(a) You are provided with a set of masses, a metre rule, a thread, two retort stands and clamps, a stop watch, a knife edge and split corks.
Carry out the following instructions using the diagram above as a guide.
(i) Determine the centre of gravity, C, of the metre rule using the knife edge.
(ii) Read and record the mass, M, of the metre rule written on the reverse side of it.
(iii) Suspend the metre rule by means of two parallel threads of equal length, h= 70 cm with one at the 10 cm mark and the other at 90 cm mark of the metre rule.
(iv) Attach a mass m = 30g firmly to the metre rule at C. Ensure that the graduated face of the metre rule is facing upwards and that d= 80 cm throughout the experiment. (v) Set the metre rule into small angular oscillations about the vertical axis through its centre of gravity by displacing its ends in opposite directions.
(VI) Determine the time,i, for 20 oscillations and evaluate the period T, T\(^2\) and T\(^{-2}\).
(vii) Repeat the procedure for four other values of m =40 g, 50 g, 60 g and 70 g n each case, determine I and evaluate T\(^2\) and T\(^{-2}\).
(viii) Plot a graph of T on the vertical axis and m on the horizontal axis.
(ix) Determine the slope, s, of the graph.
(x) Evaluate Q= 0.68 / s.
(xi) State two precautions taken to ensure accurate results.
(b) (i) Give two examples of simple harmonic motion other than the motion of a simple pendulum.
(ii) Explain the term centre of gravity of a body.
Nature of the experiment. The metre rule hangs horizontally from two equal parallel threads (a bifilar suspension) and is set into small angular (torsional) oscillations about the vertical axis through its centre of gravity. The period T is found for several attached masses m.
Procedure summary. For each mass m (30, 40, 50, 60, 70 g attached at C), the time t for 20 complete oscillations is measured, and the period and its powers are evaluated:
\[ T = \frac{t}{20}, \qquad T^{2}, \qquad T^{-2} \]
Sample table (headings).
m (g)
t for 20 osc. (s)
T (s)
\(T^{2}\)
\(T^{-2}\)
30
-
-
-
-
40
-
-
-
-
50
-
-
-
-
60
-
-
-
-
70
-
-
-
-
Graph and evaluation. Plot T (vertical) against m (horizontal). Determine the slope s from \(s = \dfrac{\Delta T}{\Delta m}\), then evaluate
\[ Q = \frac{0.68}{s} \]
Two precautions.
The oscillations were kept small (small angular displacement) and purely rotational, and timing was taken over 20 swings to reduce reaction-time error.
The two threads were of equal length and vertical, the mass firmly attached at C, and the eye kept steady to avoid parallax/miscounting; timing started and stopped at an end position of the swing.
(b)(i) Two examples of simple harmonic motion (other than the simple pendulum): the vertical oscillation of a mass on a spiral spring; the vibration of a tuning-fork prong; the up-and-down motion of a loaded test-tube (or hydrometer) floating in a liquid; the oscillation of liquid in a U-tube.
(b)(ii) Centre of gravity. The centre of gravity of a body is the single point through which the whole weight of the body appears to act, irrespective of the position of the body.
Nature of the experiment. The metre rule hangs horizontally from two equal parallel threads (a bifilar suspension) and is set into small angular (torsional) oscillations about the vertical axis through its centre of gravity. The period T is found for several attached masses m.
Procedure summary. For each mass m (30, 40, 50, 60, 70 g attached at C), the time t for 20 complete oscillations is measured, and the period and its powers are evaluated:
\[ T = \frac{t}{20}, \qquad T^{2}, \qquad T^{-2} \]
Sample table (headings).
m (g)
t for 20 osc. (s)
T (s)
\(T^{2}\)
\(T^{-2}\)
30
-
-
-
-
40
-
-
-
-
50
-
-
-
-
60
-
-
-
-
70
-
-
-
-
Graph and evaluation. Plot T (vertical) against m (horizontal). Determine the slope s from \(s = \dfrac{\Delta T}{\Delta m}\), then evaluate
\[ Q = \frac{0.68}{s} \]
Two precautions.
