ENGINEERING PHYSICS- I QP CODE: 199

SOLUTIONS WITH DETAILED ALLOCATION OF MARKS OCT/NOV-2019

Answer key

Q.No Part-A Marks

1.

Scalar quantity with Example:

Physical quantities which have magnitude only are called scalar quantities.

(e.g) Mass, Volume, Speed

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2. Principle of moments:

The principle of moment states that if a body is in equilibrium under the action of forces, the sum

of the clockwise moments equal to the sum of anti-clockwise moments.

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3. Co efficient of viscosity:

Coefficient of viscosity of a liquid is defined as the viscous force acting between two layers of a

liquid having unit area of layers and unit velocity gradient normal to the direction of flow of the liquid.

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4.

Trajectory:

The path described by the projectile from the point of projection to the point where the projectile

reaches the horizontal plane passing through the point of projection. 2

5.

Normal acceleration:

When a particle is moving along a circular path, the particle is accelerated towards the centre of

the circular path in a direction normal to the direction of its motion.

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6. Laws of conservation of angular momentum:

When there is no external torque acting on a rotating body, the angular momentum of that body

remains a constant. Iω = constant

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(or)

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7. Wavelength:

The wavelength is the distance between two consecutive particles of the medium which are in

the same state of vibration.

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8.

Magnetic moment:

The moment of the couple acting on a magnet when it is placed perpendicular to a uniform

magnetic field of unit intensity.

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Q.No Part-B

9. Dimensional formula:

(i) Force: MLT-2

(ii) Work: ML2T

-2

(iii) Power: ML2T

-3

(3x1=3)

10. Vector quantity can be resolved into two rectangular components:

The two component forces are mutually perpendicular, they are called rectangular

components. Let a force R act at a point O at an angle θ with x- axis. This force can be resolved

into two rectangular components along x and y axis. Two lines CA and CB are drawn perpendicular

to x and y axis as shown in the figure.

In the right angled triangle OAC,

OA Cos q =

OC

OA = OC cosθ = R cosθ

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Therefore the horizontal component of the force R is R cosθ. In the right angled triangle OBC,

OB Cos(90 - q) =

OC

OB Sin q =

OC

OB = OC sin θ = R sin θ

The vertical component of the force R is R sin θ.

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11.

(i) Young’s modulus: It is defined as the ratio of linear stress to linear

strain.

(ii) Bulk modulus: It is defined as the ratio of bulk stress to bulk strain.

(iii) Rigidity modulus: The ratio of the shearing stress to the shearing

strain.

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12. Streamline motion: (any 3 points)

i.Streamline flow will happen only when the driving pressure is small.

ii.The velocity of all particles in a line of flow is a constant.

iii.The lines of flow are parallel to the axis of the tube.

iv.The velocity of flow is always less than the critical velocity.

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13. Relation between linear velocity and angular velocity:

Consider a particle moving along a circular path of radius r. Initially the particle is at a point A. After

t second the particle reaches the point B. Let q be the angle swept by the radius vector in t-second.

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14. Uses of artificial satellites: (any 3 points)

The artificial satellites are launched for many purposes by different countries. The important uses of artificial satellite are

i.Collection of scientific data

ii.Weather monitoring Military

iii.Spying

iv.Remote sensing

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v.Communication purpose – the satellite receives microwaves and TV signals from the earth and

amplifies them and transmits them back to various stations on the earth.

15. Longitudinal wave:

If the particles of the medium vibrate parallel to the direction of propagation of the wave, the wave

is known as longitudinal wave.

Examples:

1. The propagation of sound in air

2. The propagation of sound in gas

3. The propagation of sound inside the liquid

The longitudinal waves travel in a medium in the form of compressions and rarefactions. The place

where the particles of the medium crowded together are called compressions and the places where the

particles spread out are called rarefactions. The particles of the medium simply move back and forth

about their mean position parallel to the direction of the propagation of the wave.

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16.

Noise pollution:

The unwanted sounds which are dumped into the atmosphere, thus producing a nuisance and adverse

effect on the health of human beings, are known as noise pollution. Noise pollution leads to severe health

problems by creating either physical or mental problems.

