ENGINEERING PHYSICS PRACTICAL Practical Manual Book N-Scheme (For Polytechnic Colleges) I & II SEMESTER AUTHOR: Dr P.M.PONNUSAMY M.Sc., M.Phil., Ph.D., SET, Lecturer in Physics, CIT Sandwich Polytechnic College, Coimbatore- 641 014 CO-AUTHORS: M.NITHYAKALYANI M.Sc., BEd.,M.Phil, Lecturer in Physics, CIT Sandwich Polytechnic College, Coimbatore- 641 014 S.PRABAHARAN M.Sc., M.Phil, Lecturer in Physics, CIT Sandwich Polytechnic College, Coimbatore- 641 014 1
SEMESTER- I ENGINEERING PHYSICS PRACTICAL CONTENTS Expt. Name of the Experiment Page No. No. 1 Screw Gauge ( Micrometer) 2 Vernier Caliper 3 Parallelogram law 4 Lami’s theorem 5 Comparison of Viscosities 6 Stoke’s Method 7 Sonometer 8 Deflection Magnetometer 2
Expt. Date INDEX Marks Remarks Faculty No. Awarded Signature Name of the Experiment (10) 3
Screw gauge: To find the Least Count (LC): Pitch = ������������������������������������������������ ������������������������������ ������������ ������������������ ������������������������������ ������������������������������ mm ������������������������������������ ������������ ������������������������������������������������������ ������������������������������ ������������ ������������������ ������������������������ ������������������������������ = mm = mm No. of Head Scale divisions = _____________ div. LC = ������������������������������ mm ������������������������������������ ������������ ������������������������������������������������������ ������������ ������������������������ ������������������������������ = mm LC = mm To find the Zero Error (ZE) and Zero Correction (ZC): ZE = div. ZC = div. 4
Expt. No. : Date : SCREW GAUGE (MICROMETER) Aim: To determine the thickness of the given irregular glass plate using Screw gauge. To measure the area of the glass plate using a graph sheet and to calculate the volume of the glass plate. Apparatus required: 1. Screw gauge 2. Given glass plate 3. Graph sheet Formula : Volume of the glass plate = Area x Thickness of the glass plate V = A x T m3 Where, A - Area of the glass plate (m2). T - Thickness of the glass plate (m). Procedure: I ) To find the least count(LC) of the Screw gauge: To find the least count, Pitch value can be calculated by using a known number of rotations given to head scale and finding the distance moved in the pitch scale. Then the number of divisions calibrated in the head scale is also noted (100 div). Pitch = ������������������������������������������������ ������������������������������ ������������ ������������������ ������������������������������ ������������������������������ ������������������������������������ ������������ ������������������������������������������������������ ������������������������������ ������������ ������������������ ������������������������ ������������������������������ LC= ������������������������������ ������������������������������������ ������������ ������������������������������������������������������ ������������ ������������������������ ������������������������������ II ) To find the Zero Error ( ZE) & Zero Correction (ZC): The threading screw is rotated to keep the studs A & B to touch each other with the observations of 2 or 3 click sound. If The horizontal Pitch scale zero is match with the zero of rotating Head scale, then there is no error and hence no zero correction (Fig.1). 5
To find the Thickness of the glass plate (T): ZE = div LC= mm ZC = div. S. No. PSR HSC CHSC = HSC +ZC CR= PSR +(CHSC x LC) mm div. div. mm 1 2 3 4 5 Average, T= mm 5 T= mm Calculation: mm2 = x 10-6 m2 Area of the glass plate, A = mm = x 10-3 m Thickness of the glass plate, T = x 10-3 m Volume of the glass plate, V = A x T m3 = x 10-6 m2 x V = x 10-9 m3 6
If the Head scale zero placed below the horizontal line of Pitch scale, then the error is positive(+ve) and the zero correction is negative (–ve) (fig.2). Example: ZE = +7 div. ZC = -7 div. If the Head scale zero placed above the horizontal line of Pitch scale, then the error is negative (–ve) and the zero correction is positive(+ve) (fig.3). Example: ZE = - (100-93) div. ZC = +7 div. Note: If the error is positive, then the correction is negative and vise versa. III ) To find the thickness (T) of the glass plate: The glass plate is placed between the studs A&B and gently tightens the threading screw to hear 2 or 3 click sound only. Then the pitch scale reading(PSR) and the head scale coincidence(HSC) division are noted. Correct