield point is 1% or much less than that?
Answer:
Let my length = L
Let the increase in length = l
Strain
=lL=1100⇒l=L100So, the increase in length will be L100.Yes, the yield point is much less than the 1% strain because the human body consists of joints and not one uniform solid structure.
Question 5:
When rubber sheets are used in a shock absorber, what happens to the energy of vibration?
Answer:
The energy of vibration dissipates as heat from the shock absorber.
Question 6:
If a compressed spring is dissolved in acid, what happened to the elastic potential energy of the spring?
Answer:
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.
Question 7:
A steel blade placed gently on the surface of water floats on it. If the same blade is kept well inside the water, it sinks. Explain.
Answer:
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.
Question 8:
When some wax is rubbed on a cloth, it becomes waterproof. Explain.
Answer:
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.
Question 9:
The contact angle between pure water and pure silver is 90°. If a capillary tube made of silver is dipped at one end in pure water, will the water rise in the capillary?
Answer:
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 byh=2TcosθrρgHere, θ = 900⇒h =2Tcos900rρg ⇒h=0Thus, the water level neither rises nor falls.
Question 10:
It is said that a liquid rises or is depressed in capillary due to the surface tension. If a liquid neither rises nor depresses in a capillary, can we conclude that the surface tension of the liquid is zero?
Answer:
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.
Question 11:
The contact angle between water and glass is 0°. When water is poured in a glass to the maximum of its capacity, the water surface is convex upward. The angle of contact in such a situation is more than 90°. Explain.
Answer:
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.
Question 12:
A uniform vertical tube of circular cross section contains a liquid. The contact angle is 90°. Consider a diameter of the tube lying in the surface of the liquid. The surface to the right of this diameter pulls the surface on the left of it. What keeps the surface on the left in equilibrium?
Answer:
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.
Question 13:
When a glass capillary tube is dipped at one end in water, water rises in the tube. The gravitational potential energy is thus increased. Is it a violation of conservation of energy?
Answer:
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.
Question 14:
If a mosquito is dipped into water and released, it is not able to fly till it is dry again. Explain.
Answer:
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.
Question 15:
The force of surface tension acts tangentially to the surface whereas the force due to air pressure acts perpendicularly on the surface. How is then the force due to excess pressure inside a bubble balanced by the force due to the surface tension?
Answer:
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.
Question 16:
When the size of a soap bubble is increased by pushing more air in it, the surface area increases. Does it mean that the average separation between the surface molecules is increased?
Answer:
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.
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Question 17:
Frictional force between solids operates even when they do not move with respect to each other. Do we have viscous force acting between two layers even if there is no relative motion?
Answer:
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 nonexistent in a static liquid.
Question 18:
Water near the bed of a deep river is quiet while that near the surface flows. Give reasons.
Answer:
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.
Question 19:
If water in one flask and castor oil in other are violently shaken and kept on a table, which will come to rest earlier?
Answer:
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.
Question 1:
A rope 1 cm in diameter breaks if the tension in it exceeds 500 N. The maximum tension that may be given to a similar rope of diameter 2 cm is
(a) 500 N
(b) 250 N
(c) 1000 N
(d) 2000 N
Answer:
Correct option: (d) 2000 N
F1=500NLet the required breaking force on the 2 cm wire be F.Breaking stress in 1 cm wire = F1A1=500π0.0122Breaking stress in 2 cm wire = F2A2=F2π0.0222The breaking stress is the same for a material.⇒500π0.0122=F2π0.0222=>F2=2000N
Question 2:
The breaking stress of a wire depends on
(a) material of the wire
(b) length of the wire
(c) radius of the wire
(d) shape of he cross section.
Answer:
Correct option: (a)
Breaking stress depends upon the intermolecular/ interatomic forces of attraction within materials. In other words, it depends upon the material of the wire.
Question 3:
A wire can sustain the weight of 20 kg before breaking. If the wire is cut into two equal parts, each part can sustain a weight of
(a) 10 kg
(b) 20 kg
(c) 40 kg
(d) 80 kg
Answer:
Correct option: (b) 20 kg
As the wire is cut into two equal parts, both have equal crosssectional 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 crosssectional areas, both the wires can sustain a weight of 20 kg.
Question 4:
Two wires A and B are made of same material. The wire A has a length l and diameter r while the wire B has a length 2l and diameter r/2. If the two wires are stretched by the same force, the elongation in A divided by the elongation in B is
(a) 1/8
(b) 1/4
(c) 4
(d) 8
Answer:
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=2lLet 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
Question 5:
A wire elongates by 1.0 mm when a load W is hung from it. If this wire goes over a a pulley and two weights W each are hung at the two ends, he elongation of he wire will be
(a) 0.5 m
(b) 1.0 mm
(c) 2.0 mm
(d) 4.0 mm
Answer:
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=WLAxLet 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 = x2Total elongation in the wire = 2y =2×2= x = 1mm
Question 6:
A heave uniform rod is hanging vertically form a fixed support. It is stretched by its won weight. The diameter of the rod is
(a) smallest at the top and gradually increases down the rod
(b) largest at the top and gradually decreased down the rod
(c) uniform everywhere
(d) maximum in the middle.
