LAB MANUAL – BASIC PHYSICS (3300005) Expt. -1: LINEAR DIMENSION BY VERNIER CALIPERS 1. 2. 3. 4. CONCLUSION: Volume of a rectangular block by vernier calipers is ____________ cm3. QUESTIONS: The smallest measurement that can be observed by the instrument is defined as least count. Bring the jaws A and C in contact with each other if the zero mark of the vernier does not coincide with the zero mark of the main scale the vernier callipers is said to have an error. Positive error: Bring the two jaws A and C in contact with each other if the zero mark of the vernier scale stands to the right of the zero mark of the main scale then error is called positive error. Negative error: If the zero mark of the vernier scale stands to the left of the zero mark of the main scale then error is called negative error. In order to increase the least count of the vernier one should increase the total number of division on vernier scale. ππππππππππ ππππππππππ = 5. M = 1 cm 20 ππππππππππππππππ ππππππππππππππππ π¨π¨π¨π¨ π¦π¦π¦π¦π¦π¦π¦π¦ ππππππππππ (ππ) ππππππππππ π§π§π§π§π§π§π§π§π§π§π§π§ π¨π¨π¨π¨ ππππππππππππππππ π¨π¨π¨π¨ π―π―π―π―π―π―π―π―π―π―π―π―π―π― π¬π¬π¬π¬π¬π¬π¬π¬π¬π¬ (ππ) = 0.05 cm , N = 50 ππππππππππ ππππππππππ = 6. (1) M= 1 mm , N= 25 (2) ππ = 1° 2 , N = 30 Divisions ππ. ππππ ππππ = ππ. ππππππ ππππ = ππ. ππππππ × ππππ π¦π¦π¦π¦ ππππ ππππππππππ ππππππππππ = ππ. ππππ π¦π¦π¦π¦ ππ π¦π¦π¦π¦ = ππ. ππππ π¦π¦π¦π¦ ππππ ππππππππππ ππππππππππ = ππ. ππππ π¦π¦π¦π¦ ππππππππππ ππππππππππ = ππ° ππ° π³π³π³π³π³π³π³π³π³π³ πͺπͺπͺπͺπͺπͺπͺπͺπͺπͺ = ππ = = ππ′ ππππ ππππ οΏ½β΅ ππ° = ππππ′ βΉ ππ° ππππ′ ππ° = βΉ = ππ′ οΏ½ ππππ ππππ ππππ Expt. – 2: LINEAR DIMENSION BY MICROMETER SCREW CONCLUSION: Thickness of the Rectangular block with micro meter Screw is ___________ mm. QUESTIONS: 1. This instrument has least count of the order of micro meter and this instrument is used to determine the gauge of the wire so, it is named as micro meter screw. 2. (i) Pitch of the screw is the distance between two consecutive threads of the screw. ππππππππβ (ππ) (ii) Least count = ππππππππππππ ππππ ππππππππππππππππ ππππ βππππππ π π π π π π π π π π (ππ) = 1 =0.01 mm 100 3. Positive error : If the zero of the head scale stands below the base line called positive error. Negative error: If the zero of the head scale stand above the base line then error is called negative error. 4. (1) 10-6 (2) 10-3 (3) 10-9 (4) 1010 Expt. – 3: RESISTANCE BY OHM’S LAW 1. 2. 3. 4. 5. 6. CONCLUSION: The value of given resistance is _______ β¦. QUESTIONS: It is defined as a flow of electron passing through a cross section of conductor in one second. (1) 1 Ohm: In given circuit if 1V potential difference is applied and current flowing through the circuit is 1A then it has 1ohm resistance. (2) 1Volt: If the work done in passing unit positive charge from one end of conductor to another end is 1 joule then potential difference is 1 volt. (3) 1 ampere: it is defined as a flow of one coulomb current passing through unit cross section of conductor. It is measured in Ampere. At constant temperature, the current flowing through circuit is directly proportion to applied voltage and inversely proportional to the resistance. In order to measure the current with ammeter, all current should pass through ammeter so it is connected in series. Voltmeter has infinite resistance, hence no current passing through it. In order to measure voltage it must be connected in parallel with resistance. In order to control the current and desired value of current, rheostat is connected with the circuit. Expt. – 4: RESISTANCE IN SERIES AND PARALLEL CONCLUSION OF SERIES CONNECTION: Hence The Low of Series Resistance is verified. 1. CONCLUSION OF PARALLEL CONNECTION: Hence The Low of Parallel Resistance is verified. When minimum power is required in circuit or purpose of saving Battery .i.e. Power supply, series connection is used. 2. When maximum power is required in circuit series connection is used. 3. For series connection R = R1 + R 2 + R 3 and For parallel connection 1 R = 8. For series connection R = R1 + R 2 + R 3 and For parallel connection 1 R 1 1 1 = R1 + R2 + R3 4. 5. 6. 7. I = I1 + I2 + I3 I = I1 = I2 = I3 V = V1 = V2 = V3 V = V1 + V2 + V3 1 R1 + 1 R2 + 1 R3 Expt. – 8: DIVERGENCE OF HE-NE LASER BEAM 1. 2. 3. 4. 5. CONCLUSION: If we increases distance from laser to screen, divergence of laser beam also increases. QUESTIONS: (1) In Medical (2) In Industry (3) Space Technology (4) In Research Field (1) Gas type Laser (2) Chemical Laser (3) Solid Laser (4) Semiconductor Laser (5) Bio Laser (6) XeF Laser In order to measure the diameter of laser beam accurately dark room is required. The range of divergence of a laser used in laboratory is 0.5 milli radian During the practical Direct eye contact with laser should be avoided. Expt – 9: MEASUREMENT OF A.C. POWER CONCLUSION: The calculated power is slightly greater than actual power. QUESTIONS: 1. It is a work done per unit time. Its unit is Joule/s or Watt 2. ππ = οΏ½3 ππ β πΌπΌ β cos ππ 3. ππ = ππ β πΌπΌ β cos ππ 4. (a) Parallel (b) Series (c) Parallel (d) Series Expt – 11: P-N JUNCTION IN FORWARD BIAS CONCLUSION: In Forward bias, if we increases the voltage current will be approximately zero up to threshold voltage after that current will increases exponentially. QUESTIONS: 1. Yes, we can use diode as a switch. 2. Silicone = 0.7 V Germanium = 0.3 V 3. P-N Junction diode : 4. (1) P-N Junction diode as a rectifier (2) Diode gates in digital electronics 5. It is a barrier with bound charges acting as potential barrier. 6. It is a slope of I-V characteristics of diode at particular operating point. It is called dynamic because it is based on increment in voltage to increment in current. 7. Zener break down. 8. (1) Diode Clamping (2) Limiters (3) Signal Processing (4) Diode gates. Expt-12: SURFACE AREA TO VOLUME RATIO 1. 2. 3. 4. 5. 6. CONCLUSION: The SA/V ratio for intact cube is __________ cm−1 while for cut cube it is ________cm−1 . QUESTIONS: It is engineering functional system at the molecular scale by using advance techniques to make complete high performance. SA/V changes linearly as size of the object changes into particularly shape smaller the size higher the SA/V ratio. SA/V ratio changes as texture of surface changes rougher the surface higher the SA/V ratio. Rusting, Burning, Decay, Surface tension and binding Strength. SA/V ratio changes as shape of objects changes for same volume flatter and elongated shape has more SA/V ratio. The amount of surface area determines the degree of reactivity of material so if we increase SA/V ratio then this additional surface area takes part in reactivity.