Head Scale Coincidence (CHSC)= HSC + ZC. Correct Reading (CR) = PSR + (CHSC x LC). The experiment is repeated for 5 different positions of the glass plate and the readings are tabulated. The average of 5 calculated value gives the thickness of the glass plate. IV ) To find the area (A) of the given glass plate: The given glass plate is placed on the graph sheet and trace the boundary line. The number of smaller square units present inside the boundary line gives the area of the glass plate(A). The volume of the given glass plate is calculated by using the formula, Volume of the glass plate, V = A x T m3 Result: Volume of the glass plate, V = _________________ x 10 -9 m3 7
Vernier Caliper To find Least Count (LC): The value of 1 MSD = cm. div. No. of divisions in Vernier scale, n = LC = Value of 1 MSD cm = ������ cm LC = cm To find Zero Error (ZE) and Zero Correction (ZC): ZE = div. ZC = div. 8
Expt. No. : Date: VERNIER CALIPER Aim: To determine the length of the given solid cylinder using Vernier Calliper and to calculate the volume of the solid cylinder. Apparatus required: 1. Vernier Calliper 2. Given solid cylinder Formula : Volume of the solid cylinder , m3 V = π r 2l Where, r - radius of the solid cylinder ( m). l – length of the solid cylinder ( m). Description: The Vernier Calliper consists of main scale (MS) calibrated in cm and inches along the lower and upper sides of the main framerespectively. The lower jaws are used to measure the external dimension and the upper jaws are used to measure internal dimension of the object. The Vernier scale divisions are calibrated in movable jaw and can be fixed at any point using clamping screw S. Procedure: I ) To find the least count(LC) of the Vernier caliper: Least count is the least possible value that can be measured from the instrument. The value of 1 main scale division (MSD) and number of divisions (n) on the Vernier scale are noted. Then the LC of the instrument can be calculated by using the formula, LC = Value of 1 MSD cm ������ 9
To find the length of the solid cylinder (l): ZE = div. LC= cm ZC = div. S. No. MSR VSC CVSC = VSC +ZC CR= MSR +(CVSC x LC) cm div. div. cm 1 2 3 4 5 Average, l= cm 5 l= cm 10
II ) To find the Zero Error ( ZE) & Zero Correction (ZC): The instrument is to check the initial error. If the error exist, then to made suitable corrections. When the two jaws of Vernier Caliper is made to contact with each other and notice that the zero of main scale exactly coincides with the zero of vernier scale then the instrument has no error (Fig.1). If the Vernier scale zero is right of main scale zero then the instrument has positive error. Hence the correction is negative (Fig.2). If the Vernier scale zero is left of main scale zero then the instrument has negative error. Hence the correction is positive (Fig.3). To determine Correct Vernier Scale Coincidence (CVSC), note the coincidence of Vernier scale division with the main scale division (straight line image). Applying this Vernier scale coincidence in the given formula and calculate the value of CVSC. CVSC = VSC + ZE III ) To find the length (l) and radius of the solid cylinder (r): The given solid cylinder is placed in between the two lower jaws of Vernier caliper in such a way that the length of the solid cylinder is parallel to the main scale frame. The main scale reading (MSR) is noted by the position of Vernier scale zero lies in the main scale. If the Vernier zero is not exactly lies on the main scale division, then the previous division is considered as MSR. The any one of the Vernier scale division is exactly match with the main scale division and form a straight line image, then the Vernier scale division is taken as Vernier scale coincidence (VSC). The solid cylinder is rotated and take five set of readings. The corrected reading (CR) can be calculated using the formula, CR = MSR + (CVSC x LC) cm 11
To find the radius of the solid cylinder (r) ZE = div. LC= cm ZC = div. S. No. MSR VSC CVSC = VSC +ZC CR= MSR +(CVSC x LC) cm cm div. div. 1 2 3 4 5 Average, diameter = 5 cm Radius, r = cm r= 2 cm Calculation: cm = x 10-2 m Length of the solid cylinder, l = Radius of the solid cylinder, r = cm = x 10-2m Value of π = 3.14 Volume of the solid cylinder, V = π r 2l m3 V= m3 V= m3 V = x 10-6 m3 12
