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Chapter: 11th Physics : UNIT 7 : Properties of Matter

Long Questions and Answer

Physics : Properties of Matter : Book Back Important Questions, Answers, Solutions : Long Questions and Answer

Properties of Matter (Physics)

Long Answer Questions

1. State Hooke’s law and verify it with the help of an experiment.

1) For a small deformation, within the elastic limit the stress and strain are proportional to each other.

2) It can be verified by stretching a thin straight wire (stretches like spring) of length L and uniform cross-sectional area A suspended from a fixed point O.

3) A pan and a pointer are attached at the free end of the wire as shown in Figure (a).

4) The extension produced on the wire is measured using a vernier scale arrangement. The experiment shows that for a given load, the corresponding stretching force is F and the elongation produced on the wire is ∆L. 

5) It is directly proportional to the original length L and inversely proportional to the area of cross section A. A graph is plotted using F on the X-axis and ∆L on the Y-axis.

6) This graph is a straight line passing through the origin as shown in Figure (b). Therefore.

∆L = (slope) F Multiplying and dividing by volume.

V = A L

F (slope) = [AL / AL] ∆L

Rearranging, we get, F / A = (L / A(slope) ) ∆L/ L

Therefore, F/A ( ∆L/L)

Comparing with stress equation, the stress is proportional to the strain in the elastic limit.

2. Explain the different types of modulus of elasticity.

There are three types of elastic modulus.

(a) Young's modulus, (b) Rigidity modulus (or shear modulus) (c) Bulk modulus.

Young's modulus:

● When a wire is stretched or compressed, then the ratio between tensile stress (or compressive stress) and tensile strain (or compressive strain) is defined as Young's modulus.

Y = Tensile stress or compressive stress / Tensile stress or compressive strain

● The unit for young modulus has the same unit of stress because, strain has no unit. So S.I. unit of Young modulus is N m-2 or pascal.

Bulk modulus:

● Bulk modulus is defined as the ratio of volume stress to the volume strain:

● Bulk modulus, K = Normal (Perpendicular) stress or pressure / Volume strain

● The normal stress or pressure is on σn = Fn / ΔA = Δp

● The volume strain is εv = ΔV / V

● Therefore, Bulk modulus is K = − σn / εv = − (Δp) / (ΔV/V)

● The negative sign in the equation means that when pressure is applied on the body, its volume decreases. A material can be easily compressed if it has a small value of bulk modulus.

The rigidity modulus or shear modulus:

● The rigidity modulus is defined as Rigidity modulus or shear modulus.

ηR = Shearing stress / Angle of shear or shearing strain

● The shearing stress is σs = tangential force / area over which it is applied = Ft / ΔA

● The angle of shear or shearing strain εx / h = θ

● Therefore, Rigidity modulus is ηR= σ / ε

● A material can be easily twisted if it has small value of rigidity modulus.

● Consider a wire, when it is twisted through an angle , a restoring torque is developed ……..τ α θ

● For a larger torque, wire will twist by a larger amount (angle of shear θ is large). Since, rigidity modulus is inversely proportional to angle of shear, the modulus of rigidity is small.

3. Derive an expression for the elastic energy stored per unit volume of a wire.

● When a body is stretched, work is done against the restoring force (internal force).

● This work done is stored in the body in the form of elastic energy.

● Consider a wire whose un-stretch length is L and area of cross section is A.

● Let a force produce an extension 𝑙 and further assume that the elastic limit of the wire has not been exceeded and there is no loss in energy.

● Then the work done by the force F is equal to the energy gained by the wire.

● The work done in stretching the wire by d𝑙, dW = F d𝑙

● The total work done in stretching the wire from 0 to 1 is

W = ʃ10 Fd𝑙           ……….. (1)

● From Young's modulus of elasticity.

Y = F/A × L/𝑙

F = YA𝑙 / L         ………….(2)

● Substituting equation (2) in equation (1). We get W = ʃ10 YA𝑙 / L d𝑙 = YA𝑙2/ L.2 = 1/2 F𝑙

W = 1/2  F𝑙 = Elastic potential energy.

