WAEC SSCE - Physics - 2022 (Essay)
Tambaya 1 Rahoto
State:
(a) The S.I. unit of the intensity of a blackbody radiation.
(b) Two features of the intensity-wavelength graph of a perfect blackbody at different temperatures.
(a) The S.I. unit of intensity of blackbody radiation is watt per square metre, \(\mathrm{W\,m^{-2}}\).
(b) Intensity-wavelength curves for a perfect blackbody at two different temperatures are shown below.
- As the temperature increases, the maximum intensity increases. Thus, the curve for \(T_2\) has a higher peak than that for \(T_1\).
- As the temperature increases, the wavelength at which maximum intensity occurs decreases. Thus, \(\lambda_{\max}\) shifts towards shorter wavelengths, in accordance with \(\lambda_{\max}T=\text{constant}\).
Bayanin Amsa
(a) The S.I. unit of intensity of blackbody radiation is watt per square metre, \(\mathrm{W\,m^{-2}}\).
(b) Intensity-wavelength curves for a perfect blackbody at two different temperatures are shown below.
- As the temperature increases, the maximum intensity increases. Thus, the curve for \(T_2\) has a higher peak than that for \(T_1\).
- As the temperature increases, the wavelength at which maximum intensity occurs decreases. Thus, \(\lambda_{\max}\) shifts towards shorter wavelengths, in accordance with \(\lambda_{\max}T=\text{constant}\).
Tambaya 2 Rahoto
You are provided with a battery of e.m.f, E, a standard resistor, R, of resistance 2 \(\Omega\), a key, K, an ammeter, A, a jockey, J, a potentiometer, UV, and some connecting wires.
(i) Measure and record the emf, E, of the battery.
(ii) Set up the circuit as shown in the diagram above with the key open.
(iii) Place the jockey at the point, U, of the potentiometer wire. Close the key and record the reading, i, of the ammeter.
(iv) Place the jockey at a point T on the potentiometer wire UV such that d = UT = 30.0 cm.
(v) Close the circuit, read and record the current, I, on the ammeter,
(vi) Evaluate \(I^1\).
(vi) Repeat the experiment for four other values of d = 40.0 cm, 50.0 cm, 60.0 cm and 70.0 cm. In each case, record I and evaluate \(I^1\).
(vii) Tabulate the results
(ix) Plot a graph with d on the vertical axis and I on the horizontal axis stalling both axes from the origin (0,0).
(x) Determine the slope, s, of the graph.
(xi) From the graph determine the value \(I_1\), of I when d = 0. (ci) Given that=s, calculate 8.
(xii) State two precautions taken to ensure accurate results.
(xii) Given that \(\frac{E}{\delta}\) = s, calculate \(\delta\).
(b)(i) Write down the equation that connects the resistance, R, of a wire and the factors on which it depends. State the meaning of each of the symbols.
(ii) An electric fan draws a current of0.75 A in a 240 V circuit. Calculate the cost of using, the fan for 10 hours if the utility rate is $ 0.50 per kWh.
(a) Potentiometer experiment
(i) The e.m.f. of the battery is:
\[E=2.0\ \text{V}\]
With the jockey at \(U\), the ammeter reading is:
\[i=1.00\ \text{A}\]
(ii) Table of results
| Distance, \(d\) (cm) | Current, \(I\) (A) | \(I^{-1}\) (A−1) |
|---|---|---|
| 30.0 | 0.769 | 1.30 |
| 40.0 | 0.714 | 1.40 |
| 50.0 | 0.667 | 1.50 |
| 60.0 | 0.625 | 1.60 |
| 70.0 | 0.588 | 1.70 |
As \(d\) increases, \(I\) decreases.
(iii) Graph of \(d\) against \(I^{-1}\)
(iv) Slope of the graph
Using two widely separated points on the straight line, \((1.20\ \text{A}^{-1},20.0\ \text{cm})\) and \((1.70\ \text{A}^{-1},70.0\ \text{cm})\):
\[s=\frac{70.0-20.0}{1.70-1.20}=\frac{50.0}{0.50}=100\ \text{cm A}.\]
(v) Value of \(I^{-1}\) when \(d=0\)
From the intercept on the \(I^{-1}\)-axis,
\[I^{-1}=1.00\ \text{A}^{-1}.\]
Therefore,
\[I=\frac{1}{1.00}=1.00\ \text{A},\]
which agrees with the current \(i\) when the jockey is at \(U\).
(vi) Resistance per unit length, \(\delta\)
Given that
\[s=\frac{E}{\delta},\]
\[\delta=\frac{E}{s}=\frac{2.0}{100}=0.020\ \Omega\,\text{cm}^{-1}.\]
(vii) Precautions
- The key was closed only while taking a reading, to prevent heating of the potentiometer wire and consequent change in resistance.
- The jockey was touched lightly on the wire and not pressed hard, to ensure good contact without damaging the wire.
(b)
(i) The resistance of a uniform wire is given by
\[R=\rho\frac{l}{A}.\]
\(R\) is the resistance of the wire, \(\rho\) is the resistivity of the material, \(l\) is the length of the wire, and \(A\) is its cross-sectional area.
(ii)
\[P=VI=240\times0.75=180\ \text{W}=0.180\ \text{kW}.\]
Energy used in 10 hours:
\[E=Pt=0.180\times10=1.80\ \text{kWh}.\]
Cost of energy:
\[\text{Cost}=1.80\times\$0.50=\boxed{\$0.90}.\]
Bayanin Amsa
(a) Potentiometer experiment
(i) The e.m.f. of the battery is:
\[E=2.0\ \text{V}\]
With the jockey at \(U\), the ammeter reading is:
\[i=1.00\ \text{A}\]
(ii) Table of results
| Distance, \(d\) (cm) | Current, \(I\) (A) | \(I^{-1}\) (A−1) |
|---|---|---|
| 30.0 | 0.769 | 1.30 |
| 40.0 | 0.714 | 1.40 |
| 50.0 | 0.667 | 1.50 |
| 60.0 | 0.625 | 1.60 |
| 70.0 | 0.588 | 1.70 |
As \(d\) increases, \(I\) decreases.
(iii) Graph of \(d\) against \(I^{-1}\)
(iv) Slope of the graph
Using two widely separated points on the straight line, \((1.20\ \text{A}^{-1},20.0\ \text{cm})\) and \((1.70\ \text{A}^{-1},70.0\ \text{cm})\):
\[s=\frac{70.0-20.0}{1.70-1.20}=\frac{50.0}{0.50}=100\ \text{cm A}.\]
(v) Value of \(I^{-1}\) when \(d=0\)
From the intercept on the \(I^{-1}\)-axis,
\[I^{-1}=1.00\ \text{A}^{-1}.\]
Therefore,
\[I=\frac{1}{1.00}=1.00\ \text{A},\]
which agrees with the current \(i\) when the jockey is at \(U\).
(vi) Resistance per unit length, \(\delta\)
Given that
\[s=\frac{E}{\delta},\]
\[\delta=\frac{E}{s}=\frac{2.0}{100}=0.020\ \Omega\,\text{cm}^{-1}.\]
(vii) Precautions
- The key was closed only while taking a reading, to prevent heating of the potentiometer wire and consequent change in resistance.
- The jockey was touched lightly on the wire and not pressed hard, to ensure good contact without damaging the wire.
(b)
(i) The resistance of a uniform wire is given by
\[R=\rho\frac{l}{A}.\]
\(R\) is the resistance of the wire, \(\rho\) is the resistivity of the material, \(l\) is the length of the wire, and \(A\) is its cross-sectional area.
(ii)
\[P=VI=240\times0.75=180\ \text{W}=0.180\ \text{kW}.\]
Energy used in 10 hours:
\[E=Pt=0.180\times10=1.80\ \text{kWh}.\]
Cost of energy:
\[\text{Cost}=1.80\times\$0.50=\boxed{\$0.90}.\]
Tambaya 3 Rahoto
(a) State two factors that affect the rate of evaporation of a liquid.
(b) Explain the term latent heal.
(c) Explain each of the following phenomena:
(i) On a dry day, water in a clay pot is cooler than water in a rubber container:
(ii) Cooking of food is faster in a pressure cooker than in an ordinary pot.
(d) A 40 V electric heater is used to supply a current of 12 A for 1400 seconds to a body mass of 1.5 kg at its melting point. The body melts and its temperature rises by 60°C in an extra 1.2 minutes, Calculate the:
(i) Latent heat of fusion of the body
(ii) Specific heat capacity of the body.
(e) State two differences between evaporation and boiling.
