WAEC SSCE - Physics - 2020 (Essay)
Ajụjụ 1 Ripọtì
An elastic material of length 3m is to be stretched to reduce an extension three times its original length. Calculate the force required to produce the extension
Principle (Hooke's law). Within the elastic limit the applied force is directly proportional to the extension produced:
\[ F = K e \]where \(K\) is the force constant of the elastic material and \(e\) is the extension.
Given data
- Original length, \(L = 3\text{ m}\)
- Force constant, \(K = 988.3\text{ N/m}\)
- Extension is three times the original length: \(e = 3 \times L = 3 \times 3 = 9\text{ m}\)
Calculation
\[ F = K e = 988.3 \times 9 \]\[ F = 8894.7\text{ N} \]The force required to produce the extension is \(8894.7\text{ N}\).
Akọwa Nkọwa
Principle (Hooke's law). Within the elastic limit the applied force is directly proportional to the extension produced:
\[ F = K e \]where \(K\) is the force constant of the elastic material and \(e\) is the extension.
Given data
- Original length, \(L = 3\text{ m}\)
- Force constant, \(K = 988.3\text{ N/m}\)
- Extension is three times the original length: \(e = 3 \times L = 3 \times 3 = 9\text{ m}\)
Calculation
\[ F = K e = 988.3 \times 9 \]\[ F = 8894.7\text{ N} \]The force required to produce the extension is \(8894.7\text{ N}\).
Ajụjụ 2 Ripọtì
(a)Explain resonance frequency as applied in RLC series Circuit.
(ii) Sketch a diagram to illustrate the variation of frequency, f, with the resistance, R, the capacitive reactance, X\(_c\) and the inductive reactance X\(_L\), in RLC series circuit.
(iii) Using the diagram drawn in (a)(ii) state whether the current in the circuit leads, lags or is in phase with the supply voltage when: (\(\alpha\)) f = f\(_o\); (\(\beta\)) f < f\(_o\) ; (\(\gamma\))f\(_o\); when f\(_o\) is the resonant frequency.
b)(i) Define mutual inductance.
(ii) The coil of an electric generator has 500 turns and 8.0cm diameter. If it rotates in a magnetic field of density 0.25T, calculate the angular speed when its peak voltage is 480V. [\(\pi\) = 3.142].
(a)(i) Resonance in a series RLC circuit
Resonance occurs at the resonant frequency \(f_o\), when the inductive reactance equals the capacitive reactance:
\[X_L=X_C\]
At this frequency, the inductive and capacitive effects cancel. Therefore, the circuit impedance is at its minimum value and is equal to the resistance \(R\). The current is consequently maximum.
\[f_o=\frac{1}{2\pi\sqrt{LC}}\]
(a)(ii) Variation of \(R\), \(X_L\), and \(X_C\) with frequency
The resistance \(R\) is constant as frequency changes. The inductive reactance increases with frequency:
\[X_L=2\pi fL\]
The capacitive reactance decreases as frequency increases:
\[X_C=\frac{1}{2\pi fC}\]
The point where the \(X_L\) and \(X_C\) curves meet is the resonant frequency \(f_o\).
(a)(iii) Phase relationship between current and supply voltage
- When \(f=f_o\), \(X_L=X_C\). The circuit is purely resistive, so the current is in phase with the supply voltage.
- When \(f<f_o\), \(X_C>X_L\). The circuit is net capacitive, so the current leads the supply voltage.
- When \(f>f_o\), \(X_L>X_C\). The circuit is net inductive, so the current lags the supply voltage.
The supplied reference answer reverses the lead/lag relationships away from resonance. In a capacitive circuit current leads voltage; in an inductive circuit current lags voltage.
(b)(i) Mutual inductance
Mutual inductance is the production of an induced e.m.f. in one coil when the current, and hence magnetic flux, in a nearby linked coil changes. Quantitatively, it is the ratio of induced e.m.f. in one coil to the rate of change of current in the other coil.