The oscillations were kept small (small angular displacement) and purely rotational, and timing was taken over 20 swings to reduce reaction-time error.
The two threads were of equal length and vertical, the mass firmly attached at C, and the eye kept steady to avoid parallax/miscounting; timing started and stopped at an end position of the swing.
(b)(i) Two examples of simple harmonic motion (other than the simple pendulum): the vertical oscillation of a mass on a spiral spring; the vibration of a tuning-fork prong; the up-and-down motion of a loaded test-tube (or hydrometer) floating in a liquid; the oscillation of liquid in a U-tube.
(b)(ii) Centre of gravity. The centre of gravity of a body is the single point through which the whole weight of the body appears to act, irrespective of the position of the body.
You have been provided with a ray box, a converging lens, a lens holder, a screen, a metre rule, and half- metre rule. Use the diagram above as a guide to perform the experiment.
(i) Determine the approximate focal length f, of the lens by focusing a distant object on the screen.
(ii) Place the ray box and the screen such that the distance between the illuminated cross-Wire and the screen, \(D= 150\ \text{cm}\).
(iii) Place the lens at a position L where a sharp mage of the cross-Wire Is obtained on the screen Note L.
(iv) Move the lens at a position L, to 0btain another sharp image of the cross-wire on the screen. Note L
(V) Measure the distance, d. between \(L_1\) and \(L_2\).
(Vi) Evaluate \(D^2\): \(d^2\) and \(D^2 - d^2\).
(vii) Repeat the procedure for four other values of \(D = 130\text{cm}, 100\text{ cm}, 90\text{ cm}\) and \(80\text{ cm}\). in each case.evaluate \(D^2\); \(d^2\) and \(D^2 - d^2\).
(viii) Tabulate the result
(ix) Plot a graph with \(D^2 - d^2\) on the vertical axis and \(D\) on the horizontal axis.
(x) Determine the r values of D axis and Determine the slopes, S, of the graph.
(xi) Evaluate \(k = \frac{s}{4}\)
(xii) State two precautions taken to ensure accurate results.
(bi) Distinguish between a virtual image and. plain image?
(ii) With the aid of a ray diagram, explain how a converging lens produces a Virtual image
Displacement (Bessel) method for the focal length of a converging lens
(i) Approximate focal length. With the ray box removed, a distant object (a window across the room) is focused sharply on the screen. The distance from the lens to the screen is measured with the metre rule:
\[ f \approx 15\ \text{cm} \]
Principle for parts (ii)-(viii). The illuminated cross-wire (object) and the screen are kept a fixed distance \(D\) apart. For any \(D>4f\) there are two positions of the lens, \(L_1\) and \(L_2\), that each throw a sharp image on the screen. Their separation is \(d=L_2-L_1\), and
The procedure of parts (iii)-(vii) is carried out for \(D=150,\,130,\,100,\,90\) and \(80\ \text{cm}\). The lens positions \(L_1,\,L_2\) are read off the metre rule (object taken as the zero of the scale), and the quantities \(D^{2}\), \(d^{2}\) and \(D^{2}-d^{2}\) are evaluated.
(viii) Table of results
\(D\) (cm)
\(L_1\) (cm)
\(L_2\) (cm)
\(d=L_2-L_1\) (cm)
\(D^{2}\) (cm\(^2\))
\(d^{2}\) (cm\(^2\))
\(D^{2}-d^{2}\) (cm\(^2\))
150
16.91
133.09
116.19
22500
13500
9000
130
17.31
112.69
95.39
16900
9100
7800
100
18.38
81.62
63.25
10000
4000
6000
90
19.02
70.98
51.96
8100
2700
5400
80
20.00
60.00
40.00
6400
1600
4800
(ix) Graph. \(D^{2}-d^{2}\) is plotted on the vertical axis against \(D\) on the horizontal axis. The points lie on a straight line passing through the origin.
Straight line through the origin; slope S = 60 cm, so k = S/4 = 15 cm = f.
