First semester diploma Engineering physics i

OPTC Edavanna Page 1

MODULE I

1. Differentiate between dimension, dimensional formula and dimensional equation

with a suitable example

Dimension of a physical quantity is powers to which the fundamental units of mass,

length and time must be raised to represent a derived unit of the quantity.

Dimensional formula is an expression which tells us which and how the physical

quantity depends on the fundamental units.

Dimensional equation is the equation obtained when a physical quantity is equated to its

Dimensional formula.

Eg: velocity:

Dimension is zero in mass (M), +1 in length (L) and -1 in time (T).

Dimensional formula is LT-1

Dimensional equation: V = [M°L1T

-1]

2. Give the physical quantities and their units in which SI is based on ?

Quantity Unit

Length meter

Mass kg

Time S

Electric current ampere

Temperature K

Amount of substance mol

Luminous intensity cd

Plane angle rad

3. Write the advantages of SI system over other systems of unit?

Universally accepted system

Comprehensive

Coherent system

Requires no separate practical units

Orphanage Polytechnic College, Edavanna REVISION PACKAGE

Engineering Physics-I ( Revision 2015)


4. Write the limitations of dimensional analysis?

Numerical constants, trigonometric ratios, angles etc appearing in a formula

cannot be deduced from dimensional method

This fails if a physical quantity depends on more than 3 others quantities

5. Define a Newton. Write dimensional formula for force ?

Newton is the SI unit of force. One Newton is that force will cause a mass of 1 kg to

have an acceleration 1m/s2.

Force = Mass X Acceleration.

Therefore dimensional formula for force is MLT-2

.

6. Write down the equation of motion of the body moving under gravity

V = u + at

S = ut +

t

2

V2 = U

2 + 2as

When a body is moving under gravity a = g

V = u + gt

S = ut +

t

2

V2 = U

2 + 2gs

7. Analyse the statement – Newton’s first law defines force and second law provides a

means to measure force

Newton’s first law states that a body continues to be in it state of rest or of uniform

motion along a straight line until and unless it is compelled by an external force. From the

statement , force can be defined as that which changes or tends to changes the state of

rest or that of uniform motion.

Second law tells what happens when a force is actually exerted on a body. When a force

acts on a body and there occurs a change in its momentum, then according to second law,

the rate of change of momentum is proportional to the applied force and in its direction


8. When a body is thrown up, show that time of ascent is equal to the time of descent

Let a body be projected vertically up with a velocity u. Let time taken to reach the

maximum height (time of ascent) be t1.

At the highest point, velocity is zero using v = u+ at, we get

0 = u – gt1 or t1 = u/g --- (A)

Let h be the maximum height reached. V2 = U

2 + 2as

0 = U2 – 2gh

H =

------------- (1)

Let t2 be the time of ascent for downward travel, initial velocity is zero

S = ut +

t

2

h = 0 +

t2

2

Substitute (1) in above equation

U2/2g =

t2

2

t22 = U

2 / g

2

t2 = u/g ------ (B)

Comparing (A) and (B), t1 = t2

ie, time of ascent = time of descent

9. What is impulse? Derive equation for impulse.

Large force acting on a body for short time is called impulse (I).

Impulse, I = F.t

F = ma = m (v – u)/t

I =

= m (v – u) = mv - mu

Means impulse of a force is measured by the change it produces in momentum.


10. State the law of conservation of momentum. Prove it in the case of collision of two

bodies moving in the same direction

Law of conservation of momentum states that when two or more bodies collide, the

sum of their momenta before impact is equal to the sum of momenta after

impact.

Momentum of m2 before Collision = m2u2

Momentum of m2 after Collision = m2v2

Changes of momentum in t seconds = m2v2 - m2u2

Rate of change of momentum m2 =

Change of momentum of first body in t seconds = m1v1=m1u1

Rate of change of momentum of the first body =

This rate of change of first body is the reaction. Since action and reaction are

equal and opposite.

= -

m1u1 + m2u2 = m1v1 + m2v2

i.e., total momentum before collision is equal to the total momentum after

collision.

11. State Newton’s laws of motion

First law: Everybody continues in its state of rest or of uniform motion along a straight

line unless it is compelled by an external force to change its state

Second law: The rate of change of momentum of a body is proportional to the applied

force and takes place in the direction in which the force acts.


Third law: Every action has an equal and opposite reaction

12. Prove Newton’s second law of motion and derive expression for force

Consider a body of mass moving with a velocity u. When an unbalanced force F acts

on it for a time t, its velocity changes to v.

So, its initial momentum = mu

Final momentum = mv

Therefore changes in momentum = mv-mu

Rate of change of momentum =

According to the second law , the rate of change of momentum is proportional to force

F. Thus,

F ma F=k ma , k = 1

F = ma

13. A gun weighting 10 kg fires a bullet of 30g with a velocity of 330m/s. With what

velocity does the gun recoil?

