WAEC SSCE - Physics - 1991 (Essay)
Question 1 Rapport
(a)(i) By means of a labelled diagram, describe the mode uf operation of a modern X-ray tube.
(ii) State the energy transformation which takes place during the operation.
(b) Explain the terms hardness and intensity as applied to X-rays
(c)(i) State three uses of X-rays
(ii) State one hazard of over-exposure to X-rays in a radiological laboratory, indicating two safety precautions.
(a)(i) Modern X-ray (Coolidge) tube
The tube is highly evacuated. A low-voltage supply heats the tungsten filament at the cathode. The hot filament emits electrons by thermionic emission. A very high potential difference between the cathode and the anode accelerates and focuses the electrons towards the tungsten target.
When the high-speed electrons strike the tungsten target, they are suddenly decelerated. A small fraction of their kinetic energy is emitted as X-rays, which leave through the window. Most of the energy becomes heat; hence the tungsten target is mounted in copper and the anode is cooled.
(a)(ii) Energy transformation
Electrical energy is converted to kinetic energy of the electrons, and then mainly to heat energy and partly to X-ray electromagnetic radiation:
\[\text{electrical energy} \rightarrow \text{kinetic energy of electrons} \rightarrow \text{heat energy} + \text{X-ray energy}.\]
(b) Hardness and intensity of X-rays
- Hardness is the penetrating power, or quality, of an X-ray beam. Hard X-rays have high frequency, short wavelength and high photon energy, and are therefore more penetrating. Their hardness is increased by increasing the accelerating potential difference across the tube. X-rays produced at lower voltage are softer and less penetrating.
- Intensity is the quantity of X-rays emitted per second, that is, the number of X-ray photons in the beam. It is increased by increasing the filament heating current, since more electrons are then emitted and strike the target.
(c)(i) Uses of X-rays
- Radiography of the body, for example to locate fractured bones.
- Detecting internal cracks and defects in welds, metal castings and other engineering materials.
- Studying crystal structure by X-ray diffraction.
(c)(ii) Hazard and precautions
Hazard: Over-exposure damages living cells and tissues and may cause radiation burns, mutations or cancer.
Safety precautions:
- Use lead shielding, such as lead-lined screens or a lead apron, between the operator and the X-ray source.
- Keep exposure time as short as possible and operate the equipment from a safe distance, outside the direct beam.
Détails de la réponse
(a)(i) Modern X-ray (Coolidge) tube
The tube is highly evacuated. A low-voltage supply heats the tungsten filament at the cathode. The hot filament emits electrons by thermionic emission. A very high potential difference between the cathode and the anode accelerates and focuses the electrons towards the tungsten target.
When the high-speed electrons strike the tungsten target, they are suddenly decelerated. A small fraction of their kinetic energy is emitted as X-rays, which leave through the window. Most of the energy becomes heat; hence the tungsten target is mounted in copper and the anode is cooled.
(a)(ii) Energy transformation
Electrical energy is converted to kinetic energy of the electrons, and then mainly to heat energy and partly to X-ray electromagnetic radiation:
\[\text{electrical energy} \rightarrow \text{kinetic energy of electrons} \rightarrow \text{heat energy} + \text{X-ray energy}.\]
(b) Hardness and intensity of X-rays
- Hardness is the penetrating power, or quality, of an X-ray beam. Hard X-rays have high frequency, short wavelength and high photon energy, and are therefore more penetrating. Their hardness is increased by increasing the accelerating potential difference across the tube. X-rays produced at lower voltage are softer and less penetrating.
- Intensity is the quantity of X-rays emitted per second, that is, the number of X-ray photons in the beam. It is increased by increasing the filament heating current, since more electrons are then emitted and strike the target.
(c)(i) Uses of X-rays
- Radiography of the body, for example to locate fractured bones.
- Detecting internal cracks and defects in welds, metal castings and other engineering materials.
- Studying crystal structure by X-ray diffraction.
(c)(ii) Hazard and precautions
Hazard: Over-exposure damages living cells and tissues and may cause radiation burns, mutations or cancer.
Safety precautions:
- Use lead shielding, such as lead-lined screens or a lead apron, between the operator and the X-ray source.
- Keep exposure time as short as possible and operate the equipment from a safe distance, outside the direct beam.
Question 2 Rapport
(a) Explain the terms reactance and impedance in an a.c circuit.
(b) A source of e.m.f. 240 v and frequency 50 Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 10A, the potential difference across the resistor is 140V and that across the inductor is 50V. Draw the vector diagram of the potential difference across the inductor, the capacitor and the resistor.
Calculate the (i) potential difference across the capacitor; (ii) capacitance of the capacitor; (iii) inductance of the inductor.