Answer:
Correct option: (a) sâ€‹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.
Question 7:
When a metal wire is stretched by a load, the fractional change in its volume âˆ†V/V is proportional to
(a)
∆ll(b)
∆ll2(c)
∆l/l(d) none of these
Answer:
Correct option: (a) âˆ†ll
C.S.A. = ALength = lVolume of the wire V = AlAssuming 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.
Question 8:
The length of a metal wire is l1 when the tension in it T1 and is l2 when the tension is T2. The natural length of the wire is
(a)
l1+l22(b)
l1l2(c)
l1T2l2T1T2T1(d)
l1T2+l2T1T2+T1
Answer:
Correct option: (c)
)
Let the Young’s modulus be Y.C.S.A. = AActual length of the wire = LFor tension T1:Y=T1ALl1L…(1)For tension T2:Y=T2ALl2L…(2)From (1) and (2):T1ALl1L=T2ALl2L⇒T1Ll1=T2Ll2⇒L=T2l1T1l2T2T1
Question 9:
A heavy mass is attached to a thin wire and is whirled in a vertical circle. The wire is most likely to break
(a) when the mass is at the highest point
(b) when the mass is at the lowest point
(c) when the wire is horizontal
(d) at an angle of cos^{−1}(1/3) from the upward vertical.
Answer:
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.
Question 10:
When a metal wire elongates by hanging a load on it, the gravitational potential energy is decreased.
(a) This energy completely appears as the increased kinetic energy of the block.
(b) This energy completely appears as the increased elastic potential energy of the wire
(c) This energy completely appears as heat.
(d) None of these.
Answer:
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.
Question 11:
By a surface of a liquid we mean
(a) a geometrical plane like x = 0
(b) all molecules exposed to the atmosphere
(c) a layer of thickness of the order of 10^{−8}m
(d) a layer of thickness of the order of 10^{−4}m
Answer:
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 =
109m, so a diameter of 10 to 15 times would be of order â€‹
10×109=108m.
Question 12:
An ice cube is suspended in vacuum in a gravity free hall. As the ice melts it
(a) will retain its cubical shape
(b) will change its shape to spherical
(c) will fall down on the floor of the hall
(d) will fly up.
Answer:
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.
Question 13:
When water droplets merge to form a bigger drop
(a) energy is liberated
(b) energy is absorbed
(c) energy is neither liberated nor absorbed
(d) energy may either be liberated or absorbed depending on the nature of the liquid.
Answer:
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.
Question 14:
The dimension ML^{−}^{1}T^{−2} can correspond to
(a) moment of a force
(b) surface tension
(c) modulus of elasticity
(d) coefficient of viscosity
Answer:
Correct option: (c)
â€‹Dimension of modulus of elasticity:
FA∆ll=MLT2L2=ML1T2Dimension of moment of force: â€‹
FL=MLT2[L]=ML2T2Dimension of surface tension: â€‹
FL=MLT2L=MT2Dimension of coefficient of viscosity: â€‹
FLAv=MLT2LL2LT1=ML1T1
Question 15:
Air is pushed into a soap bubble of radius r to double its radius. If the surface tension of the soap solution in S, the work done in the process is
(a) 8 π r^{2} S
(b) 12 π r^{2} S
(c) 16 π r^{2} S
(d) 24 π r^{2} S
Answer:
Correct option: (d)
No. of surfaces of a soap bubble = 2Increase in surface area = 4π(2r)24π(r)2=12πr2Total increase in surface area = 2×12πr2=24πr2Work done = change in surface energy =S×24πr2 = 24πr2S
Question 16:
If more air is pushed in a soap bubble, the pressure in it
(a) decreases
(b) increases
(c) remains same
(d) becomes zero.
Answer:
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.
Question 17:
If two soap bubbles of different radii are connected by a tube,
(a) air flows from bigger bubble to the smaller bubble till the sizes become equal
(b) air flows from bigger bubble to the smaller bubble till the sizes are interchanged
(c) air flows from the smaller bubble to the bigger
(d) there is no flow of air.
Answer:
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.
Question 18:
Figure shows a capillary tube of radius r dipped into water. If the atmospheric pressure is P_{0}, the pressure at point A is
(a) P_{0}
(b)
P0+2Sr(c)
P02Sr(d)
P04SrFigure
Answer:
Correct option: (c)
Here:Radius of the tube = rNet upward force due to surface tension = Scosθ×2πrUpward pressure = Scosθ×2πrπr2=2ScosθrNet downward pressure due to atmosphere = Po⇒Net pressure at A = Po 2ScosθrSince θ is small, cosθ≈ 1.⇒Net pressure = Po 2Sr
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Question 19:
The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is
(a) 4
(b) 2
(c) 1
(d) 0.125
Answer:
Correct option: (d)
Let the excess pressure inside the second bubble be P.
â€‹∴ 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=4Sx⇒x = R2Volume of the first bubble = 43πx3Volume of the second bubble = 43πR3⇒43πx3=n43πR3⇒x3=nR3⇒R23=nR3⇒n=18=0.125
Question 20:
Which of the following graphs may represent the relation between the capillary rise h and the radius r of the capillary?
