PracticalPhysicsQuestions & Answers. 1a) Lo=125.00cm^3 Diameter of measuring cylinder=3.21cm^3 S/N:1,2,3,4,5 m(g):50.0,100.0,150.0,200.0,250.0 Lo(cm^3):125.0,125.0,125.0,125.0,125.0 L(cm? 3):135.0,145.0,165.0,185.0,205.0 y=L-Lo(cm^3):10.0,20.0,40.0,60.0,80.0 V=0.785d^2y:80.89,161.77,323.55,485.32,647.10 1aviii) Slope=(250.0-150.0)/(647.10-323.55) =100.0/323.55 =0.309 1aix) -I ensured that parallax error is avoided -I ensured that mass is gently inserted into the cylinder to avoid spillage 1b) Archimede's principle stste that when a body is partially or wholly immersed into a fluid it experiences an upthrust which is equal to weight of the body 1bii) volume of the wood = 2.0 × 10^-5m^3 mass = 5.0 × 10^-3kg g = 10.0ms^-2 density of water = 1.0 × 10^3kgm^-3 volume of object = mass/density density = mass/volume density of the object = mass of the object/volume = 5.0×10^-3/2.0×10^-5 = 2.5×10^2kgm^3 upthrust = volume 0f the object × density of liquid × g =2.0×10^-5 × 1.5×10^2×10 = 0.3N 3) TABULATE READINGS R(N) - |2.00| |4.00| |6.00| |8.00| |10.00| 1(A) - |2.85| |2.21| |2.00| |1.75| |1.65| R^-1 - |0.5| |0.25| |0.17| |0.13| |0.10| i)i ensure zero error in the terminal used ii) i ensure clean terminals iii)i ensure all wires are tightened together 3b) Ohm's law states that current flow through a conductivity wire is directly proportional to its potential difference applied across its end provided that temperature remain constant Revisit after 40 seconds for Answers Alternative 2 1) S/N - |1|2|3|4|5| M(g) - |50.0|100.0|150.0|200.0|250.0| Lv(cm) - |50.0|50.0|50.0|50.0|50.0| L1(cm) - |60.0|70.0|80.0|90.0|100.0| y=Le-v(cm^3) - |10.00|20.00|30.00|40.00|50.00| 1aix) i. i avoided error due to parallel ii. i ensured that the mass is gently dropped into the liquid (1bi) Archimedes principle states that when a body is wholly or partly immersed in a liquid the weight of the body is equal to the mass of the fluid displaced 1bii) volume of the wood = 2.0 × 10^-5m^3 mass = 5.0 × 10^-3kg g = 10.0ms^-2 density of water = 1.0 × 10^3kgm^-3 volume of object = mass/density density = mass/volume density of the object = mass of the object/volume = 5.0×10^-3/2.0×10^-5 = 2.5×10^2kgm^3 upthrust = volume 0f the object × density of liquid × g =2.0×10^-5 × 1.5×10^2×10 = 0.3N =============================== CLICK HERE FOR NUMBER 3 GRAPH 3) TABULATE READINGS R(N) - |2.00| |4.00| |6.00| |8.00| |10.00| 1(A) - |2.85| |2.21| |2.00| |1.75| |1.65| R^-1 - |0.5| |0.25| |0.17| |0.13| |0.10| PRECAUTIONS i)i ensured clean terminals ii)i avioved zero error on the ammeter used iii)i ensured tight connection of all wires 3b) ohm's law states that the current flowing through a metallic conductor is proportional to its portential difference provided that physical condition is constant i.e V x I (3 bii ) E & L E =kL E / L = k E 1 / L 1 =E 2 / L 2 l 2 =E 2 L 1 / E 1 =(1 .5 * 1 )/ 2 . 5 L 2 =0 .6 cm the new balance point when 1 .5 v cell is connected across it is 0 .6 cm 3c) slope(s) = DR^-1/DI = 0.25-0.10/2.21-1.65 =0.15/0.56 slope(s)=0.27 intercept c = -0.35 evaluation 1/5=1/0.27=3.70 evalution 1/C = 1/-0.35 =-2.86