(b)(ii) Angular speed of the generator coil
For a rotating coil generator, the peak e.m.f. is:
\[E_0=NBA\omega\]
The coil diameter is \(8.0\,\text{cm}=0.080\,\text{m}\), so its radius is:
\[r=\frac{0.080}{2}=0.040\,\text{m}\]
Its area is:
\[A=\pi r^2=3.142(0.040)^2=5.027\times10^{-3}\,\text{m}^2\]
Substitute \(E_0=480\,\text{V}\), \(N=500\), and \(B=0.25\,\text{T}\):
\[\omega=\frac{E_0}{NBA}\]
\[\omega=\frac{480}{500\times0.25\times5.027\times10^{-3}}\]
\[\omega=\frac{480}{0.6284}=7.64\times10^2\,\text{rad s}^{-1}\]
Therefore, the angular speed is \(764\,\text{rad s}^{-1}\) (approximately).
Akọwa Nkọwa
(a)(i) Resonance in a series RLC circuit
Resonance occurs at the resonant frequency \(f_o\), when the inductive reactance equals the capacitive reactance:
\[X_L=X_C\]
At this frequency, the inductive and capacitive effects cancel. Therefore, the circuit impedance is at its minimum value and is equal to the resistance \(R\). The current is consequently maximum.
\[f_o=\frac{1}{2\pi\sqrt{LC}}\]
(a)(ii) Variation of \(R\), \(X_L\), and \(X_C\) with frequency
The resistance \(R\) is constant as frequency changes. The inductive reactance increases with frequency:
\[X_L=2\pi fL\]
The capacitive reactance decreases as frequency increases:
\[X_C=\frac{1}{2\pi fC}\]
The point where the \(X_L\) and \(X_C\) curves meet is the resonant frequency \(f_o\).
(a)(iii) Phase relationship between current and supply voltage
- When \(f=f_o\), \(X_L=X_C\). The circuit is purely resistive, so the current is in phase with the supply voltage.
- When \(f<f_o\), \(X_C>X_L\). The circuit is net capacitive, so the current leads the supply voltage.
- When \(f>f_o\), \(X_L>X_C\). The circuit is net inductive, so the current lags the supply voltage.
The supplied reference answer reverses the lead/lag relationships away from resonance. In a capacitive circuit current leads voltage; in an inductive circuit current lags voltage.
(b)(i) Mutual inductance
Mutual inductance is the production of an induced e.m.f. in one coil when the current, and hence magnetic flux, in a nearby linked coil changes. Quantitatively, it is the ratio of induced e.m.f. in one coil to the rate of change of current in the other coil.
(b)(ii) Angular speed of the generator coil
For a rotating coil generator, the peak e.m.f. is:
\[E_0=NBA\omega\]
The coil diameter is \(8.0\,\text{cm}=0.080\,\text{m}\), so its radius is:
\[r=\frac{0.080}{2}=0.040\,\text{m}\]
Its area is:
\[A=\pi r^2=3.142(0.040)^2=5.027\times10^{-3}\,\text{m}^2\]
Substitute \(E_0=480\,\text{V}\), \(N=500\), and \(B=0.25\,\text{T}\):
\[\omega=\frac{E_0}{NBA}\]
\[\omega=\frac{480}{500\times0.25\times5.027\times10^{-3}}\]
\[\omega=\frac{480}{0.6284}=7.64\times10^2\,\text{rad s}^{-1}\]
Therefore, the angular speed is \(764\,\text{rad s}^{-1}\) (approximately).
Ajụjụ 3 Ripọtì
(a)i) Define a torque.
(ii)State three factors that determine a torque
(b)(i) Define free fall.
(ii) A body is thrown vertically upwards from the top of a tower 40.0m high with a velocity of 10.0ms\(^{-1}\). Calculate the time taken for the body to reach the ground. [g.= 10.0ms\(^{-2}\)]
c) A cube of wood of side 8.0cm, floats at the interface between oil and water with 2.0cm of its surface below the interface as shown in the diagram below. Given that the relative densities of oil and water are 0.72 and 1.00 respectively, calculate the mass of the wood
(a)(i) The torque (moment) of a force about a point is the turning effect of the force about that point; it is the product of the force and the perpendicular distance of its line of action from the point (the axis of rotation): \(\tau=F\times d\).
