WAEC SSCE - Physics - 2009 (Practical)

Question 1 Report

Connect the circuit as shown in the diagram above. PQ is a potentiometer wire 100 cm long and R is a standard resistor of 5\(\Omega\).
With the jockey J not making contact with PQ, close the switch. Read and record the ammeter reading I. Open the switch.
Use the jockey to make contact with PQ at the 20cm mark such that PJ = I = 20 cm. Close the switch, read and record the value I\(_{i}\) of the ammeter. Evaluate I\(^{-1}\).
Repeat the procedure for other values of I = 35, 50, 65, and 80 cm. In each case, determine the corresponding values of I\(_{i}\), and I\(^{-1}\). Tabulate your readings.
Plot a graph of I\(^{i}\) on the vertical axis and I\(_{i}\), on the horizontal axis, starting both axes from the origin (0, 0).
From your graph deduce the value, of I\(_{o}\) of I\(_{i}\), when I\(^{-1}\)= 0.
Evaluate I\(_{o}\)e
State two precautions taken to ensure accurate results.

(b)) Define the e. m.f. of a battery

ii. A cell X e.m.f. 1.00 V is balanced by a length of 40.0 cm on a potentiometer wire. Another cell Y is balanced by a length of 60.0 cm on the same wire. Calculate the e.m.f. of Y.

Answer

Answer Details

(a) Potentiometer experiment

With the jockey off the potentiometer wire, the ammeter reading is:

I = 0.20 A

The readings obtained when the jockey makes contact at the stated lengths are tabulated below.

S/N	Length, L (cm)	Ammeter reading, I_i (A)	L^-1 (cm^-1)
1	20	1.00	0.050
2	35	0.90	0.029
3	50	0.85	0.020
4	65	0.80	0.015
5	80	0.75	0.013

The graph of I_i against L^-1 is shown below.

From the intercept on the vertical axis, when L^-1 = 0,

I_o = 0.75 A.

Hence,

\[ \frac{I_o}{I}=\frac{0.75}{0.20}=3.75. \]

Precautions

The ammeter was read with the eye directly in front of the pointer to avoid parallax error.
The switch was opened after each reading to prevent heating of the potentiometer wire and cell.

(b)(i) E.m.f. of a battery

The e.m.f. of a battery is the work done, or energy supplied, by the battery in driving one coulomb of charge round the complete circuit, including its internal resistance.

(b)(ii) E.m.f. of cell Y

For the same potentiometer wire, e.m.f. is proportional to balancing length:

\[ \frac{E_Y}{E_X}=\frac{l_Y}{l_X} \] \[ E_Y=\frac{60.0}{40.0}\times1.00=1.50\ \text{V}. \]

E.m.f. of cell Y = 1.50 V.

Read lesson note on Current Electricity (JAMB)

Current Electricity

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Question 2 Report

You are provided with a measuring cylinder, two different tins labeled C and D, a thermometer, and other necessary materials.
Use the measuring cylinder provided to measure 100 cm of water and pour it into the tin label.
Heat the water in the tin almost to boiling (90°C).
Remove the tin and place it on a cork or wooden stand.
Insert the thermometer into the tin and record the temperature of the water every minute starting from 85°C until the temperature falls to 60°C.
Repeat the experiment with the tin labeled D using exactly the same volume of water and temperature range. Tabulate your readings.
On the same graph sheet and using the same axis and scales, plot two graphs of temperature on the vertical axis and time on the horizontal axis from the readings obtained using tins C and D.
Label the graphs appropriately as C and D to correspond with the tins used.
From each graph, read off the time taken to cool from 85°C to 65°C.
State two precautions taken to ensure accurate results.

(b)i. Explain how heat losses by radiation and convection are minimized in a vacuum flask.

ii. State four factors that affect the rate of evaporation of a liquid in an open container.

Answer

(a) Cooling curves for tins C and D

100 cm³ of water was used in each tin. The temperature was recorded at one-minute intervals as the water cooled.

Time, t (min)	Temperature in tin C, T (°C)	Temperature in tin D, T (°C)
0	85	85
1	83	84
2	81	83
3	79	81
4	78	79
5	75	78
6	73	76
7	71	75
8	70	73
9	68	71
10	66	69.5
11	65	68
12	63	66
13	62	64
14	60	62
15	–	61
16	–	60

The two cooling curves, plotted on the same axes, are:

From the graph:

Time for tin C to cool from 85 °C to 65 °C = 11.0 min.
Time for tin D to cool from 85 °C to 65 °C = 12.4 min.

Precautions

Use the same volume of water and the same initial temperature in both tins.
Keep the thermometer bulb fully immersed without touching the base or side of the tin; stir the water gently before taking each reading and read the thermometer at eye level.

(b)(i) Vacuum flask

The silvered surfaces of the double walls reflect thermal radiation and are poor emitters and absorbers; hence heat loss by radiation is minimized. The vacuum between the walls contains virtually no particles, so convection currents cannot occur. The insulating stopper also prevents air circulation at the neck of the flask.

