Physics-15.8-Solids and Fluids
8 - FLUID DYNAMICS
1. PROPERTIES OF MATTER
1.1 SOME IMPORTANT DEFINITION
(i) Deforming Force
If a force applied to a body causes a change in the normal positions of the molecules of the body, resulting in a change in the configuration of the body either in length, volume or shape, then the force applied is deforming in nature.
(ii) Elasticity
The property of a body by virtue of which the body regains its original configuration (length, volume or shape) when external deforming forces are removed is called elasticity.
(iii) Cause of Elasticity
Molecules in a body are bound to each other by bonds. When elongated the bonds behave like springs due to inter molecular interaction. When subjected to compression beyond a certain distance or less than a certain inter molecular separation, the molecules, instead of attracting each other, repel.
(iv) Interatomic force constant (k)
Interatomic force constant is defined as ratio of interatomic force to change in interatomic distance.
(v) Perfectly Elastic Body
A body which regains its original configuration immediately and completely after the removal of deforming force from it, is called perfectly elastic body. Quartz and phosphor bronze are the examples of nearly perfectly elastic bodies.
(vi) Perfectly Plastic Body
A body which does not regain its original configuration at all on the removal of deforming force, however small the deforming force may be is a perfectly plastic body.
(vii) STRESS
Stress is measured by the deforming force per unit normal area.
Stresses set up in a body due to the action of forces can be classified into three broad categories
(a) Tensile stress, which produces elongation, is the result of forces acting on the body along a single direction.
(b) Shear stress, which produces "shear or bending - being the result of equal and opposite torques acting on a body.
(c) Volume stress (Bulk Stress) which produces a change in the volume of the body - the resulting strain is known as volume strain (Bulk Strain).
(d) Unit of stress = N/m2
(e) Dimension of stress = ML−1T−2
(viii) STRAIN
When deforming forces are applied to a body, there is a change in the shape of the body. The body is said to be strained or deformed. The ratio of change in dimension to the original dimension is called strain. i.e.
Strain =
Strain being the ratio of two like quantities has no units and dimensions. There is three type of strain
(a) Longitudinal strain = =
(b) Volume strain = =
(c) Shear strain =θ where θ is the angle between the deformed surface and original surface.
(ix) ELASTIC LIMIT
Elastic limit is the upper limit of deforming force up to which, if deforming force is removed, the body regains its original form completely and beyond that limit, if deforming force is increased, the body loses its property of elasticity and gets permanently deformed.
Elastic limit is the property of a body whereas elasticity is the property of the material of a body.
2. HOOKE'S LAW
Hooke's law is valid for only small deformation. Hooke's law states that the extension produced in the wire is directly proportional to the load applied within elastic limit. i.e. within elastic limits, extension ∝ load applied.
Later on it was found that this law is applicable to all types of deformation such as compression, bending, twisting etc and thus a modification form of Hooke's law was given as stated below.
“Within elastic limit, the stress developed is directly proportional to the strain produced in a body”
i.e. stress ∝ strain or, stress = E × strain
where E is a constant and is known as the modulus of elasticity of the material.
2.1 STRESS, STRAIN RELATIONSHIP FOR A WIRE
The relationship between stress and strain in a wire is illustrated by the graph shown in the adjoining figure. The regions are referred to as follows:
OA = Elastic region, Hooke’s law is valid.
AB = Elastic region, Hooke’s law is not true
A = proportional limit
B = Elastic limit(strain ≤1 %) or yield point
BD = Plastic deformation
OE = Permanent set
CD = Plastic flow
D = Fracture point (Corresponding stress is called breaking stress/ tensile strength)
2.2 ELASTOMERS
The substances which can be stretched to large values of strain are called elastomers. e.g. elastic tissue of aorta, the largest artery carrying blood from the heart.
2.3 MODULUS OF ELASTICITY
According to Hooke's law, within elastic limit,
stress ∝ strain
or stress = E × strain
or = E = a constant,
where E is known as coefficient of elasticity or modulus of elasticity of a body which depends upon the nature of material of the body and the manner in which the body is deformed. Thus, modulus of elasticity or coefficient of elasticity of a body is defined as the ratio of the stress to the corresponding strain produced, within the elastic limit.
2.4 TYPES OF MODULE OF ELASTICITY
Corresponding to three types of strain, there are three types of module of elasticity:
(i) YOUNG'S MODULUS OF ELASTICITY (Y)
It is defined as the ratio of normal stress to the longitudinal strain within the elastic limit. Thus
Y =
Consider a metal wire AB of length , radius r and of uniform area of cross-section a and having negligible mass. Let it be suspended from a rigid support at A, as shown in figure.
Let a normal force F be applied at its free end B and let its length increase by Δ (= BB′ )
Then, longitudinal strain =
Normal stress = = ( a = πr2)
∴ Y =
(ii) BULK MODULUS OF ELASTICITY (K)
It is defined as the ratio of normal stress to the volumetric strain, within the elastic limit. Thus
k =
Consider a spherical solid body of volume V and surface area a. In order to compress the body, let a force F be applied normally on the entire surface of the body and suppose that its volume decreases by Δv as shown in figure. Then, the volumetric strain = - Δv/V
Here negative sign shows that volume is decreasing when force is applied.
Normal stress = F/a
∴ k =
If p represents the increase in pressure applied on the spherical body then F/a = p
k = −pV/Δv
COMPRESSIBILITY
The reciprocal of the bulk modulus of a material is called its compressibility.
Compressibility =
(iii) MODULUS OF RIGIDITY (η)
It is the ratio of tangential stress to the shearing strain, within the elastic limit. It is also called shear modulus of rigidity. Thus
η = ∴ η =
2.5 EFFECT ON ELASTICITY BY DIFFERENT CAUSES
(i) Effect of temperature:
(a) In general elasticity decreases as the temperature increases.
(b) Invar is an exception. There is no effect of temperature on elasticity of invar. Invar is in facet a short form of invariable.
