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Swali 1 Ripoti
2a. State Faraday's law of electromagnetic induction.
b. Draw a labelled diagram of an induction coil and explain how it works.
c. How is the effect of eddy currents minimized in the coil?
d. State two reasons why a capacitor should be included in the primary circuit of the coil.
e. State three uses of an induction coil.
a. Faraday's law of electromagnetic induction states that the induced e.m.f. in a circuit is directly proportional to the rate of change of the magnetic flux, or field lines, linking the circuit i.e e ∝ \(\frac{dφ}{\text{dt}}\).
b. See diagram above. How it works: it has a primary coil of a few turns and a secondary coil of many turns. When the primary circuit is made, a current flows in P. A is then attracted, so breaking the circuit at the contact point. This makes the large magnetic flux in the secondary fall very quickly. A high voltage is then produced at its terminals X and Y, and a spark appears across the air gap as the circuit is made and broken repeatedly.
c. The eddy current can be minimized by laminating the core to reduce energy losses.
d. The spark which produce across the gap of the make and break device is reduced by the capacitor. Thus, it should be included.
e. (i). It is used in the operation of X-ray tubes. (ii). It is used in the generation of high voltages. (iii). It is used in morse radio transmitter.
Maelezo ya Majibu
a. Faraday's law of electromagnetic induction states that the induced e.m.f. in a circuit is directly proportional to the rate of change of the magnetic flux, or field lines, linking the circuit i.e e ∝ \(\frac{dφ}{\text{dt}}\).
b. See diagram above. How it works: it has a primary coil of a few turns and a secondary coil of many turns. When the primary circuit is made, a current flows in P. A is then attracted, so breaking the circuit at the contact point. This makes the large magnetic flux in the secondary fall very quickly. A high voltage is then produced at its terminals X and Y, and a spark appears across the air gap as the circuit is made and broken repeatedly.
c. The eddy current can be minimized by laminating the core to reduce energy losses.
d. The spark which produce across the gap of the make and break device is reduced by the capacitor. Thus, it should be included.
e. (i). It is used in the operation of X-ray tubes. (ii). It is used in the generation of high voltages. (iii). It is used in morse radio transmitter.
Swali 2 Ripoti
1a. Explain what is meant by the acceleration of free fall due to gravity, g.
b. State two reasons why g varies on the surface of the Earth.
c. A stone is projected upwards at an angle of 30° to the horizontal from the top of a tower of height 100 m and it hits the ground at a point Q. If the initial velocity of projection is 100 ms\(^{-1}\), calculate the:
i. maximum height of the stone above the ground (neglect air resistance and take g as 10 ms\(^{-2}\))
ii. time it takes to reach this height
iii. time of flight.
iv. horizontal distance from the foot of the tower to the point Q (neglect air resistance and take g as 10 ms\(^{-2}\))
1a. Acceleration of a free fall due to gravity is the force of attraction of the Earth on a unit mass.
b. g varies on the surface of the earth because of the shape of the earth and the rotation of the earth about its polar axis.
ci. θ = 30°; H = 100 m; u = 100 m/s; h = ?
Total height = H + h
=> h = \(\frac{u^2 sin^2 θ }{2g}\) = \(\frac{100^2 sin^2 30°}{ 2 \times 10}\)
= \(\frac{10000 \times 0.25}{ 20}\) = \(\frac{2500}{ 20}\) = 125 m
. Height from the ground = height of tower + the maximum height the stone reached = 100 + 125 = 225 m
ii. t = \(\frac{u sin θ }{\text{g}}\) = \(\frac{100 \times sin 30°}{10}\) = \(\frac{100 \times 0.5 }{10}\) = 10 × 0.5 = 5 s
iii. Time of flight = time taken for a particle which is projected to return to its original level T = \(\frac{2u sin θ}{\text{g}}\) = \(\frac{2 \times 100 \times sin 30°}{10}\) = \(\frac{200 \times 0.5}{10}\) = 10 s
Time of flight = 10 s
But the total time taken to land on the ground = T + time used to return from the top of the tower to the ground level. Since it's falling S\(_y\) = 100 m
S\(_y\) = ut sin θ - \(\frac{1}{2}\) g t\(^2\) => -100
= 100t sin 30° - \(\frac{1}{2}\) (10) t\(^2\) =>
-100 = 100t × 0.5 - 5 t\(^2\) => -100 = 50t - 5 t\(^2\) => 5 t\(^2\) - 50t - 100 = 0 => t\(^2\) - 10t - 20 = 0
t = \(\frac{10 \pm \sqrt{10^2 - 4 \times -20 \times 1}}{2}\) = \(\frac{10 \pm \sqrt{100 + 80}}{2}\)
t = \(\frac{10 + 13.416}{2}\) = \(\frac{23.416}{2}\) ≈ 11.7s (ignore the -ve part)
t = 11.7 s => Total time = 10 + 11.7 = 21.7 seconds
iv. Horizontal distance = the range = \(\frac{u^2 sin 2θ}{\text{g}}\) = \(\frac{100^2 sin 60°}{10}\) = \(\frac{10000 \times 0.866}{10}\) = 1000 × 0.866 = 866 m
Maelezo ya Majibu
1a. Acceleration of a free fall due to gravity is the force of attraction of the Earth on a unit mass.
