WAEC SSCE - Physics - 2011 (Essay)
Ajụjụ 1 Ripọtì
An electron of charge 1.60 x 10\(^{-19}\) C is accelerated under a potential difference of 1.0 x 10\(^5\) V. Calculate the energy of the electron in joules.
The energy gained by a charge \(Q\) accelerated through a potential difference \(V\) is the work done on it:
\[ E = QV. \]
Substituting \(Q = 1.60\times10^{-19}\,\text{C}\) and \(V = 1.0\times10^{5}\,\text{V}\):
\[ E = (1.60\times10^{-19})\times(1.0\times10^{5}) = 1.60\times10^{-14}\,\text{J}. \]
The energy of the electron is \(1.60\times10^{-14}\,\text{J}\).
Akọwa Nkọwa
The energy gained by a charge \(Q\) accelerated through a potential difference \(V\) is the work done on it:
\[ E = QV. \]
Substituting \(Q = 1.60\times10^{-19}\,\text{C}\) and \(V = 1.0\times10^{5}\,\text{V}\):
\[ E = (1.60\times10^{-19})\times(1.0\times10^{5}) = 1.60\times10^{-14}\,\text{J}. \]
The energy of the electron is \(1.60\times10^{-14}\,\text{J}\).
Ajụjụ 2 Ripọtì
The uncertainty in determining the duration during which an electron remains in a particular energy level before returning to the ground state is 2.0 x 10\(^{-9}\)s. Calculate the uncertainty in determining its energy at that level. (\(\frac{h}{2 \pi}\) = h = 1.054 x 10\(^{-34}\)Js)
By Heisenberg's uncertainty principle for energy and time,
\[ \Delta E \,\Delta t \geq \frac{h}{2\pi} = \hbar. \]
Taking the limiting (minimum) case, the uncertainty in energy is
\[ \Delta E = \frac{\hbar}{\Delta t} = \frac{1.054\times10^{-34}}{2.0\times10^{-9}}. \]
\[ \Delta E = 5.27\times10^{-26}\,\text{J}. \]
The uncertainty in determining the energy of that level is about \(5.27\times10^{-26}\,\text{J}\).
Akọwa Nkọwa
By Heisenberg's uncertainty principle for energy and time,
\[ \Delta E \,\Delta t \geq \frac{h}{2\pi} = \hbar. \]
Taking the limiting (minimum) case, the uncertainty in energy is
\[ \Delta E = \frac{\hbar}{\Delta t} = \frac{1.054\times10^{-34}}{2.0\times10^{-9}}. \]
\[ \Delta E = 5.27\times10^{-26}\,\text{J}. \]
The uncertainty in determining the energy of that level is about \(5.27\times10^{-26}\,\text{J}\).
Ajụjụ 3 Ripọtì
State three methods of polarizing an unpolarized light
Three methods of polarizing an unpolarized beam of light are:
- Polarization by reflection - when ordinary light is reflected from a transparent surface (for example water or glass) at the polarizing (Brewster) angle, the reflected light becomes plane polarized.
- Polarization by selective absorption (using a polaroid/dichroic material) - a polaroid sheet transmits vibrations in only one plane and absorbs those in all other planes, producing plane polarized light.
- Polarization by double refraction (using a doubly-refracting crystal such as calcite/Nicol prism) - the crystal splits the light into two plane polarized rays vibrating in mutually perpendicular planes.
(Polarization by scattering, as occurs with sunlight in the atmosphere, is also acceptable.)
Akọwa Nkọwa
Three methods of polarizing an unpolarized beam of light are:
- Polarization by reflection - when ordinary light is reflected from a transparent surface (for example water or glass) at the polarizing (Brewster) angle, the reflected light becomes plane polarized.
- Polarization by selective absorption (using a polaroid/dichroic material) - a polaroid sheet transmits vibrations in only one plane and absorbs those in all other planes, producing plane polarized light.
- Polarization by double refraction (using a doubly-refracting crystal such as calcite/Nicol prism) - the crystal splits the light into two plane polarized rays vibrating in mutually perpendicular planes.
(Polarization by scattering, as occurs with sunlight in the atmosphere, is also acceptable.)
Ajụjụ 4 Ripọtì
(a) Explain diffusion.
(b) Give one reason why the rate of diffusion is higher in gases than in liquids at the same temperature.
(a) Diffusion: Diffusion is the net movement of the molecules of a substance from a region where they are more concentrated to a region where they are less concentrated, arising from the continuous random motion of the molecules, until they become uniformly distributed.
(b) Why diffusion is faster in gases than in liquids (at the same temperature): In a gas the molecules are much farther apart and the intermolecular forces are very weak, so the molecules move freely at high speeds over large distances between collisions. In a liquid the molecules are close together with strong intermolecular forces, so their movement is slower and more restricted. Hence, at the same temperature, molecules of a gas diffuse much faster than those of a liquid.
Akọwa Nkọwa
(a) Diffusion: Diffusion is the net movement of the molecules of a substance from a region where they are more concentrated to a region where they are less concentrated, arising from the continuous random motion of the molecules, until they become uniformly distributed.