(a) Two factors that affect the rate of evaporation of a liquid are: 1. Temperature: Higher temperatures generally result in faster rates of evaporation, as the heat energy increases the kinetic energy of the liquid molecules, causing them to move faster and escape into the air more readily. 2. Surface area: A larger surface area of the liquid exposed to the air can increase the rate of evaporation, as there are more liquid molecules available to escape into the air. (b) Latent heat refers to the amount of heat energy that is required to change the phase of a substance without changing its temperature. For example, when ice is heated, it melts and changes from a solid to a liquid, but its temperature remains constant at 0°C until all the ice has melted. The energy absorbed during this phase change is known as the latent heat of fusion. (c) (i) On a dry day, water in a clay pot is cooler than water in a rubber container because the water in the clay pot evaporates faster due to the porous nature of the clay, which allows air to circulate through it. As the water evaporates, it removes heat from the remaining water, causing it to cool down. (ii) Cooking of food is faster in a pressure cooker than in an ordinary pot because the pressure inside the cooker increases the boiling point of water. This means that the food can be cooked at a higher temperature than in an ordinary pot, resulting in faster cooking times. (d) (i) Latent heat of fusion can be calculated using the formula Q = mL, where Q is the heat energy absorbed, m is the mass of the substance, and L is the latent heat of fusion. In this case, Q = (40 V x 12 A x 1400 s) = 672,000 J, and m = 1.5 kg. Assuming that the temperature of the substance was constant during melting, L can be calculated as L = Q/m = 672,000 J / 1.5 kg = 448,000 J/kg. (ii) Specific heat capacity can be calculated using the formula c = Q/(mΔT), where c is the specific heat capacity, ΔT is the change in temperature, and all other variables are the same as in part (i). In this case, Q = (40 V x 12 A x 72 s) = 34,560 J and ΔT = 60°C. Substituting these values into the formula, we get c = 34,560 J / (1.5 kg x 60°C) = 384 J/(kg·°C). (e) Two differences between evaporation and boiling are: 1. Evaporation occurs at the surface of a liquid, while boiling occurs throughout the liquid. 2. Evaporation occurs at a range of temperatures below the boiling point of the liquid, while boiling occurs only at the boiling point. Evaporation is a slower process than boiling, as it only occurs at the surface of the liquid, while boiling involves the rapid formation of bubbles throughout the liquid.
Bayanin Amsa
(a) Two factors that affect the rate of evaporation of a liquid are: 1. Temperature: Higher temperatures generally result in faster rates of evaporation, as the heat energy increases the kinetic energy of the liquid molecules, causing them to move faster and escape into the air more readily. 2. Surface area: A larger surface area of the liquid exposed to the air can increase the rate of evaporation, as there are more liquid molecules available to escape into the air. (b) Latent heat refers to the amount of heat energy that is required to change the phase of a substance without changing its temperature. For example, when ice is heated, it melts and changes from a solid to a liquid, but its temperature remains constant at 0°C until all the ice has melted. The energy absorbed during this phase change is known as the latent heat of fusion. (c) (i) On a dry day, water in a clay pot is cooler than water in a rubber container because the water in the clay pot evaporates faster due to the porous nature of the clay, which allows air to circulate through it. As the water evaporates, it removes heat from the remaining water, causing it to cool down. (ii) Cooking of food is faster in a pressure cooker than in an ordinary pot because the pressure inside the cooker increases the boiling point of water. This means that the food can be cooked at a higher temperature than in an ordinary pot, resulting in faster cooking times. (d) (i) Latent heat of fusion can be calculated using the formula Q = mL, where Q is the heat energy absorbed, m is the mass of the substance, and L is the latent heat of fusion. In this case, Q = (40 V x 12 A x 1400 s) = 672,000 J, and m = 1.5 kg. Assuming that the temperature of the substance was constant during melting, L can be calculated as L = Q/m = 672,000 J / 1.5 kg = 448,000 J/kg. (ii) Specific heat capacity can be calculated using the formula c = Q/(mΔT), where c is the specific heat capacity, ΔT is the change in temperature, and all other variables are the same as in part (i). In this case, Q = (40 V x 12 A x 72 s) = 34,560 J and ΔT = 60°C. Substituting these values into the formula, we get c = 34,560 J / (1.5 kg x 60°C) = 384 J/(kg·°C). (e) Two differences between evaporation and boiling are: 1. Evaporation occurs at the surface of a liquid, while boiling occurs throughout the liquid. 2. Evaporation occurs at a range of temperatures below the boiling point of the liquid, while boiling occurs only at the boiling point. Evaporation is a slower process than boiling, as it only occurs at the surface of the liquid, while boiling involves the rapid formation of bubbles throughout the liquid.
Tambaya 4 Rahoto
The diagram above illustrates the trajectory of a fired missile from point P at 250 ms-1
If the missile point Q after 40 s, calculate the distance |PQ|
|PQ| = Range = Ux.T
= UT Cosθ
= 250 x 40 Cos 42º
= 7431 m (7.4 km)
Bayanin Amsa
|PQ| = Range = Ux.T
= UT Cosθ
= 250 x 40 Cos 42º
= 7431 m (7.4 km)
Tambaya 5 Rahoto
State three properties of a semiconductor.
Semiconductors are materials that have an electrical conductivity between that of a conductor and an insulator. Three properties of a semiconductor are:
- Conductivity: Semiconductors have intermediate electrical conductivity, which means they are not good conductors like metals but not as poor as insulators. They can conduct electricity under certain conditions, such as when exposed to light or heat, or when impurities are added to them (a process called doping).
- Energy gap: Semiconductors have a small energy gap between their valence band (the highest energy level occupied by electrons) and their conduction band (the lowest energy level that electrons can occupy and still be free to move). This energy gap determines the conductivity of the material and can be manipulated through doping or other processes.
- Temperature sensitivity: Semiconductors are sensitive to changes in temperature, which can affect their electrical conductivity. As the temperature of a semiconductor increases, the energy of its electrons increases and they become more likely to move into the conduction band, increasing the material's conductivity. However, if the temperature gets too high, the semiconductor may begin to behave like a conductor and lose its unique properties.
Overall, these properties make semiconductors useful in a wide range of electronic devices, including transistors, diodes, and solar cells.
Bayanin Amsa
Semiconductors are materials that have an electrical conductivity between that of a conductor and an insulator. Three properties of a semiconductor are:
- Conductivity: Semiconductors have intermediate electrical conductivity, which means they are not good conductors like metals but not as poor as insulators. They can conduct electricity under certain conditions, such as when exposed to light or heat, or when impurities are added to them (a process called doping).
- Energy gap: Semiconductors have a small energy gap between their valence band (the highest energy level occupied by electrons) and their conduction band (the lowest energy level that electrons can occupy and still be free to move). This energy gap determines the conductivity of the material and can be manipulated through doping or other processes.
- Temperature sensitivity: Semiconductors are sensitive to changes in temperature, which can affect their electrical conductivity. As the temperature of a semiconductor increases, the energy of its electrons increases and they become more likely to move into the conduction band, increasing the material's conductivity. However, if the temperature gets too high, the semiconductor may begin to behave like a conductor and lose its unique properties.
Overall, these properties make semiconductors useful in a wide range of electronic devices, including transistors, diodes, and solar cells.
Tambaya 6 Rahoto
You are provided with a loaded boiling tube with a centimeter scale fixed inside it, a transparent vessel filled with water, standard masses 2 g, 5g and 10 g, and a slide vernier caliper. Use the diagram above as a guide to perform the experiment.
(i) Use the slide vernier caliper to measure and record the external diameter, D, of the boiling tube.
(ii) Evaluated \( A = 0.25\pi D^2 \), where \( \pi = 3.14 \).
(iii) Place the loaded boiling tube gently in the water in the transparent vessel such that it floats vertically.
(iv) Read and record the depth of immersion, y, from the zero mark of the scale fixed inside, the boiling tube.
(v) Add a mass, m = 2g, to the boiling tube. Read and record the new depth of immersion, y, from the zero mark of the scale.
(vi) Evaluate \( h = (y - y_0) \), log h and log m.
(vii) Repeat the experiment for four other values of m = 5g, 7g, 10 g, and 12g. In each case, record y and evaluate h, log h, and log m.
(viii) Tabulate the results.
(ix) Plot a graph with log m on the vertical axis and log h on the horizontal axis starting both axes from the origin (0,0).
(b)(i) State in full the law on which the experiment in (a) is based.
(ii) A uniform cylindrical rod is 0.63 m long and it has a cross-sectional area of \( 0.1\ \mathrm{m}^2 \). Calculate the depth of immersion of the rod if it floats vertically in a liquid of relative density 1.26. [density of rod =\(720\ \mathrm{kg\ m}^{-3}\), g = \(10\ \mathrm{m\ s}^{-2}\)].