(a) Reactance and impedance
Reactance is the opposition offered to alternating current by an inductor or a capacitor. It is measured in ohms, \(\Omega\).
For an inductor, \(X_L=2\pi fL\), while for a capacitor, \(X_C=\dfrac{1}{2\pi fC}\).
Impedance is the total opposition offered to alternating current by a circuit containing resistance and reactance. For a series RLC circuit,
\[Z=\sqrt{R^2+(X_L-X_C)^2}.\]
(b) Vector diagram
Take the current, and hence \(V_R\), as the horizontal reference. \(V_L\) leads the current by \(90^\circ\), while \(V_C\) lags the current by \(90^\circ\).
From the voltage phasor diagram,
\[V^2=V_R^2+(V_L-V_C)^2.\]
Given \(V=240\text{ V}\), \(V_R=140\text{ V}\), and \(V_L=50\text{ V}\),
\[240^2=140^2+(50-V_C)^2\]
\[(50-V_C)^2=57600-19600=38000\]
\[50-V_C=-\sqrt{38000}=-194.94\]
\[V_C=244.94\text{ V}.\]
(i) Potential difference across the capacitor:
\[\boxed{V_C\approx245\text{ V}}\]
(ii) Capacitance of the capacitor
\[X_C=\frac{V_C}{I}=\frac{244.94}{10}=24.49\ \Omega\]
\[C=\frac{1}{2\pi fX_C}=\frac{1}{2\pi\times50\times24.49}\]
\[\boxed{C=1.30\times10^{-4}\text{ F}=130\ \mu\text{F}}\]
(iii) Inductance of the inductor
\[X_L=\frac{V_L}{I}=\frac{50}{10}=5.0\ \Omega\]
\[L=\frac{X_L}{2\pi f}=\frac{5.0}{2\pi\times50}\]
\[\boxed{L=1.59\times10^{-2}\text{ H}\approx0.016\text{ H}}\]
Détails de la réponse
(a) Reactance and impedance
Reactance is the opposition offered to alternating current by an inductor or a capacitor. It is measured in ohms, \(\Omega\).
For an inductor, \(X_L=2\pi fL\), while for a capacitor, \(X_C=\dfrac{1}{2\pi fC}\).
Impedance is the total opposition offered to alternating current by a circuit containing resistance and reactance. For a series RLC circuit,
\[Z=\sqrt{R^2+(X_L-X_C)^2}.\]
(b) Vector diagram
Take the current, and hence \(V_R\), as the horizontal reference. \(V_L\) leads the current by \(90^\circ\), while \(V_C\) lags the current by \(90^\circ\).
From the voltage phasor diagram,
\[V^2=V_R^2+(V_L-V_C)^2.\]
Given \(V=240\text{ V}\), \(V_R=140\text{ V}\), and \(V_L=50\text{ V}\),
\[240^2=140^2+(50-V_C)^2\]
\[(50-V_C)^2=57600-19600=38000\]
\[50-V_C=-\sqrt{38000}=-194.94\]
\[V_C=244.94\text{ V}.\]
(i) Potential difference across the capacitor:
\[\boxed{V_C\approx245\text{ V}}\]
(ii) Capacitance of the capacitor
\[X_C=\frac{V_C}{I}=\frac{244.94}{10}=24.49\ \Omega\]
\[C=\frac{1}{2\pi fX_C}=\frac{1}{2\pi\times50\times24.49}\]
\[\boxed{C=1.30\times10^{-4}\text{ F}=130\ \mu\text{F}}\]
(iii) Inductance of the inductor
\[X_L=\frac{V_L}{I}=\frac{50}{10}=5.0\ \Omega\]
\[L=\frac{X_L}{2\pi f}=\frac{5.0}{2\pi\times50}\]
\[\boxed{L=1.59\times10^{-2}\text{ H}\approx0.016\text{ H}}\]
Question 3 Rapport
(a) Distinguish between temperature and heat. State the units in which they are measured
(ii) State two physical properties used for measuring temperature.
(b) (i) Describe, with the aid of a diagram, how the upper fixed point is determined for a mecury-in-glass thermometer. State one precaution to ensure accurate results
(ii) State one advantage which a constant-volume gas thermometer has over other thermometers and one reason why it is seldom used as an everyday laboratory instrument.
(c) Using the kinetic theory of matter explain why evaporation causes cooling.
(a)(i) Temperature and heat
| Temperature | Heat |
|---|---|
| Temperature is the degree of hotness or coldness of a body. It determines the direction in which heat flows. | Heat is energy transferred from a body at a higher temperature to a body at a lower temperature because of a temperature difference. |
| It is measured in kelvin, K, the SI unit. Degree Celsius, b0C, is also commonly used. | Its SI unit is the joule, J. |
(a)(ii) Two physical properties used in thermometry are:
- change in volume of a liquid with temperature;
- change in electrical resistance of a conductor with temperature.
