WAEC SSCE - Physics - 2012 (Practical)

Question 1 Report

You are provided with a potentiometer, an ammeter, a voltmeter, a standard resistor, and other necessary apparatus. Using the circuit diagram above as a guide carry out the following instructions.

Set up a circuit as illustrated in the diagram above.
Close the key, K.
Read and record the ammeter reading l\(_{o}\) and the voltmeter reading V\(_{o}\) when jockey J is not making contact with the potentiometer wire OQ.
Using J, make a contact with the potentiometer wire OQ at a point P such that OP = 1Ocm.
Read and record the current and the corresponding value of the voltage V.
Repeat the procedure for other values of OP= 20cm, 30cm, 40cm, 50cm, and 60cm.
Tabulate your readings.
Plot a graph with V on the vertical axis and I on the horizontal axis, starting both axes from the origin (0, 0).
Determine the slope, s, of the graph.
Determine the value of V when I = 0.
State two precautions taken to obtain accurate results.

(b) i. State two advantages of a lead-acid accumulator over a dry Leclanche cell.

ii. A cell of emf 2V and internal resistance of 1\(\Omega\) passes current through an external load of 9\(\Omega\). Calculate the potential drop across the cell.

Answer

Answer Details

(a) The potentiometer experiment

Close K with the jockey J off the wire and record the open-circuit ammeter reading \(I_o\) and voltmeter reading \(V_o\). Make contact at P for each length OP (10, 20, 30, 40, 50, 60 cm). As OP increases, more of the potentiometer wire is placed in parallel with (or in circuit with) the load, changing the current I and terminal voltage V. Record I and V for each length.

Specimen table

OP /cm	I /A	V /V
10	I₁	V₁
20	...	...
60	...	...

Graph and slope. A plot of V (vertical) against I (horizontal) gives a straight line of negative gradient. The slope \(s=\dfrac{\Delta V}{\Delta I}\) has the unit of resistance (ohms) and represents the internal resistance of the source (magnitude). The intercept on the V-axis (value of V when I = 0) is the e.m.f. E of the cell, since \(V=E-Ir\).

Two precautions:

Ensure all connections and the jockey make firm, clean contact to avoid contact resistance.
Open the key immediately after each reading to prevent heating of the wire and running down the cell.

(b)(i) Advantages of a lead-acid accumulator over a dry Leclanche cell:

It can be recharged and reused, whereas the Leclanche cell cannot.
It has a much lower internal resistance and can deliver a large current.

(b)(ii) E = 2 V, r = 1 \(\Omega\), R = 9 \(\Omega\).

Current \(I=\dfrac{E}{R+r}=\dfrac{2}{9+1}=0.2\,\text{A}\).

Potential drop across the cell (terminal p.d.) \(V=E-Ir=2-(0.2)(1)=1.8\,\text{V}\).

Read lesson note on Current Electricity (JAMB)

Read lesson note on Capacitors (JAMB)

Current Electricity

Capacitors

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Question 2 Report

Study the diagrams above and use them as guides in carrying out the following instructions.

Using the spring balance provided, determine the weight of the object of mass M = \(50.0\text{g}\) in air. Record this weight as \(W_{1}\).
Determine the weight of the object when it is completely immersed in water contained in a beaker as shown in the diagram above. Record the weight as \(W_{2}\).
Determine the weight of the object when it is completely immersed in a liquid labeled L. Record the weight as \(W_{3}\).
Evaluate \(U = (W_{1} - W_{2})\) and \(V = (W_{1} - W_{3})\).
Repeat the procedure with the objects of masses M = \(100\text{g}\), \(150\text{g}\), \(200\text{g}\), and \(250\text{g}\)
In each case, evaluate \(U = (W_{1} - W_{2})\) and \(V = (W_{1} - W_{3})\).
Tabulate your readings.
Plot a graph with V on the vertical axis and U on the horizontal axis.
Determine the slope, s, of the horizontal graph.
State two precautions taken to ensure accurate results.

(b)i. State Archimedes' principle.

ii. A piece of brass of mass \(20.0\text{g}\) is hung on a spring balance from a rigid support and completely immersed in kerosene from of density \(8.0 \times 10^{2}\text{kgm}^{-3}\). Determine the readings of the spring balance \((g = 10\text{ms}^{-2}\), density of brass \(8.0 \times 10^{3}\text{kgm}^{-3})\)

Answer

Test of practical knowledge: measurement of upthrust with a spring balance

For each object of mass M the spring balance is read three times: the weight in air \(W_1\), the weight when the object is completely immersed in water \(W_2\), and the weight when it is completely immersed in the liquid labelled L, \(W_3\). The two upthrusts are then evaluated from

\[ U = W_1 - W_2 \qquad\text{and}\qquad V = W_1 - W_3 . \]

The apparatus is set up as shown below.