(b) Why diffusion is faster in gases than in liquids (at the same temperature): In a gas the molecules are much farther apart and the intermolecular forces are very weak, so the molecules move freely at high speeds over large distances between collisions. In a liquid the molecules are close together with strong intermolecular forces, so their movement is slower and more restricted. Hence, at the same temperature, molecules of a gas diffuse much faster than those of a liquid.
Ajụjụ 5 Ripọtì
State: (a) the difference between plane polarized light and ordinary light;
(b) two uses of polaroids.
(a) Difference between plane polarized light and ordinary light: In ordinary (unpolarized) light the vibrations of the electric field take place in all directions (all planes) perpendicular to the direction of propagation. In plane polarized light the vibrations are restricted to a single plane containing the direction of propagation.
(b) Two uses of polaroids:
- In polaroid sunglasses, to reduce glare from light reflected off surfaces such as water and roads.
- In the control of light intensity and in photography (polarizing filters), and in liquid-crystal displays (LCDs) and stress analysis of transparent materials.
Akọwa Nkọwa
(a) Difference between plane polarized light and ordinary light: In ordinary (unpolarized) light the vibrations of the electric field take place in all directions (all planes) perpendicular to the direction of propagation. In plane polarized light the vibrations are restricted to a single plane containing the direction of propagation.
(b) Two uses of polaroids:
- In polaroid sunglasses, to reduce glare from light reflected off surfaces such as water and roads.
- In the control of light intensity and in photography (polarizing filters), and in liquid-crystal displays (LCDs) and stress analysis of transparent materials.
Ajụjụ 6 Ripọtì
A stone projected horizontally from the top of a tower with a speed of 4 ms\(^{-1}\) lands on the level ground at a horizontal distance of 25 m from the foot of the tower. Calculate the height of the tower. = 10 ms\(^{-2}\)]
For a projectile launched horizontally, the horizontal and vertical motions are independent.
Step 1 - find the time of flight from the horizontal motion. Horizontally the speed is constant at \(u = 4\,\text{m s}^{-1}\):
\[ R = u\,t \;\Rightarrow\; 25 = 4t \;\Rightarrow\; t = 6.25\,\text{s}. \]
Step 2 - find the height from the vertical motion. The stone starts with zero vertical velocity and falls freely:
\[ h = \tfrac{1}{2} g t^{2} = \tfrac{1}{2}\times10\times(6.25)^{2}. \]
\[ h = 5\times39.06 \approx 195.3\,\text{m}. \]
The height of the tower is approximately 195.3 m.
Akọwa Nkọwa
For a projectile launched horizontally, the horizontal and vertical motions are independent.
Step 1 - find the time of flight from the horizontal motion. Horizontally the speed is constant at \(u = 4\,\text{m s}^{-1}\):
\[ R = u\,t \;\Rightarrow\; 25 = 4t \;\Rightarrow\; t = 6.25\,\text{s}. \]
Step 2 - find the height from the vertical motion. The stone starts with zero vertical velocity and falls freely:
\[ h = \tfrac{1}{2} g t^{2} = \tfrac{1}{2}\times10\times(6.25)^{2}. \]
\[ h = 5\times39.06 \approx 195.3\,\text{m}. \]
The height of the tower is approximately 195.3 m.
Ajụjụ 7 Ripọtì
Explain why water in a narrow glass tube has a concave meniscus while mercury, in the same tube, has a convex meniscus.
The shape of a meniscus depends on the relative strengths of two forces: cohesion (the attraction between molecules of the same liquid) and adhesion (the attraction between the liquid molecules and the molecules of the glass).
Water (concave meniscus): For water in a glass tube, the adhesive force between water and glass is greater than the cohesive force between the water molecules. The water is therefore pulled up the walls of the tube, wetting the glass, and its surface curves upward at the edges to give a concave meniscus.
Mercury (convex meniscus): For mercury, the cohesive force between the mercury molecules is greater than the adhesive force between mercury and glass. The mercury therefore does not wet the glass; it is pulled together and depressed at the walls, so its surface bulges upward in the middle to give a convex meniscus.
Akọwa Nkọwa
The shape of a meniscus depends on the relative strengths of two forces: cohesion (the attraction between molecules of the same liquid) and adhesion (the attraction between the liquid molecules and the molecules of the glass).
Water (concave meniscus): For water in a glass tube, the adhesive force between water and glass is greater than the cohesive force between the water molecules. The water is therefore pulled up the walls of the tube, wetting the glass, and its surface curves upward at the edges to give a concave meniscus.
Mercury (convex meniscus): For mercury, the cohesive force between the mercury molecules is greater than the adhesive force between mercury and glass. The mercury therefore does not wet the glass; it is pulled together and depressed at the walls, so its surface bulges upward in the middle to give a convex meniscus.
Ajụjụ 8 Ripọtì
(a)(i) State two advantages of alcohol over mercury as a thermometric liquid.
(ii) When the bulb of a thermometer is placed in a beaker of hot water, the level of the mercury first falls and then rises gradually. Explain this observation.
(b) List two factors, other than temperature, that affect the rate of evaporation of a liquid.
