WAEC SSCE - Physics - 2016 (Practical)

Question 1 Report

You are provided with a variable d.c. power supply E, a \(2\propto\) standard resistor, a key, an ammeter, a voltmeter and other necessary materials.

i. Set up a circuit as shown in the diagram above with E= 1.5V

ii. Close the key k.

iii. Take and record the voltmeter reading V.

iv. Take and record the corresponding ammeter reading l.

v. Evaluate V\(^{-1}\) and l\(^{-1}\)

vi. Repeat the procedure for four other values of E= 3.0V, 4.5V, 6.0V, and 7.4V.

vii. Tabulate your readings.

viii. Plot a graph with V\(^{-1}\) on the vertical axis and l\(^{-1}\) on the horizontal axis starting both axes from the origin (0, 0).

ix. Determine the slope, s, of the graph.

x. Also determine the intercept, e, on the vertical axis.

xi. State two precautions taken to obtain accurate results.

(b)i. State two methods by which an electric current can be produced.

ii

Calculate the value of R in the circuit diagram shown above, given that the effective resistance of the circuit is \(4.0\Omega\) and the internal resistance of the cell is negligible.

Answer

(a) Table of readings

For each value of \(E\) the voltmeter reading \(V\) and the corresponding ammeter reading \(I\) are recorded, and the reciprocals \(V^{-1}\) and \(I^{-1}\) are evaluated:

S/N	\(E\)/V	\(I\)/A	\(V\)/V	\(V^{-1}\)/V\(^{-1}\)	\(I^{-1}\)/A\(^{-1}\)
1	1.5	0.30	0.80	1.250	3.333
2	3.0	0.50	1.40	0.717	2.000
3	4.5	0.70	2.00	0.500	1.429
4	6.0	0.90	2.50	0.400	1.111
5	7.5	1.10	3.00	0.333	0.909

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

\(V^{-1}\) is plotted on the vertical axis and \(I^{-1}\) on the horizontal axis, both axes starting from the origin \((0,0)\). A best straight line is drawn through the plotted points:

(ix) Slope of the graph

Two widely-separated points are read off the best line, \((I^{-1}_1, V^{-1}_1) = (0.26,\ 0.800)\) and \((I^{-1}_2, V^{-1}_2) = (1.24,\ 3.333)\):

\[ s = \frac{\Delta V^{-1}}{\Delta I^{-1}} = \frac{V^{-1}_2 - V^{-1}_1}{I^{-1}_2 - I^{-1}_1} = \frac{3.333 - 0.800}{1.24 - 0.26} = \frac{2.53}{0.98} \approx 2.6. \]

(x) Intercept on the vertical axis

Where the best line meets the \(V^{-1}\) axis (at \(I^{-1} = 0\)):

\[ e \approx 0.04\ \text{V}^{-1}. \]

(xi) Two precautions

Ensured all terminals were clean and every connection was tight, and checked for and corrected any zero error on the voltmeter and ammeter.
Opened the key immediately after each pair of readings was taken to avoid heating the resistor and running down the cell, and read both meters with the eye directly in line with the pointer to avoid parallax error.

(b)(i) Two methods of producing an electric current

Chemical action in cells, accumulators or fuel cells (chemical energy converted to electrical energy).
Electromagnetic induction using a generator or dynamo (mechanical energy converted to electrical energy). Solar/photocells and thermocouples are also acceptable.

(b)(ii) Value of \(R\)

The two resistors \(R\) are connected in parallel:

\[ \frac{1}{R_e} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \quad\Rightarrow\quad R_e = \frac{R}{2}. \]

This parallel combination is in series with the \(2\ \Omega\) standard resistor, and the effective (total) resistance of the circuit is \(4.0\ \Omega\) with negligible internal resistance:

\[ \frac{R}{2} + 2 = 4 \quad\Rightarrow\quad \frac{R}{2} = 2 \quad\Rightarrow\quad R = 4.0\ \Omega. \]

Read lesson note on Current Electricity (JAMB)

Read lesson note on Electric Cells (JAMB)

Answer Details

(a) Table of readings

For each value of \(E\) the voltmeter reading \(V\) and the corresponding ammeter reading \(I\) are recorded, and the reciprocals \(V^{-1}\) and \(I^{-1}\) are evaluated:

S/N	\(E\)/V	\(I\)/A	\(V\)/V	\(V^{-1}\)/V\(^{-1}\)	\(I^{-1}\)/A\(^{-1}\)
1	1.5	0.30	0.80	1.250	3.333
2	3.0	0.50	1.40	0.717	2.000
3	4.5	0.70	2.00	0.500	1.429
4	6.0	0.90	2.50	0.400	1.111
5	7.5	1.10	3.00	0.333	0.909

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

\(V^{-1}\) is plotted on the vertical axis and \(I^{-1}\) on the horizontal axis, both axes starting from the origin \((0,0)\). A best straight line is drawn through the plotted points:

(ix) Slope of the graph

Two widely-separated points are read off the best line, \((I^{-1}_1, V^{-1}_1) = (0.26,\ 0.800)\) and \((I^{-1}_2, V^{-1}_2) = (1.24,\ 3.333)\):

\[ s = \frac{\Delta V^{-1}}{\Delta I^{-1}} = \frac{V^{-1}_2 - V^{-1}_1}{I^{-1}_2 - I^{-1}_1} = \frac{3.333 - 0.800}{1.24 - 0.26} = \frac{2.53}{0.98} \approx 2.6. \]

(x) Intercept on the vertical axis

Where the best line meets the \(V^{-1}\) axis (at \(I^{-1} = 0\)):

\[ e \approx 0.04\ \text{V}^{-1}. \]

(xi) Two precautions

Ensured all terminals were clean and every connection was tight, and checked for and corrected any zero error on the voltmeter and ammeter.
Opened the key immediately after each pair of readings was taken to avoid heating the resistor and running down the cell, and read both meters with the eye directly in line with the pointer to avoid parallax error.

(b)(i) Two methods of producing an electric current

Chemical action in cells, accumulators or fuel cells (chemical energy converted to electrical energy).
Electromagnetic induction using a generator or dynamo (mechanical energy converted to electrical energy). Solar/photocells and thermocouples are also acceptable.

(b)(ii) Value of \(R\)

The two resistors \(R\) are connected in parallel:

\[ \frac{1}{R_e} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \quad\Rightarrow\quad R_e = \frac{R}{2}. \]

This parallel combination is in series with the \(2\ \Omega\) standard resistor, and the effective (total) resistance of the circuit is \(4.0\ \Omega\) with negligible internal resistance:

\[ \frac{R}{2} + 2 = 4 \quad\Rightarrow\quad \frac{R}{2} = 2 \quad\Rightarrow\quad R = 4.0\ \Omega. \]

Read lesson note on Current Electricity (JAMB)

Read lesson note on Electric Cells (JAMB)

Current Electricity

Electric Cells

View Answer

Question 2 Report

You are provided with a beaker, a thermometer, a stirrer, a measuring cylinder, a bunsen burner, a wire gauze, a 50g mass, a pair of tongs, water, tripod stand, and other necessary materials.

i. Using the measuring cylinder, measure \(150cm^{3}\) of water into the beaker.

ii. Record the volume \(v\) of the water in the beaker

iii. Calculate the mass \(m\) of the water, given that \(m = pv\) and; \(p = 1gcm_{-3}\).

iv. Measure and record the initial temperature \(\theta_{0}\) of the water in the beaker.

v. Hold the 50g mass with the pair of tongs in the flame of the bunsen burner for 2 minutes.

vi. Quickly transfer the 50g mass to water in the beaker.

vii. Stir gently and record the highest temperature \(\theta_{1}\), attained

viii. Evaluate \(\theta = (\theta_{1} - \theta_{0})\).

ix. Empty the content of the beaker and repeat the procedures above for the values of \(v = 200cm^{3}\), \(250cm^{3}\), \(300cm^{3}\), and \(350cm^{3}\).

x. Tabulate your readings.

xi. Plot a graph with \(m\) on the vertical axis and \(\theta\) on the horizontal axis.

xii. Determine the slope, \(s\), of the graph.

xiii. Evaluate \(k = \frac{50}{s}\).

xiv. State two precautions taken to obtain accurate results.