(a) Observations, table and graph
The external diameter of the boiling tube is
\[D=1.20\ \text{cm}\]
Hence, the cross-sectional area is
\[A=\frac{\pi D^2}{4}=\frac{3.14(1.20)^2}{4}=1.13\ \text{cm}^2.\]
The initial depth of immersion of the loaded boiling tube is
\[y_0=3.0\ \text{cm}.\]
| Mass added, \(m\) (g) | Depth of immersion, \(y\) (cm) | \(h=y-y_0\) (cm) | \(\log h\) | \(\log m\) |
|---|---|---|---|---|
| 2 | 4.8 | 1.8 | 0.255 | 0.301 |
| 5 | 7.4 | 4.4 | 0.643 | 0.699 |
| 7 | 9.2 | 6.2 | 0.792 | 0.845 |
| 10 | 11.9 | 8.9 | 0.949 | 1.000 |
| 12 | 13.6 | 10.6 | 1.025 | 1.079 |
The graph of \(\log m\) against \(\log h\) is shown below.
From the line of best fit,
\[\text{gradient}=\frac{\Delta(\log m)}{\Delta(\log h)}\approx1.01\approx1.\]
Thus, \(m\propto h\). This agrees with the flotation relation \(m=\rho Ah\), where \(\rho\) and \(A\) are constant.
(b)(i) Law of flotation
A floating body displaces a weight of fluid equal to its own weight.
(b)(ii) Depth of immersion of the rod
Density of liquid:
\[\rho_l=1.26\times1000=1260\ \text{kg m}^{-3}.\]
For equilibrium of the floating rod,
\[\rho_rALg=\rho_lAyg,\]
where \(y\) is the immersed length. Therefore,
\[y=\frac{\rho_rL}{\rho_l}=\frac{720\times0.63}{1260}=0.36\ \text{m}.\]
Depth of immersion = \(0.36\ \text{m}\).
Bayanin Amsa
(a) Observations, table and graph
The external diameter of the boiling tube is
\[D=1.20\ \text{cm}\]
Hence, the cross-sectional area is
\[A=\frac{\pi D^2}{4}=\frac{3.14(1.20)^2}{4}=1.13\ \text{cm}^2.\]
The initial depth of immersion of the loaded boiling tube is
\[y_0=3.0\ \text{cm}.\]
| Mass added, \(m\) (g) | Depth of immersion, \(y\) (cm) | \(h=y-y_0\) (cm) | \(\log h\) | \(\log m\) |
|---|---|---|---|---|
| 2 | 4.8 | 1.8 | 0.255 | 0.301 |
| 5 | 7.4 | 4.4 | 0.643 | 0.699 |
| 7 | 9.2 | 6.2 | 0.792 | 0.845 |
| 10 | 11.9 | 8.9 | 0.949 | 1.000 |
| 12 | 13.6 | 10.6 | 1.025 | 1.079 |
The graph of \(\log m\) against \(\log h\) is shown below.
From the line of best fit,
\[\text{gradient}=\frac{\Delta(\log m)}{\Delta(\log h)}\approx1.01\approx1.\]
Thus, \(m\propto h\). This agrees with the flotation relation \(m=\rho Ah\), where \(\rho\) and \(A\) are constant.
(b)(i) Law of flotation
A floating body displaces a weight of fluid equal to its own weight.
(b)(ii) Depth of immersion of the rod
Density of liquid:
\[\rho_l=1.26\times1000=1260\ \text{kg m}^{-3}.\]
For equilibrium of the floating rod,
\[\rho_rALg=\rho_lAyg,\]
where \(y\) is the immersed length. Therefore,
\[y=\frac{\rho_rL}{\rho_l}=\frac{720\times0.63}{1260}=0.36\ \text{m}.\]
Depth of immersion = \(0.36\ \text{m}\).
Tambaya 7 Rahoto
(a)(i) State the law of inertia.
(ii) Use the law stated in (a)(I) to explain how wearing a safety belt in a moving vehicle could reduce the possibilities of severe injuries when the vehicle is involved in a collision.
(b)(I) Define the term moment of a force.
(ii) A uniform plank measures \(2\,\text{m}\) long from its ends point A to point B. If the weight of the plank is \(54\,\text{N}\) and it rests on a knife edge \(0.50\,\text{m}\) from end B and point A is supported by a vertical string so that AB balances horizontally:
i. Draw a force diagram for the arrangement;
ii. Determine the tension, T, in the string.
iii. Determine the force, F, acting on the knife edge
(c) State three differences between solid friction and viscosity
(d) State one method of increasing the velocity ration of a pulley system.
(a)(i) The law of inertia states that an object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity unless acted upon by an external force.
(a)(ii) Wearing a safety belt in a moving vehicle could reduce the possibilities of severe injuries when the vehicle is involved in a collision because of the law of inertia. When a car suddenly stops in a collision, the passengers in the car tend to continue moving forward at the same speed the car was moving before the collision. If they are not restrained by a seat belt, they will keep moving forward and could collide with the dashboard, steering wheel, or windshield. However, if they are wearing a seat belt, the belt applies a force to the passenger in the opposite direction, keeping them from moving forward, and reducing the risk of injury.
(b)(i) The moment of a force is the turning effect of the force about a pivot and is given by the product of the force and the perpendicular distance from the pivot to the line of action of the force.
(b)(ii)
i. The force diagram for the arrangement is a diagram that shows all the forces acting on an object in a particular situation. For this arrangement, the force diagram will show the weight of the plank acting downwards at the center, the tension T acting upwards at point A, and the reaction force F acting upwards at the knife edge.
ii. To determine the tension, T, in the string, we can use the principle of moments. The sum of the clockwise moments about the knife edge must be equal to the sum of the anticlockwise moments about the knife edge. Thus,
54 N x 0.50 m = T x 1.50 m.
Therefore, T = 18 N.
iii. To determine the force, F, acting on the knife edge, we can use the principle of moments. The sum of the clockwise moments about the knife edge must be equal to the sum of the anticlockwise moments about the knife edge and at equilibrium, the sum of upward forces = the sum of forces acting downward
T + F = 54
T = 18 N
18 + F = 54
F = 36 N
(c) Three differences between solid friction and viscosity are:
1. Solid friction is the force that opposes the motion of an object along a solid surface, whereas viscosity is the force that opposes the motion of an object through a fluid.
2. Solid friction depends on the nature of the surfaces in contact, whereas viscosity depends on the properties of the fluid, such as its density and viscosity.
3. Solid friction is independent of velocity, whereas viscosity is directly proportional to velocity.
(d) One method of increasing the velocity ratio of a pulley system is to increase the number of pulleys in the system. The velocity ratio of a pulley system is the ratio of the distance moved by the effort to the distance moved by the load. The more pulleys there are in the system, the greater the distance the effort can move for a given distance moved by the load, thus increasing the velocity ratio.
Bayanin Amsa
(a)(i) The law of inertia states that an object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity unless acted upon by an external force.
(a)(ii) Wearing a safety belt in a moving vehicle could reduce the possibilities of severe injuries when the vehicle is involved in a collision because of the law of inertia. When a car suddenly stops in a collision, the passengers in the car tend to continue moving forward at the same speed the car was moving before the collision. If they are not restrained by a seat belt, they will keep moving forward and could collide with the dashboard, steering wheel, or windshield. However, if they are wearing a seat belt, the belt applies a force to the passenger in the opposite direction, keeping them from moving forward, and reducing the risk of injury.
(b)(i) The moment of a force is the turning effect of the force about a pivot and is given by the product of the force and the perpendicular distance from the pivot to the line of action of the force.
(b)(ii)
i. The force diagram for the arrangement is a diagram that shows all the forces acting on an object in a particular situation. For this arrangement, the force diagram will show the weight of the plank acting downwards at the center, the tension T acting upwards at point A, and the reaction force F acting upwards at the knife edge.
ii. To determine the tension, T, in the string, we can use the principle of moments. The sum of the clockwise moments about the knife edge must be equal to the sum of the anticlockwise moments about the knife edge. Thus,
54 N x 0.50 m = T x 1.50 m.
Therefore, T = 18 N.
iii. To determine the force, F, acting on the knife edge, we can use the principle of moments. The sum of the clockwise moments about the knife edge must be equal to the sum of the anticlockwise moments about the knife edge and at equilibrium, the sum of upward forces = the sum of forces acting downward
T + F = 54
T = 18 N
18 + F = 54
F = 36 N
(c) Three differences between solid friction and viscosity are:
1. Solid friction is the force that opposes the motion of an object along a solid surface, whereas viscosity is the force that opposes the motion of an object through a fluid.
2. Solid friction depends on the nature of the surfaces in contact, whereas viscosity depends on the properties of the fluid, such as its density and viscosity.
3. Solid friction is independent of velocity, whereas viscosity is directly proportional to velocity.
(d) One method of increasing the velocity ratio of a pulley system is to increase the number of pulleys in the system. The velocity ratio of a pulley system is the ratio of the distance moved by the effort to the distance moved by the load. The more pulleys there are in the system, the greater the distance the effort can move for a given distance moved by the load, thus increasing the velocity ratio.