Apparatus: object suspended from a spring balance on a retort stand and completely immersed in the liquid in a beaker.

Table of readings (all weights in newtons N; \(g = 10\ \text{m s}^{-2}\)):

M /g	\(W_1\) /N	\(W_2\) /N	\(W_3\) /N	\(U=W_1-W_2\) /N	\(V=W_1-W_3\) /N
50.0	0.50	0.42	0.43	0.08	0.07
100.0	1.00	0.84	0.86	0.16	0.14
150.0	1.50	1.26	1.29	0.24	0.21
200.0	2.00	1.68	1.72	0.32	0.28
250.0	2.50	2.10	2.15	0.40	0.35

Graph of V against U

V (vertical axis) is plotted against U (horizontal axis). The points lie on a straight line passing through the origin.

Slope of the graph

Taking two widely separated points on the line of best fit, \((U_1,\,V_1) = (0.08,\,0.07)\) and \((U_2,\,V_2) = (0.40,\,0.35)\):

\[ s = \frac{V_2 - V_1}{U_2 - U_1} = \frac{0.35 - 0.07}{0.40 - 0.08} = \frac{0.28}{0.32} = 0.88 . \]

The slope \(s = 0.88\) has no unit. Since \(U\) is the upthrust in water and \(V\) is the upthrust in liquid L for the same object, the slope equals the relative density of liquid L, \(s = \dfrac{\rho_L}{\rho_{water}} = 0.88\).

Two precautions

I read the spring balance pointer with my eye level with the scale to avoid parallax error, and allowed the pointer to come to rest before reading.
I ensured that the object was completely immersed in the liquid without touching the sides or bottom of the beaker, and that no air bubbles clung to it.

(b)(i) Archimedes' principle

When a body is wholly or partially immersed in a fluid (a liquid or a gas), it experiences an upthrust that is equal to the weight of the fluid displaced by the body.

(b)(ii) Reading of the spring balance

Data: mass of brass \(m = 20.0\ \text{g} = 0.020\ \text{kg}\); density of brass \(\rho_b = 8.0\times10^{3}\ \text{kg m}^{-3}\); density of kerosene \(\rho_k = 8.0\times10^{2}\ \text{kg m}^{-3}\); \(g = 10\ \text{m s}^{-2}\).

Weight of the brass in air:

\[ W = mg = 0.020 \times 10 = 0.20\ \text{N} . \]

Volume of the brass:

\[ V_b = \frac{m}{\rho_b} = \frac{0.020}{8.0\times10^{3}} = 2.5\times10^{-6}\ \text{m}^3 . \]

Upthrust from the kerosene (weight of kerosene displaced):

\[ F_u = \rho_k\, V_b\, g = (8.0\times10^{2})(2.5\times10^{-6})(10) = 0.020\ \text{N} . \]

The spring balance reads the apparent weight, which is the weight in air minus the upthrust:

\[ R = W - F_u = 0.20 - 0.020 = 0.18\ \text{N} . \]

The reading of the spring balance is 0.18 N.

Read lesson note on Heat Energy (WAEC)

Read lesson note on Pressure (JAMB)

Read lesson note on Elasticity (JAMB)

Read lesson note on Measurements And Units (JAMB)

Answer Details

Test of practical knowledge: measurement of upthrust with a spring balance

For each object of mass M the spring balance is read three times: the weight in air \(W_1\), the weight when the object is completely immersed in water \(W_2\), and the weight when it is completely immersed in the liquid labelled L, \(W_3\). The two upthrusts are then evaluated from

\[ U = W_1 - W_2 \qquad\text{and}\qquad V = W_1 - W_3 . \]

The apparatus is set up as shown below.

Table of readings (all weights in newtons N; \(g = 10\ \text{m s}^{-2}\)):

M /g	\(W_1\) /N	\(W_2\) /N	\(W_3\) /N	\(U=W_1-W_2\) /N	\(V=W_1-W_3\) /N
50.0	0.50	0.42	0.43	0.08	0.07
100.0	1.00	0.84	0.86	0.16	0.14
150.0	1.50	1.26	1.29	0.24	0.21
200.0	2.00	1.68	1.72	0.32	0.28
250.0	2.50	2.10	2.15	0.40	0.35

Graph of V against U

V (vertical axis) is plotted against U (horizontal axis). The points lie on a straight line passing through the origin.