b. g varies on the surface of the earth because of the shape of the earth and the rotation of the earth about its polar axis.
ci. θ = 30°; H = 100 m; u = 100 m/s; h = ?
Total height = H + h
=> h = \(\frac{u^2 sin^2 θ }{2g}\) = \(\frac{100^2 sin^2 30°}{ 2 \times 10}\)
= \(\frac{10000 \times 0.25}{ 20}\) = \(\frac{2500}{ 20}\) = 125 m
. Height from the ground = height of tower + the maximum height the stone reached = 100 + 125 = 225 m
ii. t = \(\frac{u sin θ }{\text{g}}\) = \(\frac{100 \times sin 30°}{10}\) = \(\frac{100 \times 0.5 }{10}\) = 10 × 0.5 = 5 s
iii. Time of flight = time taken for a particle which is projected to return to its original level T = \(\frac{2u sin θ}{\text{g}}\) = \(\frac{2 \times 100 \times sin 30°}{10}\) = \(\frac{200 \times 0.5}{10}\) = 10 s
Time of flight = 10 s
But the total time taken to land on the ground = T + time used to return from the top of the tower to the ground level. Since it's falling S\(_y\) = 100 m
S\(_y\) = ut sin θ - \(\frac{1}{2}\) g t\(^2\) => -100
= 100t sin 30° - \(\frac{1}{2}\) (10) t\(^2\) =>
-100 = 100t × 0.5 - 5 t\(^2\) => -100 = 50t - 5 t\(^2\) => 5 t\(^2\) - 50t - 100 = 0 => t\(^2\) - 10t - 20 = 0
t = \(\frac{10 \pm \sqrt{10^2 - 4 \times -20 \times 1}}{2}\) = \(\frac{10 \pm \sqrt{100 + 80}}{2}\)
t = \(\frac{10 + 13.416}{2}\) = \(\frac{23.416}{2}\) ≈ 11.7s (ignore the -ve part)
t = 11.7 s => Total time = 10 + 11.7 = 21.7 seconds
iv. Horizontal distance = the range = \(\frac{u^2 sin 2θ}{\text{g}}\) = \(\frac{100^2 sin 60°}{10}\) = \(\frac{10000 \times 0.866}{10}\) = 1000 × 0.866 = 866 m
Swali 3 Ripoti
4a. Define the boiling point of a liquid.
b. Describe with the aid of a labelled diagram, an experiment to determine the boiling point of a small quantity of a liquid.
c. State two factors that may affect the boiling point of a liquid.
d. Using the kinetic theory of matter, explain why pure water changes to steam at S.T.P without any change in temperature, although heat is being supplied to the water.
a. The boiling point of a liquid is the temperature at which a (pure) liquid changes from the liquid to the gaseous state without a change in temperature when the atmospheric pressure is normal (760 mmHg)
b. Aim: To determine the boiling point of a small quantity of a liquid
Apparatus: J-tube, thermometer, water, beaker & source of heat.
Procedure: The J-tube, as shown in the diagram, is heated in a beaker of water, and the water in the J-tube is monitored. It remains trapped until the water in the beaker is boiling. Then the water in the tube comes to the same level in each limb, showing that the pressure of the vapour in the closed limb is equal to the pressure of air outside. Conclusion: Equal levels in the two limb shows that the vapour pressure is equal to the atmospheric pressure, which is the boiling point of the liquid. The experiment is repeated, and the mean is taken.
c. (i). Presence of impurities (ii). Pressure on the liquid (iii) Nature of liquid.
d. According to the kinetic theory of matter, temperature depends on the average kinetic energy of the molecules. When pure water boils, the heat supplied is not used to increase the kinetic energy of the molecules. Instead, it is used to overcome the intermolecular forces holding the water molecules together in the liquid state.
This energy, called latent heat of vaporization, separates the molecules to form steam. Since the kinetic energy of the molecules does not increase, the temperature remains constant until all the water has changed to steam.