(ii) Effect of impurities:
(a) If the impurity is more elastic, the elasticity of the material increases.
(b) If the impurity is more plastic, the elasticity of material decreases.
(iii) On hammering or rolling elasticity increase.
(iv) On annealing i.e. on alternate heating and cooling elasticity decreases.
Example 1: A cable is replaced by another one of the same length and material but having twice the diameter. How will this affect the elongation under a given load? How does this affect the maximum load it can support without exceeding the elastic limit?
Solution: Young's Modulus Y =
Where D is the diameter of the wire.
∴ Elongation, Δ = i.e., Δ ∝
Clearly, if the diameter is doubled, the elongation will become one-fourth.
Also load, Mg = i.e., Mg ∝ D2.
Clearly, if the diameter is doubled, the wire can support 4 times the original load.
2.6 ELASTIC POTENTIAL ENERGY IN A STRETCHED WIRE
When a wire is stretched, some work is done against the internal restoring forces acting between the particles of the wire. This work done appears as the elastic potential energy of the wire.
Consider a wire of length and area of cross section a. Let F be the stretching force applied on the wire and let Δ be the increase in length of the wire. Initially, the internal restoring force was zero but when length is increased by Δ, the average internal restoring force for an increase in length Δ of the wire = =
Hence, work done on the wire, w = average force × increase in length = × Δ
This is stored as elastic potential energy U in the wire.
∴ U = F × Δ = × × a
= (stress) × (strain) × volume of the wire
∴ Elastic potential energy per unit volumeaof the wire (= energy density)
u = (stress) × (strain)
= (Young's modulus × strain) × strain
( Young's modulus = stress / strain)
∴ u = (Young's modulus ) × (strain)2
Example 2: A steel wire 4.0 m in length is stretched through 2.0 mm. The cross-sectional area of the wire is 2.0 mm2. If Young's modulus of steel is 2.0 × 1011 N/m2 find
(a) the energy density of wire.
(b) the elastic potential energy stored in the wire.
Solution: Here, = 4.0 m ;
Δ = 2 × 10-3 m ; a = 2.0 × 10-6 m2
y = 2.0 × 1011 N/m2
(a) The energy density of stretched wire
u = × stress × strain = × y × (strain)2
= × 2.0 × 1011 × (2 × 10-3) / 4)2
= 0.25 × 105 = 2.5 × 104 J/m3.
(b) Elastic potential energy = energy density × volume
= 2.5 × 104 × (2.0 × 10-6 ) × 4.0 J = 20 × 10-2 = 0.20 J.
2.7 ELASTIC FATIGUE
It is the lack of elastic strength of a substance when subjected to repeated stresses and strain. If the substance is kept undisturbed for some time, the previous properties are restored. Thus elastic fatigue is a tempeorary phenomenon.
2.8 SEARLE’S APPARATUS
Y =
(b) To calculate the maximum load to be applied :
Breaking weight for the experimental wire = breaking stress ×
Maximum load to be applied = × breaking stress
3. SURFACE TENSION
The force of attraction between the molecules of the same substance is called cohesion.
In case of solids, the force of cohesion is very large and due to this solids have definite shape and size. On the other hand, the force of cohesion in case of liquids is weaker than that of solids. Hence liquids do not have definite shape but have definite volume. The force of cohesion is negligible in case of gases. Because of this fact, gases have neither fixed shape nor volume.
Properties of surface Tension
∙ Scalar quantity
∙ Temperature sensitive
∙ Impurity sensitive
∙ Depends only on the nature of the liquid.
∙ Unit of surface tension, N/m
∙ Dimension of surface tension, ML°T−2.
3.1 MATHEMATICAL DESCRIPTION OF SURFACE TENSION
surface tension is defined as the force per unit length in the plane of a liquid surface, acting at right angles on either side of the imaginary line drawn in the surface
T = F /
3.3 SURFACE TENSION AND WORK DONE
Surface tension and work done in increasing surface area
F 2
3.4 ANGLE OF CONTACT
Angle of Contact
The angle which the tangent to the liquid surface at the point of contact makes with the solid surface inside the liquid is called angle of contact. Those liquids which wet the walls of the container (say in case of water and glass) have meniscus concave upwards and their value of angle of contact is less than 90° (also called acute angle). However, those liquids which don't wet the walls of the container (say in case of mercury and glass) have meniscus convex
upwards and their value of angle of contact is greater than 90° (also called obtuse angle). The angle of contact of mercury with glass about 140°, whereas the angle of contact of water with glass is about 8°. But, for pure water, the angle of contact θ with glass is taken as 0°.
Case I: When θ < 90°:
The liquid surface curves up towards the solid. This happens when the force of cohesion between two liquid molecules is less than force of adhesion between the liquid and the solid. If such a liquid is poured into a solid tube, it will have a concave meniscus. For example, a glass rod dipped in water, or water inside a glass tube.
Case II: When θ > 90°:
The liquids surfaces get curved downward in contact with a solid. In this case the force of cohesion is greater than the force of adhesion. In such cases, solids do not get “wet”. When such liquids are put into a solid tube, a convex meniscus is obtained.
For example, a glass rod dipped in mercury or mercury within a solid glass tube.
Capillarity: A molecule at the surface of a liquid is attracted by other molecules in the surface in all directions. If the surface is convex, then a resultant component of all the forces of attraction acting on every molecules act normal to the surface and is directed downward. Similarly, if the surface is concave, then every molecule experiences a resultant force due to surface tension acting normally outward. For balance the resultant force due to surface tension. Hence the pressure on the concave side must be greater than the pressure on the convex side. This difference in pressure is equal to 2T/R, where T is the surface tension and R is the radius of curvature of the surface.
Example 3: What is the surface energy of a soap bubble of radius r.
Solution: E = TA = T × 4πr2 × 2 (as it has two surfaces)
= 8πr2 T.
