WAEC SSCE - Physics - 2000 (Practical)

(a) Determination of the relative density of liquid L by upthrust

The object is hung from the spring balance and its weight in air \(W_1\) is read. It is then fully immersed in water and the reading \(W_2\) taken, and finally fully immersed in liquid L and the reading \(W_3\) taken. The upthrust in water is \(U=(W_1-W_2)\) and the upthrust in L is \(V=(W_1-W_3)\). The procedure is repeated for the five masses.

The diagram of the apparatus is shown below.

Table of readings

M /g	\(W_1\) /g	\(W_2\) /g	\(W_3\) /g	\(U=(W_1-W_2)\) /g	\(V=(W_1-W_3)\) /g
5.0	5.00	4.00	4.20	1.00	0.80
10.0	10.00	8.00	8.40	2.00	1.60
15.0	15.00	12.00	12.60	3.00	2.40
20.0	20.00	16.00	16.80	4.00	3.20
25.0	25.00	20.00	21.00	5.00	4.00

Graph of \(V=(W_1-W_3)\) against \(U=(W_1-W_2)\)

Slope of the graph

Taking two points on the line of best fit, \((U_1,V_1)=(1.00,\,0.80)\) and \((U_2,V_2)=(5.00,\,4.00)\):

\[ s=\frac{V_2-V_1}{U_2-U_1}=\frac{4.00-0.80}{5.00-1.00}=\frac{3.20}{4.00}=0.80 \]

Since the slope \(s=\dfrac{W_1-W_3}{W_1-W_2}=\dfrac{\text{upthrust in L}}{\text{upthrust in water}}\), the slope is the relative density of liquid L:

\[ \boxed{s=0.80} \]

(vii) Two precautions

• I ensured the object was completely immersed in the liquid and did not touch the sides or bottom of the beaker.
• I read the spring balance at eye level with the object at rest to avoid parallax error, and I noted and corrected the zero error of the balance.

(b)(i) Reading of the spring balance for the brass

Weight of the brass in air:

\[ W=mg=20\times10^{-3}\times10=0.2\ \text{N} \]

Volume of the brass:

\[ V=\frac{m}{\rho_{\text{brass}}}=\frac{20\times10^{-3}}{8.0\times10^{3}}=2.5\times10^{-6}\ \text{m}^3 \]

Upthrust = weight of kerosine displaced:

\[ U=\rho_{\text{k}}\,V\,g=8.0\times10^{2}\times2.5\times10^{-6}\times10=0.02\ \text{N} \]

Reading on the spring balance = tension in the spring = weight in air \(-\) upthrust:

\[ =0.2-0.02=\boxed{0.18\ \text{N}} \]

(b)(ii) Statements

Archimedes' principle: When a body is wholly or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid it displaces.

Law of flotation: A floating body displaces its own weight of the fluid in which it floats.

(a) E.m.f. of the accumulator and the R against \(I^{-1}\) experiment

E.m.f. of the accumulator, \(E = 1.5\ \text{V}\).

Table of readings

\(R\ (\Omega)\)	\(I\ (\text{A})\)	\(I^{-1}\ (\text{A}^{-1})\)
0	0.78	1.28
1	0.50	2.00
2	0.38	2.63
3	0.30	3.33
4	0.25	4.00
5	0.22	4.55

Graph of \(R\) against \(I^{-1}\)

Slope of the graph

Taking two points on the line of best fit, \((I^{-1}_1, R_1) = (2.00\ \text{A}^{-1},\ 1.1\ \Omega)\) and \((I^{-1}_2, R_2) = (4.00\ \text{A}^{-1},\ 4.1\ \Omega)\):

\[ s = \frac{R_2 - R_1}{I^{-1}_2 - I^{-1}_1} = \frac{4.1 - 1.1}{4.00 - 2.00} = \frac{3.0}{2.00} = 1.5\ \text{V}. \]

Intercept on the vertical axis

Producing the line of best fit back to \(I^{-1} = 0\), it cuts the vertical (\(R\)) axis at

\[ C = -1.9\ \Omega. \]

Interpretation. For this circuit the current is \(I = \dfrac{E}{R + S + r}\), which rearranges to \(R = E\,(I^{-1}) - (S + r)\). Comparing with \(R = s\,(I^{-1}) + C\): the slope \(s = E = 1.5\ \text{V}\) (equal to the measured e.m.f.), and the intercept \(C = -(S + r) = -1.9\ \Omega\), giving \(S + r = 1.9\ \Omega\).