Tambaya 8 Rahoto
State three differences between magnetic and non-magnetic materials.
Magnetic and non-magnetic materials are two distinct categories of materials. Here are three differences between these two: 1. Magnetic materials are attracted to a magnet, while non-magnetic materials are not. This means that if you hold a magnet near a magnetic material, it will be pulled towards the magnet, but the same won't happen with a non-magnetic material. 2. Magnetic materials can be magnetized, whereas non-magnetic materials cannot. Magnetization is the process of aligning the magnetic domains of a material, and this is only possible with magnetic materials. 3. Magnetic materials have magnetic domains, while non-magnetic materials do not. Magnetic domains are regions within a material where the magnetic moments of the atoms are aligned in the same direction, creating a magnetic field. In non-magnetic materials, the magnetic moments are randomly oriented, resulting in no net magnetic field. These are the three main differences between magnetic and non-magnetic materials.
Bayanin Amsa
Magnetic and non-magnetic materials are two distinct categories of materials. Here are three differences between these two: 1. Magnetic materials are attracted to a magnet, while non-magnetic materials are not. This means that if you hold a magnet near a magnetic material, it will be pulled towards the magnet, but the same won't happen with a non-magnetic material. 2. Magnetic materials can be magnetized, whereas non-magnetic materials cannot. Magnetization is the process of aligning the magnetic domains of a material, and this is only possible with magnetic materials. 3. Magnetic materials have magnetic domains, while non-magnetic materials do not. Magnetic domains are regions within a material where the magnetic moments of the atoms are aligned in the same direction, creating a magnetic field. In non-magnetic materials, the magnetic moments are randomly oriented, resulting in no net magnetic field. These are the three main differences between magnetic and non-magnetic materials.
Tambaya 9 Rahoto
You are provided with a metre rule, a weight hanger, slotted masses, M, a piece (if string, a weighing balance and a knife edge. Use the diagram above as a guide to perform the experiment.
(i) Using the weighing balance, determine and record the mass, \(M_o\), of the unloaded metre rule.
(ii) Determine and record the mass, m, of the weight hanger.
(ii) Suspend the metre rule horizontally on the knife edge. Adjust the knife edge to a point G on the metre rule where it balances horizontally.
(iv) Record the distance, d = AG.
(v) Suspend the weight hanger securely at a point, P, on the metre rule such that AP = 5 cm. Keep the hanger at this point throughout the experiment
(vi) Add a mass, M = 20 g to the hanger, adjust the knife edge to a point K on the metre rule such that it balances horizontally as shown in the diagram above.
(vii) Determine and record the distance z = AK.
(vii) Record M and evaluate y - (z - 5), x - (d - z] and v = \(\frac{x}{y}\)
(ix) Repeat the experiment for M = 40 g, 60 g, 80 g and 100 g. In each case, evaluate y, x and v.
(x) Tabulate the results.
(xi) Plot a graph with M on the vertical axis and v on the horizontal axis, sinning both axes from the origin (0,0).
(xii) Determine the slope, s, of the graph.
(xii) Determine the intercept, c, on the vertical axis.
(xiv) State two precautions taken to ensure accurate results.
(b) (i) Under what condition is an object said to be in a stable equilibrium
(ii) Auniform beam of weight 50 N has a body of weight 100 N hung at one end of it. If the beam is 12 m long, determine the distance of a support from a 100 N body for it to balance horizontally.
(a) Results and graph
Mass of unloaded metre rule, \(M_0=75.0\text{ g}\).
Mass of weight hanger, \(m=20.0\text{ g}\).
Balance point of unloaded metre rule: \(d=AG=50.0\text{ cm}\).
For each load, \(y=z-5\), \(x=d-z\), and \(v=\dfrac{x}{y}\).
| \(M\) (g) | \(z=AK\) (cm) | \(y=z-5\) (cm) | \(x=d-z\) (cm) | \(v=x/y\) |
|---|---|---|---|---|
| 20 | 34.3 | 29.3 | 15.7 | 0.536 |
| 40 | 30.0 | 25.0 | 20.0 | 0.800 |
| 60 | 26.8 | 21.8 | 23.2 | 1.064 |
| 80 | 24.3 | 19.3 | 25.7 | 1.332 |
| 100 | 22.3 | 17.3 | 27.7 | 1.601 |
The plotted graph of \(M\) against \(v\), with both axes beginning at the origin, is shown below.
Using two widely separated points on the line of best fit, \((v_1,M_1)=(0.536,20)\) and \((v_2,M_2)=(1.601,100)\):
\[s=\frac{M_2-M_1}{v_2-v_1}=\frac{100-20}{1.601-0.536}=75.1\text{ g}\approx75.0\text{ g}.\]
The vertical intercept is \(c\approx-20.0\text{ g}\).
Thus, within experimental accuracy, \(s=M_0\) and \(c=-m\).
Precautions
- The metre rule was allowed to come to rest and was balanced horizontally before each reading was taken.
- All scale readings were taken with the eye vertically above the mark to avoid parallax error.
(b)
(i) An object is in stable equilibrium if, when slightly displaced, its centre of gravity rises and a restoring moment acts to return it to its original position.
(ii) Let \(y\) be the distance of the support from the \(100\text{ N}\) body. Taking moments about the support:
\[100y=50(6-y)\]
\[100y=300-50y\]
\[150y=300\]
\[y=2.0\text{ m}.\]
Therefore, the support should be placed \(\boxed{2.0\text{ m}}\) from the \(100\text{ N}\) body.
Bayanin Amsa
(a) Results and graph
Mass of unloaded metre rule, \(M_0=75.0\text{ g}\).
Mass of weight hanger, \(m=20.0\text{ g}\).
Balance point of unloaded metre rule: \(d=AG=50.0\text{ cm}\).
For each load, \(y=z-5\), \(x=d-z\), and \(v=\dfrac{x}{y}\).
| \(M\) (g) | \(z=AK\) (cm) | \(y=z-5\) (cm) | \(x=d-z\) (cm) | \(v=x/y\) |
|---|---|---|---|---|
| 20 | 34.3 | 29.3 | 15.7 | 0.536 |
| 40 | 30.0 | 25.0 | 20.0 | 0.800 |
| 60 | 26.8 | 21.8 | 23.2 | 1.064 |
| 80 | 24.3 | 19.3 | 25.7 | 1.332 |
| 100 | 22.3 | 17.3 | 27.7 | 1.601 |
The plotted graph of \(M\) against \(v\), with both axes beginning at the origin, is shown below.
Using two widely separated points on the line of best fit, \((v_1,M_1)=(0.536,20)\) and \((v_2,M_2)=(1.601,100)\):
\[s=\frac{M_2-M_1}{v_2-v_1}=\frac{100-20}{1.601-0.536}=75.1\text{ g}\approx75.0\text{ g}.\]
The vertical intercept is \(c\approx-20.0\text{ g}\).
Thus, within experimental accuracy, \(s=M_0\) and \(c=-m\).
Precautions
- The metre rule was allowed to come to rest and was balanced horizontally before each reading was taken.
- All scale readings were taken with the eye vertically above the mark to avoid parallax error.
(b)
(i) An object is in stable equilibrium if, when slightly displaced, its centre of gravity rises and a restoring moment acts to return it to its original position.
(ii) Let \(y\) be the distance of the support from the \(100\text{ N}\) body. Taking moments about the support:
\[100y=50(6-y)\]
\[100y=300-50y\]
\[150y=300\]
\[y=2.0\text{ m}.\]
Therefore, the support should be placed \(\boxed{2.0\text{ m}}\) from the \(100\text{ N}\) body.
Tambaya 10 Rahoto
The effective potential energy, E, of a lunar satellite of mass, m, moving in an. elliptical orbit around the moon of mass, m, is given by
\[ E = \frac{K^2}{2m_1r^2} - \frac{Gm_1m_2}{r} \] where r is the distance of the satellite from the mooń and G is the universal gravitational constant of dimensions, \(M^{-1}L^3T^2\).
Ďetermine the dimensions of the angular momentum, K, of the satellite using dimensional analysis.
Dimensions of the angular momentum, K =
[K]2[m1][r]2=[G][m1][m2][r]
[k]2ML2=M−1L3T−2M2L
[K] = (M2 L4T−2 )½
∴ [k] = ML2T−1
Bayanin Amsa
Dimensions of the angular momentum, K =
[K]2[m1][r]2=[G][m1][m2][r]
[k]2ML2=M−1L3T−2M2L
[K] = (M2 L4T−2 )½
∴ [k] = ML2T−1
Tambaya 11 Rahoto
(a) What is a geostationary satellite?
(b) Name two types of Lasers.