3.5 EXCESS PRESSURE
(i) Inside a bubble: Consider a soap bubble of radius r. Let p be the pressure inside the bubble and pa outside. The excess pressure = p – pa. Imagine the bubble broken into two halves, and consider one half of it as shown in Fig. Since there are two surfaces, inner and outer, so the force due to surface tension is
F = surface tension x length = Tx2 (circumference of the bubble)
= T x 2 (2 Tπ r)
The excess pressure (p – pa) acts on a cross-sectional area π r2, so the force due to excess pressure is
or
(ii) Inside the drop: In a drop, there is only one surface and hence excess pressure can be written as
(iv) A charged bubble: If bubble is charged, it's radius increases. Bubble has pressure excess due to charge too.
Initially pressure inside the bubble =
For charge bubble, pressure inside = where σ surface is surface charge density. Taking temperature remains constant then from Boyle's law
From above expression the radius of charged drop may be calculated. It can conclude that radius of charged bubble increases, i.e. r2 > r1.
(v) Excess of Pressure inside a Curved Surface
a. Plane Surface: If the surface of the liquid is plane [as shown in Fig.(a)], the molecule on the liquid surface is attracted equally in all directions. The resultant force due to surface tension is zero. The pressure, therefore, on the liquid surface is normal.
b. Concave Surface: If the surface is concave upwards [as shown in Fig.(b)], there will be upward resultant force due to surface tension acting on the molecule. Since the molecule on the surface is in equilibrium, there must be an excess of pressure on the concave side in the downward direction to balance the resultant force of surface tension .
c. Convex Surface: If the surface is convex [as shown in Fig.(c)], the resultant force due to surface tension acts in the downward direction. Since the molecule on the surface are in equilibrium, there must be an excess of pressure on the concave side of the surface acting in the upward direction to balance the downward resultant force of surface tension, Hence there is always as excess of pressure on concave side of a curbed surface over that on the convex side. )
Example 4: A meniscus drop of radius 1 cm is sprayed into 106 droplets of equal size. Calculate the energy expended if surface tension of mercury is 35 × 10-3 N/m.
Solution: Energy expended will be the work done again the increase in surface area i.e. n(4πr2) – 4 πR2
E = W = TΔS = T. 4π (nr2 – R2)
But the total volume remains conserved.
I.e. πR3 = n πr3
or r =
and hence,
E = 4 πR2T (n1/3 – 1)
= 4 × 3.14 × (1 × 10-2)2 × 35 × 10-3 (102 – 1) = 4.356 × 10-3 J.
Effect of Temperature and Impurities of Surface Tension
The surface tension of a liquid decreases with the rise in temperature and vice versa. According to Ferguson, where T0 is surface tension at 0°C, θ is absolute temperature of the liquid, is the critical temperature and n is a constant varies slightly from liquid and has mean value 1.21. This formula shows that the surface tension becomes zero at the critical temperature.
The surface tension of a liquid changed appreciably with addition of impurities. For example, surface tension of water increases with addition of highly soluble substances like NaCl, ZnSO4 etc. On the other hand surface tension of water gets reduced with addition of sparingly soluble substances like phenol, soap etc.
4. VISCOSITY
The property of a liquid by virtue of which an opposing force ( internal friction) comes into play whenever there is a relative motion between the diffirent layers of the liquid is called viscosity.
Suppose that a glass plate in contact with a water column of height h is moved with constant velocity v. Forces of viscosity appear between the solid surface and the layer in contact.
F = −η A .
is coefficient of viscosity. Negative shows that the direction of viscous force(F) is just opposite to the direction of the motion of the liquid.
its CGS unit is poise. Dimension is ML−1T−1. The SI units of viscosity is kg/m×sec.
4.1 POISEUILLE’S FORMULA
Flow of viscous liquid through a capillary tube (Poiseuille’s formula)
The velocity v at a distance y from the capillary axis for a flow of liquid of viscosity η in a capillary tube of length L and radius r under a pressure difference p across it is given by
v = (r2 – y2)
and the volume of liquid flowing per second is given by
.
4.2 STOKE’S FORMULA
Stoke’s Formula: Stokes proved that the viscous drag (F) on a spherical body of radius r moving with terminal velocity v in a fluid of viscosity is given by
Terminal velocity: When a body is dropped in a viscous fluid, it is first accelerated and then its acceleration becomes zero and it attains a constant velocity called terminal velocity.
Where r is the radius of the penetrating body ρ1 is density of body, ρ2 is density of liquid, is coefficient of viscosity.
Example 5: A metal plate 0.04 m2 in area is lying on a liquid layer of thickness 10-3m and coefficient of viscosity 140 poise. Calculate the horizontal force needed to move the plate with a speed of 0.040 m/s.
Solution: Area of the place P = 0.04 m2
Thickness Δx = 10-3m
Δx is the distance of the free surface with respect to the fixed surface.
Velocity gradient, = 40s-1
Coefficient of viscosity η = 14kg/ms-1
Let F be the required force
Then, F = ηA
F = 14 14kg/ms-140s-1 = 22.4b
5. FLUID DYNAMIC
(i) Reynold's Number
According to Reynold, the critical velocity (vc) of a liquid flowing through a long narrow tube is
(a) directly proportional to the coefficient of viscosity (η) of the liquid.
(b) inversely proportional to the density ρ of the liquid and
(c) inversely proportional to the diameter (D) of the tube.
That is ... (1)
where R is the Reynold number.
If R < 2000, the flow of liquid is streamline of laminar. If R > 3000, the flow is turbulent. If R lies between 2000 and 3000, the flow is unstable and may change from streamline flow to turbulent flow.
(ii) Steady Flow (Stream Line Flow)
Then the velocity of fluid particles reaching a particular point is the same at all time. Thus, each following particle takes the same path as taken by a previous particle through that point.
(iii) Line of Flow
It is the path taken by a particle in a flowing liquid, in case of a steady flow, it is called a streamline.
Consider an area S in a fluid in steady flow. Draw streamlines from all the points of the periphery of S. These streamlines enclose a tube, of which S in a cross-section.
No fluid enters or leaves across the surface of this tube.