(b)i. Define heat capacity.

ii. An electric kettle rated 1.2kw is used to heat 800g of water initially at a temperature of 20 C. Neglecting heat losses, calculate the time taken for the kettle to heat the water to its boiling point. [Take the boiling point of water= 101 C specific heat capacity of water = 4200 Jkg' K'1 (odv)

Answer

Test of Practical Knowledge: heat carried into water by a hot mass

For each volume of water the mass is \(m=\rho v\) with \(\rho=1\ \text{g cm}^{-3}\), so numerically \(m/\text{g}=v/\text{cm}^3\). The temperature rise is \(\theta=\theta_1-\theta_0\), the initial temperature being \(\theta_0=34^{\circ}\text{C}\) in every trial.

(x) Table of readings

\(v\) / cm\(^3\)	\(m=\rho v\) / g	\(\theta_0\) / \(^{\circ}\)C	\(\theta_1\) / \(^{\circ}\)C	\(\theta=\theta_1-\theta_0\) / \(^{\circ}\)C
150.0	150.0	34	88	54
200.0	200.0	34	86	52
250.0	250.0	34	84	50
300.0	300.0	34	80	46
350.0	350.0	34	76	42

(xi) Graph of \(m\) against \(\theta\)

Plotting \(m\) on the vertical axis against \(\theta\) on the horizontal axis gives a straight line of best fit with a negative gradient: the larger the mass of water, the smaller the temperature rise produced by the same 50 g hot mass.

(xii) Slope of the graph

Two points read off the line of best fit are \((\theta=42^{\circ}\text{C},\,m=360\ \text{g})\) and \((\theta=54^{\circ}\text{C},\,m=166\ \text{g})\).

\[|s|=\left|\frac{\Delta m}{\Delta\theta}\right|=\frac{360-166}{54-42}=\frac{194}{12}=16.2\ \text{g }^{\circ}\text{C}^{-1}.\]

The gradient is negative, \(s=-16.2\ \text{g }^{\circ}\text{C}^{-1}\); its magnitude is used below.

(xiii) Evaluation of \(k\)

\[k=\frac{50}{|s|}=\frac{50}{16.2}=3.1.\]

(xiv) Two precautions

I stirred the water gently and avoided splashing so that heat was evenly distributed before the highest temperature was read.
I avoided parallax error by reading the thermometer and the measuring cylinder with the eye level with the meniscus, and I transferred the hot mass quickly and gently into the water.

(b)(i) Heat capacity

The heat capacity of a body is the quantity of heat required to raise the temperature of the whole body by one kelvin (one degree Celsius). Its SI unit is the joule per kelvin, \(\text{J K}^{-1}\).

(b)(ii) Time to reach boiling point

Data: \(P=1.2\ \text{kW}=1200\ \text{W}\), \(m=800\ \text{g}=0.8\ \text{kg}\), \(c=4200\ \text{J kg}^{-1}\text{K}^{-1}\), temperature rise \(\Delta\theta=101-20=81\ \text{K}\).

Neglecting heat losses, the electrical energy supplied equals the heat gained by the water:

\[Pt=mc\,\Delta\theta\] \[t=\frac{mc\,\Delta\theta}{P}=\frac{0.8\times4200\times81}{1200}=\frac{272160}{1200}=226.8\ \text{s}.\]

The kettle takes about 226.8 s (approximately 3.8 minutes) to bring the water to its boiling point.

Read lesson note on Heat Energy (WAEC)

Answer Details

Test of Practical Knowledge: heat carried into water by a hot mass

For each volume of water the mass is \(m=\rho v\) with \(\rho=1\ \text{g cm}^{-3}\), so numerically \(m/\text{g}=v/\text{cm}^3\). The temperature rise is \(\theta=\theta_1-\theta_0\), the initial temperature being \(\theta_0=34^{\circ}\text{C}\) in every trial.