Figure
Answer:
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∝1rThis equation has the characteristic of a rectangular hyperbola. Therefore, curve (c) is a rectangular hyperbola.
Question 21:
Water rises in a vertical capillary tube up to a length of 10 cm. If the tube is inclined at 45°, the length of water risen in the tube will be
(a) 10 cm
(b)
102cm
(c)
10/2cm
(d) none of these
Answer:
Correct option: (b)
Given:l = 10 cmα=450Rise in water level after the tube is tilted = h⇒l = hcos450⇒h = lcos450=1012=102 cm
Question 22:
A 20 cm long capillary tube is dipped in water. The water rises up to 8 cm. If the entire arrangement is put in a freely falling elevator, the length of water column in the capillary tube will be
(a) 8 cm
(b) 6 cm
(c) 10 cm
(d) 20 cm
Answer:
Correct option: (d)
Height of water column in capillary tube is given by:h= 2Tcosθ rρgA 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.
Question 23:
Viscosity is a property of
(a) liquids only
(b) solids only
(c) solids and liquids only
(d) liquids and gases only.
Answer:
Correct option: (d)
Viscosity is one property of fluids. Fluids include both liquids and gases.
Question 24:
The force of viscosity is
(a) electromagnetic
(b) gravitational
(c) nuclear
(d) weak.
Answer:
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.
Question 25:
The viscous force acting between two layers of a liquid is given by
FA=ηdvdz. This F/A may be called
(a) pressure
(b) longitudinal stress
(c) tangential stress
(d) volume stress
Answer:
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.
Question 26:
A raindrop falls near the surface of the earth with almost uniform velocity because
(a) its weight is negligible
(b) the force of surface tension balances its weight
(c) the force of viscosity of air balance its weight
(d) the drops are charged and atmospheric electric field balances its weight.
Answer:
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.
Question 27:
A piece of wood is taken deep inside a long column of water and released. It will move up
(a) with a constant upward acceleration
(b) with a decreasing upward acceleration
(c) with a deceleration
(d) with a uniform velocity
Answer:
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.
Question 28:
A solid sphere falls with a terminal velocity of 20 m s^{−1} in air. If it is allowed to fall in vacuum,
(a) terminal velocity will be 20 m s^{−1}
(b) terminal velocity will be less than 20 m s^{−1}
(c) terminal velocity will be more than 20 m s^{−1}
(d) there will be no terminal velocity
Answer:
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.
Question 29:
A spherical ball is dropped in a long column of a viscous liquid. The speed of the ball as a function of time may be best represented by the graph
(a) A
(b) B
(c) C
(d) D
Figure
Answer:
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.
Question 1:
A student plots a graph from his reading on the determination of Young modulus of a metal wire but forgets to put the labels. the quantities on X and Yaxes may be respectively
(a) weight hung and length increased
(b) stress applied and length increased
(c) stress applied and strain developed
(d) length increased and the weight hung.
Figure
Answer:
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.
Question 2:
The properties of a surface are different from those of the bulk liquid because the surface molecules
(a) are smaller than other molecules
(b) acquire charge due to collision from air molecules
(c) find different type of molecules in their range of influence
(d) feel a net force in one direction.
Answer:
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.
Question 3:
The rise of a liquid in a capillary tube depends on
(a) the material
(b) the length
(c) the outer radius
(d) the inner radius of the tube
Answer:
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.
Question 4:
The contact angle between a solid and a liquid is a property of
(a) the material of the solid
(b) the material of the liquid
(c) the shape of the solid
(d) the mass of the solid
Answer:
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.
Question 5:
A liquid is contained in a vertical tube of semicircular cross section. The contact angle is zero. The force of surface tension on the curved part and on the flat part are in ratio
(a) 1:1
(b) 1:2
(c) π:3
(d) 2:π
Answer:
Correct option: (c)
Let the height of the liquidfilled column be L.
Let the radius be denoted by R.
Total perimeter of the curved part=semicircumference of upper area= πr Total surface tension force =πRSTotal perimeter of the flat part =2R Total surface tension force = 2RSRatio of curved surface force to flat surface force = πRS2RS=π2
Question 6:
When a capillary tube is dipped into a liquid, the liquid neither rises nor falls in the capillary.
(a) The surface tension of the liquid must be zero.
(b) The contact angle must be 90°.
(c) The surface tension may be zero.
(d) The contact angle may be 90°.
Answer:
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°.
Question 7:
A solid sphere moves at a terminal velocity of 20 m s^{−1} in air at a place where g = 9.8 m s^{−2}. The sphere is taken in a gravityfree hall having air at the same pressure and pushed down at a speed of 20 m s^{−1}.
(a) Its initial acceleration will be 9.8 m s^{−2} downward.
(b) It initial acceleration will be 9.8 m s^{−2} upward.
(c) The magnitude of acceleration will decrease as the time passes.
(d) It will eventually stop
Answer:
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 = 6πηrvTmComparing equations (1) and (2), we find that:a = gThus, 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.