(a)(ii) Factors that determine a torque:
- The magnitude of the applied force.
- The perpendicular distance between the line of action of the force and the pivot (axis).
- The angle at which the force is applied relative to the arm (line of action direction).
(b)(i) Free fall is the motion of a body falling only under the action of gravity, with no other force (such as air resistance) acting on it; its acceleration is g.
(b)(ii) Take upward positive, u = 10.0 m/s, tower height 40.0 m so displacement to the ground = -40.0 m, g = 10.0 m/s\(^2\).
\[s=ut-\tfrac{1}{2}gt^{2}\Rightarrow-40=10t-5t^{2}.\]
\[5t^{2}-10t-40=0\Rightarrow t^{2}-2t-8=0\Rightarrow(t-4)(t+2)=0.\]
Taking the positive root, \(t=4\,\text{s}\).
(c) Cube side 8.0 cm, so face area \(A=8\times8=64\,\text{cm}^2\). It floats with 2.0 cm below the oil-water interface (in water) and \(8-2=6\,\text{cm}\) in the oil. By the principle of flotation, the weight of the wood equals the total upthrust from oil and water.
Mass \(=\rho_{water}V_{water}+\rho_{oil}V_{oil}\) (using densities in g/cm\(^3\): water 1.00, oil 0.72).
\(V_{water}=64\times2=128\,\text{cm}^3\); \(V_{oil}=64\times6=384\,\text{cm}^3\).
\[m=(1.00\times128)+(0.72\times384)=128+276.48=404.48\,\text{g}\approx404.5\,\text{g}.\]
Akọwa Nkọwa
(a)(i) The torque (moment) of a force about a point is the turning effect of the force about that point; it is the product of the force and the perpendicular distance of its line of action from the point (the axis of rotation): \(\tau=F\times d\).
(a)(ii) Factors that determine a torque:
- The magnitude of the applied force.
- The perpendicular distance between the line of action of the force and the pivot (axis).
- The angle at which the force is applied relative to the arm (line of action direction).
(b)(i) Free fall is the motion of a body falling only under the action of gravity, with no other force (such as air resistance) acting on it; its acceleration is g.
(b)(ii) Take upward positive, u = 10.0 m/s, tower height 40.0 m so displacement to the ground = -40.0 m, g = 10.0 m/s\(^2\).
\[s=ut-\tfrac{1}{2}gt^{2}\Rightarrow-40=10t-5t^{2}.\]
\[5t^{2}-10t-40=0\Rightarrow t^{2}-2t-8=0\Rightarrow(t-4)(t+2)=0.\]
Taking the positive root, \(t=4\,\text{s}\).
(c) Cube side 8.0 cm, so face area \(A=8\times8=64\,\text{cm}^2\). It floats with 2.0 cm below the oil-water interface (in water) and \(8-2=6\,\text{cm}\) in the oil. By the principle of flotation, the weight of the wood equals the total upthrust from oil and water.
Mass \(=\rho_{water}V_{water}+\rho_{oil}V_{oil}\) (using densities in g/cm\(^3\): water 1.00, oil 0.72).
\(V_{water}=64\times2=128\,\text{cm}^3\); \(V_{oil}=64\times6=384\,\text{cm}^3\).
\[m=(1.00\times128)+(0.72\times384)=128+276.48=404.48\,\text{g}\approx404.5\,\text{g}.\]
Ajụjụ 4 Ripọtì
(a) In the design of an optical fibre, what type of material is most suitable for the design of the core?
(b) State one condition necessary to confine signals to the core of an optical fibre.
(a) The core of an optical fibre is best made of a transparent material of high refractive index, such as very pure (high-quality) glass or silica. It must be optically denser than the surrounding cladding so that light can be totally internally reflected within it.
(b) For signals to be confined to the core, the light must strike the core-cladding boundary at an angle of incidence greater than the critical angle, so that total internal reflection occurs at every point along the boundary. (This requires the core to be optically denser than the cladding.)
Akọwa Nkọwa
(a) The core of an optical fibre is best made of a transparent material of high refractive index, such as very pure (high-quality) glass or silica. It must be optically denser than the surrounding cladding so that light can be totally internally reflected within it.