(x) Table of readings

\(v\) / cm\(^3\)	\(m=\rho v\) / g	\(\theta_0\) / \(^{\circ}\)C	\(\theta_1\) / \(^{\circ}\)C	\(\theta=\theta_1-\theta_0\) / \(^{\circ}\)C
150.0	150.0	34	88	54
200.0	200.0	34	86	52
250.0	250.0	34	84	50
300.0	300.0	34	80	46
350.0	350.0	34	76	42

(xi) Graph of \(m\) against \(\theta\)

Plotting \(m\) on the vertical axis against \(\theta\) on the horizontal axis gives a straight line of best fit with a negative gradient: the larger the mass of water, the smaller the temperature rise produced by the same 50 g hot mass.

(xii) Slope of the graph

Two points read off the line of best fit are \((\theta=42^{\circ}\text{C},\,m=360\ \text{g})\) and \((\theta=54^{\circ}\text{C},\,m=166\ \text{g})\).

\[|s|=\left|\frac{\Delta m}{\Delta\theta}\right|=\frac{360-166}{54-42}=\frac{194}{12}=16.2\ \text{g }^{\circ}\text{C}^{-1}.\]

The gradient is negative, \(s=-16.2\ \text{g }^{\circ}\text{C}^{-1}\); its magnitude is used below.

(xiii) Evaluation of \(k\)

\[k=\frac{50}{|s|}=\frac{50}{16.2}=3.1.\]

(xiv) Two precautions

I stirred the water gently and avoided splashing so that heat was evenly distributed before the highest temperature was read.
I avoided parallax error by reading the thermometer and the measuring cylinder with the eye level with the meniscus, and I transferred the hot mass quickly and gently into the water.

(b)(i) Heat capacity

The heat capacity of a body is the quantity of heat required to raise the temperature of the whole body by one kelvin (one degree Celsius). Its SI unit is the joule per kelvin, \(\text{J K}^{-1}\).

(b)(ii) Time to reach boiling point

Data: \(P=1.2\ \text{kW}=1200\ \text{W}\), \(m=800\ \text{g}=0.8\ \text{kg}\), \(c=4200\ \text{J kg}^{-1}\text{K}^{-1}\), temperature rise \(\Delta\theta=101-20=81\ \text{K}\).

Neglecting heat losses, the electrical energy supplied equals the heat gained by the water:

\[Pt=mc\,\Delta\theta\] \[t=\frac{mc\,\Delta\theta}{P}=\frac{0.8\times4200\times81}{1200}=\frac{272160}{1200}=226.8\ \text{s}.\]

The kettle takes about 226.8 s (approximately 3.8 minutes) to bring the water to its boiling point.

Read lesson note on Heat Energy (WAEC)

Heat Energy

View Answer

Question 3 Report

You are provided with a uniform metre rule, a knife-edge, some masses and other necessary materials.

i. Determine and record the centre of gravity of the metre rule.

ii. Fix the 100g mass marked N at a point Y, the 80cm mark of the rule using a sellotape.

iii. Suspend another 50g mass marked M at X, a distance A = 1Ocm from the 0cm mark of the rule.

iv. Balance the arrangement horizontally on the knife edge as illustrated in the diagram above.

v. Measure and record the distance B of a knife-edge from the 0cm mark of the rule.

vi. Repeat the procedure for four other values of A =15cm, 20cm, 25cm and 30cm.

vii. Tabulate your readings.

viii. Plot a graph with B on the vertical axis and A on the horizontal axis.

ix. Determine the slope, s, of the graph.

x. Also determine the intercept, c, on the vertical axis.

xi. Evaluate:
\(\propto\)) = k\(_{1}\) = (\(\frac{1 - 2s}{s}\))100

(\(\beta\)) = k\(_{2}\) = \(\frac{2c}{s}\) = 160

xii. State two precautions taken to obtain accurate results.

(b)i. Define the moment of a force about a point.

ii. State two conditions under which a rigid body at rest remains in equilibrium when acted upon by non-parallel coplanar forces.

Answer

(i) Centre of gravity of the uniform metre rule = 50.0 cm mark (the rule balances horizontally at this point before any mass is added).

The 100 g mass N is fixed with sellotape at the 80 cm mark (point Y); the 50 g mass M is suspended at point X, a distance A from the 0 cm end. The knife-edge is moved until the rule balances horizontally, and B, the distance of the knife-edge from the 0 cm end, is recorded.

Metre rule (fixed 100 g mass N at 80 cm, 50 g mass M at distance A from the 0 cm end) balanced horizontally on the knife-edge at distance B from the 0 cm end.