Page No 300:
Question 1:
A load of 10 kg is suspended by a metal wire 3 m long and having a crosssectional area 4 mm^{2}. Find (a) the stress (b) the strain and (c) the elongation. Young modulus of the metal is 2.0 × 10^{11} N m^{−2}.
Answer:
Given:
Mass of the load (m) = 10 kg
Length of wire (L) = 3 m
Area of crosssection of the wire (A) = 4 mm^{2} = 4.0 × 10^{−6} m^{2}
Young’s modulus of the metal Y = 2.0 × 10^{11} N m^{−2}
(a) Stress = F/A
F = mg
=
10×10 = 100 N (g = 10 m/s^{2})
∴FA=1004×106 =2.5×107 N/m2(b) Strain =
ΔLLâ€‹
Or,
Strain = StressY Strain =2.5×1072×1011 =1.25×104 N/m2(c) Let the elongation in the wire be
∆L.
Strain=ΔLL⇒ΔL=Strain×L =1.25×104×3 =3.75×104 m
Question 2:
A vertical metal cylinder of radius 2 cm and length 2 m is fixed at the lower end and a load of 100 kg is put on it. Find (a) the stress (b) the strain and (c) the compression of the cylinder. Young modulus of the metal = 2 × 10^{11} N m^{−2}.
Answer:
Given:
Radius of cylinder (r) = 2 cm =
2×102 mLength 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/s^{2})
A is the area of crosssection = πr^{2} = 4π
×104 m2
⇒Stress ρ=mgA =100×104π×104 =7.96×105 N/m2(b) Strain is given by:
Strain=ρY=7.96×1052×1011 =4×106(c) Compression of the cylinder:
ΔL = strain × L
= 4 × 10^{−6} × 2 = 8 × 10^{−6} m
Question 3:
The elastic limit of steel is 8 × 10^{8} N m^{−2} and its Young modulus 2 × 10^{11} N m^{−2}. Find the maximum elongation of a halfmetre steel wire that can be given without exceeding the elastic limit.
Answer:
Given:
Elastic limit of steel FA=8×105 N/m2Young’s modulus of steel Y=2×1011 N/m2Length of steel wire L=12m=0.5 mThe 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 =2×103 m=2 mmHence, the required elongation of steel wire is 2 mm.
Question 4:
A steel wire and a copper wire of equal length and equal crosssectional area are joined end to end and the combination is subjected to a tension. Find the ratio of (a) the stresses developed in the two wires and (b) the strains developed. Y of steel = 2 × 10^{11} N m^{−2}. Y of copper = 1.3 × 10 11 N m^{−2}.
Answer:
Given:
Young’s modulus of steel = 2 × 10^{11} N m^{−2}
Young’s modulus of copper = 1.3 × 10 11 N m^{−2}
Both wires are of equal length and equal crosssectional area. Also, equal tension is applied on them.
As per the question:
Lsteel=LCuAsteel=ACuFCu=FSteelHere: L_{steel} and L_{Cu} denote the lengths of steel and copper wires, respectively.
A_{steel} and A_{Cu} denote the crosssectional areas of steel and copper wires, respectively.
F_{steel} and F_{Cu} denote the tension of steel and cooper wires, respectively.
(a)Stress of CuStress of Steel=FCuACuASteelFSteel=1(b)
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 CuStrain of steel=1320⇒Strain of steelStrain of Cu=2013Hence, the required ratio is 20 : 13.
Question 5:
In figure the upper wire is made of steel and the lower of copper. The wires have equal cross section. Find the ratio of the longitudinal strains developed in the two wires.
Figure
Answer:
Given that both wires are of equal length and equal crosssectional area,
the block applies equal tension on both of them.
∴
Lsteel=LCuAsteel=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=1.54Hence, the required ratio of the longitudinal strains is 20 : 13.
Question 6:
The two wires shown in figure are made of the
Figure
same material which has a breaking stress of 8 × 10^{8} N m^{−2}. The area of cross section of the upper wire is 0.006 cm^{2} and that of the lower wire is 0.003 cm^{2}. The mass m_{1} = 10 kg, m_{2} = 20 kg and the hanger is light. (a) Find the maximum load that can be put on the hanger without breaking a wire. Which wire will break first if the load is increased? (b) Repeat the above part if m_{1} = 10 kg and m_{2} = 36 kg.
Answer:
(a) Given:
Breaking stress of wire =8×108 N/m2Area of crosssection of upper wire (Au)=0.006 cm2=6×107 mArea of crosssection of lower wire (Al)=0.003 cm2=3×107 mm1=10 kg, m2=20 kgTension in lower wire
Tl=m1g+wHere: g is the acceleration due to gravity
w is the load
∴ Stress in lower wire
=TlAl=m1g+wAl
⇒m1g+wAl=8×108⇒w=8×108×3×107100⇒w=140 N or 14 kgNow, 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 kgFor 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+wHere: g is the acceleration due to gravity
w is the load
∴ Stress in lower wire:
⇒TlAl=m1g+wAl=8×105⇒w=140 NNow, tension in upper wire
T2=m1g+m2g+w∴ Stress in upper wire:
⇒TuAu=m2g+m1g+wAu=8×105⇒w=20 NFor 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.