(b) A block of lead of mass 100 kg in a crucible and at a temperature of 40 °C was placed in an electric furnace rated 10 kW. If the melting point of lead is 320 °C, calculate the:
(i) quantity of heat required to heat the lead to its melting point;
(ii) additional heat energy required to melt the lead;
(iii) time taken to supply this additional energy. (Specific heat capacity of lead = 120 Jkg\(^{-1}\) K\(^{-1}\) (Specific Idtent heat of fusion lead = 2.5 x 10\(^{4} JK^{-1}\))
(d) State two precautions necessary in an experiment to determine the specific latent heat of vaporization of water.
(a)(i) Two advantages of alcohol over mercury as a thermometric liquid: (1) Alcohol has a much lower freezing point (about -115 °C), so it can measure very low temperatures where mercury would freeze; (2) Alcohol expands more (larger expansivity) for a given temperature rise, making the thermometer more sensitive. (It is also cheaper and can be coloured for easy reading.)
(a)(ii) Mercury level first falls then rises: When the bulb is placed in hot water, the glass bulb is heated first and expands before the heat reaches the mercury. The expansion of the bulb increases the space available, so the mercury level first falls. Shortly afterwards the heat reaches the mercury, which expands much more than the glass, and the level then rises steadily.
(b) Two factors (other than temperature) affecting the rate of evaporation: (1) Surface area exposed; (2) Draught/movement of air over the surface. (Humidity of the surrounding air and nature of the liquid are also acceptable.)
(c) Heating the lead:
(i) Heat to raise the lead to its melting point:
\[ Q = mc\,\Delta\theta = 100\times120\times(320-40) = 100\times120\times280 = 3.36\times10^{6}\,\text{J}. \]
(ii) Additional heat to melt the lead:
\[ Q = mL = 100\times2.5\times10^{4} = 2.5\times10^{6}\,\text{J}. \]
(iii) Time to supply this additional energy (furnace power \(P = 10\,\text{kW} = 10000\,\text{W}\)):
\[ t = \frac{Q}{P} = \frac{2.5\times10^{6}}{10000} = 250\,\text{s}. \]
(d) Two precautions in determining specific latent heat of vaporization of water: (1) Lag (insulate) the apparatus to minimise heat loss to the surroundings; (2) Ensure steady conditions and collect the condensed water only after the water has been boiling steadily (take readings at constant rate). (Correcting for heat losses by repeating at two heating rates is also acceptable.)
Akọwa Nkọwa
(a)(i) Two advantages of alcohol over mercury as a thermometric liquid: (1) Alcohol has a much lower freezing point (about -115 °C), so it can measure very low temperatures where mercury would freeze; (2) Alcohol expands more (larger expansivity) for a given temperature rise, making the thermometer more sensitive. (It is also cheaper and can be coloured for easy reading.)
(a)(ii) Mercury level first falls then rises: When the bulb is placed in hot water, the glass bulb is heated first and expands before the heat reaches the mercury. The expansion of the bulb increases the space available, so the mercury level first falls. Shortly afterwards the heat reaches the mercury, which expands much more than the glass, and the level then rises steadily.
(b) Two factors (other than temperature) affecting the rate of evaporation: (1) Surface area exposed; (2) Draught/movement of air over the surface. (Humidity of the surrounding air and nature of the liquid are also acceptable.)
(c) Heating the lead:
(i) Heat to raise the lead to its melting point:
\[ Q = mc\,\Delta\theta = 100\times120\times(320-40) = 100\times120\times280 = 3.36\times10^{6}\,\text{J}. \]
(ii) Additional heat to melt the lead:
\[ Q = mL = 100\times2.5\times10^{4} = 2.5\times10^{6}\,\text{J}. \]
(iii) Time to supply this additional energy (furnace power \(P = 10\,\text{kW} = 10000\,\text{W}\)):
\[ t = \frac{Q}{P} = \frac{2.5\times10^{6}}{10000} = 250\,\text{s}. \]
(d) Two precautions in determining specific latent heat of vaporization of water: (1) Lag (insulate) the apparatus to minimise heat loss to the surroundings; (2) Ensure steady conditions and collect the condensed water only after the water has been boiling steadily (take readings at constant rate). (Correcting for heat losses by repeating at two heating rates is also acceptable.)
Ajụjụ 9 Ripọtì
(a)(i) What is an eclipse?
(ii) List the three types of eclipse.
(b) A student in a lecture theatre can read from the board clearly but requires a pair of spectacles to read from a book.
(i) What eye defect has this student?
(ii) What type of lens is needed to correct the eye defect?
(iii) The focal length of the lens used to correct this defect is 10cm. Calculate the power of the lens.
(c) A car B moves towards a stationary car A. If B produces an ultrasonic sound at a point and it takes 5.6 x 10\(^{-3}\)s for a beep to be heard in B, calculate the distance between the two cars at that instant. (Speed. sound in air = 340 ms\(^-\))
(d) The image of an object is located 9 cm behind a convex mirror. If magnification produced is 0.6, calculate the focal length of the mirror.
(a)(i) Eclipse: An eclipse is the total or partial disappearance of a heavenly body (the sun or the moon) from view, caused when the shadow of one body falls on another as the sun, earth and moon come to lie nearly in a straight line.
