WAEC SSCE - Physics - 2021 (Essay)
Tambaya 1 Rahoto
(a)(i) Define each of the following terms as it relates to converging lenses (i) focal length; (ii) optical Centre.
(iii) Draw a ray diagram to illustrate how a converging lens is used to produce a virtual image of an object.
(b)(i) Name the primary colors of light. (ii) Match each primary color to its corresponding complementary color.
(c) A ray passes symmetrically through a glass prism of angle 60° and refractive index of 1.5. Calculate the angle of: (i) incidence; (ii) minimum deviation.
(a)(i) Focal length. The focal length of a converging lens is the distance from the optical centre of the lens to its principal focus (the point on the principal axis to which rays travelling parallel to the axis converge after refraction).
(a)(ii) Optical centre. The optical centre is the point at the middle of the lens through which a ray of light passes without being deviated (it travels straight on).
(a)(iii) Ray diagram: virtual image formed by a converging lens. When the object is placed between the lens and its principal focus (\(u < f\)) the lens acts as a magnifying glass: the emergent rays diverge and, produced backwards, meet on the same side as the object to form a virtual, erect and magnified image.
Two standard construction rays are used:
- Ray 1 leaves the top of the object parallel to the principal axis and, after refraction, passes through the far principal focus \(F'\).
- Ray 2 passes through the optical centre \(C\) of the lens and continues undeviated.
After the lens the two emergent rays diverge, so no real image is formed. Extending them backwards (dashed) they intersect on the same side as the object, locating the tip of the virtual image \(I\).
(b)(i) Primary colours of light. Red, Green and Blue.
(b)(ii) Complementary pairs. Each primary colour pairs with the colour obtained by mixing the other two primaries:
| Primary colour | Complementary colour |
| Red | Cyan |
| Green | Magenta |
| Blue | Yellow |
(c) Ray passing symmetrically through a 60° prism, \(n = 1.5\). Symmetric passage means the ray traverses the prism at minimum deviation \(D_m\), so the refraction is described by
\[ n = \frac{\sin\!\left(\dfrac{A+D_m}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)}, \qquad A = 60^{\circ}. \](c)(i) Angle of incidence. At symmetric (minimum-deviation) passage the two refracting angles inside the prism are equal, each \(=\tfrac{A}{2}=30^{\circ}\), so the angle of incidence at the first face is
\[ n = \frac{\sin i}{\sin 30^{\circ}} \;\Rightarrow\; \sin i = 1.5 \times \sin 30^{\circ} = 1.5 \times 0.5 = 0.75, \] \[ i = \sin^{-1}(0.75) = 48.6^{\circ}. \](c)(ii) Angle of minimum deviation. Using the prism formula with \(i=\tfrac{A+D_m}{2}\):
\[ \sin\!\left(\frac{A+D_m}{2}\right) = n\sin\frac{A}{2} = 0.75 \;\Rightarrow\; \frac{A+D_m}{2} = 48.6^{\circ}, \] \[ A + D_m = 97.2^{\circ} \;\Rightarrow\; D_m = 97.2^{\circ} - 60^{\circ} = 37.2^{\circ}. \]Angle of incidence \(i \approx 48.6^{\circ}\); angle of minimum deviation \(D_m \approx 37.2^{\circ}\).
Bayanin Amsa
(a)(i) Focal length. The focal length of a converging lens is the distance from the optical centre of the lens to its principal focus (the point on the principal axis to which rays travelling parallel to the axis converge after refraction).
(a)(ii) Optical centre. The optical centre is the point at the middle of the lens through which a ray of light passes without being deviated (it travels straight on).
(a)(iii) Ray diagram: virtual image formed by a converging lens. When the object is placed between the lens and its principal focus (\(u < f\)) the lens acts as a magnifying glass: the emergent rays diverge and, produced backwards, meet on the same side as the object to form a virtual, erect and magnified image.
Two standard construction rays are used:
- Ray 1 leaves the top of the object parallel to the principal axis and, after refraction, passes through the far principal focus \(F'\).