(b) For signals to be confined to the core, the light must strike the core-cladding boundary at an angle of incidence greater than the critical angle, so that total internal reflection occurs at every point along the boundary. (This requires the core to be optically denser than the cladding.)
Ajụjụ 5 Ripọtì
The velocity v, of a wave in a stretched string, depends on the tension T, in the spring and the mass per unit length of the spring. Obtain an expression for v in terms of T and u, using the method of dimensions.
Dimensional derivation of the speed of a wave on a stretched string.
Assume the speed v depends on the tension T and the mass per unit length \(\mu\):
\[v=k\,T^{a}\mu^{b}\]
where k is a dimensionless constant. Writing the dimensions:
- \([v]=LT^{-1}\)
- \([T]=MLT^{-2}\) (force)
- \([\mu]=ML^{-1}\) (mass per unit length)
So \(LT^{-1}=(MLT^{-2})^{a}(ML^{-1})^{b}=M^{a+b}L^{a-b}T^{-2a}\).
Equating powers:
- M: \(a+b=0\)
- L: \(a-b=1\)
- T: \(-2a=-1\Rightarrow a=\tfrac{1}{2}\)
From these, \(b=-\tfrac{1}{2}\). Therefore
\[v=k\,T^{1/2}\mu^{-1/2}=k\sqrt{\dfrac{T}{\mu}}.\]
Thus the speed of the wave is proportional to the square root of the tension divided by the mass per unit length.
Akọwa Nkọwa
Dimensional derivation of the speed of a wave on a stretched string.
Assume the speed v depends on the tension T and the mass per unit length \(\mu\):
\[v=k\,T^{a}\mu^{b}\]
where k is a dimensionless constant. Writing the dimensions:
- \([v]=LT^{-1}\)
- \([T]=MLT^{-2}\) (force)
- \([\mu]=ML^{-1}\) (mass per unit length)
So \(LT^{-1}=(MLT^{-2})^{a}(ML^{-1})^{b}=M^{a+b}L^{a-b}T^{-2a}\).
Equating powers:
- M: \(a+b=0\)
- L: \(a-b=1\)
- T: \(-2a=-1\Rightarrow a=\tfrac{1}{2}\)
From these, \(b=-\tfrac{1}{2}\). Therefore
\[v=k\,T^{1/2}\mu^{-1/2}=k\sqrt{\dfrac{T}{\mu}}.\]
Thus the speed of the wave is proportional to the square root of the tension divided by the mass per unit length.
Ajụjụ 6 Ripọtì
(a) Define isotopes.
(b) Mention two uses of radioactive tracers in each of the following areas:
(\(\alpha\)) medicine
(\(\beta\)) industry
(\(\gamma\)) agriculture.
(c) State three features of electromagnetic waves.
(d)) Mention four components of the nuclear reactor.
(i) State the functions of each of the components stated in (d)(i).
(a) Isotopes are atoms of the same element that have the same atomic number (same number of protons) but different mass numbers (different numbers of neutrons).
(b) Uses of radioactive tracers:
- (\(\alpha\)) Medicine: to locate tumours or study the functioning of organs (e.g. iodine-131 for the thyroid); to trace blood flow and detect blood clots or circulatory disorders.
- (\(\beta\)) Industry: to detect leaks in buried pipes carrying oil, water or gas; to measure the thickness of sheets (metal, paper, plastic) and monitor wear in engine parts.
- (\(\gamma\)) Agriculture: to study the uptake and movement of fertilisers by plants (e.g. phosphorus-32); to study how plants absorb water and nutrients and to induce useful mutations.
(c) Features of electromagnetic waves:
- They are transverse waves and travel through a vacuum at the speed of light (\(3\times10^{8}\,\text{m s}^{-1}\)).
- They consist of oscillating electric and magnetic fields at right angles to each other and to the direction of travel.
- They carry energy, are not deflected by electric or magnetic fields (uncharged), and can be reflected, refracted, diffracted and polarised.
(d) Components of a nuclear reactor and their functions:
- Fuel rods (e.g. uranium-235): provide the fissile material that undergoes fission to release energy.