The types of sources of noise are generally classified into two types.

(i) Industrial noises and (ii) Non- industrial noises

The noises which are produced due to the movements of car, motor truck, motor cycle, aircraft,

machines, explosions, etc. are known as industrial noises. The noises which are produced due to the

loudspeaker, construction work, radio, etc. are known as non-industrial noises.

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Q.No Part-C

17

a)

Conventions to be followed in S.I units: (any 5 points)

1. When we write a unit in full, the first letter should not be in capital letter.

(eg) metre and not as Metre kilogram and

not as Kilogram

2. The symbols of unit in the name of scientist should be in upper case (capital) letter

(eg) N for newton, J for joule.

3. Only the singular form of the unit is to

be used.

(eg) km and not as kms

Kg and not as kgs

4. There should be no full stop at the end of the abbreviations.

(eg) mm and not as mm.

5. When temperature is expressed in Kelvin, the degree sign is omitted.

(eg ) 273K not as 2730K

6. Only accepted symbols should be used.

(eg) ampere is represented as 'A' and not 'amp' second is represented as 's'

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and not 'sec'

7. Use of the solidus or slash is to be avoided, but when used, not more than one solidus be

employed.

(eg) J/k.mol or Jk.mol-1 but not J/k/mol

J kg-1 K-1 but not J/kg/K

8. One letter space is always to be left between the number and the

symbol of the unit. (eg) 2.3 m and not as 2.3m

Kg m s-2 and not kg ms-2

9. The numerical value of any physical quantity should be expressed as p= a × 10m. Here 'a' is a

number between 1 and 10 and 'm' is the appropriate power of 10.

(eg) velocity of light c = 2.997 × 108 m s-1

17.b)

Experiment to verify the parallelogram law of forces:

A drawing board is fixed vertically on the wall. Two frictionless pulleys are fixed at the top corners

of the drawing board. Weight hangers P, Q and R are tied at the ends of the strings as shown in the

figure. The weights P, Q and R are adjusted suitably such that the point O is at rest. The point O and the

directions of the strings are marked on the paper. The points A, B and C are marked. OA, OB and OC

represent the forces P, Q and R respectively. The parallelogram OADB is drawn. Then the diagonal OD

is drawn. The length of the diagonal OD

and ∠COD are measured and tabulated.

Sl. No P Q R OA OB OC OD ∠COD

The experiment is repeated for different values of P, Q and R. It is found that OC = OD and

∠COD = 180º. Thus the parallelogram law of forces is verified experimentally.

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17.c)

Find the magnitude and direction of the resultant of two forces 60N and 80N acting at a point with

an angle 90o between them.

Given; P = 60N, Q= 80N, θ = 90o

Formula : R = √ P2 + 2PQ cosθ + Q

2

α = tan-1

Q sinθ

P+Q cos θ

R = √ (60)2 + 2x60x80x cos 90

o + (80)

2

= √10000

R = 100N

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α = tan-1

80 sin 90

o

60 + 80 cos 90o

α = tan-1

80 = tan

-1 ( 1.333)

60

α = 53o 7 ’

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18.

a) Experimental determination of the coefficient of viscosity of a high viscous liquid by

Stokes' method:

Procedure:

The pure, transparent, highly viscous liquid (castor oil) is taken in the tall glass jar. On the outer

surface of the jar, two markings B and C or C and D are made at a distance of say 50 cm. Then only

the sphere can attain terminal velocity when travelling from B to C or C to D. Then using the screw

gauge, the diameters of the given glass beads are measured. The radii of the beads are calculated.

Values of r and s are taken from standard tables.

The sphere is placed gently on the surface of the highly viscous liquid and dropped. When it crosses

the marking B, a stop clock is started and when it crosses the marking C, the clock is stopped. The time

taken (t) and the distance between B and C is noted as 'h'. The same procedure is repeated for all the

given spheres. Then the coefficient of viscosity of highly viscous liquid (castor oil) is calculated, using

the formula.