Mass of gun, M=10 kg

Mass of bullet, m=30 kg

Velocity of bullet, v = 330 m/s

Recoil velocity of gun, V =

= 9.90m/s

14. Derive the equation for the displacement of body during the nth

second of its motion

S1 = un +

an

2

S2 = u(n-1) +

(n-1)

2

Distance Sn covered in nth seconds, Sn = S1 – S2

= un +

an

2 – [ u(n-1) +

(n-1)

2 ]

= u + an -

a

= u + a(n -


15. Explain the term ‘Elastic fatigue’.

The property of an elastic body due to which its behaviour becomes less elastic under the

action of repeated alternating deforming forces

16. A body covers 120m in the 4thsecond.If it travels 240m in 8s ,calculate its acceleration,

initial velocity and velocity at the end of 8th second

S = u + a(n -

120 = u + a(4 -

S = ut +

at

2

240 = 8u +

x a x 8

2

The equations get simplified as 240 = 2u + 7a and 480 = 16u + 64a

Solving, we get a= -180 m/s2, u = 750 m/s

By 8 seconds, the distance covered by the body be S=ut+ ½ at2

S = 750 x 8 + ½ (-180) x 82 = 240 m

Using v2 = u2 + 2as = 7502 – 2 x 180 x 240 = 476100

Therefore, v = 690 m/s

I.e. by the end of 8 seconds, the body will have a velocity 690 m/s

17. A body moving with uniform acceleration describes 10m in the 2nd second and 20m

in the 40th second of its motion. Calculate the distance moved by it in the 5th second

of its motion

S = u + a (n -

10 = u + a (2 -

20 = u +

20 = 2u+3a --------- (1)

20 = u + a (4 -

20 = u +


40 = 2u+7a………….. (2)

Solving, a = 5m/s2, u = 2.5 m/s

S5 = 2.5 + 5 (5 -1/2 ) = 25m

18. Find the initial velocity and acceleration of a particle travelling with a uniform

acceleration in a straight line, if it traverses 55m in the 8th second and 85m in the

13th second of its motion.

S = u + a (n -

55 = u + a (8 -

20 = u +

110 = 2u + 15a ------ (1)

85 = u + a (13 -

85 = u +

170 = 2u + 25a ------ (2)

Solving, a = 6 cm/s2, u = 10 m/s

19. A stone is dropped into water from a bridge 44.1 m above the water level. Another

stone is thrown vertically downward one second later. Both stones reach water

surface simultaneously. Find the downward velocity of the second stone (g = 9.8

m/s2)

Time taken for the first stone to fall freely from the bridge from 44.1 m height will be t,

S = ut +

at

2

44.1 = 0 +

X 9.8 X t

2

t = √

= 3 s


20. A neutron having a mass of 1.67 x 10-27

kg and moving at 108 m/s collides with a

deuteron at rest and sticks to it . Calculate the speed of the combination (mass of

deutron=3.34*10-27kg)

m1u1 + m2u2 = m1v1 + m2v2

Here v1 = v2 = v , u2 = 0

Thus, m1u1+0=(m1+m2) v

Therefore v =

=

= 3.33x107m/s


MODULE II

1. What is mean by concurrent forces?

Forces whose lines of action pass through a common point

2. Explain analytical method to find the resultant of two forces?

OC2 = OD

2 + CD

2

R2 = (OA + AD)

2 + CD

2

= OA2 + 2.OA.AD + AD

2 + CD

2, But AD

2 + CD

2 = AC

2

R2 = OA

2 + 2.OA.AD + AC

2

Since OA = P and AC = OB = Q

R2 = OA

2 + 2.P.AD + Q

2

From right angled triangle ADC,

cos = AD/AC

AD = AC.cos = Q. cos

R2 = P

2 + 2.P.Q.cos + Q

2

Resultant, R = √

tan =

From triangle ADC,

CD = Q.sin


AD = Q.cos

tan =

=

3. What is mean by Lami’s theorem ?

If three forces acting on a body keep the body in equilibrium, each force is

proportional to the sine of the angle between the other two.

4. A body of mass 1 kg is suspended by two strings AC and BC of length 5 cm

and 12 cm respectively from two hooks separated by 13 cm. Calculate the

tension in the strings

Downward force,F = mg = 1 x 9.8 = 9.8 N

Applying Lami’s theorem, we get,


T1 = 9.8 sinӨ

T2 = 9.8 cosӨ

But, sinӨ = 12/13 and cosӨ = 5/13

Substituting above, we get, T1 = 9.05 N and T2 = 3.77 N

5. Derive expression for work done by a couple.

FF – couple acting on a body

Moment of the couple C, C = F.AB

W = F.AA1 + F.BB1

Ө =

AA1 = AO. Ө

BB1 = BO. Ө

W = F. AO. Ө + F. BO. Ө

= F (AO + BO ). Ө

= F.AB. Ө

= C Ө

6. What is mean by power ?

Work done per second is called power. P = 2 NC


MODULE III

1. Distinguish between stress and strain. Deduce the expression for young’s

modulus

Stress is the restoring force setup per unit area inside the body. It is measured by the

applied force per unit area.