(x) Intercept and slope. The line passes through the origin, so its intercept on the \(D\)-axis is
\[ D\text{-intercept}=0. \]
Taking two well-separated points on the line of best fit, \((80,\,4800)\) and \((150,\,9000)\):
(xi) Evaluate \(k\). Since \(D^{2}-d^{2}=4fD\), the slope \(S=4f\), hence
\[ k=\frac{S}{4}=\frac{60}{4}=15\ \text{cm}. \]
This equals the focal length of the lens and agrees with the value \(f\approx 15\ \text{cm}\) obtained in part (i).
(xii) Two precautions.
The lens was mounted vertically and its centre kept in line with the cross-wire and the screen, all at the same height on the bench.
Each image was focused as sharply as possible and the eye positioned squarely over the metre rule to avoid parallax and zero error when reading the lens positions.
(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 come from when produced backwards.
Can be caught (focused) on a screen.
Cannot be caught on a screen.
Inverted relative to the object.
Erect (upright) relative to the object.
(b)(ii) How a converging lens produces a virtual image
The object is placed between the lens and its principal focus, that is at an object distance \(u
A ray travelling parallel to the principal axis is refracted so that it passes through the far principal focus \(F'\).
A ray directed through the optical centre \(C\) of the lens passes straight through undeviated.
After refraction these two rays diverge; they never meet on the far side. When they are produced backwards (broken lines) they meet on the same side as the object, forming an image that is virtual, erect and magnified.
Ray diagram: object inside the focal length of a converging lens gives a virtual, erect, magnified image (broken lines show the rays produced backwards).
Displacement (Bessel) method for the focal length of a converging lens
(i) Approximate focal length. With the ray box removed, a distant object (a window across the room) is focused sharply on the screen. The distance from the lens to the screen is measured with the metre rule:
\[ f \approx 15\ \text{cm} \]
Principle for parts (ii)-(viii). The illuminated cross-wire (object) and the screen are kept a fixed distance \(D\) apart. For any \(D>4f\) there are two positions of the lens, \(L_1\) and \(L_2\), that each throw a sharp image on the screen. Their separation is \(d=L_2-L_1\), and
The procedure of parts (iii)-(vii) is carried out for \(D=150,\,130,\,100,\,90\) and \(80\ \text{cm}\). The lens positions \(L_1,\,L_2\) are read off the metre rule (object taken as the zero of the scale), and the quantities \(D^{2}\), \(d^{2}\) and \(D^{2}-d^{2}\) are evaluated.
(viii) Table of results
\(D\) (cm)
\(L_1\) (cm)
\(L_2\) (cm)
\(d=L_2-L_1\) (cm)
\(D^{2}\) (cm\(^2\))
\(d^{2}\) (cm\(^2\))
\(D^{2}-d^{2}\) (cm\(^2\))
150
16.91
133.09
116.19
22500
13500
9000
130
17.31
112.69
95.39
16900
9100
7800
100
18.38
81.62
63.25
10000
4000
6000
90
19.02
70.98
51.96
8100
2700
5400
80
20.00
60.00
40.00
6400
1600
4800
(ix) Graph. \(D^{2}-d^{2}\) is plotted on the vertical axis against \(D\) on the horizontal axis. The points lie on a straight line passing through the origin.
Straight line through the origin; slope S = 60 cm, so k = S/4 = 15 cm = f.
(x) Intercept and slope. The line passes through the origin, so its intercept on the \(D\)-axis is
\[ D\text{-intercept}=0. \]
Taking two well-separated points on the line of best fit, \((80,\,4800)\) and \((150,\,9000)\):
(xi) Evaluate \(k\). Since \(D^{2}-d^{2}=4fD\), the slope \(S=4f\), hence
\[ k=\frac{S}{4}=\frac{60}{4}=15\ \text{cm}. \]
This equals the focal length of the lens and agrees with the value \(f\approx 15\ \text{cm}\) obtained in part (i).
(xii) Two precautions.
The lens was mounted vertically and its centre kept in line with the cross-wire and the screen, all at the same height on the bench.
Each image was focused as sharply as possible and the eye positioned squarely over the metre rule to avoid parallax and zero error when reading the lens positions.