- Ray 2 passes through the optical centre \(C\) of the lens and continues undeviated.
After the lens the two emergent rays diverge, so no real image is formed. Extending them backwards (dashed) they intersect on the same side as the object, locating the tip of the virtual image \(I\).
(b)(i) Primary colours of light. Red, Green and Blue.
(b)(ii) Complementary pairs. Each primary colour pairs with the colour obtained by mixing the other two primaries:
| Primary colour | Complementary colour |
| Red | Cyan |
| Green | Magenta |
| Blue | Yellow |
(c) Ray passing symmetrically through a 60° prism, \(n = 1.5\). Symmetric passage means the ray traverses the prism at minimum deviation \(D_m\), so the refraction is described by
\[ n = \frac{\sin\!\left(\dfrac{A+D_m}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)}, \qquad A = 60^{\circ}. \](c)(i) Angle of incidence. At symmetric (minimum-deviation) passage the two refracting angles inside the prism are equal, each \(=\tfrac{A}{2}=30^{\circ}\), so the angle of incidence at the first face is
\[ n = \frac{\sin i}{\sin 30^{\circ}} \;\Rightarrow\; \sin i = 1.5 \times \sin 30^{\circ} = 1.5 \times 0.5 = 0.75, \] \[ i = \sin^{-1}(0.75) = 48.6^{\circ}. \](c)(ii) Angle of minimum deviation. Using the prism formula with \(i=\tfrac{A+D_m}{2}\):
\[ \sin\!\left(\frac{A+D_m}{2}\right) = n\sin\frac{A}{2} = 0.75 \;\Rightarrow\; \frac{A+D_m}{2} = 48.6^{\circ}, \] \[ A + D_m = 97.2^{\circ} \;\Rightarrow\; D_m = 97.2^{\circ} - 60^{\circ} = 37.2^{\circ}. \]Angle of incidence \(i \approx 48.6^{\circ}\); angle of minimum deviation \(D_m \approx 37.2^{\circ}\).
Tambaya 2 Rahoto
(a)(i) State the principal factor that determines the relative stability of a radioactive nucleus.
(ii) Arrange the following radioactive nucleus in decreasing order of stability. Justify your answer: X,W and Y:
\(\displaystyle {}^{40}_{20}X \quad {}^{920}_{36}Y \text{ and } {}^{95}_{42}Z\)
(b)(i) Explain the term ionization potential.
(ii)
The diagram above illustrates energy levels in the hydrogen atom. E, is the energy of the \(E_0\) ground state.
(i) When an electron makes a transition from level n = 3 to level n = 1, it emits a photon of wavelength \(1.02 \times 10^{-7}\,\text{m}\). Calculate \(E_0\).
(ii) Calculate the ionization potential of the hydrogen atom.
(c)(i) Explain the statement, the work function of sodium is 2.0 eV. (ii) Light of wavelength 160 mm is shone on the surface of a sodium metal of work function 2.0 eV. Determine whether photoelectrons will be emitted. [\(h = 6.6 \times 10^{-34}\,\text{Js}\), \(e = 3.0 \times 10^{8}\,\text{m/s}\), I eV = \(1.6 \times 10^{-19}\,\text{J}\)]
(a)(i) Principal factor for nuclear stability. The relative stability of a nucleus is determined principally by its neutron-to-proton (n/p) ratio (equivalently, by its binding energy per nucleon). Nuclei whose n/p ratio lies within the stability band are stable; those far from it are unstable.
(a)(ii) Order of stability. Compare the neutron-to-proton ratios of the given nuclides. A nucleus whose n/p ratio is closest to 1 (for light nuclei) or lies nearest the stability line is the most stable, and stability decreases as the n/p ratio departs further from it. Arrange the nuclides so that the one with the n/p ratio nearest the stability band comes first and the one furthest from it comes last (for example the light nuclide with n/p close to 1 is the most stable, and the heavy neutron-rich nuclide is the least stable). Justify each placement by quoting its computed n/p ratio.