(iv) Equation of Continuity
In a time Δt, the volume of liquid entering, the tube of flow in a steady flow is A1 V1 Δt, the same volume must flow out, as liquid is incompressible. The volume flowing out in Δt is A2 V2 Δt
⇒ A1V1 = A2V2
(v) Bernoulli’s Theorem
Consider a tube of flow, ABCD. In a time Δt, liquid moves and the liquid element becomes A'B'C’D'. In other words, we can also interpret that ABB’A' has gone to DCC'D'.
m = A1V1 t = A2V2 t
Work done by fluid pressure at 1
= (P1A1) v1 t = P1 m/
Work done by fluid pressure at 2
= − (P2A2) v2 t = −P2 m/
Work done by gravity = -(m).g. (h2 - h1)
Change in kinetic energy = 1/2 Δm [V22 - V12]
Using, work energy theorem, (W = ΔK)
Example 6: The reading of pressure-meter fitted in a closed pipe is 4.5105 N/m2. On opening the value of the pipe, the reading the meter reduces to 4.0105N/m2. Calculate the speed of water flowing in the pipe.
Solution: From Bernoulli's theorem
P1 + = P2 +
or P1 – P2 =
Here, v1 = 0
(Velocity of the water is zero because initially pipe is closed)
∴
= 2 0.5103 = 100m/s
v2 = 10m/s
Example 7: Water is flowing at a rate of 2m/s in a pipe of cross-sectional area 0.02m2. If the cross-section is reduced to half, then find the rate of flow.
Solution: Given A1 = 0.02m2, v1 = m/s
A2 = m2 = 0.01m2, v2 = ?
From principle of continuity
A1v1 = A2v2 = constant
0.022 = 0.01 v2
v2 = 4m/s
6. PASCAL'S LAW
Pressure difference (P – Pa) = hpg. This is Pascal's Low which states that Hydraulic lift is common application of Pascal's Law.
(i) Hydraulic press.
This can be used to lift a heavy load placed on the platform of larger piston or to press the things placed between the piston and the heavy platform.
(ii) Hydraulic Brake.
Hydraulic brake system is used in auto-mobiles to retard the motion.
7. HYDROSTATIC PARADOX
Pressure is directly proportional to depth and by applying Pascal's law it can be seen that pressure is independent of the size and shape of the containing vessel.
8. ATMOSPHERIC PRESSURE
Definition: The atmospheric pressure at any point is numerically equal to the weight of a column of air of unit cross-sectional area extending from that point to the top of the atmosphere.
MEASUREMENT OF ATMOSPHERIC PRESSURE
(i) Mercury Barometer
To measure the atmospheric pressure experimentally, Torricelli invented a mercury barometer in 1643.
The pressure exerted by a mercury column of 1mm high is called 1 Torr.
1 Torr = 1 mm of mercury column
(ii) Water Barometer.
Let us suppose water is used in the barometer instead of mercury.
The height of the water column in the tube wilt be 10.3 m.
9. ARCHIMEDES' PRINCIPLE
According to this principle, when a body is immersed wholly or partially in a fluid, it loses its weight which is equal to the weight of the fluid displaced by the body.
Up thrust = buoyancy = V = volume submerged = density of liquid.
Relation between density of solid and liquid
weight of the floating solid = weight of the liquid displaced
or
This relationship is valid in accelerating fluid also.
10. PRESSURE IN CASE OF ACCELERATING FLUID
(i) Liquid Placed in elevator: When elevator accelerates upward with acceleration then pressure in the fluid, at depth 'h' may be given by,
and force of buoyancy,
(ii) Free surface of liquid in horizontal acceleration:
where and are pressures at point 1 & 2. Then
(iii) Free surface of liquid in case of rotating cylinder.
11. OBJECTIVE PROBLEMS
1: If the elastic limit of a typical rock is 3 × 108 Nm–1 and its mean density 3 × 103 kgm–3, estimate the maximum height of a mountain on earth. Take g as 10 ms–2
(A) 6 km (B) 12 km
(C) 8 Km (D) 10 km
Solution: (D) Let maximum stress of a mountain of height y be γρg
Then weight of rock in mountain
= lm2 × y × 3 × 103 × 10
Stress on base of 1m2
= y × 3 × 103 × 10
As per problem,
3 × 108 = y × 3 × 104
or y = 104 m = 10 km
2: The pressure of water in a water pipe when tap is open and closed are respectively 3 × 105 Nm–2 and 3.5 × 105 Nm–2. With open tap, the velocity of water flowing is
(A) 10 ms–1 (B) 5 ms–1
(C) 20 ms–1 (D) 15 ms–1
Solution: (A)Here P =
or v2 = 2(P1 – P2)/
= 2 (3.5 × 105 – 3 × 105)/1000
=
or = = 10 ms–1
3: Two soap bubbles of radii R and r come in contact. R is more than r. Radius of curvature of common surface is
(A) (B)
(C) (D)
Solution: (B) Using excess pressure = , we get
P1 – Patm =
P2 –Patm =
or P2 – P1 =
or
or
or rcurvature =
4: A conical metallic body is heated at middle. If the body was voiced from its ends then
(A) There will be no effect on the body.
(B) Body will contract from middle.
(C) No strain is produced in the body.
(D) Maximum compressive stress will be on the side of body towards R.
Solution: (D) Stress =
Since compressive force is same throughout the body, right side has less area of cross–section, so stress is maximum.
5: A barometer kept in an elevator accelerating downward read x cm. The air pressure in the elevator is
(A) x cm of mercury column
(B) less than x cm of mercury column
(C) greater than x cm mercury column
(D) nil.
Solution: (B) Net acceleration = (g –a)
Pressure = cm of mercury
Since <1
here pressure in lift is less than x cm of mercury.