Question 7:
Two persons pull a rope towards themselves. Each person exerts a force of 100 N on the rope. Find the Young modulus of the material of the rope if it extends in length by 1 cm. Original length of the rope = 2 m and the area of cross section = 2 cm^{2}.
Answer:
Given:
Force (F) applied by two persons on the rope = 100 N
Original length of rope L=2 mExtension in the rope ∆L=0.01 mArea of crosssection of the rope A=2×104We know that:Young’s modulus Y=FA×L∆L =1002×104×20.01⇒Y=1×108 N/m2Hence, the required Young’s modulus for the rope is
1×108 N/m2.
Question 8:
A steel rod of crosssectional area 4 cm^{2} and 2 m shrinks by 0.1 cm as the temperature decreases in night. If the rod is clamped at both ends during the day hours, find the tension developed in it during night hours. Young modulus of steel = 1.9 × 10^{11} N m^{−2}.
Answer:
Given:
Crosssectional area of steel rod A = 4 cm^{2} = 4 × 10^{−4} m^{2}
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 × 10^{11} N m^{−2}
Let the tension developed at night be F.
Y=FA×L∆L⇒F=YA∆LL =1.9×1011×4×104×1032 =3.8×104 N∴ Required tension developed in steel rod during night hours = 3.8 × 10^{4} N.
Question 9:
Consider the situation shown in figure. The force F is equal to the m_{2}g/2. If the area of cross section of the string is A and its Young modulus Y, find the strain developed in it. The string is light and there is no friction anywhere.
Figure
Answer:
Given:
Force (F) = m_{2}g/2
Area of crosssection of the string = A
Young’s modulus = Y
Let a be the acceleration produced in block m_{2} in the downward direction and T be the tension in the string.
From the free body diagram:
m2gT=m2a …iTF=m1a …(ii)
From equations (i) and (ii), we get:
a=m2gFm1+m2Applying F=m2g 2⇒a=m2g2m1+m2Again, T = F + m_{1}a
On applying the values of F and a, we get:
⇒T=m2g2+m1m2g2m1+m2We 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.
Question 10:
A sphere of mass 20 kg is suspended by a metal wire of unstretched length 4 m and diameter 1 mm. When in equilibrium, there is a clear gap of 2 mm between the sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle θ with the vertical and is released. Find the maximum value of θ so that the sphere does not rub the floor. Young modulus of the metal of the wire is 2.0 × 10^{11} N m^{−2}. Make appropriate approximations.
Answer:
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 × 10^{11} 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‘â€‹.
⇒T’=mg+mv2rThe change in tension is due to the centrifugal force.
∴â€‹
∆T=T’T
∆T=mv2r …iNow, using work energy principle:
12mv20=mgr1cosθ⇒v2=2gr1cosθ …2Applying the value of v^{2} in (i):
∆T=m 2gr1cosθr =2mg1cosθ Now, F=∆TAlso, F=YA∆LL⇒YA∆LL=2mg1cosθ⇒cosθ=1YA∆LL2mg⇒cosθ=12×1011×4×3.14×52×108×2×1034×2×20×10⇒cosθ=0.80Or, θ=36.4°Hence, the required maximum value of θ is 35.4Ëš.
Page No 301:
Question 11:
A steel wire of original length 1 m and crosssectional area 4.00 mm^{2} is clamped at the two ends so that it lies horizontally and without tensions. If a load of 2.16 kg is suspended from the middle point of the wire, what would be its vertical depression?
Y of the steel = 2.0 × 10^{11} N m^{−2}. Take g = 10 m s^{−2}.
Answer:
Given:
Original length of steel wire (L) = 1 m
Area of crosssection (A) = 4.00 mm^{2} = 4 × 10^{−2} cm^{2}
Load = 2.16 kg
Young’s modulus of steel (Y) = 2 × 10^{11} N/m^{2}
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=xx2+l2=xl1+x2l21/2Expanding the above equation using the binomial theorem:
cosθ=xl112x2l2 neglecting the higher order termsSince x<<l, x2l2 can be neglected.⇒cosθ=xlIncrease in length:
ΔL = (AC + CB) − AB
AC = (l^{2} + x^{2})^{1/2}
ΔL = 2 (l^{2} + x^{2})^{1/2} − 2l
We know that:
Y=FAL∆L⇒2×1012=T×1004×102×2502+x21/2100From the free body diagram:
2Tcosθ=mg2Tx50=2.16×103×980⇒2×2×1012×4×102×2502+x2 12100×100×50=2.16×103×980On solving the above equation, we get x = 1.5 cm.
Hence, the required vertical depression is 1.5 cm.
Question 12:
A copper wire of crosssectional area 0.01 cm^{2} is under a tension of 20N. Find the decrease in the crosssectional area. Young modulus of copper = 1.1 × 10^{11} N m^{−2} and Poisson ratio = 0.32.