Energy per unit volume is called energy density.

 u = Elastic potential energy / Volume

= 1/ 2 (Stress × Srain)

4. Derive an equation for the total pressure at a depth ‘h’ below the liquid surface.

● Consider a water sample of cross sectional area in the form of a cylinder. Let h1 and h2 be the depths from the air-water interface to level 1 and level 2 of the cylinder, respectively as shown in Figure (a).

● Let F1 be the force acting downwards on level 1 and F2 be the force acting upwards on level 2.

● Such that, F1 = P1A and F2 = P2A . Let us assume the mass of the sample to be m.

● Under equilibrium condition the total upward force (F2) is balanced by the total downward force (F1 + mg). The gravitational force will act downward which is being exactly balanced by the difference between the force. F2 – F1.

F2 – F1. = mg = FG

● Where m is the mass of the water available in the sample element. Let ρ be the density of the water then, the mass of water available in the sample element is m = ρV = ρA(h2 – h1) Hence, gravitational force.

FG = ρA(h2 – h1) g

● On substituting the value of FG in equation

F2 = F1 + mg

 => P2A = P1A + ρA (h2 – h1) g

Cancelling out A on both sides, P2 = P1 + ρ (h2 – h1) g

● If we choose the level 1 at the surface of the liquid (i.e., air-water interface) and the level 2 at a depth 'h' below the surface (as shown in Figure (b), then the value of h1 becomes zero (h1=0) and in turn P1 assumes the value of atmospheric pressure (Pa).

● The pressure (P2) at a depth becomes P. Substituting these values in equation, we get P = Pa + ρgh

● The pressure at a depth h is greater than the pressure on the surface on the liquid, where Pa is the atmospheric pressure which is equal to 1.013 × 105 Pa.

● If the atmospheric pressure is neglected or ignored then P = ρgh


5. State and prove Pascal’s law in fluids.

● Hydraulic lift which is used to lift a heavy load with a small force.

It is a force multiplier.

● It consists of two cylinders A and B connected to each other by a horizontal pipe, filled with a liquid.

● They are fitted with frictionless pistons of cross sectional areas A1 and A2 (A2 > A1).

● A downward force F is applied on the smaller piston.

● The pressure of the liquid under this piston increases to P, where P = F1 / A1

● But according to Pascal's law, this increased pressure P is transmitted undiminished in all directions.

● So a pressure is exerted on piston B.

● Upward force on piston B is F2 = P × A2 = [ F1/A1 ] × A2 F2 = [ A2/A1 ] × F1

● Therefore by changing the force on the smaller piston A, the force on the piston B has been increased by the factor A2/A1 and this factor is called the mechanical advantage of the lift.


6. State and prove Archimedes principle.

It states that when a body is partially or wholly immersed in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it and its up-thrust acts through the centre of gravity of the liquid displaced.

Up-thrust or buoyant force = weight of liquid displaced.

7. Derive the expression for the terminal velocity of a sphere moving in a high viscous fluid using stokes force.

● Expression for terminal velocity.

● Consider a sphere of radius r which falls freely through a highly viscous liquid of coefficient of viscosity η. Let the density of the material of the sphere be ρ and the density of the fluid be σ.

● Gravitational force acting on the sphere.

FG = mg = 4/3 πr3 ρg (downward force) Up thrust, U = 4/3 πr3 σg (upward force) viscous force

● At terminal velocity u,

● downward force = upward force

FG – U = F

4/3 πr3 ρg – 4/3 πr3 σg = 6πηrvt

vt = [2/9] × [r2 (ρ – σ) / η]g vt r2

● The terminal speed of the sphere is directly proportional to the square of its radius. If σ is greater than ρ. then the term (ρ - σ) becomes negative leading to a negative terminal velocity.


8. Derive Poiseuille’s formula for the volume of a liquid flowing per second through a pipe under streamlined flow.

● Consider a liquid flowing steadily through a horizontal capillary tube.

● Let v = ( V / t ) be the volume of the liquid flowing out per second through a capillary tube.

● It depends on (1) coefficient of viscosity (η) of the liquid. (2) radius of the tube (r), and (3) the pressure gradient (P / l).

● Then, v∞ηarb(P / l)c ; v = kηarb(P/l)c           ………..(1)

Where k is a dimensionless constant.