The average value of CR is taken as the length (l) of the given solid cylinder. The above same procedure is used to find the diameter of the solid cylinder. The half of the diameter value is taken as radius (r) of the solid cylinder. IV ) To find the volume of the solid cylinder: Using the value of length (l) and radius (r) of the solid cylinder, the volume of the given solid cylinder can be calculated by using the formula, Volume of the solid cylinder, V = π r 2l m3 Result: Volume of the solid cylinder, V = _________________ x 10 -6 m3 13
Parallelogram law of forces: 14
Expt. No. : Date: PARALLELOGRAM LAW Aim: To verify Parallelogram law of force Apparatus required: 1. Drawing board 2. Smooth pulleys with clamp 3. Three sets of slotted weights 4. Thread 5. Drawing paper 6. Scale 7. Protractor 8. Compass Formula : Parallelogram law of forces: Equilibrant (OC) = Resultant (OD) ∟COD = 180º Where, P, Q and R are the applied forces Description: A drawing board is supported vertically on a wall. Two smooth freely movable pulleys are fixed at the two top corners of the board. The two weight hangers P and Q are attached to the two end of a long thread passing through the pulleys. The third weight hanger R is attached to the middle of the thread as shown in the figure. 15
Tabulation for Parallelogram law of forces: S.No. Forces Sides x10-2 m ∟COD x 10-3 Kg OB OC OD P Q R OA 1 2 3 1) The equilibrant = The resultant i.e., OC = OD, Hence the diagonal OD represent the magnitude of the resultant. 2) ∟COD = 180º, Hence the diagonal OD represent the direction of the resultant. 16
Procedure: Three threads are knotted together at a point O and two of them are passed over two smooth frictionless pulleys attached to the top corners of vertical drawing board. The third thread is freely hanging downwards. The free ends of the threads carry weight P, Q and R. Three threads are knotted together at a point O and two of them are passed over two smooth frictionless pulleys attached to the top corners of vertical drawing board. The third thread is freely hanging downwards. The free ends of the threads carry weight P, Q and R. Parallelogram law of forces: Mark a point A and B such that OA is equal to OB. Using compass, draw a line AD parallel to OB and BD parallel to OA. The parallelogram ADBO is obtained. The value of resultant OD and the angle ∟COD is noted and tabulated. Result: Equilibrant = Resultant, i.e., OC = OD and ∟COD = 180º Hence Parallelogram law of forces is verified. 17
18
Expt. No. : Date: LAMI’S THEOREM Aim: To verify Lami’s theorem. Apparatus required: 1. Drawing board 2. Smooth pulleys with clamp 3. Three sets of slotted weights 4. Thread 5. Drawing paper 6. Scale 7. Protractor 8. Compass Formula : Lami’s theorem: ������ ������ ������ ������������������ ������ = ������������������������ = ������������������ ������ Where, P, Q and R are the applied forces α, β and γ are the angles between the forces Description: A drawing board is supported vertically on a wall. Two smooth freely movable pulleys are fixed at the two top corners of the board. The two weight hangers P and Q are attached to the two end of a long thread passing through the pulleys. The third weight hanger R is attached to the middle of the thread as shown in the figure. 19
Tabulation for Lami’s theorem: Forces Angles ������ ������ ������ x 10-3 Kg S.No. P Q R α β γ ������������������ ������ ������������������ ������ ������������������ ������ Calculation: ������ = sin ������ 1) ������ = ������ = sin ������ sin ������ ������ = ������ = ������������������ ������ sin ������ ������ = sin ������ 2) ������ = sin ������ ������ = ������������������ ������ ������ = sin ������ 20
Procedure: Three threads are knotted together at a point O and two of them are passed over two smooth frictionless pulleys attached to the top corners of vertical drawing board. The third thread is freely hanging downwards. The free ends of the threads carry weight P, Q and R. Three threads are knotted together at a point O and two of them are passed over two smooth frictionless pulleys attached to the top corners of vertical drawing board. The third thread is freely hanging downwards. The free ends of the threads carry weight P, Q and R. Lami’s theorem: Using protractor, the angle between QOR, POR and POQ are measured and noted as α, β and γ. The value of ������ , ������ and ������ are calculated and tabulated. In all cases the values of ������������������ ������ ������������������ ������ ������������������ ������ ������ , ������ and ������ are all equal. ������������������ ������ ������������������ ������ ������������������ ������ Lami’s theorem is verified by using the formula, ������ ������ ������ ������������������ ������ = ������������������������ = ������������������ ������ Result: ������ ������ ������ ������������������ ������ = ������������������������ = ������������������ ������ Hence Lami’s theorem is verified. 21