(a)(ii) Three types of eclipse:
- Eclipse of the sun (solar eclipse)
- Eclipse of the moon (lunar eclipse)
- Annular eclipse
(b)(i) Eye defect: The student sees distant objects (the board) clearly but cannot focus on near objects (the book), so the defect is long sight (hypermetropia).
(b)(ii) Correcting lens: A convex (converging) lens.
(b)(iii) Power of the lens: \(f = 10\,\text{cm} = 0.10\,\text{m}\).
\[ P = \frac{1}{f} = \frac{1}{0.10} = +10\,\text{D (dioptres)}. \]
(c) Distance between the cars: The ultrasonic beep travels from B to A and is reflected back to B, so it covers a total path of \(2d\) in \(t = 5.6\times10^{-3}\,\text{s}\).
\[ 2d = vt = 340\times5.6\times10^{-3} = 1.904\,\text{m}, \]
\[ d = \frac{1.904}{2} = 0.952\,\text{m}. \]
The two cars are 0.952 m apart at that instant.
(d) Focal length of the convex mirror: The image is virtual, formed \(9\,\text{cm}\) behind the mirror, so \(v = -9\,\text{cm}\). The magnification \(m = \dfrac{-v}{u} = 0.6\) gives \(u = \dfrac{9}{0.6} = 15\,\text{cm}\).
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-9} + \frac{1}{15} = -0.1111 + 0.0667 = -0.0444, \]
\[ f = -22.5\,\text{cm}. \]
The focal length of the convex mirror is 22.5 cm, the negative sign confirming that it is a diverging (convex) mirror.
Akọwa Nkọwa
(a)(i) Eclipse: An eclipse is the total or partial disappearance of a heavenly body (the sun or the moon) from view, caused when the shadow of one body falls on another as the sun, earth and moon come to lie nearly in a straight line.
(a)(ii) Three types of eclipse:
- Eclipse of the sun (solar eclipse)
- Eclipse of the moon (lunar eclipse)
- Annular eclipse
(b)(i) Eye defect: The student sees distant objects (the board) clearly but cannot focus on near objects (the book), so the defect is long sight (hypermetropia).
(b)(ii) Correcting lens: A convex (converging) lens.
(b)(iii) Power of the lens: \(f = 10\,\text{cm} = 0.10\,\text{m}\).
\[ P = \frac{1}{f} = \frac{1}{0.10} = +10\,\text{D (dioptres)}. \]
(c) Distance between the cars: The ultrasonic beep travels from B to A and is reflected back to B, so it covers a total path of \(2d\) in \(t = 5.6\times10^{-3}\,\text{s}\).
\[ 2d = vt = 340\times5.6\times10^{-3} = 1.904\,\text{m}, \]
\[ d = \frac{1.904}{2} = 0.952\,\text{m}. \]
The two cars are 0.952 m apart at that instant.
(d) Focal length of the convex mirror: The image is virtual, formed \(9\,\text{cm}\) behind the mirror, so \(v = -9\,\text{cm}\). The magnification \(m = \dfrac{-v}{u} = 0.6\) gives \(u = \dfrac{9}{0.6} = 15\,\text{cm}\).
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-9} + \frac{1}{15} = -0.1111 + 0.0667 = -0.0444, \]
\[ f = -22.5\,\text{cm}. \]
The focal length of the convex mirror is 22.5 cm, the negative sign confirming that it is a diverging (convex) mirror.
Ajụjụ 10 Ripọtì
(a). State: (a) two applications of electrolysis in an industry
(b) one application of electrolysis in a school laboratory.
(a) Two applications of electrolysis in industry:
- Electroplating - coating objects with a thin layer of metal (for example chromium, silver or gold plating) to protect them from corrosion or to improve appearance.
- Extraction and purification (refining) of metals - such as the extraction of aluminium from molten alumina and the purification of copper by electrolytic refining. (Electrolytic production of chemicals such as caustic soda and chlorine is also acceptable.)
(b) One application of electrolysis in a school laboratory: Electrolysis of acidified water to prepare and collect hydrogen and oxygen gases (for example in a Hofmann voltameter). (Electroplating of a small object or determining the chemical equivalent of a metal are also acceptable.)
Akọwa Nkọwa
(a) Two applications of electrolysis in industry:
- Electroplating - coating objects with a thin layer of metal (for example chromium, silver or gold plating) to protect them from corrosion or to improve appearance.
- Extraction and purification (refining) of metals - such as the extraction of aluminium from molten alumina and the purification of copper by electrolytic refining. (Electrolytic production of chemicals such as caustic soda and chlorine is also acceptable.)
(b) One application of electrolysis in a school laboratory: Electrolysis of acidified water to prepare and collect hydrogen and oxygen gases (for example in a Hofmann voltameter). (Electroplating of a small object or determining the chemical equivalent of a metal are also acceptable.)
Ajụjụ 11 Ripọtì
(a) What is a vector quantity?
(b) Three vectors \(3\ \text{ms}^{-1}\) N \(45^\circ\) W, \(12\ \text{ms}^{-1}\) W and \(5\ \text{ms}^{-1}\) S act at a point.
(i) Sketch a vector diagram to illustrate the given information.
(ii) Calculate the resultant of the vectors.