Answer:
Given:
Crosssectional area of copper wire A = 0.01 cm^{2} = 10^{−6} m^{2}
Applied tension T = 20 N
Young modulus of copper Y = 1.1 × 10^{11} N/m^{2}
Poisson ratio σ = 0.32
We know that:
Y=FLA∆L ⇒∆LL=FAY =20106×1.1×1011=18.18×105 Poisson’s ratio,σ=∆dd∆LL=0.32Where d is the transverse lengthSo, ∆dd=0.32×∆LL =0.32×18.18×105=5.81×105Again, ∆AA=2∆rr=2∆dd⇒∆A=2∆ddA⇒∆A=2×5.8×105×0.01 =1.164×106 cm2Hence, the required decrease in the cross sectional area is
1.164×106 cm2.
Question 13:
Find the increase in pressure required to decrease the volume of a water sample by 0.01%. Bulk modulus of water = 2.1 × 10^{9} N m^{−2}.
Answer:
Given:
Bulk modulus of water (B) =
2.1×109 Nm2In 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=104From B=PV∆V, we have:⇒P=B∆VV =2.1×109×104 =2.1×105 N/m2Hence, the required increase in pressure is
2.1×105 Nm2.
Question 14:
Estimate the change in the density of water in ocean at a depth of 400 m below the surface. The density of water at the surface = 1030 kg m^{−3} and the bulk modulus of water = 2 × 10^{9} N m^{−2}.
Answer:
Given:
Bulk modulus of water B=2×109 N/m2Depth (d) = 400 m
Density of water at the surface (ρ_{0}) = 1030 kg/m^{3}
We know that:
Density at surface ρ0=mV0Density at depth ρd=mVd⇒ ρdρ0=V0Vd …iHere: ρ_{d }= density of water at a depth
m = mass
V_{0} = volume at the surface
V_{d} = volume at a depth
Pressure at a depth d=ρ0gd Acceleration due to gravity g = 10 ms2Volume strain = V0VdV0B=PressureVolume strain⇒B=ρ0gdV0VdV0⇒1VdV0=ρ0gdB⇒VdV0=1p0 gdB …iiUsing equations (i) and (ii), we get:
ρdρ0=11ρ0gdB⇒ρd=11 ρ0ghBρ0⇒ ρd=1030 11030×10×4002×109 ≈1032 kg/m3Change in density =ρdρ0 =10321030=2 kg/m3Hence, the required density at a depth of 400 m below the surface is 2 kg/m^{3}.â€‹
Question 15:
A steel plate of face area 4 cm^{2} and thickness 0.5 cm is fixed rigidly at the lower surface. A tangential force of 10 N is applied on the upper surface. Find the lateral displacement of the upper surface with respect to the lower surface. Rigidity modulus of steel = 8.4 × 10^{10} N m^{−2}.
Answer:
Given:
Face area of steel plate A = 4 cm^{2} = 4 × 10^{−4} m^{2}
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 × 10^{10} N m^{−2}
Let θ be the angular displacement.
Rigidity modulus
m=FAθ
⇒m=104×104θ⇒θ=104×104×8.4×1010 =0.297×106∴ 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.
Question 16:
A 5.0 cm long straight piece of thread is kept on the surface of water. Find the force with which the surface on one side of the thread pulls it. Surface tension of water = 0.076 N m^{−1}.
Answer:
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.
Question 17:
Find the excess pressure inside (a) a drop of mercury of radius 2 mm (b) a soap bubble of radius 4 mm and (c) an air bubble of radius 4 mm formed inside a tank of water. Surface tension of mercury, soap solution and water are 0.465 N m^{−1}, 0.03 N m^{−1} and 0.076 N m^{−1} respectively.
Answer:
Given:
Radius of mercury drop r=2 mm=2×103 mRadius of soap bubble r=4 mm=4×103 mRadius of air bubble r=4 mm=4×103 mSurface tension of mercury THg=0.465 N/mSurface tension of soap solution Ts=0.03 N/mSurface tension of water Ta=0.076 N/m(a) Excess pressure inside mercury drop:
P=2THgr =0.465×22×103=465 N/m2(b) Excess pressure inside the soap bubble:
P=4Tsr =4×0.034×103=30 N/m2(c) Excess pressure inside the air bubble:
P=2Tar =2×0.0764×103=38 N/m2
Question 18:
Consider a small surface area of 1 mm^{2} at the top of a mercury drop of radius 4.0 mm. Find the force exerted on this area (a) by the air above it (b) by the mercury below it and (c) by the mercury surface in contact with it. Atmospheric pressure = 1.0 × 10^{5} Pa and surface tension of mercury = 0.465 N m^{−1}. Neglect the effect of gravity. Assume all numbers to be exact.
Answer:
Given:
Surface area of mercury drop, A = 1 mm^{2} = 10^{−6} m^{2}
Radius of mercury drop, r = 4 mm = 4 × 10^{−3} m
Atmospheric pressure, P_{0} = 1.0 × 10^{5} P_{a}
Surface tension of mercury, T = 0.465 N/m
(a) Force exerted by air on the surface area:
F = P_{0}A
⇒ F= 1.0 × 10^{5} × 10^{−6} = 0.1 N
(b) Force exerted by mercury below the surface area:
Pressure P’ = P0+2TrF=P’A=P0+2TrA =0.1+2×0.4654×103×106 =0.1+0.00023=0.10023 N(c) Force exerted by mercury surface in contact with it:
P=2TrF=PA=2TrA =2×0.4654×103×106=0.00023 N
Question 19:
The capillaries shown in figure have inner radii 0.5 mm, 1.0 mm and 1.5 mm respectively. The liquid in the beaker is water. Find the heights of water level in the capillaries. The surface tension of water is 7.5 × 10^{−2} N m^{−1}.