Glass beads Radius (r) Radius2 (r2) Time taken (t) r2t Unit mm mm2 s mm2s

Bead 1

Bead 2

The average of r2t = mm2s

= × 10-6 m2s

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18.

b) Expression for the surface tension by capillary rise:

Let a capillary tube of uniform radius (r) be dipped vertically into a liquid of density ρ taken in a

beaker. Let the liquid rises through a height (h) in the capillary tube. Let θ be the angle of contact with

glass and T be the surface tension of the liquid. The surface tension is acting tangential to the liquid

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surface. The reaction R=T, offered by the walls of tube acts at an angle θ to the verticals. This reaction

T along AB can be resolved vertically as Tcosθ and horizontally as Tsinθ. The horizontal components

along the circumference are equal and opposite in direction. So they get cancelled. So the total force

due to surface tension is only due to vertical components.

The total upward force acting through the length of the meniscus 2πr

= 2 π r. T cosθ

This force balances the weight of the liquid column of height h and radius r.

The weight of the liquid column = π r2h ρ g

2πrTcosθ = πr²hρg T = rhρg/2 cos θ

In the case of water θ = 0° and hence cosθ = 1

Hence for water, Surface tension T = rhρg/2

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18.

c)

The length of a wire increases from 1.25 m to 1.2508 m when a load of 12 kg is suspended.

The radius of the wire is 0.5 mm. Find the stress, strain and young's modulus of material of the

wire.

Given :

L = 1.25m, l = 1.2508 – 1.25 = 0.0008m, M = 12kg, r = 0.5mm = 0.5x10-3

m

Formula: Stress = F / A Nm-2

Strain = l / L

Young’s modulus = Stress / Strain Nm-2

Stress = F / A = (Mg) / (π r2)

= (12 x 9.8) / (3.14 x (0.5 x 10-3

)2) = 117.6 x 10

6 / 0.785

Strain = l /L = 0.0008 / 1.25

Young’s Modulus = Stress / Strain

= 149.8 x 106 / 6.4 x 10

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19.

a) Expressions for maximum height, time of flight:

Consider a body be projected upwards from the point of projection O with the velocity u. Let α be

the angle of projection. The velocity u can be resolved into two components. Let u sinα and u cosα are

Stress = 149.8 x 106 Nm

-2

Strain = 6.4 x 10-4

E = 23.4 x 1010

Nm-2

the vertical and horizontal components of the velocity.

Maximum height reached by the projectile:

Let H be the maximum height reached by the projectile. At the highest point the vertical component

of velocity becomes zero. Considering the vertical motion of the projectile,

Taking the equation of motion

v2 = u2 + 2as

Final velocity at A, v = 0

Initial velocity at O, along vertical direction u = u sin α

Acceleration a = -g

Vertical displacement AC, s= H

Substituting these values in equation (1)

0 = u2 sin2 α + 2(-g) H

0 = u2sin2 α - 2 gH

2gH = u2sin2 α

∴Maximum height, H= u2sin2 α/2g

Time of Flight of the Projectile:

The time taken by the projectile to describe the path is called time of flight. i.e., It is the time taken

by the projectile to travel from 'O' to 'B', via A.

Let T be the time of flight. Here, the vertical displacement at B is zero.

Considering the vertical motion, Taking the equation of motion

S= ut+1/2 at2

Vertical displacement S = 0

Initial velocity at O, along vertical direction u = u sin α

Acceleration a = -g ; Time t = T; Substituting these values in equation

0 = u sinα T+1/2(-g)T2

1/2gT2 = u sinα T

∴Time of flight, T= 2u sinα/g

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19.

b) Expressions for Normal acceleration and Centripetal force:

Consider a particle of mass 'm' moving along a circular path of radius r. Let v be the velocity of the

particle. Initially the particle is at a point A. After t second the particle reaches the point B. Let q be the

angle covered by the radius vector. Let B be very close to A, so that the angle covered q is small.

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19.

c)

An electric train has to travel on a railway track with a curve of radius 120 m with a speed of

36 kmph. Calculate the angle of banking of the rails.