(c) In a laboratory experiment to determine the force constant of a spiral spring, the mass or, the spring was varied and the corresponding extensions were measured and recorded as shown in the table below.
| Mass M/g | Weight W/N | Extension e/cm |
| 50 | 6.5 | |
| 100 | 11.0 | |
| 150 | 15.0 | |
| 200 | 20.0 | |
| 250 | 25.0 |
(i) Copy and complete the table. (Take \(g = 10\ \text{ms}^{-2}\))
(ii) Plot a graph with weight, W, on the vertical axis and extension, e, on the horizontal axis.
(iii) Using the graph, determine the force constant of the spring.
(iv) Determine the natural length of the spring if its length was 38.0 cm when loaded with 250 g mass.
(a) A vector quantity is a physical quantity which has both magnitude and direction. Examples include velocity, force and displacement.
(b)(i) Vector diagram
(b)(ii) Resultant of the vectors
Take east as the positive x-direction and north as the positive y-direction.
| Vector | x-component / m s-1 | y-component / m s-1 |
|---|---|---|
| 3 m s-1, N45°W | \(-3\sin45^\circ=-2.12\) | \(+3\cos45^\circ=+2.12\) |
| 12 m s-1, W | \(-12.00\) | \(0\) |
| 5 m s-1, S | \(0\) | \(-5.00\) |
| Resultant components | \(-14.12\) | \(-2.88\) |
\[R=\sqrt{(-14.12)^2+(-2.88)^2}=14.4\ \text{m s}^{-1}\]
\[\theta=\tan^{-1}\left(\frac{2.88}{14.12}\right)=11.5^\circ\]
Hence, the resultant is \(14.4\ \text{m s}^{-1}\), \(11.5^\circ\) south of west, or equivalently S78.5°W.
(c)(i) Completed table
Since \(W=Mg\), with \(g=10\ \text{m s}^{-2}\):
| Mass, \(M\) / g | Weight, \(W\) / N | Extension, \(e\) / cm |
|---|---|---|
| 50 | 0.5 | 6.5 |
| 100 | 1.0 | 11.0 |
| 150 | 1.5 | 15.0 |
| 200 | 2.0 | 20.0 |
| 250 | 2.5 | 25.0 |
(c)(ii) Graph of weight against extension
(c)(iii) The force constant, \(k\), is the gradient of the graph:
Using two widely separated points on the line of best fit, \((10.0\ \text{cm},0.903\ \text{N})\) and \((25.0\ \text{cm},2.530\ \text{N})\),
\[k=\frac{\Delta W}{\Delta e}=\frac{2.530-0.903}{25.0-10.0}=0.1085\ \text{N cm}^{-1}\]
\[k=0.1085\times100=10.85\ \text{N m}^{-1}\approx \boxed{10.8\ \text{N m}^{-1}}\]
(c)(iv) At a mass of 250 g, the extension is 25.0 cm. Therefore,
\[\text{natural length}=38.0-25.0=\boxed{13.0\ \text{cm}}\]
Akọwa Nkọwa
(a) A vector quantity is a physical quantity which has both magnitude and direction. Examples include velocity, force and displacement.
(b)(i) Vector diagram
(b)(ii) Resultant of the vectors
Take east as the positive x-direction and north as the positive y-direction.
| Vector | x-component / m s-1 | y-component / m s-1 |
|---|---|---|
| 3 m s-1, N45°W | \(-3\sin45^\circ=-2.12\) | \(+3\cos45^\circ=+2.12\) |
| 12 m s-1, W | \(-12.00\) | \(0\) |
| 5 m s-1, S | \(0\) | \(-5.00\) |
| Resultant components | \(-14.12\) | \(-2.88\) |
\[R=\sqrt{(-14.12)^2+(-2.88)^2}=14.4\ \text{m s}^{-1}\]
\[\theta=\tan^{-1}\left(\frac{2.88}{14.12}\right)=11.5^\circ\]
Hence, the resultant is \(14.4\ \text{m s}^{-1}\), \(11.5^\circ\) south of west, or equivalently S78.5°W.
(c)(i) Completed table
Since \(W=Mg\), with \(g=10\ \text{m s}^{-2}\):
| Mass, \(M\) / g | Weight, \(W\) / N | Extension, \(e\) / cm |
|---|---|---|
| 50 | 0.5 | 6.5 |
| 100 | 1.0 | 11.0 |
| 150 | 1.5 | 15.0 |
| 200 | 2.0 | 20.0 |
| 250 | 2.5 | 25.0 |
(c)(ii) Graph of weight against extension
(c)(iii) The force constant, \(k\), is the gradient of the graph:
Using two widely separated points on the line of best fit, \((10.0\ \text{cm},0.903\ \text{N})\) and \((25.0\ \text{cm},2.530\ \text{N})\),
\[k=\frac{\Delta W}{\Delta e}=\frac{2.530-0.903}{25.0-10.0}=0.1085\ \text{N cm}^{-1}\]
\[k=0.1085\times100=10.85\ \text{N m}^{-1}\approx \boxed{10.8\ \text{N m}^{-1}}\]
(c)(iv) At a mass of 250 g, the extension is 25.0 cm. Therefore,
\[\text{natural length}=38.0-25.0=\boxed{13.0\ \text{cm}}\]
Ajụjụ 12 Ripọtì
In fig. la and fig.1 b above, Ida and Wb represent the respective loads on a spring placedtiear a 30 cm rule, when in air and when in water:
(a) Identify the force causing a shrink in the spring in fig.(b).