(b)(i) Determination of the upper fixed point, or steam point
The apparatus is arranged as shown below.
Pure water is boiled in a hypsometer. The bulb of the mercury-in-glass thermometer is surrounded by dry steam above the boiling water. The pressure of the steam is measured with the manometer and is adjusted to standard atmospheric pressure, 76 cm Hg (760 mm Hg). When the mercury thread becomes steady, its level is marked on the stem. This mark is the upper fixed point, equal to 100 b0C.
Precaution: Ensure that the thermometer bulb is completely surrounded by steam and does not touch the boiling water or the wall of the vessel.
(b)(ii)
- Advantage: A constant-volume gas thermometer gives very accurate measurements over a wide temperature range, since the expansion of the containing glass is negligible compared with the change in pressure of the gas. It is therefore suitable as a standard thermometer for calibration.
- Reason it is seldom used routinely: It is bulky, delicate and inconvenient to handle, and it gives readings slowly.
(c) Why evaporation causes cooling
Liquid molecules are in continual random motion and have different kinetic energies. At the surface, the fastest molecules, having sufficiently high kinetic energy, escape from the liquid as vapour. These molecules carry away more than the average kinetic energy. The average kinetic energy of the molecules left in the liquid therefore decreases. Since temperature is proportional to the average kinetic energy of the molecules, the temperature of the remaining liquid falls, causing cooling.
Détails de la réponse
(a)(i) Temperature and heat
| Temperature | Heat |
|---|---|
| Temperature is the degree of hotness or coldness of a body. It determines the direction in which heat flows. | Heat is energy transferred from a body at a higher temperature to a body at a lower temperature because of a temperature difference. |
| It is measured in kelvin, K, the SI unit. Degree Celsius, b0C, is also commonly used. | Its SI unit is the joule, J. |
(a)(ii) Two physical properties used in thermometry are:
- change in volume of a liquid with temperature;
- change in electrical resistance of a conductor with temperature.
(b)(i) Determination of the upper fixed point, or steam point
The apparatus is arranged as shown below.
Pure water is boiled in a hypsometer. The bulb of the mercury-in-glass thermometer is surrounded by dry steam above the boiling water. The pressure of the steam is measured with the manometer and is adjusted to standard atmospheric pressure, 76 cm Hg (760 mm Hg). When the mercury thread becomes steady, its level is marked on the stem. This mark is the upper fixed point, equal to 100 b0C.
Precaution: Ensure that the thermometer bulb is completely surrounded by steam and does not touch the boiling water or the wall of the vessel.
(b)(ii)
- Advantage: A constant-volume gas thermometer gives very accurate measurements over a wide temperature range, since the expansion of the containing glass is negligible compared with the change in pressure of the gas. It is therefore suitable as a standard thermometer for calibration.
- Reason it is seldom used routinely: It is bulky, delicate and inconvenient to handle, and it gives readings slowly.
(c) Why evaporation causes cooling
Liquid molecules are in continual random motion and have different kinetic energies. At the surface, the fastest molecules, having sufficiently high kinetic energy, escape from the liquid as vapour. These molecules carry away more than the average kinetic energy. The average kinetic energy of the molecules left in the liquid therefore decreases. Since temperature is proportional to the average kinetic energy of the molecules, the temperature of the remaining liquid falls, causing cooling.
Question 4 Rapport
(a) Explain the terms:- uniform acceleration and average speed.
(b) A body at rest is given an initial uniform acceleration of 8.0ms\(^{-2}\) for 30s after which the acceleration is reduced to 5.0ms\(^{-2}\) for the next 20s. The body maintains the speed attained for 60s after which it is brought to rest in 20s. Draw the velocity-time graph of the motion using the information given above.
(c) Using the graph, calculate the: (i) maximum speed attained during the motion; (ii) average retardation as the body is being brought to rest; (iii) total distance travelled during the first 50s; (iv) average speed during the same interval as in (ii).
(a) Definitions
Uniform acceleration means that velocity changes by equal amounts in equal time intervals. Therefore, the acceleration is constant.
\[ a=\frac{v-u}{t} \]Average speed is the total distance travelled divided by the total time taken:
\[ \text{average speed}=\frac{\text{total distance travelled}}{\text{total time taken}} \](b) Velocity–time graph
Calculate the speeds at the end of each stage:
\[ v=0+(8.0)(30)=240\ \text{m s}^{-1} \] \[ v=240+(5.0)(20)=340\ \text{m s}^{-1} \]The body then travels at \(340\ \text{m s}^{-1}\) for \(60\ \text{s}\), from \(t=50\ \text{s}\) to \(t=110\ \text{s}\), before slowing uniformly to rest by \(t=130\ \text{s}\).