Stress = F/A

Strain is the fractional deformation resulting from a stress. It is measured by the ratio

of change in dimension of a body to the original dimension.

Longitudinal strain =

young’s modulus is the ratio of longitudinal stress to the longitudinal strain

Y =

2. Write applications of Bernoulli’s principle

Atomiser

Air foil

3. Explain Hook’s law. Formulate the three elastic moduli.

Hooke’s law states that the strain produced in a body is directly proportional to the

stress provided the stress is not very large.

(1) Modulus of elasticity

= Constant.

This Constant is called Modulus of elasticity

Young’s modulus : It is ratio of the longitudinal stress to the longitudinal strain

Y =

=

(2) Modulus of Rigidity or Rigidity modulus

Ratio of shearing stress to shearing strain


Rigidity modulus,

=

(3) Bulk Modulus

Bulk modulus comes into play when a body is subjected to a uniform normal

force distributed over the whole of its surface. If V is original volume and ⱴ the

change in volume, bulk strain is ⱴ/V.

K

=

4. A cable is re placed by another of the same length and material but twice the

diameter. Analyze how it effects the elongation under a given load

Y =

=

Since both cables have same length, load and material, but diameter differs,

the eg for elongation can be given as,

l1 = FL / r12y)

l2 = FL / r22y)

Since r2 = 2r, l1/l2 = (2r1)2/r1

2 = 4

i.e. l1 = 4l2

Elongation becomes four times compared to the first cable

5. A brass wire of young’s modulus 90 GPa of length 3.14 and a diameter 0.4 mm is

stretched. Find the mass to be suspended to produce an elongation of 0.98 cm.

Y= 90 GPa = 90 X 109 Pa

L= 3.14 m

Diameter = 0.4 mm, Radius = 0.2 mm

L = 0.98 cm = 0.98 X 10-2

m.

Y =

F =

= = [90 X 10

9 x 0.98 X 10

-2 x (0.2x10-3)

2 ] / 3.14

F = mg = 35.298 N

Therefore mass to be suspended, m = 35.298/9.8 = 3.6 kg


6. Explain principle of continuity?

Relation representing conservation of mass

For a steady, non-viscous, non-uniform flow, the product of the area of cross

section and the velocity of flow remains the same at every point in the tube

Volume of liquid flowing per second at P = a1V1

Volume of liquid flowing per second at Q = a2V2

Volume entering at P = volume leaving at Q

a1V1 = a2V2 = a constant

This expression is called equation of continuity.


Module IV

1. Write examples of simple harmonic motion?

Simple pendulum

Loaded spring

U tube

Test tube floating in water

Freely suspended magnet

2. Explain characteristics of simple harmonic motion

Period (T): time taken for one complete vibration

Frequency (f): number of vibrations executed per second. F =

Amplitude (a): maximum displacement of particle in simple harmonic motion

Phase: stage of vibration through which the particle is passing

3. Define wave length?

It is the linear distance between any two nearest particles of the medium which are in

the same state of vibration.

4. Derive equation for simple harmonic motion?

The projection of a uniform circular motion along a diameter of circle is called simple

harmonic motion.

y = asin t

Differentiating,

= - 2 y [y = asinwt ]

This is the differential equation for simple harmonic motion

5. Derive the relation v = f

T - time period

– distance travelled by the wave

Velocity , v =

= (since f = 1/T )


6. What is mean by resonance?

A phenomenon occurring in vibratory bodies having the same natural

frequency

At resonance the body vibrates with very large amplitude

7. What are applications of ultrasonic waves

Sonar (echo sounding)

Ultrasonic signalling

Emulsification of immiscible liquids

Metal testing

Under water communication

Metallurgical uses

Medical and biological uses

Ultrasonic cleaners

8. Explain second mode of vibration? What is first overtone?

Two nodes and two anti-nodes

Fundamental frequency, f =

Frequency of vibration in this mode, f1 =

----- (1)

Vibrating length, L =

+

+

L =

1=

Substitute value of 1 in (1), we get,

f1 =

= 3 x

= 3f

This frequency is called first overtone or third harmonic


9. Calculate the velocity of sound using resonance column experiment.

A tuning fork is sounded on upper end of tube

Tube raised slowly until resonance occurs.

At this point, frequency of tuning fork = frequency of air column

Length of air column above water level is measured using a meter scale

( l1 + e ) =

-------------- (1)

= 4 ( l1 + e )

e – end correction = 0.3d

Velocity of sound in air, v = f = f x 4 ( l1 + e )

End correction can be eliminated if the second resonant length l2 is measured

( l2 + e ) =

-------------- (2)

(2) – (1) (l2 – l1) = or = 2 (l2 – l1)

= f = 2 f (l2 – l1)