(b) Given that the force constant of the spring is 2.0 x 10\(^{11}\) Nm\(^{-1}\), calculate the work done by the force in causing the shrink.
(a) The force causing the spring to shrink in fig.(b)
When the load is lowered into the water it experiences an upthrust (buoyant force) from the water, acting vertically upwards. This upthrust partly supports the weight of the load, so the tension pulling on the spring is reduced and the spring contracts (shrinks). The force responsible for the shrink is therefore the upthrust of the water on the load.
(b) Work done by the force in causing the shrink
Reading the contraction from the metre rule. The pointer attached to the load marks the position of the lower end of the spring against the fixed rule:
- In air (fig. a) the pointer reads \(20\ \text{cm}\) (the spring is fully stretched by the whole weight \(W_a\)).
- In water (fig. b) the upthrust lifts the load, so the pointer rises to \(18\ \text{cm}\).
The contraction (shrink) of the spring is the difference of the two readings:
\[x = 20\ \text{cm}-18\ \text{cm}=2\ \text{cm}=2\times10^{-2}\ \text{m}.\]
Work done. For a spring obeying Hooke's law, the work done in changing its length by \(x\) against a force constant \(k\) is
\[W=\tfrac{1}{2}kx^{2}.\]
With \(k=2.0\times10^{11}\ \text{N m}^{-1}\) and \(x=2\times10^{-2}\ \text{m}\):
\[W=\tfrac{1}{2}\times(2.0\times10^{11})\times(2\times10^{-2})^{2}\]
\[W=\tfrac{1}{2}\times(2.0\times10^{11})\times(4\times10^{-4})\]
\[W=\mathbf{4.0\times10^{7}\ \text{J}}.\]
The key step is to read \(x\) as the difference of the two rule readings (\(20\ \text{cm}\) in air and \(18\ \text{cm}\) in water), giving \(x = 2\ \text{cm}\), and only then substitute into \(W=\tfrac12 kx^{2}\).
Akọwa Nkọwa
(a) The force causing the spring to shrink in fig.(b)
When the load is lowered into the water it experiences an upthrust (buoyant force) from the water, acting vertically upwards. This upthrust partly supports the weight of the load, so the tension pulling on the spring is reduced and the spring contracts (shrinks). The force responsible for the shrink is therefore the upthrust of the water on the load.
(b) Work done by the force in causing the shrink
Reading the contraction from the metre rule. The pointer attached to the load marks the position of the lower end of the spring against the fixed rule:
- In air (fig. a) the pointer reads \(20\ \text{cm}\) (the spring is fully stretched by the whole weight \(W_a\)).
- In water (fig. b) the upthrust lifts the load, so the pointer rises to \(18\ \text{cm}\).
The contraction (shrink) of the spring is the difference of the two readings:
\[x = 20\ \text{cm}-18\ \text{cm}=2\ \text{cm}=2\times10^{-2}\ \text{m}.\]
Work done. For a spring obeying Hooke's law, the work done in changing its length by \(x\) against a force constant \(k\) is
\[W=\tfrac{1}{2}kx^{2}.\]
With \(k=2.0\times10^{11}\ \text{N m}^{-1}\) and \(x=2\times10^{-2}\ \text{m}\):
\[W=\tfrac{1}{2}\times(2.0\times10^{11})\times(2\times10^{-2})^{2}\]
\[W=\tfrac{1}{2}\times(2.0\times10^{11})\times(4\times10^{-4})\]
\[W=\mathbf{4.0\times10^{7}\ \text{J}}.\]
The key step is to read \(x\) as the difference of the two rule readings (\(20\ \text{cm}\) in air and \(18\ \text{cm}\) in water), giving \(x = 2\ \text{cm}\), and only then substitute into \(W=\tfrac12 kx^{2}\).
Ajụjụ 13 Ripọtì
State three properties of cathode rays which suggest the particle nature of matter
Three properties of cathode rays which suggest the particle (corpuscular) nature of matter are:
- They travel in straight lines and cast sharp shadows of objects placed in their path, as would a stream of moving particles.
- They are deflected by electric and magnetic fields, showing that they consist of charged particles (electrons carrying negative charge) that possess mass and charge.
- They possess momentum and kinetic energy - they can exert a mechanical force, for example turning a small paddle wheel placed in their path, which shows the moving stream carries the properties of particles.
(They also produce heat on striking a target and can be deflected/behave as though made of discrete massive particles.)
Akọwa Nkọwa
Three properties of cathode rays which suggest the particle (corpuscular) nature of matter are:
- They travel in straight lines and cast sharp shadows of objects placed in their path, as would a stream of moving particles.
- They are deflected by electric and magnetic fields, showing that they consist of charged particles (electrons carrying negative charge) that possess mass and charge.
- They possess momentum and kinetic energy - they can exert a mechanical force, for example turning a small paddle wheel placed in their path, which shows the moving stream carries the properties of particles.
(They also produce heat on striking a target and can be deflected/behave as though made of discrete massive particles.)
