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 σ* _{n}* =

●
The volume strain is ε_{v }= Δ*V* / *V*

●
Therefore, Bulk modulus is K = − σ* _{n}* / ε

●
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.

η_{R}
= Shearing stress / Angle of shear or shearing strain

●
The shearing stress is σ_{s} = tangential force / area over which it is
applied =* F _{t}* / Δ

● The angle of shear or shearing strain εs = *x* / *h* = θ

● Therefore, Rigidity modulus is ηR= σ / εs

● 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
= ʃ^{1}_{0 }*Fd*𝑙 ……….. (1)

●
From Young's modulus of elasticity.

Y
= *F/A* × *L*/𝑙

⇒ F =* YA*𝑙 / *L* ………….(2)

●
Substituting equation (2) in equation (1). We get W = ʃ^{1}_{0 }*YA*𝑙 / *L* *d*𝑙 = *YA*𝑙^{2}/ *L*.2
= 1/2 *F*𝑙

W
= 1/2* F*𝑙 =
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 h_{1
}and h_{2 }be the depths from the air-water interface to level 1
and level 2 of the cylinder, respectively as shown in Figure (a).

●
Let F_{1} be the force acting downwards on level 1 and F_{2} be
the force acting upwards on level 2.

●
Such that, F_{1 }= P_{1}A and F_{2} = P_{2}A . Let
us assume the mass of the sample to be m.

●
Under equilibrium condition the total upward force (F_{2}) is balanced
by the total downward force (F_{1 }+ mg). The gravitational force will
act downward which is being exactly balanced by the difference between the
force. F_{2} – F_{1}.

F_{2}
– F_{1}. = mg = F_{G}

●
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(h_{2} – h_{1}) Hence, gravitational force.

F_{G
}= ρA(h_{2} – h_{1}) g

●
On substituting the value of F_{G} in equation

F_{2}
= F_{1} + mg

=> P_{2}A = P_{1}A + ρA (h_{2}
– h_{1}) g

Cancelling
out A on both sides, P_{2} = P_{1} + ρ (h_{2} – h_{1})
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 h_{1} becomes zero (h_{1}=0) and in turn
P_{1} assumes the value of atmospheric pressure (P_{a}).

●
The pressure (P_{2}) at a depth becomes P. Substituting these values in
equation, we get P = P_{a} + ρgh

●
The pressure at a depth h is greater than the pressure on the surface on the
liquid, where P_{a} is the atmospheric pressure which is equal to 1.013
× 10^{5} P_{a}.

● 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 A_{1}
and A_{2} (A_{2} > A_{1}).

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

●
The pressure of the liquid under this piston increases to *P,* *where*
*P* = *F*_{1} / *A*_{1}

●
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 *F*_{2} = *P* × *A*_{2}
= [ *F*_{1}/*A*_{1 }] × *A*_{2 }⇒ *F*_{2} = [ *A*_{2}/*A*_{1}*
*] *× F*_{1}

●
Therefore by changing the force on the smaller piston A, the force on the
piston B has been increased by the factor *A*_{2}/*A*_{1}*
*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.

●
*F*_{G} = *mg* = 4/3 π*r*^{3} ρ*g*
(downward force) Up thrust, U = 4/3 π*r*^{3} σ*g* (*upward
force*) viscous force

●
At terminal velocity u,

●
downward force = upward force

*F*_{G }– U = *F*

⇒ 4/3 π*r*^{3} ρ*g*
– 4/3 π*r*^{3} σ*g* = 6πηrv_{t}

v_{t}
= [2/9] × [*r*^{2 }(ρ – σ) / η]*g *⇒ v_{t }∞ *r*^{2}

● 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∞η* ^{a}r^{b}*(

Where
k is a dimensionless constant.

Therefore,
[*v*] = *volume* / *time* = [ *L*^{3}*T*^{–1}],

[
*dP* / *dx* ] = *Presence* / *distance *= [*ML*^{–2}*T*^{–2}],

[η]=[*
ML*^{–1}*T*^{–1}] *and* [*r*] = [*L*]

Substituting
in equation (1) we get,

[L^{3}T^{−1}]
= [ML^{−1}T^{−1}]^{a} [L]^{b} [ML^{−2}T^{2}]^{c}

[M^{0}L^{3}T^{−1}]
= M^{a+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^{−1}*r*^{4} (*P*/*l*)^{1}

The
value of k = π / 8

∴ v = π*r*^{4}*P
*/ 8η*l* …………(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 F_{T} = 2πRT towards right

ii)
Force due to outside pressure F_{P1} = P_{1}πR^{2}
towards right

iii)
Force due to inside pressure F_{P2} = P_{2}πR^{2}
towards left

As
the drop is in equilibrium, F_{P2} = F_{T }+ Fp_{1}

P_{2}πR^{2}
= 2πRT + P_{1}πR^{2 }

=>
(P_{2} – P_{1} )πR^{2} = 2πRT

Excess
pressure is ∆P = P_{2} – P_{1 }= 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 F_{T} = 4πRT towards right