(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 come from when produced backwards.
Can be caught (focused) on a screen.
Cannot be caught on a screen.
Inverted relative to the object.
Erect (upright) relative to the object.
(b)(ii) How a converging lens produces a virtual image
The object is placed between the lens and its principal focus, that is at an object distance \(u
A ray travelling parallel to the principal axis is refracted so that it passes through the far principal focus \(F'\).
A ray directed through the optical centre \(C\) of the lens passes straight through undeviated.
After refraction these two rays diverge; they never meet on the far side. When they are produced backwards (broken lines) they meet on the same side as the object, forming an image that is virtual, erect and magnified.
Ray diagram: object inside the focal length of a converging lens gives a virtual, erect, magnified image (broken lines show the rays produced backwards).
(a) You are provided with a battery, an ammeter, a voltmeter, a resistance box, a key and connection wires.
(i) Set up circuit as shown in the diagram above.
(ii) With the key opened, measure and record the e.m.f. \(E_o\) of the battery
(iii) With the key closed, select the resistance \(R=1\) on the resistance box. Read and record the current, 1Ω.
(iv) Evaluate \(I^{-1}\).
(v) Repeat the procedure for five other values of \(R=2\Omega\), \(3\Omega\), \(4\Omega\), \(5\Omega\), and \(6\Omega\).
In each case, record \(I\) and evaluate \(I^{-1}\) results.
(vii) Plot a graph with \(R\) on the vertical axis and \(I^{-1}\) on the horizontal axis.
(Viii) Determine the slope, \(s\), of the graph.
(ix) Determine the intercept \(C\), On the vertical axis.
(x) State two precautions taken to ensure accurate results.
(b)(i) Define potential difference in an electric field.
(ii) A piece of resistance wire of diameter 0.2 mm and length 25 cm has a resistance of 7Ω. Calculate the resistivity of the wire of the battery. [\(\pi = \frac{22}{7}\)]
Theory. For a cell of e.m.f. E and internal resistance r driving current I through an external resistance R,
\[ E = I(R + r) \;\Rightarrow\; R = \frac{E}{I} - r = E\,I^{-1} - r \]
Expected graph. A plot of R (vertical) against \(I^{-1}\) (horizontal) is a straight line of
\[ \text{slope } s = E \quad\text{and}\quad \text{intercept on the } R\text{-axis } C = -r \]
So the slope gives the e.m.f. of the battery and the magnitude of the (negative) intercept gives its internal resistance.
Sample table (headings).
R (Ω)
I (A)
\(I^{-1}\) (A\(^{-1}\))
1
-
-
2
-
-
3
-
-
4
-
-
5
-
-
6
-
-
Two precautions.
The key was closed only while taking a reading, to prevent the battery running down and its e.m.f. drifting.
Clean, tight connections were made and the eye placed squarely in front of the meter to avoid parallax when reading the ammeter.
(b)(i) Potential difference. The potential difference between two points in an electric field is the work done in moving one coulomb of positive charge from one point to the other.
Theory. For a cell of e.m.f. E and internal resistance r driving current I through an external resistance R,
\[ E = I(R + r) \;\Rightarrow\; R = \frac{E}{I} - r = E\,I^{-1} - r \]
Expected graph. A plot of R (vertical) against \(I^{-1}\) (horizontal) is a straight line of
\[ \text{slope } s = E \quad\text{and}\quad \text{intercept on the } R\text{-axis } C = -r \]
So the slope gives the e.m.f. of the battery and the magnitude of the (negative) intercept gives its internal resistance.
Sample table (headings).
R (Ω)
I (A)
\(I^{-1}\) (A\(^{-1}\))
1
-
-
2
-
-
3
-
-
4
-
-
5
-
-
6
-
-
Two precautions.
The key was closed only while taking a reading, to prevent the battery running down and its e.m.f. drifting.
Clean, tight connections were made and the eye placed squarely in front of the meter to avoid parallax when reading the ammeter.
(b)(i) Potential difference. The potential difference between two points in an electric field is the work done in moving one coulomb of positive charge from one point to the other.