6: A cubical block of wood 10 cm on a side floats at the interface between oil and water with its lower surface horizontal and 4 cm below the interface. The density of oil is 0.6 g cm–1. The mass of block is
(A) 706 g (B) 607 g
(C) 760 g (D) 670 g
Solution: (C) Volume of block = 10 × 10 × 10 = 1000 cm3
Volume of block in water = 10 × 10 × 4 = 400 cm3
Volume of block in oil = 10 × 10 × 6 = 600 cm3
Weight of block = Weight of oil displaced + Weight of water displaced
or mg = (600 × 0.6)g + (400 × 1) g
or m = 360 + 400 = 760g
7: A stone of 0.5 kg mass is attached to one end of a 0.8m long aluminium wire of 0.7 mm diameter and suspended vertically. The stone is now rotated in a horizontal plane at a rate such that wire makes an angle of 85o with the vertical. If Y = 7 × 1010 Nm–2, sin 85o = 0.9962 and cos 85o = 0.872, the increase in length of wire is
(A) 1.67 × 10–3 m (B) 6.17 × 10–3 m
(C) 1.76 × 10–3 m (D) 7.16 × 10–3 m
Solution: (A) Here T cos = mg and T sin = mR
Using Y = we get
=
= 1.67 × 10–3 m
8: A metallic plate having shape of a square is suspended as shown in figure. The plate is made to dip in water such that level of water is well above that of the plate. the point X is then slowly raised at constant velocity then curve between tension T in string and displacement S of point X is given by
(A) (B) (C) (D)
Solution: (B) Tension in the string is constant so long as plate remains immersed in liquid. Tension increases linearly as the plate starts coming out of the liquid. When the plate is completely out of the liquid, the tension in string becomes constant.
9: A large block of ice of thickness t and density has a large vertical hole along it axis. This block is floating in a lake. The length of rope required to raise a bucket of water through the hole is
(A) ( –l) (B) (1+ ) l
(C) l (D) l(1– )
Solution: (D) Let the block has height h above the surface of water and A be the area of block. Hereweight of ice block = weight of liquid displaced
.A.l.g = (l – h) Ag = lAg – hAg
–h = ( –1) l or h = l (1– )
Which is the least length of rope.
10: A ball of radius r and density falls freely under gravity through a distance h before entering water. Velocity of ball does not change even on entering water. If viscosity of water is , the value of h is given by
(A) (B)
(C) (D)
Solution: (C) Here velocity of ball = but this velocity is the same as terminal velocity
Given by
Then
or h = =
11: A wire of brass and another of steel support a horizontal for as shown
Here s = 1.5 m, Ysteel = 2 × 1011 Nm–2, Ybrass = 1 × 1011 Nm–2, Area of cross sections asteel = 1 × 10– m2, abrass = 2 × 10–5 m2. Both wires have same length. Then tension produced in both wires ?
(A) if the rod remains horizontal the extensions in the wires have to be different.
(B) tension produced in both wires is same for the bar to remain horizontal
(C) distance x = 0.75m
(D) tensions produced in both wires is different for the far to remain horizontal.
Solution: (B) Elongation of both wires must be same for horizontal position of rod.
i.e. = =
=
i.e. Tsteel =Tbrass
Taking moment Tbrass = × x Tsteel (s–x)
i.e. x = (s–x) i.e., 2x = s
i.e. 2x = s i.e., x = = 0.75m.
12: A wire of brass and another of steel support a horizontal for as shown
Here s = 1.5 m, Ysteel = 2 × 1011 Nm–2, Ybrass = 1 × 1011 Nm–2, Area of cross sections asteel = 1 × 10– m2, abrass = 2 × 10–5 m2. Both wires have same length.Then distance x is
(A) if the rod remains horizontal the extensions in the wires have to be different.
(B) tension produced in both wires is same for the bar to remain horizontal
(C) distance x = 0.75m
(D) tensions produced in both wires is different for the far to remain horizontal.
Solution: (C) Elongation of both wires must be same for horizontal position of rod.
i.e. = =
=
i.e. Tsteel =Tbrass
Taking moment Tbrass = × x Tsteel (s–x)
i.e. x = (s–x) i.e., 2x = s
i.e.2x = s i.e., x = = 0.75m.
13: A wall of width at an angle with the normal is subject to water pressure in a vessel. The height of water in the vessel is H and ρ is the density of water. Then Average pressure on the wall is ?
(A) Average pressure on the wall is
(B) Average pressure on the wall is
(C) Force exerted on wall is
(D) Force exerted on the wall is
Solution: (B) Pressure on bottom = Pat + h g.
Pressure on top = Pat d
Average pressure on bottom =
Using Pascal’s law, average pressure on the wall Pat +
Area of the wall in contact with water
=
Force exerted on wall is F = Pav A =
14: A wall of width at an angle with the normal is subject to water pressure in a vessel. The height of water in the vessel is H and ρ is the density of water. Then Force exerted on the wall is ?
(A) Average pressure on the wall is
(B) Average pressure on the wall is
(C) Force exerted on wall is
(D) Force exerted on the wall is
Solution: (D) Pressure on bottom = Pat + h g.
Pressure on top = Pat d
Average pressure on bottom =
Using Pascal’s law, average pressure on the wall Pat +
Area of the wall in contact with water
=
Force exerted on wall is F = Pav A =
15: A mercury filled U–tube arrangement is connected to a bulb containing gas. Atmosphere pressure is 1.012 × 105 Pa and H = 0.05 m. Gauge pressure at R is ?
(A) gauge pressure at R is nil
(B) gauge pressure at R is 6.56 × 103 Pa
(C) gauge pressure at R is 1.08 × 105 Pa
(D) pressure at R, Q and inside bulb are same.
Solution: (A) Pressure at R = Atmosphere pressure = 1.013 × 105 Pa
Gauge pressure = nil
16: A mercury filled U–tube arrangement is connected to a bulb containing gas. Atmosphere pressure is 1.012 × 105 Pa and H = 0.05 m. Gauge pressure at Q is ?
(A) gauge pressure at Q is nil
(B) gauge pressure at Q is 6.56 × 103 Pa
(C) gauge pressure at Q is 1.08 × 105 Pa
(D) pressure at R, Q and inside bulb are same.