Ajụjụ 14 Ripọtì
(a) Explain mutual induction
(b) State four use of electromagnets
The diagram above illustrates two coils X an arranged so that their axes are collinear. X is connected to an a.c. supply and has an amme in series with it while Y is connected to a lamp Explain the following observations. The (i) lamp is lit when the a.c. supply is switched on;
(ii) brightness of the light from the lamp increases when distance between X. and Y is decreased;
(iii) filament of the lamp glows brighter whet bundle of insulatttriron wires is placed along ccmmon axis of the coils.
(d) Two cells, one have an emf of 2.0 V and an internal resistance of 0.4 and the other having an emf of 2.0 V and an internal resistance of 0.1 \(\Omega\), are connected in parallel. The combination is then connected in series with a resistor.
(i) Draw a circuit diagram of the arrangement.
(ii) Calculate the current through 5\(\Omega\) resistor.
(a) Mutual induction is the production of an e.m.f. in one coil or circuit when the current in a nearby coil or circuit is changing. The changing current produces a changing magnetic flux which links the second coil.
(b) Uses of electromagnets
- Electric bells.
- Electromagnetic relays and switches.
- Electromagnetic cranes for lifting iron and steel.
- Magnetic separation of iron scraps from non-magnetic materials.
(c)
- The alternating current in coil X produces a continuously changing magnetic flux. Some of this flux links coil Y and induces an alternating e.m.f. in it. A current therefore flows through the lamp and it lights.
- When the coils are brought closer, more of the magnetic flux from X links coil Y. The induced e.m.f. and current in Y increase, so the lamp becomes brighter.
- The bundle of insulated soft-iron wires has high magnetic permeability and provides a path that concentrates the magnetic flux through both coils. The flux linkage and induced e.m.f. in Y increase, producing a larger current and a brighter lamp. The wires are insulated to reduce eddy currents.
(d)(i) Circuit diagram
(d)(ii) Current through the \(5\,\Omega\) resistor
Since the two cells have equal e.m.f.s and are connected in parallel, the resultant e.m.f. is
\[E = 2.0\text{ V}\]
The equivalent internal resistance is
\[ \frac{1}{r}=\frac{1}{0.4}+\frac{1}{0.1}=2.5+10=12.5 \]
\[ r=\frac{1}{12.5}=0.08\,\Omega \]
Total resistance of the circuit:
\[ R_{\text{total}}=5.0+0.08=5.08\,\Omega \]
Hence,
\[ I=\frac{E}{R_{\text{total}}}=\frac{2.0}{5.08}=0.394\text{ A} \]
Current through the \(5\,\Omega\) resistor \(=0.39\text{ A}\) (approximately).
Akọwa Nkọwa
(a) Mutual induction is the production of an e.m.f. in one coil or circuit when the current in a nearby coil or circuit is changing. The changing current produces a changing magnetic flux which links the second coil.
(b) Uses of electromagnets
- Electric bells.
- Electromagnetic relays and switches.
- Electromagnetic cranes for lifting iron and steel.
- Magnetic separation of iron scraps from non-magnetic materials.
(c)
- The alternating current in coil X produces a continuously changing magnetic flux. Some of this flux links coil Y and induces an alternating e.m.f. in it. A current therefore flows through the lamp and it lights.
- When the coils are brought closer, more of the magnetic flux from X links coil Y. The induced e.m.f. and current in Y increase, so the lamp becomes brighter.
- The bundle of insulated soft-iron wires has high magnetic permeability and provides a path that concentrates the magnetic flux through both coils. The flux linkage and induced e.m.f. in Y increase, producing a larger current and a brighter lamp. The wires are insulated to reduce eddy currents.
(d)(i) Circuit diagram
(d)(ii) Current through the \(5\,\Omega\) resistor
Since the two cells have equal e.m.f.s and are connected in parallel, the resultant e.m.f. is
\[E = 2.0\text{ V}\]
The equivalent internal resistance is
\[ \frac{1}{r}=\frac{1}{0.4}+\frac{1}{0.1}=2.5+10=12.5 \]
\[ r=\frac{1}{12.5}=0.08\,\Omega \]
Total resistance of the circuit:
\[ R_{\text{total}}=5.0+0.08=5.08\,\Omega \]
Hence,
\[ I=\frac{E}{R_{\text{total}}}=\frac{2.0}{5.08}=0.394\text{ A} \]
Current through the \(5\,\Omega\) resistor \(=0.39\text{ A}\) (approximately).
Ajụjụ 15 Ripọtì
(a) Explain the terms:
(i) transmutation as it relates to radioactivity; (ii) stopping potential.
(b) \(^{23}_{11}A + ^2_1B\) —> \(^p_qC\) + proton
\(^p_qC\) —> \(^r_sE\) + beta
A nucleus C, formed artificially from A and B radioactive and quickly decays to another nucli E as indicated in the nuclear equations abc Datermine the values of p, q, r and s.