Maelezo ya Majibu
a. The boiling point of a liquid is the temperature at which a (pure) liquid changes from the liquid to the gaseous state without a change in temperature when the atmospheric pressure is normal (760 mmHg)
b. Aim: To determine the boiling point of a small quantity of a liquid
Apparatus: J-tube, thermometer, water, beaker & source of heat.
Procedure: The J-tube, as shown in the diagram, is heated in a beaker of water, and the water in the J-tube is monitored. It remains trapped until the water in the beaker is boiling. Then the water in the tube comes to the same level in each limb, showing that the pressure of the vapour in the closed limb is equal to the pressure of air outside. Conclusion: Equal levels in the two limb shows that the vapour pressure is equal to the atmospheric pressure, which is the boiling point of the liquid. The experiment is repeated, and the mean is taken.
c. (i). Presence of impurities (ii). Pressure on the liquid (iii) Nature of liquid.
d. According to the kinetic theory of matter, temperature depends on the average kinetic energy of the molecules. When pure water boils, the heat supplied is not used to increase the kinetic energy of the molecules. Instead, it is used to overcome the intermolecular forces holding the water molecules together in the liquid state.
This energy, called latent heat of vaporization, separates the molecules to form steam. Since the kinetic energy of the molecules does not increase, the temperature remains constant until all the water has changed to steam.
Swali 4 Ripoti
3a. What is surface tension? Explain the phenomenon in terms of intermolecular forces.
b. Describe a simple experiment to demonstrate the surface tension of a liquid.
c. State three examples to illustrate the effects of surface tension.
d. Why does water wet a clean glass surface, whereas mercury does not?
e. State two methods by which the surface tension of a liquid may be reduced.
a. Surface tension: Surface tension is a property of a liquid whereby its surface behaves as if it were covered by an elastic skin. Or it is the tangential force of a liquid.
b. The surface “skin” of a liquid exists because of the attractive forces between its molecules. A molecule inside the liquid is pulled equally in all directions by surrounding molecules, so the net force on it is zero. However, a molecule at the surface has more neighboring molecules below than above, creating a net inward (downward) pull. This inward force tends to minimize the surface area, causing the surface to behave like a stretched elastic skin. This effect is known as surface tension.
See diagram above.
Aim: To demonstrate surface tension in a liquid
Apparatus: A needle, water, a beaker, and a blotting paper
Procedure: The beaker of water is filled as shown above. A needle is now placed on a blotting paper and then pushed gently to the middle of the water surface. The paper wets and sinks, but the needle remains floating on the water's surface.
Conclusion: The surface acts like a skin covering the liquid, which can support the weight of the needle.
c. (i). When mercury spills on a glass sheet, it forms a spherical droplet (ii). A flowing needle on water (iii). A toy duck on liquid.
d. The cohesion force between molecules of water is less than its adhesion to glass, therefore when water is allowed to spill on glass, it wets. Conversely, the cohesion between the molecules of mercury is greater than its adhesion to glass. Hence, it does not wet glass.
e. (i). By adding detergents, (ii). By heating the liquid (ii). By addition of camphor
Maelezo ya Majibu
a. Surface tension: Surface tension is a property of a liquid whereby its surface behaves as if it were covered by an elastic skin. Or it is the tangential force of a liquid.
b. The surface “skin” of a liquid exists because of the attractive forces between its molecules. A molecule inside the liquid is pulled equally in all directions by surrounding molecules, so the net force on it is zero. However, a molecule at the surface has more neighboring molecules below than above, creating a net inward (downward) pull. This inward force tends to minimize the surface area, causing the surface to behave like a stretched elastic skin. This effect is known as surface tension.
See diagram above.
Aim: To demonstrate surface tension in a liquid
Apparatus: A needle, water, a beaker, and a blotting paper
Procedure: The beaker of water is filled as shown above. A needle is now placed on a blotting paper and then pushed gently to the middle of the water surface. The paper wets and sinks, but the needle remains floating on the water's surface.
Conclusion: The surface acts like a skin covering the liquid, which can support the weight of the needle.
c. (i). When mercury spills on a glass sheet, it forms a spherical droplet (ii). A flowing needle on water (iii). A toy duck on liquid.
d. The cohesion force between molecules of water is less than its adhesion to glass, therefore when water is allowed to spill on glass, it wets. Conversely, the cohesion between the molecules of mercury is greater than its adhesion to glass. Hence, it does not wet glass.
e. (i). By adding detergents, (ii). By heating the liquid (ii). By addition of camphor
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