ii)
Force due to outside pressure F_{p1} = P_{1}πR^{2}
towards right

iii)
Force due to inside pressure Fp_{2} = P_{2}πR^{2}
towards left

As
the bubble is in equilibrium, F_{p2} = F_{T} + F_{p1}

P_{2}πR^{2}
= 4πRT + P_{1}πR^{2}

=>
(P_{2 }– P_{1} ) πR^{2} = 4πRT

Excess
pressure is ∆P = P_{2} − P_{1} = 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 P_{1} and P_{2} be the pressure outside and inside the air bubble, respectively. Now, the
excess pressure inside the air bubble is ∆P = P_{1} − P_{2}.
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 F_{T} = 2πRT

ii)
The force due to outside pressure P_{1} is to the right acting across a
cross sectional area of πR^{2} is Fp_{1 }= P_{1} πR^{2}

iii)
The force due to pressure P_{2} inside the bubble, acting to the left
is Fp_{2 }= P_{2} πR^{2}. As the air bubble is in
equilibrium under the action of these forces, Fp_{2 }= F_{T} +
Fp_{1}

P_{2}
πR^{2 }= 2πRT + P_{1} πR^{2 }

⇒ (P_{2 }- P_{1})
πR^{2} = 2πRT

Excess
pressure is ∆P = P_{2 }− P_{1} = 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 F_{T} 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
= π*r*^{2}*h* + (π*r*^{2}×*r* – 2/3 π*r*^{3}
)

⇒ V = πr^{2}*h* + 1/3
π*r*^{3}

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

●
Therefore,

2π*rT
cos*θ = π*r*^{2} ( *h*+1/3*r*) ρ*g*

⇒ *T* = [ *r*(*h* +
1/3 *r*) ρ*g* ] / 2*cos*θ

●
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* = *r*ρ*gh* / 2cosθ

*h* = 2*T *cosθ
/ *r*ρ*g *

⇒ *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 a_{1 }and a_{2}
such that a_{1} > a_{2 }.

●
A non-viscous and incompressible liquid flows steadily through the pipe, with
velocities v_{1} and v_{2} in area a_{1} and a_{2}
respectively.

●
Let m_{1} be the mass of fluid flowing through section A in time ∆t, m_{1}
= (a_{1}v_{1} ∆t)ρ

●
Let m_{2} be the mass of fluid flowing through section B in time ∆t, m_{2}
= (a_{2}v_{2} ∆t)ρ . For an incompressible liquid, mass is
conserved m_{1} = m_{2}

a_{1}v_{1}
∆tρ = a_{2}v_{2 }∆t ρ

a_{1}v_{1}
= a_{2}v_{2 }

=>
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
v^{2} + gh = constant, this is known as Bernoulli's equation.

**Proof:**

●
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 a_{A}, v_{A} and P_{A} 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 F_{A}= P_{A}a_{A}

●
Distance travelled by the liquid in time t is d = v_{A} t

Therefore,
the work done is W = F_{A}d = P_{A}a_{A} v_{A}
t

But,
a_{A}v_{A} t = a_{A}d = V, volume of the liquid
entering at A.

Thus,
the work done is the pressure energy (at A), W = F_{A}d = P_{A}V

Pressure
energy per unit volume at *A *= *Pressure energy* / *Volume* = *P _{A}V*
/

Pressure
energy per unit mass at *A* = *Pressure energy* / *Mass* = *P _{A}V*
/

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

*E _{PA}*

Potential
energy of the liquid at A, P_{EA }= mg h_{A},

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

*KE _{A
}*= 1/2

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

E_{A
}= E_{PA} + KE_{A} + PE_{A}

*E _{A}* =

Similarly,
let a_{B}, v_{B}, and P_{B} 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

E_{B}
= *m *P_{B}/ρ + 1/2 mv^{2}_{B} + mg* *h_{B}

From
the law of conservation of energy, E_{A} = E_{B}

*m* P_{A}/ρ
+ 1/2 mv^{2}_{A} + mgh_{A }= *m* P_{B}/ρ +
1/2 mv^{2}_{B} + mg h_{B}

P_{A}/ρ
+ 1/2 v^{2}_{A} + gh_{A }= P_{B}/ρ + 1/2 v^{2}_{B}
+ g h_{B} = constant

Thus,
the above equation can be written as

P/ρg
+ 1/2 v^{2}/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.**

**Venturimeter.**

●
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 P_{1} 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 'v_{1}' and
into the narrow region, its speed increases to 'v_{2}'

●
According to the Bernoulli's equation, this increase in speed is accompanied by
a decreases in the fluid pressure P_{2} 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=P_{1}−P_{2}) between the
surfaces of the monometer liquid.

●
From the equation of continuity, Av_{1} = a v_{2}

v_{2}
= [ *A*/*a* ] v_{1}

●
Using Bernoulli's equation, *P*_{1 }+ ρ (v_{1}^{2}/
2) = *P*_{2} + ρ(v_{2}^{2}/2) = *P*_{2}
+ ρ 1/2( (*A* / *a*)
v_{1})^{2}

●
From the above equation, the pressure difference, Δ*P* = *P*_{1}*
*- *P*_{2} = ρ v_{1}^{2}/ 2 ( [*A*^{2}
– *a*^{2}] / *a*^{2} )

●
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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