- Moderator (e.g. graphite or heavy water): slows down the fast neutrons so that they can cause further fission.
- Control rods (e.g. boron or cadmium): absorb excess neutrons to control the rate of the chain reaction (and shut it down).
- Coolant (e.g. water, carbon dioxide gas): removes the heat produced in the core and carries it to the heat exchanger to raise steam. (A thick concrete/lead shield protects workers from radiation.)
Akọwa Nkọwa
(a) Isotopes are atoms of the same element that have the same atomic number (same number of protons) but different mass numbers (different numbers of neutrons).
(b) Uses of radioactive tracers:
- (\(\alpha\)) Medicine: to locate tumours or study the functioning of organs (e.g. iodine-131 for the thyroid); to trace blood flow and detect blood clots or circulatory disorders.
- (\(\beta\)) Industry: to detect leaks in buried pipes carrying oil, water or gas; to measure the thickness of sheets (metal, paper, plastic) and monitor wear in engine parts.
- (\(\gamma\)) Agriculture: to study the uptake and movement of fertilisers by plants (e.g. phosphorus-32); to study how plants absorb water and nutrients and to induce useful mutations.
(c) Features of electromagnetic waves:
- They are transverse waves and travel through a vacuum at the speed of light (\(3\times10^{8}\,\text{m s}^{-1}\)).
- They consist of oscillating electric and magnetic fields at right angles to each other and to the direction of travel.
- They carry energy, are not deflected by electric or magnetic fields (uncharged), and can be reflected, refracted, diffracted and polarised.
(d) Components of a nuclear reactor and their functions:
- Fuel rods (e.g. uranium-235): provide the fissile material that undergoes fission to release energy.
- Moderator (e.g. graphite or heavy water): slows down the fast neutrons so that they can cause further fission.
- Control rods (e.g. boron or cadmium): absorb excess neutrons to control the rate of the chain reaction (and shut it down).
- Coolant (e.g. water, carbon dioxide gas): removes the heat produced in the core and carries it to the heat exchanger to raise steam. (A thick concrete/lead shield protects workers from radiation.)
Ajụjụ 7 Ripọtì
State three properties of lasers that make them preferable to ordinary light beam.
Properties of lasers that make them preferable to an ordinary light beam:
- Monochromatic: laser light is of a single (or very narrow band of) wavelength/colour, unlike ordinary light which is a mixture of many wavelengths.
- Coherent: the waves are all in phase with one another, both in space and time.
- Highly directional and narrow (collimated): the beam is very parallel and spreads very little over long distances, giving high intensity concentrated in a small area (it is also very intense/bright).
Akọwa Nkọwa
Properties of lasers that make them preferable to an ordinary light beam:
- Monochromatic: laser light is of a single (or very narrow band of) wavelength/colour, unlike ordinary light which is a mixture of many wavelengths.
- Coherent: the waves are all in phase with one another, both in space and time.
- Highly directional and narrow (collimated): the beam is very parallel and spreads very little over long distances, giving high intensity concentrated in a small area (it is also very intense/bright).
Ajụjụ 8 Ripọtì
A bullet is fired from a gun at 30\(^o\) to the horizontal. The bullet remains in flight for 25s before touching the ground. Calculating the velocity of projection. [g = 10ms\(^{-2}\)]
Projectile fired at an angle. For a projectile launched with speed u at angle \(\theta\) to the horizontal, the time of flight is
\[T=\dfrac{2u\sin\theta}{g}.\]
Given \(\theta=30^{\circ}\), \(T=25\,\text{s}\) and \(g=10\,\text{m s}^{-2}\):
\[25=\dfrac{2u\sin30^{\circ}}{10}=\dfrac{2u(0.5)}{10}=\dfrac{u}{10}.\]
Therefore \(u=25\times10=250\,\text{m s}^{-1}\).
The velocity of projection is \(250\,\text{m s}^{-1}\) at \(30^{\circ}\) to the horizontal.