(b)(i) Ionization potential. The ionization potential is the minimum energy (or the potential difference through which an electron must be accelerated to acquire that energy) required to completely remove the most loosely bound electron from an isolated atom in its ground state.
(b)(ii) Ground-state energy \(E_0\). The photon emitted in the transition \(n=3 \to n=1\) carries energy \(hc/\lambda\). (Here \(c = 3.0\times10^{8}\ \text{m/s}\).)
\[ E_{3}-E_{1} = \frac{hc}{\lambda} = \frac{6.6\times10^{-34}\times3.0\times10^{8}}{1.02\times10^{-7}} = 1.94\times10^{-18}\ \text{J} \]For hydrogen \(E_n = \dfrac{E_0}{n^{2}}\) with \(E_0\) the ground-state energy, so the emitted energy is \(|E_0|\left(1-\tfrac{1}{9}\right)=\tfrac{8}{9}|E_0|\):
\[ \tfrac{8}{9}|E_0| = 1.94\times10^{-18} \;\Rightarrow\; |E_0| = 2.18\times10^{-18}\ \text{J} \]So \(E_0 \approx -2.18\times10^{-18}\ \text{J}\) (about \(-13.6\ \text{eV}\)).
Ionization potential of hydrogen. The energy to remove the electron from the ground state is \(|E_0| = 2.18\times10^{-18}\ \text{J}\); in volts,
\[ V = \frac{|E_0|}{e} = \frac{2.18\times10^{-18}}{1.6\times10^{-19}} \approx 13.6\ \text{V} \](c)(i) Work function of sodium is 2.0 eV. This means the minimum energy needed to just free an electron from the surface of sodium metal is 2.0 electron-volts \((= 2.0\times1.6\times10^{-19} = 3.2\times10^{-19}\ \text{J})\).
(c)(ii) Photoelectron emission (\(\lambda = 160\ \text{nm} = 1.6\times10^{-7}\ \text{m}\)). Energy of the incident photon:
\[ E = \frac{hc}{\lambda} = \frac{6.6\times10^{-34}\times3.0\times10^{8}}{1.6\times10^{-7}} = 1.24\times10^{-18}\ \text{J} = 7.7\ \text{eV} \]Since the photon energy (7.7 eV) is greater than the work function (2.0 eV), photoelectrons will be emitted, each with maximum kinetic energy \(7.7 - 2.0 = 5.7\ \text{eV}\).
Bayanin Amsa
(a)(i) Principal factor for nuclear stability. The relative stability of a nucleus is determined principally by its neutron-to-proton (n/p) ratio (equivalently, by its binding energy per nucleon). Nuclei whose n/p ratio lies within the stability band are stable; those far from it are unstable.
(a)(ii) Order of stability. Compare the neutron-to-proton ratios of the given nuclides. A nucleus whose n/p ratio is closest to 1 (for light nuclei) or lies nearest the stability line is the most stable, and stability decreases as the n/p ratio departs further from it. Arrange the nuclides so that the one with the n/p ratio nearest the stability band comes first and the one furthest from it comes last (for example the light nuclide with n/p close to 1 is the most stable, and the heavy neutron-rich nuclide is the least stable). Justify each placement by quoting its computed n/p ratio.
(b)(i) Ionization potential. The ionization potential is the minimum energy (or the potential difference through which an electron must be accelerated to acquire that energy) required to completely remove the most loosely bound electron from an isolated atom in its ground state.