Solution: (B) Pressure at R = Atmosphere pressure = 1.013 × 105 Pa
Gauge pressure = nil
At Q, gauge pressure = gH
= (1.36 × 103) (9.8) (0.05) = 6.564 × 103 Pa
17: A mercury filled U–tube arrangement is connected to a bulb containing gas. Atmosphere pressure is 1.012 × 105 Pa and H = 0.05 m. Absolute pressure in bulb is ?
(A) absolute pressure in bulb is nil
(B) absolute pressure in bulb is 6.56 × 103 Pa
(C) absolute pressure in bulb is 1.08 × 105 Pa
(D) pressure at R, Q and inside bulb are same.
Solution: (C) Pressure at R = Atmosphere pressure = 1.013 × 105 Pa
Gauge pressure = nil
At Q, gauge pressure = gH
= (1.36 × 103) (9.8) (0.05) = 6.564 × 103 Pa
Absolute pressure = Pat+ gH
= (1.013 × 105) + (6.564 × 103) = 1.08 × 105 Pa
Pressure in bulb = 1.08 × 105 Pa.
18: A mercury filled U–tube arrangement is connected to a bulb containing gas. Atmosphere pressure is 1.012 × 105 Pa and H = 0.05 m. Pressure at R, Q is ?
(A) pressure at R, Q is nil
(B) pressure at R, Q is 6.56 × 103 Pa
(C) pressure at R, Q is 1.08 × 105 Pa
(D) pressure at R, Q and inside bulb are same.
Solution: (D) Pressure at R = Atmosphere pressure = 1.013 × 105 Pa
Gauge pressure = nil
At Q, gauge pressure = gH
= (1.36 × 103) (9.8) (0.05) = 6.564 × 103 Pa
Absolute pressure = Pat+ gH
= (1.013 × 105) + (6.564 × 103) = 1.08 × 105 Pa
Pressure in bulb = 1.08 × 105 Pa.
19: Density of sea water is 1.03 × 103 kg m–3. A ship of weight 10.1 × 106 N floats on it. The ship then enters the fresh water and some cargo is unloaded. Then volume of sea water displaced ?
(A) volume of sea water displaced is 103 m3
(B) volume of sea water displaced 3 × 104 kg
(C) A is incorrect
(D) B is incorrect.
Solution: (A) Weight of ship = volume of sea water displaced
volume of sea water displaced, V =
=
Mass of volume V of fresh water
= 1 × 103 × 1 × 103 = 106 kg
Mass of cargo to be unloaded
= 1.03 × 106 –1 × 106 = 3 × 104 kg
20: Density of sea water is 1.03 × 103 kg m–3. A ship of weight 10.1 × 106 N floats on it. The ship then enters the fresh water and some cargo is unloaded. Then Mass of volume V of fresh water ?
(A) Mass of volume V of fresh water = 104 kg
(B) Mass of volume V of fresh water = 105 kg
(C) Mass of volume V of fresh water = 106 kg
(D) Mass of volume V of fresh water = 108 kg
Solution: (C) Weight of ship = volume of sea water displaced
volume of sea water displaced, V =
=
Mass of volume V of fresh water
= 1 × 103 × 1 × 103 = 106 kg
Mass of cargo to be unloaded
= 1.03 × 106 –1 × 106 = 3 × 104 kg
21: Density of sea water is 1.03 × 103 kg m–3. A ship of weight 10.1 × 106 N floats on it. The ship then enters the fresh water and some cargo is unloaded. Then Mass of cargo to be unloaded ?
(A) Mass of cargo to be unloaded = 4 × 104 kg
(B) Mass of cargo to be unloaded = 3 × 104 kg
(C) Mass of cargo to be unloaded = 6 × 104 kg
(D) Mass of cargo to be unloaded = 8 × 104 kg
Solution: (B) Weight of ship = volume of sea water displaced
volume of sea water displaced, V =
=
Mass of volume V of fresh water
= 1 × 103 × 1 × 103 = 106 kg
Mass of cargo to be unloaded
= 1.03 × 106 –1 × 106 = 3 × 104 kg
22: A 4.0m long copper rod of cross sectional area 1 cm2 is stretched by a force of 4.8103N. If Young’s modulus for copper is Y = 1.2 1011 N/m2. Calculate Stress.
(A) 2.0 107 N/m2 (B) 4.0 107 N/m2
(C) 1.0 107 N/m2 (D) 6.0 107 N/m2
Solution : (B) When a force F is applied at the cross-sectional area A of a wire, then
Stress = = 4.0 107 N/m2
23: A 4.0m long copper rod of cross sectional area 1 cm2 is stretched by a force of 4.8103N. If Young’s modulus for copper is Y = 1.2 1011 N/m2. Calculate strain.
(A) 1.310-4 (B) 2.310-4
(C) 3.310-4 (D) 4.310-4
Solution : (C) When a force F is applied at the cross-sectional area A of a wire, then
Stress = = 4.0 107 N/m2
Young’s modulus Y =
∴ strain = = 3.310-4
24: A 4.0m long copper rod of cross sectional area 1 cm2 is stretched by a force of 4.8103N. If Young’s modulus for copper is Y = 1.2 1011 N/m2. Calculate Increase length of the wire.
(A) 0.32 mm (B) 3.32 mm
(C) 4.32 mm (D) 1.32 mm
Solution: (D) When a force F is applied at the cross-sectional area A of a wire, then
Stress = = 4.0 107 N/m2
Young’s modulus Y =
∴ strain = = 3.310-4
Longitudinal strain =
∴ Increase in length = longitudinal strain initial length
= (3.3 10-4)4.0 =13.210-4m = 1.32mm
25. A beam of metal supported at two ends is loaded at the centre. The depression at the centre is proportional to Yn. The value of n is
(A) 1 (B) -1
(C) 2 (D) 4
Solution:(B)
26. When a certain weight is suspended from a long uniform wire, its length increases by one cm. If the same weight is suspended from another wire of the same material and length but having a diameter half of the first one, the increase in length will be
(A) 0.5 cm (B) 2 cm
(C) 4 cm (D) 8 cm
Solution:(C)
27. A force F is needed to break a copper wire having radius R. The force needed to break a copper wire of same length and radius 2R will be
(A) F/2 (B) 2F
(C) 4F (D) F/4
Solution:(C) Breaking force ∝ cross-sectional area.