Therefore, [v] = volume / time = [ L3T–1],

[ dP / dx ] = Presence / distance = [ML–2T–2],

[η]=[ ML–1T–1] and [r] = [L]

Substituting in equation (1) we get,

[L3T−1] = [ML−1T−1]a [L]b [ML−2T2]c

[M0L3T−1] = Ma+c L−a+b−2c T−a−2c

So, equating the powers of M,L, and T on both sides, we get a+c=0, −a+b−2c = 3, and −a−2c = −1.

● Solving these equations, we get a = −1, b = 4 and c = 1.

● Therefore, equation becomes, v = kη−1r4 (P/l)1

The value of k = π / 8

v = πr4P / 8ηl               …………(2)

9. Obtain an expression for the excess of pressure inside a i) liquid drop ii) liquid bubble iii) air bubble.

i) Excess pressure inside the liquid drop

Consider a liquid drop of radius R and the surface tension of the liquid is T as shown in Figure.

The various forces acting on the liquid drop are,

i) Force due to surface tension FT = 2πRT towards right

ii) Force due to outside pressure FP1 = P1πR2 towards right

iii) Force due to inside pressure FP2 = P2πR2 towards left

As the drop is in equilibrium, FP2 = FT + Fp1

P2πR2 = 2πRT + P1πR2

=> (P2 – P1 )πR2 = 2πRT

Excess pressure is ∆P = P2 – P1 = 2T / R

ii) Excess pressure inside a soap bubble.

Consider a soap bubble of radius R and the surface tension of the soap bubble be T. A soap bubble has two liquid surfaces in contact with air, one inside the bubble and other outside the bubble.

Therefore the force on the soap bubble due to surface tension is 2 × 2πRT. The various forces acting on the soap bubble are,

i) Force due to surface tension FT = 4πRT towards right

ii) Force due to outside pressure Fp1 = P1πR2 towards right

iii) Force due to inside pressure Fp2 = P2πR2 towards left

As the bubble is in equilibrium, Fp2 = FT + Fp1

P2πR2 = 4πRT + P1πR2

=> (P2 – P1 ) πR2 = 4πRT

Excess pressure is ∆P = P2 − P1 = 4T / R

iii) Excess of pressure inside air bubble in a liquid.

Consider an air bubble of radius R inside a liquid having surface tension T as shown in Figure. Let P1 and P2 be the pressure outside and inside the air bubble, respectively. Now, the excess pressure inside the air bubble is ∆P = P1 − P2. Consider the forces acting on the air bubble are

i) The force due to surface tension acting towards right around the rim of length 2πR is FT = 2πRT

ii) The force due to outside pressure P1 is to the right acting across a cross sectional area of πR2 is Fp1 = P1 πR2

iii) The force due to pressure P2 inside the bubble, acting to the left is Fp2 = P2 πR2. As the air bubble is in equilibrium under the action of these forces, Fp2 = FT + Fp1

P2 πR2 = 2πRT + P1 πR2

(P2 - P1) πR2 = 2πRT

Excess pressure is ∆P = P2 − P1 = 2T / R

10. What is capillarity? Obtain an expression for the surface tension of a liquid by capillary rise method.

● Consider a capillary tube which is held vertically in a beaker containing water; the water rises in the capillary tube to a height due to surface tension.

● The surface tension force FT acts along the tangent at the point of contact downwards and its reaction force upwards.

● Surface tension T, is resolved into two components i) Horizontal component T sin θ and

ii) Vertical component T cos θ acting upwards, all along the whole circumference of the meniscus.

● Total upward force = (T cos θ)( 2πr) = 2πr T cos θ

● Where θ is the angle of contact; r is the radius of the tube.

● Let p be the density of water and h be the height to which the liquid rises inside the tube. Then,

(the volume of liquid column in the tube, V ) = (volume of the liquid column of radius r height h ) + (volume of liquid of radius r and height r-volume of the hemisphere of radius r )

V = πr2h + (πr2×r – 2/3 πr3 )

V = πr2h + 1/3 πr3

● The upward force supports the weight of the liquid column above the free surface.

● Therefore,

rT cosθ = πr2 ( h+1/3r) ρg

T = [ r(h + 1/3 r) ρg ] / 2cosθ

● If the capillary is a very fine tube of radius (i.e., radius is very small) then r/3 can be neglected when it is compared to the height h.