Figure
Answer:
Given:
Surface tension of water T = 7.5 × 10^{−2} N/m
Taking cos θ = 1:
Radius of capillary A (r_{A}) = 0.5 mm = 0.5 × 10^{−3} m
Height of water level in capillary A:
hA=2T cos θrAρg =2×7.5×1020.5×103×1000×10 =3×102 m=3 cmRadius of capillary B (r_{B}) = 1 mm = 1 × 10^{−3} m
Height of water level in capillary B:
hB=2Tcos θrBρg =2×7.5×1021×103×103×10 =15×103 m=1.5 cmRadius of capillary C (r_{C}) = 1.5 mm = 1.5 × 10^{−3} m
Height of water level in capillary C:
hC=2T cos θrCρg=2×7.5×1021.5×103×103×10=151.5×103 m=1 cm
Question 20:
The lower end of a capillary tube is immersed in mercury. The level of mercury in the tube is found to be 2 cm below the outer level. If the same tube is immersed in water, up to what height will the water rise in the capillary?
Answer:
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=2TcosθrρgNow, 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 =2.19Height of the water level:
h_{w} = 2 × 2.19 = 4.38 cm
Hence, the required rise in the water level in the capillary tube is 4.38 cm.
Question 21:
A barometer is constructed with its tube having radius 1.0 mm. Assume that the surface of mercury in the tube is spherical in shape. If the atmospheric pressure is equal to 76 cm of mercury, what will be the height raised in the barometer tube? The contact angle of mercury with glass = 135° and surface tension of mercury = 0.465 N m^{−1}. Density of mercury = 13600 kg m^{−3}.
Answer:
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 =2×465×1/2103×13600×10=0.0048 m =0.48 cm∴ Net rise in level in the barometer tube = H − h
= 76 − 0.48
= 75.52 cm
Question 22:
A capillary tube of radius 0.50 mm is dipped vertically in a pot of water. Find the difference between the pressure of the water in the tube 5.0 cm below the surface and the atmospheric pressure. Surface tension of water = 0.075 N m^{−1}.
Answer:
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/m^{2}
Excess pressure at the surface is given by:
P0=2Tr=2×0.755×104 =300 N/m2Difference in pressure: P_{0} − P
=490300=190 N/m2Hence, the required difference in pressure is 190 N/m^{2}.
Question 23:
Find the surface energy of water kept in a cylindrical vessel of radius 6.0 cm. Surface tension of water = 0.075 J m^{−2}.
Answer:
Given:
Radius of cylindrical vessel, r = 6.0 cm = 0.06 m
Surface tension of water, T = 0.075 J/m^{2}
Area, A = πr^{2} = 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.
Question 24:
A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury = 0.465 J m^{−2}.
Answer:
Given:
Initial radius of mercury drop R = 2 mm = 2 × 10^{−3} m
Surface tension of mercury T = 0.465 J/m^{2}
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=43πr3×8Taking 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 × 4πr^{2} − 4πR^{2}) T
=4πT8×R24R2=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.
Question 25:
A capillary tube of radius 1 mm is kept vertical with the lower end in water. (a) Find the height of water raised in the capillary. (b) If the length of the capillary tube is half the answer of part (a), find the angle θ made by the water surface in the capillary with the wall.
Answer:
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=2×0.076103×10×100 =1.52 cm =1.52×102 m =1.52 cm
(b) Let the new length of the tube be h’.
h’=2Tcos θrρgcos θ=h’rρg2TUsing equation i, we get:cos θ=h’h=12 Because h’=h2⇒ θ=cos112=60°The water surface in the capillary makes an angle of 60^{âˆ˜}with the wall.
Question 26:
The lower end of a capillary tube of radius 1 mm is dipped vertically into mercury. (a) Find the depression of mercury column in the capillary. (b) If the length dipped inside is half the answer of part (a), find the angle made by the mercury surface at the end of the capillary with the vertical. Surface tension of mercury = 0.465 N m^{−1} and the contact angle of mercury with glass −135°.
Answer:
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°103×13600×9.8 =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°
Question 27:
Two large glass plates are placed vertically and parallel to each other inside a tank of water with separation between the plates equal to 1 mm. Find the rise of water in the space between the plates. Surface tension of water = 0.075 Nm^{−1}.
Answer:
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 = 10^{3} kg/m^{3}
Applying law of conservation of energy:
T (2L) = [1 × (10^{−3}) × h] ρg
⇒h=2×0.075103×103×10 =0.015 m = 1.5 cmTherefore, the rise of water in the space between the plates is 1.5 cm.
Question 28:
Consider an ice cube of edge 1.0 cm kept in a gravityfree hall. Find the surface area of the water when the ice melts. Neglect the difference in densities of ice and water.