(c)(i) Maximum speed
The maximum speed is the highest value on the graph:
\[ \boxed{340\ \text{m s}^{-1}} \](c)(ii) Average retardation while coming to rest
The body slows from \(340\ \text{m s}^{-1}\) to \(0\ \text{m s}^{-1}\) in \(20\ \text{s}\).
\[ a=\frac{v-u}{t}=\frac{0-340}{20}=-17\ \text{m s}^{-2} \]The acceleration is negative because the velocity is decreasing. The retardation, stated as a positive magnitude, is:
\[ \boxed{17\ \text{m s}^{-2}} \](c)(iii) Total distance travelled during the first \(50\ \text{s}\)
Distance is the area under a velocity–time graph.
From \(0\) to \(30\ \text{s}\), the area is a triangle:
\[ s_1=\frac{1}{2}\times30\times240=3600\ \text{m} \]From \(30\) to \(50\ \text{s}\), the area is a trapezium:
\[ s_2=\frac{1}{2}(240+340)\times20=5800\ \text{m} \] \[ \text{Total distance}=3600+5800=\boxed{9400\ \text{m}} \](c)(iv) Average speed during the first \(50\ \text{s}\)
\[ \text{average speed}=\frac{9400}{50}=\boxed{188\ \text{m s}^{-1}} \]Important correction: The values \(240\ \text{m s}^{-1}\), \(200\ \text{m s}^{-1}\), and \(160\ \text{m s}^{-1}\) in the supplied reference answer do not follow from the stated motion. After reaching \(240\ \text{m s}^{-1}\) at \(30\ \text{s}\), the body continues to accelerate at \(5.0\ \text{m s}^{-2}\) for \(20\ \text{s}\), so its speed must increase by \(100\ \text{m s}^{-1}\) to \(340\ \text{m s}^{-1}\).
Détails de la réponse
(a) Definitions
Uniform acceleration means that velocity changes by equal amounts in equal time intervals. Therefore, the acceleration is constant.
\[ a=\frac{v-u}{t} \]Average speed is the total distance travelled divided by the total time taken:
\[ \text{average speed}=\frac{\text{total distance travelled}}{\text{total time taken}} \](b) Velocity–time graph
Calculate the speeds at the end of each stage:
\[ v=0+(8.0)(30)=240\ \text{m s}^{-1} \] \[ v=240+(5.0)(20)=340\ \text{m s}^{-1} \]The body then travels at \(340\ \text{m s}^{-1}\) for \(60\ \text{s}\), from \(t=50\ \text{s}\) to \(t=110\ \text{s}\), before slowing uniformly to rest by \(t=130\ \text{s}\).
(c)(i) Maximum speed
The maximum speed is the highest value on the graph:
\[ \boxed{340\ \text{m s}^{-1}} \](c)(ii) Average retardation while coming to rest
The body slows from \(340\ \text{m s}^{-1}\) to \(0\ \text{m s}^{-1}\) in \(20\ \text{s}\).
\[ a=\frac{v-u}{t}=\frac{0-340}{20}=-17\ \text{m s}^{-2} \]The acceleration is negative because the velocity is decreasing. The retardation, stated as a positive magnitude, is:
\[ \boxed{17\ \text{m s}^{-2}} \](c)(iii) Total distance travelled during the first \(50\ \text{s}\)
Distance is the area under a velocity–time graph.
From \(0\) to \(30\ \text{s}\), the area is a triangle:
\[ s_1=\frac{1}{2}\times30\times240=3600\ \text{m} \]From \(30\) to \(50\ \text{s}\), the area is a trapezium:
\[ s_2=\frac{1}{2}(240+340)\times20=5800\ \text{m} \] \[ \text{Total distance}=3600+5800=\boxed{9400\ \text{m}} \](c)(iv) Average speed during the first \(50\ \text{s}\)
\[ \text{average speed}=\frac{9400}{50}=\boxed{188\ \text{m s}^{-1}} \]Important correction: The values \(240\ \text{m s}^{-1}\), \(200\ \text{m s}^{-1}\), and \(160\ \text{m s}^{-1}\) in the supplied reference answer do not follow from the stated motion. After reaching \(240\ \text{m s}^{-1}\) at \(30\ \text{s}\), the body continues to accelerate at \(5.0\ \text{m s}^{-2}\) for \(20\ \text{s}\), so its speed must increase by \(100\ \text{m s}^{-1}\) to \(340\ \text{m s}^{-1}\).