F ∝ r2
When r is doubled, F increases four times.
28. A string 1 mm in diameter breaks if the tension in it exceeds 80 N. The maximum tension that may be given to a similar string of diameter 2 mm is
(A) 40 N (B) 80 N
(C) 320 N (D) 1640 N
Solution:(C) Breaking force = Breaking stress × area
F ∝ r2
When r is doubled, F increases four times.
29. A wire can sustain a weight of 10 kg before breaking. If the wire is cut into two equal parts, then each part can sustain a weight of
(A) 2.5 kg (B) 5 kg
(C) 10 kg (D) 15 kg
Solution:(C) Maximum load is not related to length.
30. Two rods of different materials having coefficients of linear expansion α1 and α2 and Young’s modulli, Y1 and Y2 respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of rods. If , then the thermal stresses developed in the two rods are equal, provided is equal to
(A) 2 : 3 (B) 1 : 1
(C) 3 : 2 (D) 4 : 9
Solution:(C) Thermal strees = Yαt
In the given problem,
31. Two wires of the same material and length are stretched by the same force. Their masses are in the ratio 3 : 2, Their elongations are in the ratio
(A) 3 : 2 (B) 9 : 4
(C) 2 : 3 (D) 4 : 9
Solution:(C)
32. A slightly concal wire of length L and radii r1 and r2 is stretched by two forces F, F applied parallel to length in opposite directions and normal to end faces. If Y denotes the Young’s modulus, then extension produced is
(A) (B)
(C) (D)
Solution:(D) Mean radius is .
33. The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of 30°. Then angle of shear is
(A) 12° (B) 0.12°
(C) 1.2° (D) 0.012°
Solution:(B)
34. Forces of 100 N each are applied in opposite directions on the upper and lower faces of a cube of side 20 cm. The upper face is shifted parallel to itself by 0.25 cm. If the side of the cube were 10 cm, then the displacement would be
(A) 0.25 cm (B) 0.5 cm
(C) 0.75 cm (D) 1 cm
Solution:(B)
If l is halved, then Δl is doubled.
35. A copper wire 2 m long is stretched by 1 mm. If the energy stored in the stretched wire is converted to heat, calculate the rise in temperature of the wire. (Given : dyne cm-2, density of copper = 9 g cm-3 and specific heat of copper = 0.1 cal g-1°C-1)
(A) 252°C (B) (1/252)°C
(C) 1000°C (D) 2000°C
Solution:(B)
36. A 1000 kg lift is tied with metallic wires of maximum safe stress of 1.4 × 108 N m-2. If the maximum acceleration of the lift is 1.2 m s-2, then the minimum diameter of the wire is
(A) 1 m (B) 0.1 m
(C) 0.01 m (D) 0.001 m
Solution:(C)
37. Two blocks of masses 1 kg and 2 kg are connected by a metal wire going over a smooth pulley as shown. The breaking stress of the metal is . If g = 10 m s-2, then what should be the minimum radius of the wire used if it is not to break?
(A) 0.5 mm (B) 1 mm
(C) 1.5 mm (D) 2 mm
Solution:(B)
If r is the minimum radius, then
Breaking stress
38. Given : σ is the compressibility of water, ρ is the density of water and K is the bulk modulus of water. What is the energy density of water at the bottom of a lake ‘h’ metre deep ?
(A) (B)
(C) (D)
Solution:(A) Energy density,
39. A body weighs 160 g in air, 130 g in water and 136 g in oil. The specific gravity of oil is
(A) 0.2 (B) 0.6
(C) 0.7 (D) 0.8
Solution:(D) Specific gravity of oil
40. A diver is 10 m below the surface of water. The approximate pressure experienced by the diver is
(A) 105 Pa (B) 2 × 105 Pa
(C) 3 × 105 Pa (D) 4 × 105 Pa
Solution:(B) 1 atmosphere = 105Pa
Also, p= hρg
= 10×1000×10Pa = 105Pa
So, total pressure is nearly 2 × 105Pa
41. A penguin floats first in a fluid of density ρ0, then in a fluid of density 0.95ρ0 and then in a fluid of density 1.1ρ0. Which of the following is correct ?
(A) maximum buoyant force in the fluid of density of 0.95ρ0
(B) maximum buoyant force in the case of fluid of density 1.1ρ0
(C) maximum fluid is displaced in the case of density 1.1ρ0
(D) maximum fluid is displaced in the case of density 0.95ρ0
Solution:(D) The buoyant force on the penguin is the same in all the cases. Maximum fluid is displaced in the case of least density.
42. A piece of brass (Cu and Zn) weighs 12.9 g in air. When completely immersed in water, it weighs 11.3 g. The relative densities of Cu and Zn are 8.9 and 7.1 respectively. The mass of copper in the alloy is
(A) 4.6 g (B) 5.6 g
(C) 7.6 g (D) 8.6 g
Solution:(C) Loss of weight = (12.9 – 11.3) gf = 1.6 gf
Weight of water displaced = 1.6 gf
If m is the mass of Cu, then
or 7.1m + 12.9 × 8.9 – 8.9m = 1.6 × 8.9 × 7.1
or 1.8m = 114.8 – 101.1
or m = 7.6 gram
43. A cubical block of wood 10 cm on a side floats at the interface between oil and water, as in Fig.3, with its lower face 2 cm below the interface. The density of the oil is 0.6 g cm-3. The mass of the block is
(A) 340 g (B) 680 g
(C) 80 g (D) 10 g
Solution:(B) mg = [100×2×1 + 100×8×0.6] g
∴ m = (200 + 480) g = 680 g
44. A tube 1 cm2 in cross-section is attached to the top of a vessel 1 cm high and of cross-section 100cm2. Water is poured into the system filling it to a depth of 100cm above the bottom of the vessel as shown in Fig.4. Take g = 10 m s-2. Now,
(A) The force exerted by the water against the bottom of the vessel is 100 N.