(a) A geostationary satellite is an artificial satellite that orbits the Earth at the same rate as the Earth's rotation. This means that it stays fixed at a particular point in the sky over the Earth's equator. From the ground, it appears as though the satellite is stationary. (b) Two types of lasers are solid-state lasers and gas lasers. Solid-state lasers use a solid material as the active medium, while gas lasers use a gas or a mixture of gases as the active medium. Solid-state lasers are commonly used in industry, medicine, and research, while gas lasers are used for applications such as cutting, welding, and laser printers.
Bayanin Amsa
(a) A geostationary satellite is an artificial satellite that orbits the Earth at the same rate as the Earth's rotation. This means that it stays fixed at a particular point in the sky over the Earth's equator. From the ground, it appears as though the satellite is stationary. (b) Two types of lasers are solid-state lasers and gas lasers. Solid-state lasers use a solid material as the active medium, while gas lasers use a gas or a mixture of gases as the active medium. Solid-state lasers are commonly used in industry, medicine, and research, while gas lasers are used for applications such as cutting, welding, and laser printers.
Tambaya 12 Rahoto
(a)(i) State Coulomb's law of electrostatics.
(ii) The electron and proton of a hydrogen atom are separated by a mean distance of \(5.2 \times 10^{-11}\,\text{m}\).
Calculate the magnitude of the electrostatic force between the particles.
[\(e = 1.6 \times 10^{-19}\,\text{C}\), \((4\pi \mathcal{E}_0)^{-1} = 9.0 \times 10^9\,\text{mF}^{-1}\)]
(b) The diagram below shows a potential divider circuit.
%IMG%
i. Show that \(V_{\text{out}} = V_{\text{in}}\left(\frac{R_1}{R_1 + R_2}\right)\)
ii. If \(\frac{V_{\text{in}}}{V_{\text{out}}} = 2.5\) and \(R_2 = 30\,\Omega\), calculate \(R_1\)
iii. Define the volt.
(c) Explain why wood is not suitable for use as the core of transformers.
(d) State one application for the cathode ray tube.
(a)(i) Coulomb's law of electrostatics states that the force of attraction or repulsion between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Mathematically, F = kQ1Q2/r^2, where F is the electrostatic force, Q1 and Q2 are the charges on the objects, r is the distance between them, and k is Coulomb's constant.
(ii) Using Coulomb's law, the magnitude of the electrostatic force between the electron and proton of a hydrogen atom can be calculated as follows:
F = k(Q1Q2/r^2) = (9.0 x 10^9 Nm^2/C^2)(1.6 x 10^-19 C)(1.6 x 10^-19 C)/(5.2 x 10^-11 m)^2 = 8.2 x 10^-8 N.
(b)(I) Proof of Vout
Let the current through the circuit be I
Vin
= I(R1 + R2)
OR
I = VinR1+R2
Vout = IR2
II. Calculation of R1
;
Vout
= Vin(R1R1+R2)
OR
VoutVin=R1R1+R2
12.5=R1R1+R30R1 = 20Ω
(ii) The Volt The unit of potential difference between two points when one joule of work is done in taking one coulomb of charge between the points
(c) Wood is not suitable for use as the core of transformers because it is not a good conductor of electricity. Transformers require a material with a high magnetic permeability, which allows for efficient transfer of energy between coils. Wood does not have the necessary magnetic properties and can also be damaged by the heat generated during operation.
(d) One application for the cathode ray tube (CRT) is in television and computer displays. A CRT works by producing a beam of electrons that is directed towards a phosphorescent screen, causing it to emit light and create an image. By controlling the path of the electron beam, the image can be manipulated and displayed on the screen. Although CRT technology has largely been replaced by LCD and LED displays, it is still used in some specialized applications such as medical imaging and oscilloscopes.
Bayanin Amsa
(a)(i) Coulomb's law of electrostatics states that the force of attraction or repulsion between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Mathematically, F = kQ1Q2/r^2, where F is the electrostatic force, Q1 and Q2 are the charges on the objects, r is the distance between them, and k is Coulomb's constant.
(ii) Using Coulomb's law, the magnitude of the electrostatic force between the electron and proton of a hydrogen atom can be calculated as follows:
F = k(Q1Q2/r^2) = (9.0 x 10^9 Nm^2/C^2)(1.6 x 10^-19 C)(1.6 x 10^-19 C)/(5.2 x 10^-11 m)^2 = 8.2 x 10^-8 N.
(b)(I) Proof of Vout
Let the current through the circuit be I
Vin
= I(R1 + R2)
OR
I = VinR1+R2
Vout = IR2
II. Calculation of R1
;
Vout
= Vin(R1R1+R2)
OR
VoutVin=R1R1+R2
12.5=R1R1+R30R1 = 20Ω
(ii) The Volt The unit of potential difference between two points when one joule of work is done in taking one coulomb of charge between the points
(c) Wood is not suitable for use as the core of transformers because it is not a good conductor of electricity. Transformers require a material with a high magnetic permeability, which allows for efficient transfer of energy between coils. Wood does not have the necessary magnetic properties and can also be damaged by the heat generated during operation.
(d) One application for the cathode ray tube (CRT) is in television and computer displays. A CRT works by producing a beam of electrons that is directed towards a phosphorescent screen, causing it to emit light and create an image. By controlling the path of the electron beam, the image can be manipulated and displayed on the screen. Although CRT technology has largely been replaced by LCD and LED displays, it is still used in some specialized applications such as medical imaging and oscilloscopes.
Tambaya 13 Rahoto
A \(50\ \mathrm{N}\) force is applied to the free end of a spiral spring of force constant, \(100\ \mathrm{N\,m^{-1}}\). Calculate the work done by the force to stretch the spring.
Calculation of workdone by the force to scratch the string
W = ½ ke\(^{2}\) = \(\frac{f^{2}}{2k}\)
W = \(\frac{50^{2}}{2 \times 100}\)
W = 12.5 J
OR
e = \(\frac{F}{K} = \frac{50}{100}\) = 0.5m
w = ½ = ke\(^{2}\)
w = ½ x 100 x 0.5\(^{2}\)
W = 12.5 J
Bayanin Amsa
Calculation of workdone by the force to scratch the string
W = ½ ke\(^{2}\) = \(\frac{f^{2}}{2k}\)
W = \(\frac{50^{2}}{2 \times 100}\)
W = 12.5 J
OR
e = \(\frac{F}{K} = \frac{50}{100}\) = 0.5m
w = ½ = ke\(^{2}\)
w = ½ x 100 x 0.5\(^{2}\)
W = 12.5 J
Tambaya 14 Rahoto
(a)(i) What is meant by the term artificial radioactivity?
(ii) Complete the table below
| Emission | Nature | Charge | Ionizing |
| High speed electron | Moderately ionizing | ||
| Neutral | Negligible ionizing ability | ||
| Alpha particles | Positive |
(b) In an x-ray tube, an electron is accelerated from rest towards a metal target by a 30 kV source. Calculate the kinetic energy of the electron. [e=1.6 x \(10^{-19}\) C]
(c) The table below shows the frequencies of radiations incident on a certain metal and the corresponding kinetic energies of the photoelectrons.
| Frequency x \(10^{14}\) (Hz) | 6.8 | 8.0 | 9.2 | 10.0 | 11.0 |
| Kinetic energy x \(10^{-19}\) (j) | 0.8 | 1.6 | 2.4 | 2.9 | 3.8 |
(i) Plot a graph of kinetic energy, K.E, on the vertical axis and frequency, f, on the horizontal axis starting both axes from the origin (0,0).
(ii) From the graph, determine the:
i. Planck's constant;
ii. Threshold frequency of radiations;
iii. Work function of the metal.
(a)() Meaning of artificial radioactivity;
The process by which a stable nucleus is bombarded with a neutron to make it unstable and so disintegrates/decays with the emission of particles/radiation and. energy.
| Emission | Nature | Charge | Ionizing |
| Beta (\(\beta\)) | High speed electron | Negative | Moderately ionizing |
| Gamma (\(\gamma\)) | Electro-magnetic radiation | Neutral | Negligible ionizing ability |
| Alpha particles | Helium nucleus | Positive | Highly ionizing |
(b) The kinetic energy of the electron can be calculated using the formula: KE = qV, where q is the charge of the electron and V is the potential difference. Substituting the given values, we get:
K.E = eV
KE = (1.6 x 10^-19 C)(30,000 V)
KE = 4.8 x 10^-15 J (c)
(i)
To calculate the slope of the graph, you need to determine the change in the dependent variable (kinetic energy) divided by the change in the independent variable (frequency). In this case, you can choose any two points on the graph and calculate the slope using the following formula:
slope = (kinetic_energy2 - kinetic_energy1) / (frequency2 - frequency1)
Let's take two points from the given data, for example:
Point 1: (frequency1, kinetic_energy1) = (6.8 x 10^14 Hz, 0.8 x 10^-19 J)
Point 2: (frequency2, kinetic_energy2) = (8.0 x 10^14 Hz, 1.6 x 10^-19 J)
Now, we can calculate the slope:
slope = (1.6 x 10^-19 J - 0.8 x 10^-19 J) / (8.0 x 10^14 Hz - 6.8 x 10^14 Hz)
slope = 1 x 10^-5 J Hz^(-1).