(b)(ii) Ground-state energy \(E_0\). The photon emitted in the transition \(n=3 \to n=1\) carries energy \(hc/\lambda\). (Here \(c = 3.0\times10^{8}\ \text{m/s}\).)
\[ E_{3}-E_{1} = \frac{hc}{\lambda} = \frac{6.6\times10^{-34}\times3.0\times10^{8}}{1.02\times10^{-7}} = 1.94\times10^{-18}\ \text{J} \]For hydrogen \(E_n = \dfrac{E_0}{n^{2}}\) with \(E_0\) the ground-state energy, so the emitted energy is \(|E_0|\left(1-\tfrac{1}{9}\right)=\tfrac{8}{9}|E_0|\):
\[ \tfrac{8}{9}|E_0| = 1.94\times10^{-18} \;\Rightarrow\; |E_0| = 2.18\times10^{-18}\ \text{J} \]So \(E_0 \approx -2.18\times10^{-18}\ \text{J}\) (about \(-13.6\ \text{eV}\)).
Ionization potential of hydrogen. The energy to remove the electron from the ground state is \(|E_0| = 2.18\times10^{-18}\ \text{J}\); in volts,
\[ V = \frac{|E_0|}{e} = \frac{2.18\times10^{-18}}{1.6\times10^{-19}} \approx 13.6\ \text{V} \](c)(i) Work function of sodium is 2.0 eV. This means the minimum energy needed to just free an electron from the surface of sodium metal is 2.0 electron-volts \((= 2.0\times1.6\times10^{-19} = 3.2\times10^{-19}\ \text{J})\).
(c)(ii) Photoelectron emission (\(\lambda = 160\ \text{nm} = 1.6\times10^{-7}\ \text{m}\)). Energy of the incident photon:
\[ E = \frac{hc}{\lambda} = \frac{6.6\times10^{-34}\times3.0\times10^{8}}{1.6\times10^{-7}} = 1.24\times10^{-18}\ \text{J} = 7.7\ \text{eV} \]Since the photon energy (7.7 eV) is greater than the work function (2.0 eV), photoelectrons will be emitted, each with maximum kinetic energy \(7.7 - 2.0 = 5.7\ \text{eV}\).
Tambaya 3 Rahoto
The circuit diagram below is a simple current rectifier circuit. Use it to answer the questions that follow:
(a) State the function of each of the parts labelled A and B.
b) Sketch the output signal produces.
A is the a.c. supply (source). It produces an alternating potential difference, so its polarity reverses continuously.
B is the smoothing capacitor. It charges when the rectified voltage rises and discharges through the load when the rectified voltage falls. This reduces the variation in the output, producing a smoother d.c. voltage with a small ripple.
The output is therefore direct current with ripple: it stays positive and does not reverse direction, but is not perfectly constant.
Examination point: do not draw negative half-cycles after the rectifier. The capacitor makes the positive output flatter, but some ripple remains because it discharges between voltage peaks.
Bayanin Amsa
A is the a.c. supply (source). It produces an alternating potential difference, so its polarity reverses continuously.
B is the smoothing capacitor. It charges when the rectified voltage rises and discharges through the load when the rectified voltage falls. This reduces the variation in the output, producing a smoother d.c. voltage with a small ripple.
The output is therefore direct current with ripple: it stays positive and does not reverse direction, but is not perfectly constant.
Examination point: do not draw negative half-cycles after the rectifier. The capacitor makes the positive output flatter, but some ripple remains because it discharges between voltage peaks.
Tambaya 4 Rahoto
A projectile is fired at an angle of 30° to the horizontal with a velocity of 40 m/s Calculate the velocity attained after 1 s. [g = 10 m/s\(^2\)]
Resolve the initial velocity. With \(u = 40\ \text{m/s}\) at \(30^{\circ}\) to the horizontal:
\[ u_x = 40\cos30^{\circ} = 40\times0.866 = 34.6\ \text{m/s} \] \[ u_y = 40\sin30^{\circ} = 40\times0.5 = 20\ \text{m/s} \]After t = 1 s. The horizontal component is unchanged (no horizontal force):
\[ v_x = 34.6\ \text{m/s} \]The vertical component is reduced by gravity:
\[ v_y = u_y - g t = 20 - 10\times1 = 10\ \text{m/s} \]Resultant velocity.
\[ v = \sqrt{v_x^{2}+v_y^{2}} = \sqrt{34.6^{2}+10^{2}} = \sqrt{1197+100} = \sqrt{1297} \approx 36.0\ \text{m/s} \]Direction above the horizontal.
\[ \tan\theta = \frac{v_y}{v_x} = \frac{10}{34.6} = 0.289 \;\Rightarrow\; \theta \approx 16.1^{\circ} \]The velocity after 1 s is about 36.0 m/s directed at about 16° above the horizontal.