Answer:
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=3a34π1/3Surface area of spherical water surface = 4πr^{2}
=4π3a34π2/3=36π1/3 cm2
Question 29:
A wire forming a loop is dipped into soap solution and taken out so that a film of soap solution is formed. A loop of 6.28 cm long thread is gently put on the film and the film is pricked with a needle inside the loop. The thread loop takes the shape of a circle. Find the tension the the thread. Surface tension of soap solution = 0.030 N m^{−1}.
Answer:
Given:
Surface tension of soap solution T = 0.030 N m^{−1}
Let the radius of the thread loop be r.
⇒2πr=6.28 cm⇒r=6.282×3.14=1 cm
The excess pressure inside the loop is expressed as follows:
∆P=4TrTension in the thread:
T’=∆P×area of loop
⇒T’=4Tr×πr2⇒T’=4πTr =4×0.030×3.14×102 N =3.8×103 N
Question 30:
A metal sphere of radius 1 mm and mass 50 mg falls vertically in glycerine. Find (a) the viscous force exerted by the glycerine on the sphere when the speed of the sphere is 1 cm s^{−}^{1}, (b) the hydrostatic force exerted by the glycerine on the sphere and (c) the terminal velocity with which the sphere will move down without acceleration. Density of glycerine = 1260 kg m^{−3} and its coefficient of viscosity at room temperature = 8.0 poise.
Answer:
Radius of metallic sphere r = 1 mm = 10^{−3} m
Speed of the sphere v = 10^{−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/m^{3}
(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
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=mg43πr2σg6πηr =50×10343×3.14×106×1260×106×3.14×0.8×103 =50043×3.14×103×1260×106×3.14×0.8 =2.3 cm/s
Question 31:
Estimate the speed of vertically falling raindrops from the following data. Radius of the drops = 0.02 cm, viscosity of air = 1.8 × 10^{−4} poise, g= 9.9 × 10 ms^{−}^{2} and density of water = 1000 kg m^{−3}.
Answer:
Given:
Radius of the drops r = 0.02 cm = 2 × 10^{−4} m
Viscosity of air η = 1.8 × 10^{−4} poise = 1.8 × 10^{−5} decapoise
Acceleration due to gravity g = 9.9 m/s^{2}
Density of water ρ = 1000 kg/m^{3}
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
=2×0.02×1022×1000×9.99×1.8×105= 5 m/s.Hence, the required vertical speed of the falling raindrops is 5 m/s.
Page No 302:
Question 32:
Water flows at a speed of 6 cm s^{−}^{1} through a tube of radius 1 cm. Coefficient of viscosity of water at room temperature is 0.01 poise. Calculate the Reynolds number. Is it a steady flow?
Answer:
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 =6×103×103×2×102102 =120Here, the Reynolds number is less than 2000. Therefore, it is a steady flow.
HC Verma Solutions for Class 11 Physics – Part 1
 Chapter 1 – Introduction to Physics
 Chapter 2 – Physics and Mathematics
 Chapter 3 – Rest and Motion: Kinematics
 Chapter 4 – The Forces
 Chapter 5 – Newton’s Laws of Motion
 Chapter 6 – Friction
 Chapter 7 – Circular Motion
 Chapter 8 – Work and Energy
 Chapter 9 – Center of Mass, Linear Momentum, Collision
 Chapter 10 – Rotational Mechanics
 Chapter 11 – Gravitation
 Chapter 12 – Simple Harmonic Motion
 Chapter 13 – Fluid Mechanics
 Chapter 14 – Some Mechanical Properties of Matter
 Chapter 15 – Wave Motion and Wave on a String
 Chapter 16 – Sound Wave
 Chapter 17 – Light Waves
 Chapter 18 – Geometrical Optics
 Chapter 19 – Optical Instruments
 Chapter 20 – Dispersion and Spectra
 Chapter 21 – Speed of Light
 Chapter 22 – Photometry
HC Verma Solutions for Class 12 Physics – Part 2
 Chapter 23 – Heat and Temperature
 Chapter 24 – Kinetic Theory of Gases
 Chapter 25 – Calorimetry
 Chapter 26 – Laws of Thermodynamics
 Chapter 27 – Specific Heat Capacities of Gases
 Chapter 28 – Heat Transfer
 Chapter 29 – Electric Field and Potential
 Chapter 30 – Gauss’s Law
 Chapter 31 – Capacitors
 Chapter 32 – Electric Current in Conductors
 Chapter 33 – Thermal and Chemical Effects of Electric Current
 Chapter 34 – Magnetic Field
 Chapter 35 – Magnetic Field due to a Current
 Chapter 36 – Permanent Magnets
 Chapter 37 – Magnetic Properties of Matter
 Chapter 38 – Electromagnetic Induction
 Chapter 39 – Alternating Current
 Chapter 40 – Electromagnetic Waves
 Chapter 41 – Electric Current through Gases
 Chapter 42 – Photoelectric Effect and Wave Particle Duality
 Chapter 43 – Bohr’s Model and Physics of the Atom
 Chapter 44 – Xrays
 Chapter 45 – Semiconductors and Semiconductor Devices
 Chapter 46 – The Nucleus
 Chapter 47 – The Special Theory of Relativity
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