Akọwa Nkọwa
Projectile fired at an angle. For a projectile launched with speed u at angle \(\theta\) to the horizontal, the time of flight is
\[T=\dfrac{2u\sin\theta}{g}.\]
Given \(\theta=30^{\circ}\), \(T=25\,\text{s}\) and \(g=10\,\text{m s}^{-2}\):
\[25=\dfrac{2u\sin30^{\circ}}{10}=\dfrac{2u(0.5)}{10}=\dfrac{u}{10}.\]
Therefore \(u=25\times10=250\,\text{m s}^{-1}\).
The velocity of projection is \(250\,\text{m s}^{-1}\) at \(30^{\circ}\) to the horizontal.
Ajụjụ 9 Ripọtì
(a) (i) Define Optical angle.
(ii) Explain two conditions necessary for total internal reflection to occur.
(iii) List three practical applications of total internal reflection.
(b) State two effects of refraction.
(c)(i) Define progressive waves.
(ii) A plane progressive wave is represented by the equation y = 0.5 sin(1000\(\pi\)r = \(\frac{100 \pi \lambda}{17}\)) where y is in millimetres, t in seconds and x in metres. Calculate the: (\(\alpha\)) frequency of the wave; (\(\beta\))of the wave; (\(\gamma\)) speed of the wave
(a)(i) The (critical) optical angle is the angle of incidence in the denser medium for which the angle of refraction in the less dense medium is exactly 90 degrees; at this angle the refracted ray grazes the surface, and beyond it total internal reflection occurs.
(a)(ii) Conditions for total internal reflection:
- Light must travel from an optically denser medium to an optically less dense medium (from high to low refractive index).
- The angle of incidence in the denser medium must be greater than the critical angle for the two media.
(a)(iii) Applications: optical fibres (communication/endoscopes); totally reflecting prisms in periscopes and binoculars; the shining/sparkle of diamonds; the formation of mirages.
(b) Effects of refraction: a stick partly immersed in water appears bent; a pool of water or swimming pool appears shallower than it really is (real depth greater than apparent depth). (Also the twinkling of stars and apparent raising of the sun at sunset.)
(c)(i) A progressive (travelling) wave is one that moves outward from a source, transferring energy from one point to another through a medium, with the disturbance advancing continuously.
(c)(ii) Comparing with \(y=A\sin(\omega t-kx)\), the angular frequency is \(\omega=1000\pi\,\text{rad s}^{-1}\) and the wave number is \(k=\dfrac{100\pi}{17}\,\text{m}^{-1}\) (from the given equation).
(\(\alpha\)) Frequency: \(f=\dfrac{\omega}{2\pi}=\dfrac{1000\pi}{2\pi}=500\,\text{Hz}\).
(\(\beta\)) Wavelength: \(\lambda=\dfrac{2\pi}{k}=\dfrac{2\pi\times17}{100\pi}=0.34\,\text{m}\).
(\(\gamma\)) Speed: \(v=f\lambda=500\times0.34=170\,\text{m s}^{-1}\).
Akọwa Nkọwa
(a)(i) The (critical) optical angle is the angle of incidence in the denser medium for which the angle of refraction in the less dense medium is exactly 90 degrees; at this angle the refracted ray grazes the surface, and beyond it total internal reflection occurs.
(a)(ii) Conditions for total internal reflection:
- Light must travel from an optically denser medium to an optically less dense medium (from high to low refractive index).
- The angle of incidence in the denser medium must be greater than the critical angle for the two media.
(a)(iii) Applications: optical fibres (communication/endoscopes); totally reflecting prisms in periscopes and binoculars; the shining/sparkle of diamonds; the formation of mirages.
(b) Effects of refraction: a stick partly immersed in water appears bent; a pool of water or swimming pool appears shallower than it really is (real depth greater than apparent depth). (Also the twinkling of stars and apparent raising of the sun at sunset.)
(c)(i) A progressive (travelling) wave is one that moves outward from a source, transferring energy from one point to another through a medium, with the disturbance advancing continuously.
(c)(ii) Comparing with \(y=A\sin(\omega t-kx)\), the angular frequency is \(\omega=1000\pi\,\text{rad s}^{-1}\) and the wave number is \(k=\dfrac{100\pi}{17}\,\text{m}^{-1}\) (from the given equation).
