PHYSIC PRATICAL QUESTION AND ANSWERS FOR WAEC 2013

QUESTION 2: (a.) Measure and record the thickness of the glass block provided. Trace the outline ABCD of the glass block on the sheetof paper as shown. Remove the 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 other pins atpoint 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 axesfrom 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): 0.0216,0.0204,0.0188,0.0164,0.01 SinPdegree: 0.04226,0.5736,0.7071,0.8191,0.9 Sin^2Pdegree: 0.1786,0.3290,0.5000,0.6710,0.82 *************************** 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 4 in a table form. Therefore, S/N has 1,2,3,4,5 under it, in a table 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 aconstantan wire, a 2-ohm standard resistor, an accumulatorE, 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^-1means 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 4 in a table form. Therefore, S/N has 1,2,3,4,5 under it, in a table 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 beingtaken. (b.) I ensured that the terminals were clean.

CHEMISTRY WAEC QUESTION AND ANSWERS FOR 2013

1. VOLUMENTRIC ANALYSIS QUESTION 1. 'A' is a solution of tetraoxosulphate(vi). B is a solution coomtaining !.4g of potassiumb hydroxide per 25.0cm3. (a.) Put A into burette and titrate with 20.0cm3 or 25.0cm3 portion into of B using Methyl Orange as an indicator. Record the volume of your pipette. Tabulate your burette readings and calculate the average volume of acid used. (b.) From your result and information provided. Calculate the (i.) Concentration of B in mol/ dm3 (ii.) Concentration of A inmol/ dm3 (iii.) Number of Hydrogenions in 1.0dm^2 of A. The equation for the reaction is: H2SO4 + 2KOH --> k2SO4 + 2H2O. {H=1, O=16, K=39} SOLUTION Burrete reading(cm3)| Rough | 1st | 2nd | 3rd Final(cm3) | 18.60 | 18.00| 20.30| 28.00 Initial(cm3) | 0.00 | 0.00 |2.30 | 10.00 Vol. Acid used (cm3)| 18.60 |18.00| 18.00| 18.00 The average volume of acid used = 18.00 + 18.00 + 18.00/3 = 18.00cm3. (bi.) Mass of KOH per dm3 of B = 39+16+1= 56. Conc. of B = 5.6/56 = 0.01mol/ dm3. (bii.) The equation for the reaction is: H2SO4 + 2KOH --> k2SO4 + 2H2O. From the equation ==> Conc of A * Volume of A* Mole ratio = CA*18.00/0.10*2 = 1/2.'. CA= 0.10 * 25/2*18' CA= 0.0694mol/ dm3. (biii.) 1dm3 of 1.0mol/ dm3H2SO4 contains (2*6.0*10^23). Hydrogen atom ==> 1dm3 of 0.694mol/ dm3 H2SO4 contains 2*6.0*10^23 = 8.328x10^22.