(vii) Table of readings

S/N	A / cm	B / cm
1	10.0	54.0
2	15.0	55.0
3	20.0	56.0
4	25.0	57.0
5	30.0	57.0
6	35.0	58.0

(viii) Graph of B against A

B is plotted on the vertical axis against A on the horizontal axis and the best straight line is drawn through the points.

(ix) Slope, s

Using two widely-spaced points on the best line, \((A_1,B_1)=(0,\,53.0)\) and \((A_2,B_2)=(30.0,\,59.0)\):

\[s=\frac{B_2-B_1}{A_2-A_1}=\frac{59.0-53.0}{30.0-0}=\frac{6.0}{30.0}=0.2\]

(x) Intercept, c, on the vertical axis (where the line cuts A = 0):

\[c=53.0\ \text{cm}\]

(xi) Evaluation

\[k_1=\left(\frac{1-2s}{s}\right)100=\left(\frac{1-2(0.2)}{0.2}\right)100=\left(\frac{1-0.4}{0.2}\right)100=\left(\frac{0.6}{0.2}\right)100=300\]\[k_2=\frac{2c}{s}-160=\frac{2(53.0)}{0.2}-160=\frac{106.0}{0.2}-160=530-160=370\]

(xii) Two precautions

The rule was made to balance exactly horizontal on the knife-edge before each reading of B was taken.
Error due to parallax was avoided by reading the metre rule with the eye placed vertically above the mark being read.

(b)(i) Moment of a force about a point

The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force. Its unit is the newton-metre (N m).

(b)(ii) Two conditions for equilibrium under non-parallel coplanar forces

The vector sum of all the forces is zero (i.e. the algebraic sum of their resolved components in any given direction is zero).
The algebraic sum of the moments of the forces about any point is zero.

Read lesson note on Equilibrium Of Forces (WAEC)

Answer Details

(i) Centre of gravity of the uniform metre rule = 50.0 cm mark (the rule balances horizontally at this point before any mass is added).

The 100 g mass N is fixed with sellotape at the 80 cm mark (point Y); the 50 g mass M is suspended at point X, a distance A from the 0 cm end. The knife-edge is moved until the rule balances horizontally, and B, the distance of the knife-edge from the 0 cm end, is recorded.

(vii) Table of readings

S/N	A / cm	B / cm
1	10.0	54.0
2	15.0	55.0
3	20.0	56.0
4	25.0	57.0
5	30.0	57.0
6	35.0	58.0

(viii) Graph of B against A

B is plotted on the vertical axis against A on the horizontal axis and the best straight line is drawn through the points.

(ix) Slope, s

Using two widely-spaced points on the best line, \((A_1,B_1)=(0,\,53.0)\) and \((A_2,B_2)=(30.0,\,59.0)\):

\[s=\frac{B_2-B_1}{A_2-A_1}=\frac{59.0-53.0}{30.0-0}=\frac{6.0}{30.0}=0.2\]

(x) Intercept, c, on the vertical axis (where the line cuts A = 0):

\[c=53.0\ \text{cm}\]

(xi) Evaluation

\[k_1=\left(\frac{1-2s}{s}\right)100=\left(\frac{1-2(0.2)}{0.2}\right)100=\left(\frac{1-0.4}{0.2}\right)100=\left(\frac{0.6}{0.2}\right)100=300\]\[k_2=\frac{2c}{s}-160=\frac{2(53.0)}{0.2}-160=\frac{106.0}{0.2}-160=530-160=370\]

(xii) Two precautions

The rule was made to balance exactly horizontal on the knife-edge before each reading of B was taken.
Error due to parallax was avoided by reading the metre rule with the eye placed vertically above the mark being read.

(b)(i) Moment of a force about a point

The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force. Its unit is the newton-metre (N m).

(b)(ii) Two conditions for equilibrium under non-parallel coplanar forces

The vector sum of all the forces is zero (i.e. the algebraic sum of their resolved components in any given direction is zero).
The algebraic sum of the moments of the forces about any point is zero.

Read lesson note on Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

View Answer






Press & Hold to Drag Around
Click Here to Close