Diagram: 22
Expt. No. : Date: COMPARISON OF VISCOSITIES OF TWO LIQUIDS Aim: To compare the co-efficient of viscosity of the given two liquids. Apparatus required: 1. Capillary tube 2. Burette 3. Stand with clamp 4. Beaker 5. Given two liquids 6. Stop watch 7. Rubber tube with pinch cock 8.Funnel Formula : The ratio of co-efficient of viscosity of two liquids, ������������ = ������������ ������ ������������ ������������ ������������ ������������ Where, η1 – co-efficient of viscosity of first liquid ( Nsm-2). η2 – co-efficient of viscosity of second liquid ( Nsm-2). t1 – time taken for the flow of first liquid (s). t2 – time taken for the flow of second liquid (s). ρ1 –density of first liquid ( 1000 kgm-3 for water). ρ2 –density of second liquid (kgm-3). Procedure: A fine bore capillary tube is connected to the graduated burette through a long rubber tube and liquid flow is closed by pinch cock. The burette is fixed vertically on a stand and capillary tube is placed horizontally by using small stand. The given first liquid is fully loaded in the burette. Pinch cock is open and allows the liquid to flow through the capillary tube freely (Avoid control flow). 23
Tabulation First liquid = ____________________, Second liquid = ____________________, Burette Time taken to cross Burette Time of flow ������1 reading the level reading ������2 S.No. Range For first For second (ml) First Second liquid t1 liquid t2 (s) liquid liquid (ml) (s) (s) (s) 1 0 0 0 0-5 25 5-10 3 10 4 15 10-15 5 20 6 25 15-20 20-25 Mean , ������������ = = ������������ 5 Calculation: The density of first liquid, ρ1 = 1000 kgm-3 kgm-3 The density of second liquid, ρ2 = The mean value of ������������ = ___________ ������������ The ratio of co-efficient of viscosity of two liquids, x ������������ = ������������ ������ ������������ ������������ ������������ ������������ = = ������������ = ________________ ������������ 24
Switch on the stop watch and the time is noted for cross the liquid level 0ml, 5ml, 10ml up to 25ml. The time taken for the range (0-5)ml, (5-10)ml, (10-15)ml, (15-20)ml and (20-25)ml are calculated and the time of flow for the particular range is taken as (t1). The experiment is repeated for the second liquid and the time of flow is taken as (t2). Let ρ1 and ρ2 be density of two liquids. Then the ratio of co-efficient of viscosity of two liquids ( ������������ ) can be calculated by using the formula, ������������ ������������ = ������������ ������ ������������ ������������ ������������ ������������ Result: The ratio of co-efficient of viscosity of two liquids, ������������ = ________________ ������������ 25
Diagram: To find Least Count (LC): Pitch = ������������������������������������������������ ������������������������������ ������������ ������������������ ������������������������������ ������������������������������ mm ������������������������������������ ������������ ������������������������������������������������������ ������������������������������ ������������ ������������������ ������������������������ ������������������������������ = mm = mm No. of Head Scale divisions = _____________ div. LC= ������������������������������ mm ������������������������������������ ������������ ������������������������������������������������������ ������������ ������������������������ ������������������������������ = mm LC = mm 26
Expt. No. : Date: CO-EFFICIENT OF VISCOSITY- STOKE’S METHOD Aim: To find the co-efficient of viscosity of high viscous liquid (castor oil) using Stoke’s method. Apparatus required: 1. Measuring jar 2. Spherical balls of various radius 3.Castor oil 4. Screw gauge 5. Rubber band 6. Stop watch Formula : The co-efficient of viscosity of high viscous liquid, ������ = ������ (������ − ������)������ ������ ������������ ������ ������������������−������ ������������ Where, r – the radius of the spherical ball (m). h – distance travel by the ball of height “h” (m). t – time taken by the ball to travel the distance of height h (s). ρ –density of the ball ( kgm-3). σ –density of the castor oil (970 kgm-3). g –acceleration due to gravity (9.8 ms-2). Procedure: 1) To find the radius of the spherical ball ‘r’ : Using screw gauge, the radii of the given spherical balls are determined preciously and marked as r1, r2 and r3. 27