(b)(ii) Factors affecting evaporation

Temperature of the liquid.
Area of the liquid surface exposed.
Humidity of the surrounding air.
Speed of air movement or draught over the surface.

Read lesson note on Heat Energy (WAEC)

Read lesson note on Gravitational Field (JAMB)

Read lesson note on Temperature And Its Measurement (JAMB)

Answer Details

(a) Cooling curves for tins C and D

100 cm³ of water was used in each tin. The temperature was recorded at one-minute intervals as the water cooled.

Time, t (min)	Temperature in tin C, T (°C)	Temperature in tin D, T (°C)
0	85	85
1	83	84
2	81	83
3	79	81
4	78	79
5	75	78
6	73	76
7	71	75
8	70	73
9	68	71
10	66	69.5
11	65	68
12	63	66
13	62	64
14	60	62
15	–	61
16	–	60

The two cooling curves, plotted on the same axes, are:

From the graph:

Time for tin C to cool from 85 °C to 65 °C = 11.0 min.
Time for tin D to cool from 85 °C to 65 °C = 12.4 min.

Precautions

Use the same volume of water and the same initial temperature in both tins.
Keep the thermometer bulb fully immersed without touching the base or side of the tin; stir the water gently before taking each reading and read the thermometer at eye level.

(b)(i) Vacuum flask

The silvered surfaces of the double walls reflect thermal radiation and are poor emitters and absorbers; hence heat loss by radiation is minimized. The vacuum between the walls contains virtually no particles, so convection currents cannot occur. The insulating stopper also prevents air circulation at the neck of the flask.

(b)(ii) Factors affecting evaporation

Temperature of the liquid.
Area of the liquid surface exposed.
Humidity of the surrounding air.
Speed of air movement or draught over the surface.

Read lesson note on Heat Energy (WAEC)

Read lesson note on Gravitational Field (JAMB)

Read lesson note on Temperature And Its Measurement (JAMB)

Heat Energy

Gravitational Field

Temperature And Its Measurement

View Answer

Question 3 Report

You are provided with two meter rules and other necessary apparatus.
Place one of the rules on a knife edge and determine its centre of gravity C. Mark this pos with a piece of chalk.
Read and record the mass M\(_{R}\) of the metre rule written on the reverse side of it.
Attach the mass M= 100g firmly to the rule AB at C using sellotape.
Suspend the metre rule by two parallel threads of length h = 40 cm each at the 10 cm marks. Ensure that the graduated face to the metre rule is facing upwards.
Set the rule AB into a small angular oscillation about the vertical axis through its centre of gravity.
Determine the time, t for 20 complete oscillations. Evaluate the period T and T\(^{2}\)
Read and record the value of d in meters.
Keeping d constant throughout the experiment, repeat the procedure for other values of h = 50, 60, 70, and 80 cm. In each case determine the corresponding values of f T and T. Tabulate your reading. (x) Plot a graph of T on the vertical axis and h on the horizontal axis.
Determine the slope S, of the graph. Evaluate k =s, where Q =2 S, Q 250P
State two precautions taken to ensure accurate results.

(b)i. Define the term couple as it relates to rotational or oscillatory systems.

ii. Give two practical application of a couple in everyday life.

Answer

(a) Torsional oscillation of a suspended metre rule

The metre rule is balanced on the knife edge and its centre of gravity is located at the 49.5 cm mark, which is marked with chalk. The mass printed on the reverse of the rule is M_R = 135 g. The 100 g mass is fixed at C, the rule is suspended by two parallel threads of length \(h\) attached at the 10 cm marks, and the separation of the threads is kept constant at d = 80 cm = 0.80 m. The rule is twisted through a small angle about the vertical axis through C and released; the time \(t\) for 20 complete oscillations is taken with a stopwatch. The period is \(T=\dfrac{t}{20}\) and \(T^{2}\) is evaluated. The procedure is repeated for \(h = 40, 50, 60, 70\) and \(80\) cm.

Table of readings

\(h\) (cm)	\(d\) (m)	\(t\) (s)	\(T=\dfrac{t}{20}\) (s)	\(T^{2}\) (s\(^2\))
40	0.80	28.0	1.40	1.9600
50	0.80	29.0	1.45	2.1025
60	0.80	31.0	1.55	2.4025
70	0.80	34.0	1.70	2.8900
80	0.80	37.0	1.85	3.4225

Graph of \(T^{2}\) against \(h\)

Slope of the graph

Two points are taken on the line of best fit: \((h_1, T^2_1) = (40\ \text{cm}, 1.81\ \text{s}^2)\) and \((h_2, T^2_2) = (80\ \text{cm}, 3.30\ \text{s}^2)\).

\[ S=\frac{\Delta T^{2}}{\Delta h}=\frac{(3.30-1.81)\ \text{s}^2}{(80-40)\ \text{cm}}=\frac{1.49\ \text{s}^2}{40\ \text{cm}} \] \[ S = 0.0371\ \text{s}^2\,\text{cm}^{-1} \]