Given v = 36 kmph = (36 ×1000)/3600

v =10 ms

-1, r = 120 m , g = 9.8 ms-2

formula: tanθ = v

2/rg

θ = tan-1

(v2/rg)

θ = tan

-1( (10)

2/120 x 9.8)

θ= tan-1

(0.085) = 40 51'

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20.

a) Expression for Kinetic Energy of a Rigid body rotating about an axis:

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20.

b)

Expression for orbital velocity of a satellite :

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20.

c) The escape velocity of the surface of the earth is 11.2 kms-1. Find the radius of the earth.

Given: Ve = 11.2 kms-1

=> 11200ms-1

, g = 9.8 ms-2

Formula: Ve = √2gR

Ve2 = 2gR

R =

Ve2

=

(11200)2

2g 2 x 9.8

R = 125440000

19.6

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21.

a) Experimental determination of frequency of a tuning fork using sonometer:

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R = 6400km (or) 6400 x 103 m

The sonometer consists of a hollow wooden box. A nail is fixed at one end and a smooth pulley is

fixed at the other end of the box. One end of a sonometer string is tied to the nail and other end of the

string passes over the smooth pulley. The free end is attached to the weight hanger. A, B and C are three

knife edges placed on the box under the string. A, B are fixed and C is movable.

A suitable tension (T = Mg) is applied to the string. A small paper rider is placed on the string in

between A and C. The tuning fork of frequency ‘n’ is kept on the sonometer box, the string is made to

vibrate. The movable knife-edge C is adjusted and the vibrating length of the string AC= l is

measured. Let r be the radius of the string measured by using screw gauge and 'ρ' be the density of the

materialof the wire, then the linear density, m =πr2⍴

The experiment is repeated by changing the value of tension (load). The readings are tabulated.

S.No Load M kg Vibrating length l

cm

l2 cm

2 M/l

2

Mean :

2

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21.

b) Acoustics of Buildings (Architectural acoustics):

The term 'Acoustics of building' describe the production, transmission and reception of sound inside the building. The following points should be taken into consideration for the acoustic design of the buildings, auditorium, cinema theatres etc.

1. The sound heard by the audience should be sufficiently loud in any part of the hall.

2. The quality of the speech and music should not be changed any where inside the hall.

3. There should not be focusing of sound due to walls and ceiling, in any part of the hall.

4. There should not be any vibrations due to resonance.

5. There should not be any other noise from other sources, both from outside the hall and from

with in the hall.

Two more important factors should also be considered for good acoustic design of the buildings.

1. Echo , 2. Reverberation

1. Echo :

The direct sound from the source and the reflected sound (echo) from the walls produce confusion

in certain buildings. A hall with large number of open windows is free from the defect. Echoes can be

eliminated by making the walls rough. But in the case of musical hall, however echoes are desirable, to

a certain extent.

2. Reverberation :

The Reverberation is the prolonged reflections of sound from the walls, floor and ceiling of a hall.

As a result the clarity of the successive speech is affected. Due to this reason, the reverberation time

should be optimum to hear the clear sound in the hall.To minimize this defect, Sabine derived an

equation for the reverberation time,

Where V is the volume of the hall, α is the coefficient of absorption of each reflecting surface present

in the hall and A is the area of the each sound absorbing surface present in the hall.

T= 0.16 V/ A α

In the above Sabine's equation, T will be minimized by increasing the value of α. So the

reverberation is minimized by covering the inner surfaces of the hall using absorption materials like

carpets, wall screen, porous tiles etc.

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21.

c)

The length ,breadth and thickness of a bar magnet are 30cm, 2cm and 1 cm respectively.

calculate the intensity of magnetisation ,if its magnetic moments is 6×10-6 Am2.

Given: l =30 cm or 0.30 m ,b = 2 cm or 0.02 m and h = 1 cm or 0.01 m

m = 6×10-6 Am2

Formula:

Volume = lbh

Intensity of magnetization(M) = m/v

Volume = lbh = 0.30 × 0.02 × 0.01 =0.00006

Intensity of magnetization(M) = m/v = 6×10-6 / 0.00006

M = 0.1 Am-1

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V.KAILASAPATHY M.Sc., M.Phil. B. Ed.

LECTURER- PHYSICS

215, SAKTHI POLYTECHNIC COLLEGE,

SAKTHI NAGAR-638315