To determine Planck's constant from the given graph and slope, we can use the equation:
slope = h / e
where h is Planck's constant and e is the elementary charge (1.602176634 x 10^-19 C).
From the previous calculation, the slope of the graph is 1 x 10^-5 J Hz^(-1).
Let's substitute the values into the equation to solve for Planck's constant:
1 x 10^-5 J Hz^(-1) = h / (1.602176634 x 10^-19 C)
To isolate h, we can rearrange the equation:
h = slope * e
Substituting the values:
h = (1 x 10^-5 J Hz^(-1)) * (1.602176634 x 10^-19 C)
Evaluating the expression:
h ≈ 1.602176634 x 10^-24 J·s
Therefore, from the given graph and slope, the approximate value of Planck's constant is 1.602176634 x 10^-24 J·s.
(ii) To determine the threshold frequency of radiation from the given information, we need to use the concept of the photoelectric effect and the relationship between the kinetic energy of photoelectrons and the frequency of incident radiation.
According to the photoelectric effect, electrons are ejected from a metal surface when illuminated by electromagnetic radiation of sufficient energy. The minimum frequency of radiation required to eject electrons is known as the threshold frequency.
The relationship between the kinetic energy of photoelectrons and the frequency of incident radiation is given by the equation:
K.E. = h * (frequency - threshold_frequency)
where K.E. is the kinetic energy of the photoelectrons, h is Planck's constant, frequency is the frequency of incident radiation, and threshold_frequency is the threshold frequency.
From the graph, we have the slope, which is equal to h, and the kinetic energy corresponding to each frequency. We can select any point on the graph where the kinetic energy is non-zero and solve for the threshold frequency.
Let's choose the point (frequency, kinetic energy) = (6.8 x 10^14 Hz, 0.8 x 10^-19 J) from the given data.
0.8 x 10^-19 J = slope * (6.8 x 10^14 Hz - threshold_frequency)
Substituting the slope value:
0.8 x 10^-19 J = 1.602176634 x 10^-24 J·s * (6.8 x 10^14 Hz - threshold_frequency)
To solve for the threshold frequency, we can rearrange the equation:
threshold_frequency = 6.8 x 10^14 Hz - (0.8 x 10^-19 J / (1.602176634 x 10^-24 J·s))
Calculating the threshold frequency:
threshold_frequency = 6.8 x 10^14 Hz - 4.992706701 x 10^4 Hz
threshold_frequency ≈ 6.799500729 x 10^14 Hz
Therefore, the threshold frequency of radiation is approximately 6.799500729 x 10^14 Hz.
(iii)
To determine the work function of the metal, we can use the equation:
Work function = h * threshold_frequency
where h is Planck's constant and threshold_frequency is the threshold frequency of radiation.
From the previous calculations, the approximate value of Planck's constant is 1.602176634 x 10^-24 J·s and the threshold frequency is approximately 6.799500729 x 10^14 Hz.
Substituting these values into the equation, we can calculate the work function:
Work function = (1.602176634 x 10^-24 J·s) * (6.799500729 x 10^14 Hz)
Work function ≈ 1.090589631 x 10^-9 J
Therefore, based on the given information, the approximate value of the work function of the metal is 1.090589631 x 10^-9 J.
Bayanin Amsa
(a)() Meaning of artificial radioactivity;
The process by which a stable nucleus is bombarded with a neutron to make it unstable and so disintegrates/decays with the emission of particles/radiation and. energy.
| Emission | Nature | Charge | Ionizing |
| Beta (\(\beta\)) | High speed electron | Negative | Moderately ionizing |
| Gamma (\(\gamma\)) | Electro-magnetic radiation | Neutral | Negligible ionizing ability |
| Alpha particles | Helium nucleus | Positive | Highly ionizing |
(b) The kinetic energy of the electron can be calculated using the formula: KE = qV, where q is the charge of the electron and V is the potential difference. Substituting the given values, we get:
K.E = eV
KE = (1.6 x 10^-19 C)(30,000 V)
KE = 4.8 x 10^-15 J (c)
(i)
To calculate the slope of the graph, you need to determine the change in the dependent variable (kinetic energy) divided by the change in the independent variable (frequency). In this case, you can choose any two points on the graph and calculate the slope using the following formula:
slope = (kinetic_energy2 - kinetic_energy1) / (frequency2 - frequency1)
Let's take two points from the given data, for example:
Point 1: (frequency1, kinetic_energy1) = (6.8 x 10^14 Hz, 0.8 x 10^-19 J)
Point 2: (frequency2, kinetic_energy2) = (8.0 x 10^14 Hz, 1.6 x 10^-19 J)
Now, we can calculate the slope:
slope = (1.6 x 10^-19 J - 0.8 x 10^-19 J) / (8.0 x 10^14 Hz - 6.8 x 10^14 Hz)
slope = 1 x 10^-5 J Hz^(-1).
To determine Planck's constant from the given graph and slope, we can use the equation:
slope = h / e
where h is Planck's constant and e is the elementary charge (1.602176634 x 10^-19 C).
From the previous calculation, the slope of the graph is 1 x 10^-5 J Hz^(-1).
Let's substitute the values into the equation to solve for Planck's constant:
1 x 10^-5 J Hz^(-1) = h / (1.602176634 x 10^-19 C)
To isolate h, we can rearrange the equation:
h = slope * e
Substituting the values:
h = (1 x 10^-5 J Hz^(-1)) * (1.602176634 x 10^-19 C)
Evaluating the expression:
h ≈ 1.602176634 x 10^-24 J·s
Therefore, from the given graph and slope, the approximate value of Planck's constant is 1.602176634 x 10^-24 J·s.
(ii) To determine the threshold frequency of radiation from the given information, we need to use the concept of the photoelectric effect and the relationship between the kinetic energy of photoelectrons and the frequency of incident radiation.
According to the photoelectric effect, electrons are ejected from a metal surface when illuminated by electromagnetic radiation of sufficient energy. The minimum frequency of radiation required to eject electrons is known as the threshold frequency.
The relationship between the kinetic energy of photoelectrons and the frequency of incident radiation is given by the equation:
K.E. = h * (frequency - threshold_frequency)
where K.E. is the kinetic energy of the photoelectrons, h is Planck's constant, frequency is the frequency of incident radiation, and threshold_frequency is the threshold frequency.
From the graph, we have the slope, which is equal to h, and the kinetic energy corresponding to each frequency. We can select any point on the graph where the kinetic energy is non-zero and solve for the threshold frequency.
Let's choose the point (frequency, kinetic energy) = (6.8 x 10^14 Hz, 0.8 x 10^-19 J) from the given data.
0.8 x 10^-19 J = slope * (6.8 x 10^14 Hz - threshold_frequency)
Substituting the slope value:
0.8 x 10^-19 J = 1.602176634 x 10^-24 J·s * (6.8 x 10^14 Hz - threshold_frequency)
To solve for the threshold frequency, we can rearrange the equation:
threshold_frequency = 6.8 x 10^14 Hz - (0.8 x 10^-19 J / (1.602176634 x 10^-24 J·s))
Calculating the threshold frequency:
threshold_frequency = 6.8 x 10^14 Hz - 4.992706701 x 10^4 Hz
threshold_frequency ≈ 6.799500729 x 10^14 Hz
Therefore, the threshold frequency of radiation is approximately 6.799500729 x 10^14 Hz.
(iii)
To determine the work function of the metal, we can use the equation:
Work function = h * threshold_frequency
where h is Planck's constant and threshold_frequency is the threshold frequency of radiation.
From the previous calculations, the approximate value of Planck's constant is 1.602176634 x 10^-24 J·s and the threshold frequency is approximately 6.799500729 x 10^14 Hz.
Substituting these values into the equation, we can calculate the work function:
Work function = (1.602176634 x 10^-24 J·s) * (6.799500729 x 10^14 Hz)
Work function ≈ 1.090589631 x 10^-9 J
Therefore, based on the given information, the approximate value of the work function of the metal is 1.090589631 x 10^-9 J.
Tambaya 15 Rahoto
You are provided with a loaded boiling tube with a centimeter scale fixed inside it, a transparent vessel filled with water, standard masses 2 g, 5g and 10 g, and a slide vernier caliper. Use the diagram above as a guide to perform the experiment.
(i) Use the slide vernier caliper to measure and record the external diameter, D, of the boiling tube.
(ii) Evaluate \(A = 0.25\pi D^2\), where \(\pi = 3.14\).