Evaluation of \(k\)

Given \(k=\dfrac{S}{Q}\) where \(Q=\dfrac{2}{25\,d^{2}}\), then

\[ k=\frac{S}{Q}=\frac{S\times 25\,d^{2}}{2} \]

With \(S = 0.0371\ \text{s}^2\,\text{cm}^{-1}\) and \(d = 0.80\ \text{m}\) so that \(d^{2}=0.64\ \text{m}^2\):

\[ k=\frac{0.0371\times 25\times 0.64}{2} \] \[ k=\frac{0.5936}{2}=0.297\ \text{s}^2\,\text{cm}^{-1}\,\text{m}^{2} \]

Two precautions

The two suspension threads were kept exactly equal in length, vertical and parallel so that the rule hung horizontally and oscillated smoothly in a horizontal plane.
Only a small angular twist was given, and the stopwatch was read at eye level to avoid parallax error while timing 20 complete oscillations.

(b)(i) Couple

A couple is a pair of two forces that are equal in magnitude, parallel and opposite in direction, but whose lines of action do not pass through the same point. A couple produces a turning (rotational) effect only, with no resultant translational force. Its moment (torque) is:

\[ \tau = F\times d \]

where \(F\) is the magnitude of one of the forces and \(d\) is the perpendicular distance between their lines of action.

(b)(ii) Two practical applications of a couple

Turning a tap or a water valve on and off with the fingers.
Turning a spanner or a screwdriver, and turning the steering wheel of a vehicle with both hands.

Read lesson note on Equilibrium Of Forces (WAEC)

Answer Details

(a) Torsional oscillation of a suspended metre rule

The metre rule is balanced on the knife edge and its centre of gravity is located at the 49.5 cm mark, which is marked with chalk. The mass printed on the reverse of the rule is M_R = 135 g. The 100 g mass is fixed at C, the rule is suspended by two parallel threads of length \(h\) attached at the 10 cm marks, and the separation of the threads is kept constant at d = 80 cm = 0.80 m. The rule is twisted through a small angle about the vertical axis through C and released; the time \(t\) for 20 complete oscillations is taken with a stopwatch. The period is \(T=\dfrac{t}{20}\) and \(T^{2}\) is evaluated. The procedure is repeated for \(h = 40, 50, 60, 70\) and \(80\) cm.

Table of readings

\(h\) (cm)	\(d\) (m)	\(t\) (s)	\(T=\dfrac{t}{20}\) (s)	\(T^{2}\) (s\(^2\))
40	0.80	28.0	1.40	1.9600
50	0.80	29.0	1.45	2.1025
60	0.80	31.0	1.55	2.4025
70	0.80	34.0	1.70	2.8900
80	0.80	37.0	1.85	3.4225

Graph of \(T^{2}\) against \(h\)

Slope of the graph

Two points are taken on the line of best fit: \((h_1, T^2_1) = (40\ \text{cm}, 1.81\ \text{s}^2)\) and \((h_2, T^2_2) = (80\ \text{cm}, 3.30\ \text{s}^2)\).

\[ S=\frac{\Delta T^{2}}{\Delta h}=\frac{(3.30-1.81)\ \text{s}^2}{(80-40)\ \text{cm}}=\frac{1.49\ \text{s}^2}{40\ \text{cm}} \] \[ S = 0.0371\ \text{s}^2\,\text{cm}^{-1} \]

Evaluation of \(k\)

Given \(k=\dfrac{S}{Q}\) where \(Q=\dfrac{2}{25\,d^{2}}\), then

\[ k=\frac{S}{Q}=\frac{S\times 25\,d^{2}}{2} \]

With \(S = 0.0371\ \text{s}^2\,\text{cm}^{-1}\) and \(d = 0.80\ \text{m}\) so that \(d^{2}=0.64\ \text{m}^2\):

\[ k=\frac{0.0371\times 25\times 0.64}{2} \] \[ k=\frac{0.5936}{2}=0.297\ \text{s}^2\,\text{cm}^{-1}\,\text{m}^{2} \]

Two precautions

The two suspension threads were kept exactly equal in length, vertical and parallel so that the rule hung horizontally and oscillated smoothly in a horizontal plane.
Only a small angular twist was given, and the stopwatch was read at eye level to avoid parallax error while timing 20 complete oscillations.

(b)(i) Couple

A couple is a pair of two forces that are equal in magnitude, parallel and opposite in direction, but whose lines of action do not pass through the same point. A couple produces a turning (rotational) effect only, with no resultant translational force. Its moment (torque) is:

\[ \tau = F\times d \]

where \(F\) is the magnitude of one of the forces and \(d\) is the perpendicular distance between their lines of action.

(b)(ii) Two practical applications of a couple

Turning a tap or a water valve on and off with the fingers.
Turning a spanner or a screwdriver, and turning the steering wheel of a vehicle with both hands.

Read lesson note on Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

View Answer






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