(\(\alpha\)) Frequency: \(f=\dfrac{\omega}{2\pi}=\dfrac{1000\pi}{2\pi}=500\,\text{Hz}\).
(\(\beta\)) Wavelength: \(\lambda=\dfrac{2\pi}{k}=\dfrac{2\pi\times17}{100\pi}=0.34\,\text{m}\).
(\(\gamma\)) Speed: \(v=f\lambda=500\times0.34=170\,\text{m s}^{-1}\).
Ajụjụ 10 Ripọtì
In a solar panel for heat supply, state the function of each of the following parts: (a) metal flat plate; (b) thermal insulator: (c) tubes.
Functions of the parts of a solar (flat-plate) heat collector:
- (a) Metal flat plate (absorber): it is painted dull black to absorb the incident solar radiation and convert it into heat, then conducts this heat to the water flowing in the tubes attached to it.
- (b) Thermal insulator: it is placed behind and around the plate to reduce heat loss (by conduction and convection) to the surroundings, so that the collected heat is retained and transferred to the water.
- (c) Tubes: they carry the water (or fluid) in good thermal contact with the hot absorber plate, so that the water is heated and then conveyed away to the storage tank for use.
Akọwa Nkọwa
Functions of the parts of a solar (flat-plate) heat collector:
- (a) Metal flat plate (absorber): it is painted dull black to absorb the incident solar radiation and convert it into heat, then conducts this heat to the water flowing in the tubes attached to it.
- (b) Thermal insulator: it is placed behind and around the plate to reduce heat loss (by conduction and convection) to the surroundings, so that the collected heat is retained and transferred to the water.
- (c) Tubes: they carry the water (or fluid) in good thermal contact with the hot absorber plate, so that the water is heated and then conveyed away to the storage tank for use.
Ajụjụ 11 Ripọtì
A satellite launched with velocity V\(_E\) just escapes the earth's gravitational attraction. Given that the radius of the earth is R, show that V\(_E\) = \(\sqrt{20R}\) [g = 10ms\(^{-2}\)
Escape velocity. A body just escapes the earth gravitational field when its total mechanical energy is zero, i.e. its initial kinetic energy equals the work needed to move it from the earth surface to infinity (the gravitational potential energy magnitude).
\[\tfrac{1}{2}mV_E^{2}=\dfrac{GMm}{R}.\]
At the earth surface the acceleration due to gravity is \(g=\dfrac{GM}{R^{2}}\), so \(GM=gR^{2}\). Substituting:
\[\tfrac{1}{2}mV_E^{2}=\dfrac{gR^{2}m}{R}=mgR\Rightarrow V_E^{2}=2gR.\]
Therefore \(V_E=\sqrt{2gR}\). With \(g=10\,\text{m s}^{-2}\):
\[V_E=\sqrt{2\times10\times R}=\sqrt{20R}.\]
Hence \(V_E=\sqrt{20R}\) as required.
Akọwa Nkọwa
Escape velocity. A body just escapes the earth gravitational field when its total mechanical energy is zero, i.e. its initial kinetic energy equals the work needed to move it from the earth surface to infinity (the gravitational potential energy magnitude).
\[\tfrac{1}{2}mV_E^{2}=\dfrac{GMm}{R}.\]
At the earth surface the acceleration due to gravity is \(g=\dfrac{GM}{R^{2}}\), so \(GM=gR^{2}\). Substituting:
\[\tfrac{1}{2}mV_E^{2}=\dfrac{gR^{2}m}{R}=mgR\Rightarrow V_E^{2}=2gR.\]
Therefore \(V_E=\sqrt{2gR}\). With \(g=10\,\text{m s}^{-2}\):
\[V_E=\sqrt{2\times10\times R}=\sqrt{20R}.\]
Hence \(V_E=\sqrt{20R}\) as required.
Ajụjụ 12 Ripọtì
(a) In an experiment to measure the specific latent heat of vapourisation of water, a student places a heater in a beaker containing water. The beaker stands on an electronic balance so that the mass of the beaker and water could be measured. The heater is switched on and readings were taken every 100s when the water starts boiling.