Two precautions

• I made all connections clean and tight, and opened the key immediately after taking each reading so as not to run down the accumulator.
• I read the ammeter with my eye directly in front of the pointer to avoid errors due to parallax, and I noted and corrected the zero error of the ammeter.

(b)(i) Two advantages of a lead-acid accumulator over a Leclanche cell

• The lead-acid accumulator can be recharged and used again, whereas the Leclanche cell cannot be recharged.
• The lead-acid accumulator has a very low internal resistance, so it can supply a large current for a long time, which the Leclanche cell cannot.

(b)(ii) Effective resistance of the network

The \(3\ \Omega\) and \(4\ \Omega\) resistors are in parallel:

\[ \frac{1}{R_p} = \frac{1}{3} + \frac{1}{4} = \frac{4 + 3}{12} = \frac{7}{12}, \qquad R_p = \frac{12}{7} = 1.71\ \Omega. \]

This parallel section is in series with the \(4\ \Omega\) resistor (the battery has negligible internal resistance):

\[ R_{\text{eff}} = R_p + 4 = 1.71 + 4 = 5.71\ \Omega. \]

The effective resistance in the circuit \(= 5.71\ \Omega\).

(a) Lens displacement (conjugate-foci) method

For each fixed object-to-screen distance \(D\), the lens is moved to the two positions \(x_1\) and \(x_2\) that each give a sharp image on the screen. The separation of these positions is \(L = x_1 - x_2\). The readings and evaluated quantities are tabulated below.

\(D\) (cm)	\(D^{2}\) (cm\(^{2}\))	\(x_1\) (cm)	\(x_2\) (cm)	\(L=x_1-x_2\) (cm)	\(L^{2}\) (cm\(^{2}\))	\(D^{2}-L^{2}\) (cm\(^{2}\))
100	10000	80.50	18.30	62.20	3868.84	6131.16
90	8100	70.20	19.00	51.20	2621.44	5478.56
80	6400	59.10	20.00	39.10	1528.81	4871.19
70	4900	46.70	22.00	24.70	610.09	4289.91
60	3600	29.00	7.60	21.40	457.96	3142.04

Sample evaluation for the first reading (\(D = 100\ \text{cm}\)):

\[ D^{2} = 100^{2} = 10000\ \text{cm}^{2}, \quad L = 80.50 - 18.30 = 62.20\ \text{cm} \]\[ L^{2} = 62.20^{2} = 3868.84\ \text{cm}^{2}, \quad D^{2}-L^{2} = 10000 - 3868.84 = 6131.16\ \text{cm}^{2} \]

Graph of \(D^{2}-L^{2}\) against \(D\)

Plotting \(\left(D^{2}-L^{2}\right)\) on the vertical axis against \(D\) on the horizontal axis gives a straight line through the origin:

Slope and value of K

Reading the slope from the line of best fit using a large triangle, taking two clear points on the line \((D = 100\ \text{cm},\ D^{2}-L^{2} = 6000\ \text{cm}^{2})\) and \((D = 50\ \text{cm},\ D^{2}-L^{2} = 3000\ \text{cm}^{2})\):

\[ S = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6000 - 3000}{100 - 50} = \frac{3000}{50} = 60\ \text{cm} \]\[ K = \frac{S}{4} = \frac{60}{4} = 15\ \text{cm} \]

(This \(K\) is the focal length of the converging lens, since the displacement method gives \(f = \dfrac{D^{2}-L^{2}}{4D}\), so a graph of \(D^{2}-L^{2}\) against \(D\) has slope \(4f\) and \(K = \tfrac{S}{4} = f\).)

Two precautions

1. I avoided parallax error by reading the position of the lens on the metre rule with the line of sight directly above the mark.
2. I ensured that a sharp, well-focused image was formed on the screen before taking each reading of the lens position.

(b)(i) Real image versus virtual image

Real image	Virtual image
Formed by the actual intersection of refracted rays.	Formed where the refracted rays only appear to meet when produced backwards.
Can be caught (focused) on a screen.	Cannot be caught on a screen.
Inverted for a single converging lens.	Upright.

(b)(ii) Ray diagram: converging lens forming a real, diminished image

With the object placed beyond \(2F\), the converging lens forms a real, inverted and diminished image between \(F\) and \(2F\) on the far side of the lens.