Tabulation To find the radii of the spherical balls: LC= mm , ZE = div. ZC = div. Balls S.No. PSR HSC CHSC = CR= PSR Mean Mean Radius mm div. HSC +ZC +(CHSC x LC) diameter mm div. mm mm 1 Ball 2 d1= r1 = 1 3 1 Ball 2 d2= r2 = 2 3 1 d3= r3 = Ball 2 3 3 28
2)To find the co-efficient of viscosity of high viscous liquid ‘η’: The given high viscous liquid in transparent form is taken in tall measuring jar and mark two positions ‘A’ just below the top level and position ‘B’ just above the bottom level. The distance between ‘ AB’ is taken as height (h). The first spherical ball (r1) is dropped gently on the liquid and note down the time (t) taken by the ball to cross the level A to B using stop watch. The experiment is repeated for the remaining balls r2 and r3 and the readings are tabulated. From the tabulation the r2t is calculated and found to be constant for all balls. The co-efficient of viscosity of high viscous liquid ‘η’ can be calculated by applying the constant value ρ, σ & g and the measured value height(h) and r2t in the given formula, ������ = ������ (������ − ������)������ ������ ������������ ������ ������������������−������ ������������ 29
2) To find the value of r2t: Balls Radius (r) r2 Time taken to cross r2t x10-3 m x 10-6 m2 the level AB x 10-6 m2s ts 1 r1 = 2 r2 = 3 r3 = Average, r2t = x 10-6 m2s r2t = ������ x 10-6 m2s Calculation: Distance between the level A and B, h = x 10-2 m The average value, r2t = x 10-6 m2s Density of the ball, ρ= kgm-3. Density of the castor oil, σ = 970 kgm-3. Acceleration due to gravity, g = 9.8 ms-2 30
Calculation: The co-efficient of viscosity of high viscous liquid, ������ = ������ (������ − ������)������ ������ ������������ ������ ������������������−������ ������������ = η = Nsm-2 Result: Nsm-2 The co-efficient of viscosity of high viscous liquid, η= 31
Diagram: Tabulation: 1) To find the ( ������ ) value: ������������ Load Vibrating length l2 ( ������ ) x10-4m ������������ S.No Ml Kg x10-2m Kgm-2 1 2 3 Average, ( ������ )= = Kgm-2 ������������ ������ 32
Expt. No. : Date: SONOMETER Aim: To determine the frequency of the given tuning fork by using sonometer. Apparatus required: 1. Sonometer 2. Tuning fork 3. Slotted weight hanger in terms of 0.5kg 4. Paper rider 5. Screw gauge 6. Rubber hammer Formula : The frequency of tuning fork, ������ = ������ √ ������ ������ ������������ ������ ������2 ������ Where, M –mass attached to the string(kg). m – mass per unit length of the string i.e., linear density (kgm-1). l – length of the vibrating string (m). g – acceleration due to gravity (9.8 ms-2). Description: Sonometer is an instrument used to measure the unknown frequency of tuning fork. It consists of a hollow wooden box about one metre length and a string is placed over the two knife edges A &B. One end of the string is tied at one end of sonometer and the other end of string having weight hanger passed over smooth movable pully. The vibrating length of string can be adjust the position of any one of knife edge. Procedure: 1) To find the ( ������ ) value: ������������ Sonometer set up is placed in a table such a way that the weight hanger (0.5 kg) is freely suspended. Three knife edges are placed above it. The knife edge A is placed very closed to the fixed 33
To find Linear density(m): Length of the string =_______________m. Mass of the string= __________ x10-3kg. ������������������−������ ������������������������ ������������ ������������������ ������������������������������������ ������������������������������������ ������������������������������������������, ������ = ������������������������������������ ������������ ������������������ ������������������������������������ = ------------------------ ������������������−������ m = ------------------------ ������������������−������ Calculation: kg m-1 Linear density of the string, m = The average value , ( ������ ) = Kgm-2 ������������ Acceleration due to gravity, g = 9.8 ms-2 The frequency of tuning fork, ������ = ������ √ ������ ������ ������������ ������2 ������ ������ n= Hz 34