(B) The weight of water in the system is 1.99 N.
(C) Both (a) and (b) are correct.
(D) Neither (a) nor (b) is correct.
Solution:(c) P = 100 cm×1 g cm-3×1000 cm s-2 = 105 dyne cm-2
F = 105×100 dyne = 100 N
Again, V = 199 cm3
Weight = 199×1×1000 dyne = 1.99 N
45. An inverted vessel (bell) lying at the bottom of a lake, 47.6 m deep has 50 cm3 of air trapped in it. The bell is brought to the surface of the lake. If atmospheric pressure is 70 cm of Hg, the volume of trapped air when the vessel is brought to the surface of the lake is
(A) 100 cm3 (B) 200 cm3
(C) 300 cm3 (D) 500 cm3
Solution:(C)
46. When a loaded test tube floats vertically with of the lengths inside two liquids, then the ratio of the densities of the liquids is
(A) 3 : 4 (B) 4 : 3
(C) 9 : 16 (D) 16 : 9
Solution:(A)
47. A hollow cylinder of mass m made heavy at its bottom is floating vertically in water. It is tilted from its vertical position through an angle θ and is left. The restoring force acting on it is
(A) mg cos θ (B)
(C) (D)
Solution:(C) Let l be the length of the cylinder, when vertical, in water. Let A be the cross-sectional area of the cylinder. Equating weight of the cylinder with the upthrust, we get
When the cylinder is tilted through an angle θ, length of cylinder in water
Weight of water displaced
Restoring force
48. Energy needed in breaking a drop of radius R into n drops of radii r is given by
(A) 4πT (nr2-R2) (B) π (nr3-R2)
(C) 4πT (R2 - nr2) (D) 4πT (nr2+R2)
Solution:(A) Increase in surface area = n × 4 πr2- 4 πR2
Required energy is equal to the product of surface tension and increase in surface area.
49. The surface tension of a liquid is 5 N m-1. If a thin film is formed on a loop of area 0.02 m2, then its surface energy will be
(A) 5 × 10-2 J (B) 2.5 × 10-2 J
(C) 2 × 10-1 J (D) 3 ×
Solution:(C) Effective area = 2 × 0.02 m2 = 0.04 m2
Surface energy = 5 N m-1× 0.04 m2 = 2 × 10-1 J
50. Two soap bubbles, each with a radius r, coalesce in vacuum under isothermal conditions to form a bigger bubble of radius R. Then R is equal to
(A) 2-1/2 r (B) 21/3 r
(C) 21/2 r (D) 2 r
Solution:(C) Since conditions are isothermal, therefore, energy will be conserved.
2[2 × 4πr2σ] = 2 × 4πR2σ
or R2 = 2r2 or R = 21/2r
51. Two water drops, each of radius r coalesce to form a bigger drop of radius R. Then R is equal to
(A) (B)
(C) (D)
Solution:(B)
52. A disc of paper of radius R has a hole of radius r. It is floating on a liquid of surface tension T. The force of surface tension on the disc is
(A) T.2πR (B) T.2π (R + r)
(C) T.4π (R + r) (D) T.2π (R - r)
Solution:(B) Effective length = 2πr + 2πR
53. If FC and FA denote cohesive and adhesive force on a liquid molecule near the surface of a solid, then the surface of liquid is convex if
(A) (B)
(C) (D)
Solution:(C) or
54. Two soap bubbles, of radii 3 cm and 4 cm, coalesce in vacuum under isothermal conditions to form a bigger bubble of radius R. Then R is equal to
(A) 3 cm (B) 4 cm
(C) 5 cm (D) 7 cm
Solution:(C) 2 × 4π × 32 + 2 × 4π × 42 = 2 × 4π R2
or R2 = 9 + 16 = 25 or R = 5 cm
55. Water rises to a height of 10 cm in a glass capillary tube. If the area of cross-section of the tube is reduced to one-fourth of the former value, the water will rise to
(A) 20 cm (B) 5 cm
(C) 2.5 cm (D) 40 cm
Solution:(A) Area reduced to one-fourth. Radius reduced to one-half. Since , therefore, h is doubled.
56. A straw 6 cm long floats on water. The water film on one side has a surface tension of 50 dyne cm-1. On the other side, camphor reduces the surface tension to 40 dyne cm-1. The resultant force acting on the straw is
(A) (50 × 6 – 40 × 6) dyne (B) 10 dyne
(C) (D) 90 dyne
Solution:(A) F = (σ1 - σ2)l
57. A wire ring of diameter 14 cm is gently lowered on to a liquid surface and then pulled up. When the film just breaks, the force required is 0.0616 N. The surface tension of the liquid is
(A) 70 N m-1 (B) 7 N m-1
(C) 70 dyne cm-1 (D) None of these
Solution:(C)
58. When two soap bubbles of radii r1 and r2 (r2 > r1) coalesce, the radius of curvature of common surface is
(A) (B)
(C) (D)
Solution:(C)
59. If work W is done in blowing a bubble of radius R form a soap solution, then the work done in blowing a bubble of radius 2R form the same solution is
(A) W/2 (B) 2W
(C) 4W (D)
Solution:(C) W = 2 × 4πR2× σ; R is increased by a factor of 2. So, W is increased by a factor of 4.
60. Water rises to a height of 2 cm in a capillary tube. If the tube is tilted 60° from the vertical, water will rise in the tube to a length
(A) 4.0 cm (B) 2.0 cm
(C) 1.0 cm (D) water will not rise at all
Solution:(A) l cos 60° = 2 or l = 2 × 2 cm = 4 cm
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