7 April 2013

EXPECTED PHYSICS PRACTICAL SOLUTIONS (weac 2013)

QUESTION 2: (a.) Measure and record the thickness of the glass block provided. Trace the outline ABCD of the glass block on the sheet of paper as shown. Removethe block and draw Normal at N. Draw an incident ray such that the angle of incidence,i=25degree. Fix two pins at points P and Q on the incident ray. Replace the glass block and fix two otherpins at point R and Y such that the pins appear to be in a straight line with the images of the pins at P and Q when viewed through the side DC of the glass block. Remove the block and join the points at R and Y producing the line to meet DC and X. Join NX and measure its length L. Evaluate L^-2 and Sin^2i. Repeat the experiment for i=35degree, 45degree,55degree and 65degree. In each case determine the corresponding values of L, L^-2 and Sin^i. Tabulate your readings. Plot a graph of L^- on the vertical axis and Sin^2i on the horizontal axis, starting both axes from the origin. Determine the slope(s) of the graph and the intercept(I) on the vertical axis. Evaluate the expression: K=(I/S)^1/2. State two precautions taken to ensure accurate results. (i.) Using your graph deduce the value of L when i=0degree. (ii.) State Snell's law of refraction and explain why refraction occurs at the boundary between two media. ANSWER: Table of values/Observation: Width of glass block b=6.5cm. S/N: 1,2,3,4,5. P(degree): 25.00,35.00,45.00,55.00,65.00. L(cm): 6.80,7.00,7.30,7.80,8.20. L^2(cm^2): 46.24,49.00,53.29,60.84,67.24. L^-2(cm^-2): 0.0216,0.0204,0.0188,0.0164,0.0149. SinPdegree: 0.04226,0.5736,0.7071,0.8191,0.9063. Sin^2Pdegree: 0.1786,0.3290,0.5000,0.6710,0.8214. ***************************NOTE THE FOLLOWING: *Draw a table for the above values. *Note that ^ means Raise to Power. Eg: 2^-1 means Two raise to power minus One. *Note that COMMA(,) in the above table means NEXT LINE. Eg: 2,3,4,5 means dat 3 is under 2, 4 is under 3 and 5 is under 4in a table form. Therefore, S/N has 1,2,3,4,5 under it, in atable as shown in the table of values above. i.e 1 is under S/N and 2 is under 1 and 3 is under 2 and 4 is under 3 and so on..... **************************** Slope= DL^-2(cm^-2)/DSin^2i = 0.0136-0.0224/0.96-0.1 = 0.0088/0.86 = -0.01023(cm^-2) K=(1/5)^1/2 = (0.234/0.01023)^1/2 = 1.512cm, =1.5cm. Deduction intercept on vertical axis = 0.0234(cm2). PRECAUTIONS: (i.) I avoided error of parallax. (ii.) I ensured that the object pins and the image pins were erect and in a straight line. (bi) When i = OL^-2 = 0.0236cm^-2 = L^-2= 42.37cm^2. Hence L=6.5cm. QUESTION 3 - (ELECTRICITY): You are provided with a constantan wire,a 2-ohm standard resistor, an accumulator E, an ameter(A), a key(K) and other necessary apparatus. (i.) Measure and record the e.m.f of the accumulator provided. (ii.) Connect a circuit as shown in the diagram above. (iii.) Close the Key, read and record the ameter readings to when the crocodile clip is not in contact with the constantan wire. (iv.) Open the key with the key making contact with the wire, when L=90cm. Close the key. Read and record the ammeter reading I, Evaluate I^-1. (v.) Repeat the procedure for L=80,70,60 and 50cm. (vi.) In each case, read and record the ammeter reading and Evaluate I^-1(A^-1). Tabulate your readings. (vii.) Plot a graph, L on the vertical axis and I^-1 on the horizontal axis. (viii.) Determine the slope of the graph and its intercept,c on the vertical axis. (ix) Evaluate K = c/S. (x) Using your graph, determine the current(i) when L = 55cm. (xi.) State two precautions taken to ensure accurate result. SOLUTIONS: TABLE OF VALUES: S/N: 1,2,3,4,5. Io(A): 0.650,0.700,0.750,0.800,0.850. L(cm): 90.00,80.00,70.00,60.00,50.00. I^-1(A^- 1): 0.011,0.013,0.014,0.017,0.020. I(A): 0.70,0.75,0.80,0.85,0.90. **************************** NOTE THE FOLLOWING: *Draw a table for the above values. *Note that ^ means Raise to Power. Eg: 2^-1 means Two raise to power minus One. *Note that COMMA(,) in the above table means NEXT LINE. Eg: 2,3,4,5 means dat 3 is under 2, 4 is under 3 and 5 is under 4in a table form. Therefore, S/N has 1,2,3,4,5 under it, in atable as shown in the table of values above. i.e 1 is under S/N and 2 is under 1 and 3 is under 2 and 4 is under 3 and so on..... **************************** (viii.) Slope(S) = DI/DI^-1 = 0.65/0.24 = 27.1. Intercept (c) on the vertical axis = 0.43amps. (ix.) K= c/S = 0.43/27.1 = 0.016. (x.) The value of the current I when L = 55cm = 0.018cm. (xi.) Precautions: (a.) I made sure the key was removed when readings were not being taken. (b.) I ensured that the terminals were clean

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