The table below shows the readings.
| Time/s | 0 | 100 | 200 | 300 | 400 |
| Reading on balance/g | 203.22 | 201.62 | 199.79 | 198.26 | 196.50 |
| Mass of water evaporated/g | 0 | ||||
| Energy supplied by heated/J | 0 |
(I) Fill in the mass of water evaporated.
(ii) Given that the heater supplies energy at the rate of 38J/s, fill in the values of the energy supplied by the heater in 100s, 200s, 300s, and 400s.
(ii) Plot a graph of energy supplied on the vertical axis and mass of water evaporated on the horizontal axis, starting both axes from the origin(0,0).
(iv) Determine the slope of the graph.
(v) what does the value of the slope mean?
(a)(i) Completed table
The mass evaporated is the decrease in balance reading:
\[m=203.22-\text{balance reading}\]
| Time / s | 0 | 100 | 200 | 300 | 400 |
|---|---|---|---|---|---|
| Balance reading / g | 203.22 | 201.62 | 199.79 | 198.26 | 196.50 |
| Mass of water evaporated / g | 0.00 | 1.60 | 3.43 | 4.96 | 6.72 |
| Energy supplied / J | 0 | 3800 | 7600 | 11400 | 15200 |
(ii) The energy supplied is \(E=Pt\), where \(P=38\text{ J s}^{-1}\). Thus, for example, at \(100\text{ s}\), \(E=38\times100=3800\text{ J}\).
(iii) Graph of energy supplied against mass evaporated
(iv) Slope of the graph
Using two well-separated points on the line of best fit, \((0.8\text{ g},2000\text{ J})\) and \((6.1\text{ g},14000\text{ J})\):
\[\text{Slope}=\frac{\Delta E}{\Delta m}=\frac{(14000-2000)\text{ J}}{(6.1-0.8)\text{ g}}=\frac{12000}{5.3}=2.26\times10^3\text{ J g}^{-1}\]
\[\therefore\ \text{Slope}=2.26\times10^6\text{ J kg}^{-1}.\]
(v) The slope represents the specific latent heat of vapourisation of water. It is the heat energy required to change unit mass of water at its boiling point into vapour at the same temperature.
Thus, approximately \(2.26\times10^3\text{ J}\) is required to evaporate \(1\text{ g}\) of water, or \(2.26\times10^6\text{ J}\) is required for \(1\text{ kg}\) of water.
Akọwa Nkọwa
(a)(i) Completed table
The mass evaporated is the decrease in balance reading:
\[m=203.22-\text{balance reading}\]
| Time / s | 0 | 100 | 200 | 300 | 400 |
|---|---|---|---|---|---|
| Balance reading / g | 203.22 | 201.62 | 199.79 | 198.26 | 196.50 |
| Mass of water evaporated / g | 0.00 | 1.60 | 3.43 | 4.96 | 6.72 |
| Energy supplied / J | 0 | 3800 | 7600 | 11400 | 15200 |
(ii) The energy supplied is \(E=Pt\), where \(P=38\text{ J s}^{-1}\). Thus, for example, at \(100\text{ s}\), \(E=38\times100=3800\text{ J}\).
(iii) Graph of energy supplied against mass evaporated
(iv) Slope of the graph
Using two well-separated points on the line of best fit, \((0.8\text{ g},2000\text{ J})\) and \((6.1\text{ g},14000\text{ J})\):
\[\text{Slope}=\frac{\Delta E}{\Delta m}=\frac{(14000-2000)\text{ J}}{(6.1-0.8)\text{ g}}=\frac{12000}{5.3}=2.26\times10^3\text{ J g}^{-1}\]
\[\therefore\ \text{Slope}=2.26\times10^6\text{ J kg}^{-1}.\]
(v) The slope represents the specific latent heat of vapourisation of water. It is the heat energy required to change unit mass of water at its boiling point into vapour at the same temperature.
Thus, approximately \(2.26\times10^3\text{ J}\) is required to evaporate \(1\text{ g}\) of water, or \(2.26\times10^6\text{ J}\) is required for \(1\text{ kg}\) of water.