Slope of the graph

Taking two widely separated points on the line of best fit, \((U_1,\,V_1) = (0.08,\,0.07)\) and \((U_2,\,V_2) = (0.40,\,0.35)\):

\[ s = \frac{V_2 - V_1}{U_2 - U_1} = \frac{0.35 - 0.07}{0.40 - 0.08} = \frac{0.28}{0.32} = 0.88 . \]

The slope \(s = 0.88\) has no unit. Since \(U\) is the upthrust in water and \(V\) is the upthrust in liquid L for the same object, the slope equals the relative density of liquid L, \(s = \dfrac{\rho_L}{\rho_{water}} = 0.88\).

Two precautions

I read the spring balance pointer with my eye level with the scale to avoid parallax error, and allowed the pointer to come to rest before reading.
I ensured that the object was completely immersed in the liquid without touching the sides or bottom of the beaker, and that no air bubbles clung to it.

(b)(i) Archimedes' principle

When a body is wholly or partially immersed in a fluid (a liquid or a gas), it experiences an upthrust that is equal to the weight of the fluid displaced by the body.

(b)(ii) Reading of the spring balance

Data: mass of brass \(m = 20.0\ \text{g} = 0.020\ \text{kg}\); density of brass \(\rho_b = 8.0\times10^{3}\ \text{kg m}^{-3}\); density of kerosene \(\rho_k = 8.0\times10^{2}\ \text{kg m}^{-3}\); \(g = 10\ \text{m s}^{-2}\).

Weight of the brass in air:

\[ W = mg = 0.020 \times 10 = 0.20\ \text{N} . \]

Volume of the brass:

\[ V_b = \frac{m}{\rho_b} = \frac{0.020}{8.0\times10^{3}} = 2.5\times10^{-6}\ \text{m}^3 . \]

Upthrust from the kerosene (weight of kerosene displaced):

\[ F_u = \rho_k\, V_b\, g = (8.0\times10^{2})(2.5\times10^{-6})(10) = 0.020\ \text{N} . \]

The spring balance reads the apparent weight, which is the weight in air minus the upthrust:

\[ R = W - F_u = 0.20 - 0.020 = 0.18\ \text{N} . \]

The reading of the spring balance is 0.18 N.

Read lesson note on Heat Energy (WAEC)

Read lesson note on Pressure (JAMB)

Read lesson note on Elasticity (JAMB)

Read lesson note on Measurements And Units (JAMB)

Heat Energy

Pressure

Elasticity

Measurements And Units

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Question 3 Report

You are provided with a syringe, a petri-dish firmly attached to the base of the movable piston (plunger) of the syringe, a Set of weights, and other necessary apparatus.

Pull the piston of the syringe upward until it can no longer move. Read and record this position of the piston on the graduated mark on the syringe as V\(_{o}\).
Clamp the syringe and ensure that it is vertical.
Place a mass M= 500g gently at the center of the petri-dish.
Read and record the new position of the piston as V.
Evaluate V\(^{1}\).
Repeat the procedure for four other values of M= 1000g, 1500g, 2000g, and 2500g.
Tabulate your readings.
Plot a graph with V\(^{-1}\) on the vertical axis and M on the horizontal axis, starting both axes from the origin (0,0).
Determine the slope, s, of the graph.
Evaluate k = s\(^{-1}\).
State two precautions taken to ensure accurate results.

(b)i. When a weight is placed on the petri-dish, which quantities of the gas in the syringe (\(\Omega\)) increases; (\(\beta\)) decrease?

ii. What is responsible for the pressure exerted by a gas in a closed vessel?

Answer

Answer Details

(a) Table of readings

S/N	Mass, \(M\) (g)	Volume, \(V\) (cm³)	\(V^{-1}\) (cm⁻³)
1	500	9.30	0.108
2	1000	7.40	0.135
3	1500	6.40	0.156
4	2000	5.40	0.185
5	2500	4.60	0.217

For example, when \(M=500\text{ g}\),

\[V^{-1}=\frac{1}{9.30}=0.108\text{ cm}^{-3}.\]

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

Using two widely separated points on the line of best fit, approximately \((500,0.107)\) and \((2500,0.214)\):

\[s=\frac{0.214-0.107}{2500-500}=\frac{0.107}{2000}=5.35\times10^{-5}\text{ cm}^{-3}\text{g}^{-1}.\]

\[k=s^{-1}=\frac{1}{5.35\times10^{-5}}=1.87\times10^{4}\text{ g cm}^{3}.\]

Precautions

The syringe was clamped firmly in a vertical position.
Each mass was placed gently at the centre of the Petri dish, and the piston reading was taken at eye level to avoid parallax error.

(b)

(i) On placing a weight on the Petri dish, the pressure of the trapped gas increases, while its volume decreases.

(ii) Gas pressure in a closed vessel is caused by the continuous random collisions of gas molecules with the walls of the vessel.

Read lesson note on Gravitational Field (WAEC)

Gravitational Field

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