● Therefore, T = rρgh / 2cosθ

h = 2T cosθ / rρg

h α 1/r

● h is inversely proportional to the radius (r) of the tube. The smaller the radius of the tube greater will be the capillarity.

11. Obtain an equation of continuity for a flow of fluid on the basis of conservation of mass.

● Consider a pipe AB of varying cross sectional area a1 and a2 such that a1 > a2 .

● A non-viscous and incompressible liquid flows steadily through the pipe, with velocities v1 and v2 in area a1 and a2 respectively.

● Let m1 be the mass of fluid flowing through section A in time ∆t, m1 = (a1v1 ∆t)ρ

● Let m2 be the mass of fluid flowing through section B in time ∆t, m2 = (a2v2 ∆t)ρ . For an incompressible liquid, mass is conserved m1 = m2

a1v1 ∆tρ = a2v2 ∆t ρ

a1v1 = a2v2

=> a v = constant

● It is called the equation of continuity and it is a statement of conservation of mass in the flow of fluids. 

● In general, a v = constant, which means that the volume flux or flow rate remains constant throughout the pipe. In other words, the smaller the cross section, greater will be the velocity of the fluid.

12. State and prove Bernoulli’s theorem for a flow of incompressible, non-viscous, and streamlined flow of fluid.

Bernoulli's theorem:

According to Bernoulli's theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains a constant.

P/ρ + 1/2 v2 + gh = constant, this is known as Bernoulli's equation.


● Let us consider a flow of liquid through a pipe AB as shown in Figure.

● Let v be the volume of the liquid when it enters A in a time t which is equal to the volume of the liquid leaving B in the same time.

● Let aA, vA and PA be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at A respectively.

● Let the force exerted by the liquid at A is FA= PAaA

● Distance travelled by the liquid in time t is d = vA t

Therefore, the work done is W = FAd = PAaA vA t

But, aAvA t = aAd = V, volume of the liquid entering at A.

Thus, the work done is the pressure energy (at A), W = FAd = PAV

Pressure energy per unit volume at A = Pressure energy / Volume = PAV / V = PA

Pressure energy per unit mass at A = Pressure energy / Mass = PAV / m = PA / [m /V ] = PA / ρ

Since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is

EPA = PAV = PAV × (m / m) = m [PA / ρ]

Potential energy of the liquid at A, PEA = mg hA,

Due to the flow of liquid, the kinetic energy of the liquid at A,

KEA = 1/2 mv2A

Therefore, the total energy due to the flow of liquid at A,


EA =m PA/ρ + 1/2 mv2A + mghA

Similarly, let aB, vB, and PB be the area of cross section of the tube, velocity of the liquid, and pressure exerted by the liquid at B. Calculating the total energy at EB, we get

EB = m PB/ρ + 1/2 mv2B + mg hB

From the law of conservation of energy, EA = EB

m PA/ρ + 1/2 mv2A + mghA = m PB/ρ + 1/2 mv2B + mg hB

PA/ρ + 1/2 v2A + ghA = PB/ρ + 1/2 v2B + g hB = constant

Thus, the above equation can be written as

P/ρg + 1/2 v2/g + h = constant

13. Describe the construction and working of venturimeter and obtain an equation for the volume of liquid flowing per second through a wider entry of the tube.


● This device is used to measure the rate of flow (or say flow speed) of the incompressible fluid flowing through a pipe. It works on the principle of Bernoulli's theorem.

● Let P1 be the pressure of the fluid at the wider region of the tube A.

● The fluid of density 'ρ' flows from the pipe with speed 'v1' and into the narrow region, its speed increases to 'v2'

● According to the Bernoulli's equation, this increase in speed is accompanied by a decreases in the fluid pressure P2 at the narrow region of the tube B.

● Therefore, the pressure difference between the tubes A and B is noted by measuring the height difference (∆P=P1−P2) between the surfaces of the monometer liquid.

● From the equation of continuity, Av1 = a v2

v2 = [ A/a ] v1

● Using Bernoulli's equation, P1 + ρ (v12/ 2) = P2 + ρ(v22/2) = P2 + ρ 1/2( (A / a) v1)2

● From the above equation, the pressure difference, ΔP = P1 - P2 = ρ v12/ 2 ( [A2a2] / a2 )

● Thus, the speed of flow of fluid at the wide end of the tube A

● The volume of the liquid flowing out per second is

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