(c) A certain metal of work function 1.6 eV is irradiated with ultra-violet light of wavelength 3.6 x 10\(^{-7}\) Calculate the maximum
(i) kinetic energy of ejected electron in joules;
(ii) speed of an emitted electron. (1eV = 1.6 x 10\(^{-18}\) J; C = 3.0 x 10\(^{8}\) ms\(^{-1}\); m, = 9.1 x 10\(^{-31}\) kg; h = 6.6 x 10\(^{-34}\) Js)
(d) If source of the ultra-violet light in (c) above is mo away from the surface of the metal, state the of on the maximum speed of the ejected electron
(a)(i) Transmutation: Transmutation is the change of one element into another as a result of a nuclear reaction, in which the number of protons in the nucleus (the atomic number) changes, for example during radioactive decay or artificial nuclear bombardment.
(a)(ii) Stopping potential: The stopping potential is the minimum negative (retarding) potential that must be applied to the collecting electrode of a photocell to just stop the most energetic photoelectrons from reaching it, so that the photocurrent becomes zero. \(eV_s = KE_{max}\).
(b) Nuclear equations:
\[ {}^{23}_{11}A + {}^{2}_{1}B \rightarrow {}^{p}_{q}C + {}^{1}_{1}\text{(proton)}. \]
Conserving nucleon number: \(23 + 2 = p + 1 \Rightarrow p = 24\). Conserving charge: \(11 + 1 = q + 1 \Rightarrow q = 11\).
\[ {}^{24}_{11}C \rightarrow {}^{r}_{s}E + {}^{0}_{-1}\beta. \]
Conserving nucleon number: \(24 = r + 0 \Rightarrow r = 24\). Conserving charge: \(11 = s + (-1) \Rightarrow s = 12\).
Therefore \(p = 24,\; q = 11,\; r = 24,\; s = 12\).
(c) Energy of the incident photon:
\[ E = \frac{hc}{\lambda} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{3.6\times10^{-7}} = 5.5\times10^{-19}\,\text{J}. \]
Work function \(W_0 = 1.6\,\text{eV} = 1.6\times(1.6\times10^{-19}) = 2.56\times10^{-19}\,\text{J}\).
(i) Maximum kinetic energy:
\[ KE_{max} = E - W_0 = 5.5\times10^{-19} - 2.56\times10^{-19} = 2.94\times10^{-19}\,\text{J}. \]
(ii) Speed of the emitted electron: from \(KE = \tfrac{1}{2}mv^{2}\),
\[ v = \sqrt{\frac{2\,KE}{m}} = \sqrt{\frac{2\times2.94\times10^{-19}}{9.1\times10^{-31}}} = \sqrt{6.46\times10^{11}} \approx 8.04\times10^{5}\,\text{m s}^{-1}. \]
(d) Moving the source of ultra-violet light farther from the metal only reduces the intensity (number of photons per second arriving), so fewer electrons are emitted per second. It does not change the frequency/energy of each photon, so the maximum kinetic energy and hence the maximum speed of the ejected electron remain unchanged.
Akọwa Nkọwa
(a)(i) Transmutation: Transmutation is the change of one element into another as a result of a nuclear reaction, in which the number of protons in the nucleus (the atomic number) changes, for example during radioactive decay or artificial nuclear bombardment.
(a)(ii) Stopping potential: The stopping potential is the minimum negative (retarding) potential that must be applied to the collecting electrode of a photocell to just stop the most energetic photoelectrons from reaching it, so that the photocurrent becomes zero. \(eV_s = KE_{max}\).
(b) Nuclear equations:
\[ {}^{23}_{11}A + {}^{2}_{1}B \rightarrow {}^{p}_{q}C + {}^{1}_{1}\text{(proton)}. \]
Conserving nucleon number: \(23 + 2 = p + 1 \Rightarrow p = 24\). Conserving charge: \(11 + 1 = q + 1 \Rightarrow q = 11\).
\[ {}^{24}_{11}C \rightarrow {}^{r}_{s}E + {}^{0}_{-1}\beta. \]
Conserving nucleon number: \(24 = r + 0 \Rightarrow r = 24\). Conserving charge: \(11 = s + (-1) \Rightarrow s = 12\).
Therefore \(p = 24,\; q = 11,\; r = 24,\; s = 12\).
(c) Energy of the incident photon:
\[ E = \frac{hc}{\lambda} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{3.6\times10^{-7}} = 5.5\times10^{-19}\,\text{J}. \]
Work function \(W_0 = 1.6\,\text{eV} = 1.6\times(1.6\times10^{-19}) = 2.56\times10^{-19}\,\text{J}\).
(i) Maximum kinetic energy:
\[ KE_{max} = E - W_0 = 5.5\times10^{-19} - 2.56\times10^{-19} = 2.94\times10^{-19}\,\text{J}. \]
(ii) Speed of the emitted electron: from \(KE = \tfrac{1}{2}mv^{2}\),
\[ v = \sqrt{\frac{2\,KE}{m}} = \sqrt{\frac{2\times2.94\times10^{-19}}{9.1\times10^{-31}}} = \sqrt{6.46\times10^{11}} \approx 8.04\times10^{5}\,\text{m s}^{-1}. \]
(d) Moving the source of ultra-violet light farther from the metal only reduces the intensity (number of photons per second arriving), so fewer electrons are emitted per second. It does not change the frequency/energy of each photon, so the maximum kinetic energy and hence the maximum speed of the ejected electron remain unchanged.