(iii) Place the loaded boiling tube gently in the water in the transparent vessel such that it floats vertically.
(iv) Read and record the depth of immersion, y, from the zero mark of the scale fixed inside the boiling tube.
(v) Add a mass, m = 2g, to the boiling tube. Read and record the new depth of immersion, y, from the zero mark of the scale.
(vi) Evaluate \(h = (y - y_0)\), log h and log m.
(vii) Repeat the experiment for four other values of m = 5g, 7g, 10 g, and 12g. In each case, record y and evaluate h, log h, and log m.
(viii) Tabulate the results.
(ix) Plot a graph with log m on the vertical axis and log h on the horizontal axis starting both axes from the origin (0,0).
(b)(i) State in full the law on which the experiment in (a) is based.
(ii) A uniform cylindrical rod is 0.63 m long and it has a cross-sectional area of 0.1 m\(^2\). Calculate the depth of immersion of the rod if it floats vertically in a liquid of relative density 1.26. [density of rod = 720 kg m\(^{-3}\), g = 10 m s\(^{-2}\)].
(a) Results and graph
The external diameter of the boiling tube measured with the slide vernier caliper is:
\[D=2.41\ \text{cm}\]
Hence,
\[A=\frac{\pi D^2}{4}=\frac{3.14(2.41)^2}{4}=4.56\ \text{cm}^2.\]
The initial depth of immersion of the loaded boiling tube is:
\[y_0=3.000\ \text{cm}.\]
| Mass, \(m\) (g) | Depth, \(y\) (cm) | \(h=y-y_0\) (cm) | \(\log h\) | \(\log m\) |
|---|---|---|---|---|
| 2 | 3.439 | 0.439 | −0.358 | 0.301 |
| 5 | 4.097 | 1.097 | 0.040 | 0.699 |
| 7 | 4.535 | 1.535 | 0.186 | 0.845 |
| 10 | 5.193 | 2.193 | 0.341 | 1.000 |
| 12 | 5.632 | 2.632 | 0.420 | 1.079 |
The graph of \(\log m\) against \(\log h\) is shown below.
Using two widely separated points on the best-fit line, \((-0.358,\ 0.301)\) and \((0.420,\ 1.079)\),
\[\text{gradient}=\frac{1.079-0.301}{0.420-(-0.358)}=\frac{0.778}{0.778}=1.00.\]
Thus, \(\log m=\log h+0.659\), showing that \(m\propto h\).
(b)(i)
Law of flotation: A body floating in a fluid displaces a quantity of the fluid whose weight is equal to the weight of the body.
(b)(ii)
Mass of rod:
\[m=\rho V=720(0.1)(0.63)=45.36\ \text{kg}.\]
Weight of rod:
\[W=mg=45.36\times10=453.6\ \text{N}.\]
Density of liquid:
\[\rho_l=1.26\times1000=1260\ \text{kg m}^{-3}.\]
If \(y\) is the depth of immersion, then the upthrust is
\[U=\rho_l(0.1y)g=1260(0.1y)(10)=1260y.\]
For flotation, upthrust equals weight:
\[1260y=453.6.\]
\[y=\frac{453.6}{1260}=0.36\ \text{m}.\]
Depth of immersion = \(0.36\ \text{m}\).
Bayanin Amsa
(a) Results and graph
The external diameter of the boiling tube measured with the slide vernier caliper is:
\[D=2.41\ \text{cm}\]
Hence,
\[A=\frac{\pi D^2}{4}=\frac{3.14(2.41)^2}{4}=4.56\ \text{cm}^2.\]
The initial depth of immersion of the loaded boiling tube is:
\[y_0=3.000\ \text{cm}.\]
| Mass, \(m\) (g) | Depth, \(y\) (cm) | \(h=y-y_0\) (cm) | \(\log h\) | \(\log m\) |
|---|---|---|---|---|
| 2 | 3.439 | 0.439 | −0.358 | 0.301 |
| 5 | 4.097 | 1.097 | 0.040 | 0.699 |
| 7 | 4.535 | 1.535 | 0.186 | 0.845 |
| 10 | 5.193 | 2.193 | 0.341 | 1.000 |
| 12 | 5.632 | 2.632 | 0.420 | 1.079 |
The graph of \(\log m\) against \(\log h\) is shown below.
Using two widely separated points on the best-fit line, \((-0.358,\ 0.301)\) and \((0.420,\ 1.079)\),
\[\text{gradient}=\frac{1.079-0.301}{0.420-(-0.358)}=\frac{0.778}{0.778}=1.00.\]
Thus, \(\log m=\log h+0.659\), showing that \(m\propto h\).
(b)(i)
Law of flotation: A body floating in a fluid displaces a quantity of the fluid whose weight is equal to the weight of the body.
(b)(ii)
Mass of rod:
\[m=\rho V=720(0.1)(0.63)=45.36\ \text{kg}.\]
Weight of rod:
\[W=mg=45.36\times10=453.6\ \text{N}.\]
Density of liquid:
\[\rho_l=1.26\times1000=1260\ \text{kg m}^{-3}.\]
If \(y\) is the depth of immersion, then the upthrust is
\[U=\rho_l(0.1y)g=1260(0.1y)(10)=1260y.\]
For flotation, upthrust equals weight:
\[1260y=453.6.\]
\[y=\frac{453.6}{1260}=0.36\ \text{m}.\]
Depth of immersion = \(0.36\ \text{m}\).
Tambaya 16 Rahoto
(a) State one condition each necessary for the characteristics each of the following occurrences:
(i) Constructive interference of waves.
(ii) Total internal reflection.
(iii) Production of beats.
(b) In a resonance tube experiment using a tuning fork of frequency 256 Hz, the first position of resonance was 35 cm, the next position was 100 cm. Calculate the velocity of sound in air from the experiment.
(c)(i) State the three classifications of musical instalments
(ii) Give one example each of the classifications stated in (c)(i).
(d) Calculate the critical angle for light traveling from glass to air. [refractive index of glass = 1.5].
(e) The speed of sound in a medium at a temperature of 102 °C is \(240\ \mathrm{m\ s}^{-1}\). If the speed of sound in the medium is \(3\ 10\ \mathrm{m\ s}^{-1}\). Calculate its temperature
(a)
(i) One necessary condition for constructive interference of waves is that the waves have the same frequency, and the crests and troughs of the waves align with each other.
(ii) One necessary condition for total internal reflection is that the angle of incidence is greater than the critical angle for the boundary between two media, and the wave travels from a denser medium to a less dense medium.
(iii) One necessary condition for the production of beats is that two waves with slightly different frequencies interfere with each other, and their amplitudes vary periodically in time.
(b) The velocity of sound in air can be calculated as follows:
- The distance between the first and second position of resonance is 100 cm - 35 cm = 65 cm = 0.65 m.
- The wavelength of the sound wave is twice the distance between the first and second position of resonance, which is 2 x 0.65 m = 1.3 m.
- The frequency of the tuning fork is 256 Hz. - Using the equation v = fλ, where v is the velocity of sound, f is the frequency, and λ is the wavelength, we can calculate the velocity of sound as
v = 256 Hz x 1.3 m = 332.8 m/s.
(c) (i) The three classifications of musical instruments are:
- Stringed instruments
- Wind instruments
- Percussion instruments
(ii) Examples of each classification are:
- Stringed instruments: guitar, violin
- Wind instruments: flute, trumpet
- Percussion instruments: drums, xylophone
(d) The critical angle for light traveling from glass to air can be calculated as follows:
- The refractive index of glass is given as 1.5.
- Using the formula sin ?c = 1/n, where ?c is the critical angle and n is the refractive index, we can calculate the critical angle as sin θc = 1/1.5 = 0.67.
- Taking the inverse sine of 0.67, we can find the critical angle as θc = 42.3 degrees.
(e) The speed of sound in a medium is directly proportional to the square root of the temperature of the medium. Using this relationship, we can calculate the temperature of the medium as follows:
- Let T1 be the temperature of the medium where the speed of sound is 240 m/s, and T2 be the temperature of the medium where the speed of sound is 310 m/s.
- The ratio of the speeds of sound is 310/240 = 1.29.
- The ratio of the square roots of the temperatures is √(T2/T1) = 1.29. - Solving for T2, we get T2 = T1 x (1.29)^2 = T1 x 1.6641. - Substituting T1 = 102 + 273 = 375 K, we get T2 = 625 K. -
Therefore, the temperature of the medium where the speed of sound is 310 m/s is 625 - 273 = 352 °C.
Bayanin Amsa
(a)
(i) One necessary condition for constructive interference of waves is that the waves have the same frequency, and the crests and troughs of the waves align with each other.
(ii) One necessary condition for total internal reflection is that the angle of incidence is greater than the critical angle for the boundary between two media, and the wave travels from a denser medium to a less dense medium.