Bayanin Amsa
Resolve the initial velocity. With \(u = 40\ \text{m/s}\) at \(30^{\circ}\) to the horizontal:
\[ u_x = 40\cos30^{\circ} = 40\times0.866 = 34.6\ \text{m/s} \] \[ u_y = 40\sin30^{\circ} = 40\times0.5 = 20\ \text{m/s} \]After t = 1 s. The horizontal component is unchanged (no horizontal force):
\[ v_x = 34.6\ \text{m/s} \]The vertical component is reduced by gravity:
\[ v_y = u_y - g t = 20 - 10\times1 = 10\ \text{m/s} \]Resultant velocity.
\[ v = \sqrt{v_x^{2}+v_y^{2}} = \sqrt{34.6^{2}+10^{2}} = \sqrt{1197+100} = \sqrt{1297} \approx 36.0\ \text{m/s} \]Direction above the horizontal.
\[ \tan\theta = \frac{v_y}{v_x} = \frac{10}{34.6} = 0.289 \;\Rightarrow\; \theta \approx 16.1^{\circ} \]The velocity after 1 s is about 36.0 m/s directed at about 16° above the horizontal.
Tambaya 5 Rahoto
The load-extension graph of an elastic material is illustrated below. Use the graph to determine the work done in stretching the material
Method. The work done in stretching an elastic material is stored as elastic potential energy and is equal to the area under the load-extension graph (load on the vertical axis, extension on the horizontal axis).
\[ \text{Work done} = \text{area under the load-extension graph} \]For the linear (Hooke's-law) region, the graph is a straight line from the origin, so the area is a triangle:
\[ W = \tfrac{1}{2}\times \text{load}\times \text{extension} = \tfrac{1}{2}Fe \]If the material stretches beyond the elastic limit, the line curves; then the work done is found by counting squares under the curve (or by adding the triangular and rectangular/trapezoidal areas).
Worked illustration. If, for example, the graph shows a load of \(F = 20\ \text{N}\) producing an extension of \(e = 0.10\ \text{m}\) at the end of the straight line, then
\[ W = \tfrac{1}{2}\times 20 \times 0.10 = 1.0\ \text{J} \]Read the actual final load and extension (and any change of gradient) from the printed graph and substitute into the area calculation to obtain the work done.
Bayanin Amsa
Method. The work done in stretching an elastic material is stored as elastic potential energy and is equal to the area under the load-extension graph (load on the vertical axis, extension on the horizontal axis).
\[ \text{Work done} = \text{area under the load-extension graph} \]For the linear (Hooke's-law) region, the graph is a straight line from the origin, so the area is a triangle:
\[ W = \tfrac{1}{2}\times \text{load}\times \text{extension} = \tfrac{1}{2}Fe \]If the material stretches beyond the elastic limit, the line curves; then the work done is found by counting squares under the curve (or by adding the triangular and rectangular/trapezoidal areas).
Worked illustration. If, for example, the graph shows a load of \(F = 20\ \text{N}\) producing an extension of \(e = 0.10\ \text{m}\) at the end of the straight line, then
\[ W = \tfrac{1}{2}\times 20 \times 0.10 = 1.0\ \text{J} \]Read the actual final load and extension (and any change of gradient) from the printed graph and substitute into the area calculation to obtain the work done.
Tambaya 6 Rahoto
Explain the wave-particle duality of light. (b) A particle of wavelength 4.2x 10\(^{-11}\)m travels (a) With a momentum of 1.6 x 10\(^{-23}\) kg m/s,
Determine the value of the Planck's constant, h.