end of a string and the knife edge B is near to movable pulley. The movable knife edge C is placed between two knife edges A &B. A paper rider is kept on the string in-between the knife edges A and C. The unknown frequency tuning fork is made to vibrate by hitting with rubber hammer and the stem of tuning fork is kept on the sonometer box immediately. The string starts to vibrate and by adjust the vibrating length of the string by moving the knife edge C. The string vibrates vigorously and the paper rider is fall down at resonant condition. At this resonance state, the frequency of the string is exactly equal to the frequency of the tuning fork. Then the distance between the knife edges A to C is noted as the vibrating length (AC = l) of the string. The experiment is repeated with 1 and 1.5 kg load in weight hanger. The readings are tabulated and the average value of ( ������ ) is calculated. ������2 2) To find mass per unit length of vibrating string ‘m’( Linear density) The mass of the string is measured by using digital balance and by knowing the length of the string (ρ) using metre scale, the linear density can be calculated by using the formula, Mass per unit length of vibrating string ( Linear density), m = Mass of the string/ length of the string kgm-1. By applying the constant value g and the measured value ( ������ ) and ‘m’ to the below formula, the ������2 frequency of tuning fork is calculated, ������ = ������ √ ������ ������ ������������ ������ ������2 ������ Result: The frequency of tuning fork, n = Hz 35
Diagram: 36
Expt. No. : DEFLECTION MAGNETOMETER Date: Aim: To compare the magnetic moments of the given two bar magnets using deflection magnetometer in Tan-A position by equal distance method. Apparatus required: 1. Deflection magnetometer with board 2. Two bar magnets Formula : The ratio of magnetic moments of the given two bar magnets by tan-A position is , ������������ = ������������������������������ ������������ ������������������������������ Where, M 1 –magnetic moments of the first bar magnet. M 2 –magnetic moments of the second bar magnet. ������������ – the mean deflection for first magnet. ������������ – the mean deflection for second magnet. Description: The deflection magnetometer consists of magnetic needle and long aluminium pointer pivoted over a circular mirror. The circular portion is divided into four quadrant of each 0° to 90°. This deflection magnetometer is placed on the centre of the wooden board having metre scale. The magnetometer is rotated in such a way that the arms of the board and the aluminium pointer face towards east-west direction and reads 0°-0°. This position is called Tan-A position. 37
Tabulation: Distance Name of Deflections in magnetometer ������������ = ������������������������������ θ3 θ4 θ5 ������6 θ7 θ8 S.No (d) the Mean ������������ ������������������������������ θ cm Magnet θ1 θ2 First θx Magnet (M1) 1 θy Second Magnet (M2) First θx Magnet θy (M1) 2 Second Magnet (M2) First θx Magnet θy (M1) 3 Second Magnet (M2) Mean, ������������ = ������ = ������������ 38
Procedure: Equal distance method: Keep the deflection magnetometer set up in Tan-A position. The first bar magnet(M1) is placed over the scale at a distance (d) on the eastern side of magnetometer and assure that the deflection is between 30° to 60°. The distance (d) is taken as centre of the magnetometer to the centre of bar magnet and the axis of magnet parallel to east west direction and perpendicular to the magnetic meridian (North-South direction). The bar magnet axis is pass through the centre of magnetometer. The deflection produced by the magnetic needle is shown by two ends of aluminium pointer and the readings (θ1&θ2) are noted in such a way that the aluminium pointer and the image of pointer in the mirror exactly coincide with each other. Then reverse the magnet in the same position and note the deflections (θ3&θ4). Again keep the same magnet in western side of magnetometer at the same distance (d) and noted the deflection (θ5&θ6). Then reverse the magnet position in the same position and note the deflections (θ7&θ8). The averages of eight deflections are taken as θx. By keeping the same distance (d), the experiment is repeated for the second magnet(M2). And the average deflections are taken as θy. Similarly the experiment is repeated for different distance (d) valus and the deflections are tabulated. The ratio of magnetic moments of the given two bar magnets can be calculated by using the formula, ������������ = ������������������������������ ������������ ������������������������������ 39