(iii) One necessary condition for the production of beats is that two waves with slightly different frequencies interfere with each other, and their amplitudes vary periodically in time.
(b) The velocity of sound in air can be calculated as follows:
- The distance between the first and second position of resonance is 100 cm - 35 cm = 65 cm = 0.65 m.
- The wavelength of the sound wave is twice the distance between the first and second position of resonance, which is 2 x 0.65 m = 1.3 m.
- The frequency of the tuning fork is 256 Hz. - Using the equation v = fλ, where v is the velocity of sound, f is the frequency, and λ is the wavelength, we can calculate the velocity of sound as
v = 256 Hz x 1.3 m = 332.8 m/s.
(c) (i) The three classifications of musical instruments are:
- Stringed instruments
- Wind instruments
- Percussion instruments
(ii) Examples of each classification are:
- Stringed instruments: guitar, violin
- Wind instruments: flute, trumpet
- Percussion instruments: drums, xylophone
(d) The critical angle for light traveling from glass to air can be calculated as follows:
- The refractive index of glass is given as 1.5.
- Using the formula sin ?c = 1/n, where ?c is the critical angle and n is the refractive index, we can calculate the critical angle as sin θc = 1/1.5 = 0.67.
- Taking the inverse sine of 0.67, we can find the critical angle as θc = 42.3 degrees.
(e) The speed of sound in a medium is directly proportional to the square root of the temperature of the medium. Using this relationship, we can calculate the temperature of the medium as follows:
- Let T1 be the temperature of the medium where the speed of sound is 240 m/s, and T2 be the temperature of the medium where the speed of sound is 310 m/s.
- The ratio of the speeds of sound is 310/240 = 1.29.
- The ratio of the square roots of the temperatures is √(T2/T1) = 1.29. - Solving for T2, we get T2 = T1 x (1.29)^2 = T1 x 1.6641. - Substituting T1 = 102 + 273 = 375 K, we get T2 = 625 K. -
Therefore, the temperature of the medium where the speed of sound is 310 m/s is 625 - 273 = 352 °C.
Tambaya 17 Rahoto
You are provided with a battery of e.m.f, E, a standard resistor, R, of resistance 2 \( \Omega \), a key, K, an ammeter, A, a jockey, J, a potentiometer, UV, and some connecting wires.
(i) Measure and record the emf, E, of the battery.
(ii) Set up the circuit as shown in the diagram above with the key open.
(iii) Place the jockey at the point, U, of the potentiometer wire. Close the key and record the reading, i, of the ammeter.
(iv) Place the jockey at a point T on the potentiometer wire UV such that d = UT = 30.0 cm.
(v) Close the circuit, read and record the current, I, on the ammeter,
(vi) Evaluate \(I^1\).
(vi) Repeat the experiment for four other values of d = 40.0 cm, 50.0 cm, 60.0 cm and 70.0 cm. In each case, record I and evaluate \(I^1\).
(vii) Tabulate the results
(ix) Plot a graph with d on the vertical axis and I on the horizontal axis stalling both axes from the origin (0,0).
(x) Determine the slope, s, of the graph.
(xi) From the graph determine the value \(I_1\), of I when d = 0. (ci) Given that=s, calculate 8.
(xii) State two precautions taken to ensure accurate results.
(xii) Given that \( \frac{E}{\delta} = s \), calculate \( \delta \).
(b)(i) Write down the equation that connects the resistance, R, of a wire and the factors on which it depends. State the meaning of each of the symbols.
(ii) An electric fan draws a current of0.75 A in a 240 V circuit. Calculate the cost of using, the fan for 10 hours if the utility rate is $ 0.50 per kWh.
(a) Potentiometer Experiment
The e.m.f. of the battery was measured as:
\(E=3.0\text{ V}\)
With the jockey at \(U\), \(d=0\) and the ammeter reading was:
\(i=1.50\text{ A}\)
The readings obtained are shown below. The reciprocal current was evaluated from \(I^{-1}=1/I\).
| Distance, \(d\) (cm) | Current, \(I\) (A) | \(I^{-1}\) (A−1) |
|---|---|---|
| 30.0 | 1.154 | 0.867 |
| 40.0 | 1.111 | 0.900 |
| 50.0 | 1.071 | 0.933 |
| 60.0 | 1.034 | 0.967 |
| 70.0 | 1.000 | 1.000 |
A graph of \(d\) against \(I^{-1}\), with both axes beginning at the origin, is plotted below.
Using two widely separated points on the straight line, \((0.867\text{ A}^{-1},30.0\text{ cm})\) and \((1.000\text{ A}^{-1},70.0\text{ cm})\):
\[ s=\frac{70.0-30.0}{1.000-0.867} =\frac{40.0}{0.133} \approx 3.00\times10^2\text{ cm A}. \]
Hence, the slope of the graph is \(3.00\times10^2\text{ cm A}\).
Extrapolating the straight line to \(d=0\),
\(I^{-1}=0.667\text{ A}^{-1}\).
Therefore, \(I=1/0.667=1.50\text{ A}\), which agrees with the current when the jockey is at \(U\).
Given that \(s=E/\delta\),
\[ \delta=\frac{E}{s}=\frac{3.0}{3.00\times10^2} =1.00\times10^{-2}\ \Omega\text{ cm}^{-1}. \]
Precautions
- The key was closed only while a reading was being taken, to prevent heating and consequent change in resistance of the wire.
- The jockey was touched lightly on the potentiometer wire and not dragged along it, to avoid damaging the wire or altering its resistance.
(b)
(i) The resistance of a uniform wire is given by
\[R=\frac{\rho l}{A}.\]
- \(R\) is the resistance of the wire in ohms \((\Omega)\).
- \(\rho\) is the resistivity of the material in \(\Omega\,\text{m}\).
- \(l\) is the length of the wire in metres.
- \(A\) is the cross-sectional area of the wire in \(\text{m}^2\).
(ii)
\[ P=VI=240\times0.75=180\text{ W}=0.180\text{ kW}. \] \[ \text{Electrical energy}=0.180\times10=1.80\text{ kWh}. \] \[ \text{Cost}=1.80\times\$0.50=\$0.90. \]
Therefore, the cost of operating the fan for 10 hours is \(\$0.90\).
Bayanin Amsa
(a) Potentiometer Experiment
The e.m.f. of the battery was measured as:
\(E=3.0\text{ V}\)
With the jockey at \(U\), \(d=0\) and the ammeter reading was:
\(i=1.50\text{ A}\)
The readings obtained are shown below. The reciprocal current was evaluated from \(I^{-1}=1/I\).
| Distance, \(d\) (cm) | Current, \(I\) (A) | \(I^{-1}\) (A−1) |
|---|---|---|
| 30.0 | 1.154 | 0.867 |
| 40.0 | 1.111 | 0.900 |
| 50.0 | 1.071 | 0.933 |
| 60.0 | 1.034 | 0.967 |
| 70.0 | 1.000 | 1.000 |
A graph of \(d\) against \(I^{-1}\), with both axes beginning at the origin, is plotted below.
Using two widely separated points on the straight line, \((0.867\text{ A}^{-1},30.0\text{ cm})\) and \((1.000\text{ A}^{-1},70.0\text{ cm})\):
\[ s=\frac{70.0-30.0}{1.000-0.867} =\frac{40.0}{0.133} \approx 3.00\times10^2\text{ cm A}. \]
Hence, the slope of the graph is \(3.00\times10^2\text{ cm A}\).
Extrapolating the straight line to \(d=0\),
\(I^{-1}=0.667\text{ A}^{-1}\).
Therefore, \(I=1/0.667=1.50\text{ A}\), which agrees with the current when the jockey is at \(U\).
Given that \(s=E/\delta\),
\[ \delta=\frac{E}{s}=\frac{3.0}{3.00\times10^2} =1.00\times10^{-2}\ \Omega\text{ cm}^{-1}. \]
Precautions
- The key was closed only while a reading was being taken, to prevent heating and consequent change in resistance of the wire.
- The jockey was touched lightly on the potentiometer wire and not dragged along it, to avoid damaging the wire or altering its resistance.
(b)
(i) The resistance of a uniform wire is given by
\[R=\frac{\rho l}{A}.\]
- \(R\) is the resistance of the wire in ohms \((\Omega)\).
- \(\rho\) is the resistivity of the material in \(\Omega\,\text{m}\).
- \(l\) is the length of the wire in metres.
- \(A\) is the cross-sectional area of the wire in \(\text{m}^2\).
(ii)
\[ P=VI=240\times0.75=180\text{ W}=0.180\text{ kW}. \] \[ \text{Electrical energy}=0.180\times10=1.80\text{ kWh}. \] \[ \text{Cost}=1.80\times\$0.50=\$0.90. \]
Therefore, the cost of operating the fan for 10 hours is \(\$0.90\).