Calculation: The ratio of magnetic moments of the given two bar magnets by 1. ������������ = ������������������������������ = ������������������ ( )= ������������ ������������������������������ ������������������ ( ) 2. ������������ = ������������������������������ = ������������������ ( )= ������������ ������������������������������ ������������������ ( ) 3. ������������ = ������������������������������ = ������������������ ( )= ������������ ������������������������������ ������������������ ( ) Average, ������������ = ������ = ������������ 40
Result: The ratio of magnetic moments of the given two bar magnets by Equal distance method, ������������ = _____________________ ������������ 41
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SEMESTER- II ENGINEERING PHYSICS PRACTICAL CONTENTS Expt. Name of the Experiment Page No. No. 9 Refractive Index 10 Spectrometer 11 Solar Cell 12 Laws of Resistances 13 Joule’s Calorimeter 14 P-N Junction Diode 15 Copper Voltameter 16 Logic Gates 43
Reference Data 1. Density Iron = 7600kgm-3 Steel = 7800 kgm-3 Water = 1000 kgm-3 Kerosene = 830 kgm-3 Castrol oil = 970 kgm-3 2. Co-efficient of viscosity Castor oil = 452 to 986 x 10-3 Nsm-2 1.002 x 10-3 Nsm-2 Water = 2 x 10-3 Nsm-2 15.4 x 10-3 Nsm-2 Kerosene = Coconut oil = 3. Surface Tension Water = 72.75 x 10-3 Nm-1 Kerosene = 30 x 10-3 Nm-1 Coconut oil = 33 x 10-3 Nm-1 4. Refractive Index Water = 1.33 1.48 to 1.61 Crown glass = 1.53 to 1.96 Flint glass = 5. Specific Heat = 910 Jkg-1K-1 Aluminium = 373 Jkg-1K-1 Brass = 385 Jkg-1K-1 Copper = 4190 Jkg-1K-1 Water 6. ECE of copper = 3.3 x 10-7kgC-1 44
Expt. Date INDEX Marks Remarks Faculty No. Signature Name of the Experiment Awarded (10) 45
Diagram: To find Least Count (LC): The value of 1 MSD = cm. div. No. of divisions in Vernier scale, n = LC = Value of 1 MSD cm = ������ cm LC = cm 46
Expt. No. : Date : REFRACTIVE INDEX Aim: To determine the refractive index of the given transparent liquid(water) using travelling microscope. Apparatus required: 1. Travelling microscope 2. Beaker with water 3. Marked Paper 4. Coin Formula : ������������������������ ������������������������������ Refractive index of the liquid = ������������������������������������������������ ������������������������������ µ = ������������ − ������������ ������������ − ������������ Where,R1 - position of the coin placed at the bottom of the empty beaker view directly (m). R2– position of the apparent image of the coin view through water (m). R3– position of the level of water surface (m). Description: The travelling microscope consists of main scale (MS) graduated in cm and vernier scale (50 divisions). A microscope is placed in stand which moves in horizontal and vertical direction. The image of the object is focused sharply and the scale readings are noted by using magnifying lens. Procedure: A coin is placed at the bottom of empty beaker and kept below the microscope. The microscope is focused to view the sharp image of coin and the readings are noted as R1. Then add water to the beaker through particular height and focus the apparent image of the coin by adjusts the 47
Tabulation: LC= cm. CR= MSR + (VSC x LC) S.No. Coin in the Coin inside the Paper on the Real Apparent empty beaker liquid liquid surface depth depth R1 cm R2 cm R3 cm R3- R1 R3- R2 cm cm MSR VSC CR MSR VSC CR MSR VSC CR cm div cm cm div cm cm div cm 1 2 Calculation: Refractive index of the liquid, µ1 = ������������− ������������ = ������������− ������������ µ2 = ������������− ������������ = ������������− ������������ The mean value of µ = µ������+ µ������ = ������ 48
microscope height by back side screw and top fine screw. The readings are noted as R2. Drop saw dust or piece of paper over the surface of water and the surface of water can be focused and the readings are noted as R3. The experiment is repeated by changing the coin height and the refractive index of the given coin can be calculated by using the formula =Refractive index of the liquid, µ = ������������������������ ������������������������������ ������������− ������������ ������������������������������������������������ ������������������������������ ������������− ������������ Result: Volume of the solid cylinder, V = _________________ x 10 -6 m3 49
Diagram: To find Least Count (LC): The value of 1 MSD = No. of divisions in Vernier scale, n = div. LC = Value of 1 MSD = ������ LC = 50
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