(a) Wave-particle duality of light. Light shows a dual nature. In some experiments it behaves as a wave (it undergoes interference, diffraction and polarisation, as in Young's double-slit experiment), while in other experiments it behaves as a stream of particles called photons, each carrying energy \(E=hf\) (as in the photoelectric effect and the Compton effect). Light is therefore neither purely a wave nor purely a particle; it exhibits whichever behaviour the experiment probes. The two aspects are linked by the de Broglie relation \(\lambda = h/p\).
(b) Finding Planck's constant. The de Broglie relation connects wavelength and momentum:
\[ \lambda = \frac{h}{p} \quad\Rightarrow\quad h = \lambda p \]With \(\lambda = 4.2\times10^{-11}\ \text{m}\) and \(p = 1.6\times10^{-23}\ \text{kg m/s}\):
\[ h = (4.2\times10^{-11})(1.6\times10^{-23}) \] \[ h = 6.72\times10^{-34}\ \text{J s} \]Planck's constant is about \(6.72\times10^{-34}\ \text{J s}\), in good agreement with the accepted value \(6.63\times10^{-34}\ \text{J s}\).
Bayanin Amsa
(a) Wave-particle duality of light. Light shows a dual nature. In some experiments it behaves as a wave (it undergoes interference, diffraction and polarisation, as in Young's double-slit experiment), while in other experiments it behaves as a stream of particles called photons, each carrying energy \(E=hf\) (as in the photoelectric effect and the Compton effect). Light is therefore neither purely a wave nor purely a particle; it exhibits whichever behaviour the experiment probes. The two aspects are linked by the de Broglie relation \(\lambda = h/p\).
(b) Finding Planck's constant. The de Broglie relation connects wavelength and momentum:
\[ \lambda = \frac{h}{p} \quad\Rightarrow\quad h = \lambda p \]With \(\lambda = 4.2\times10^{-11}\ \text{m}\) and \(p = 1.6\times10^{-23}\ \text{kg m/s}\):
\[ h = (4.2\times10^{-11})(1.6\times10^{-23}) \] \[ h = 6.72\times10^{-34}\ \text{J s} \]Planck's constant is about \(6.72\times10^{-34}\ \text{J s}\), in good agreement with the accepted value \(6.63\times10^{-34}\ \text{J s}\).
Tambaya 7 Rahoto
(a)(i) Define dew point. (ii) Explain why dew forms more quickly on the metal parts than on the rubber parts of a bicycle placed in the open overnight.
(b)(i) Explain the statement. the specific heat capacity of copper is 400 J/kg/K. (ii) Two metals, P and Q are supplied with the same quantity of heat.
If the ratio of the specific heat capacity of P to Q is 3 : 1 and their masses are in the ratio I:2 respectively.
calculate the ratio of the temperature rise of P to Q.
(c)(i) Define coefficient of thermal conductivity of a material.
(ii)
The diagram above illustrates a composite bar of iron and copper. The bar is insulated along its sides and it has a diameter of 10 mm. The length and thermal conductivity of the iron are 0.15 m and 40 W/m/K, respectively and those of copper are 0.05 m and 360 W/m/K, respectively. If the free ends of the iron and copper are kept at 100°C and 0°C respectively. calculate the (i) temperature at the interface between the bars; (ii) rate of heat flow along the bar.
(a)(i) Dew point
Dew point is the temperature at which the water vapour present in air is just sufficient to saturate the air, so that condensation begins to occur.
(a)(ii) Why dew forms faster on metal than on rubber
Metal is a better conductor of heat than rubber. At night, the metal parts of the bicycle lose heat more rapidly and become colder faster than the rubber parts. Their temperature therefore falls below the dew point earlier, causing water vapour in the surrounding air to condense as dew on the metal surfaces. Rubber is a poor conductor of heat, so it cools more slowly and dew forms on it later.
(b)(i) Meaning of the specific heat capacity of copper being 400 J kg-1 K-1
It means that 400 J of heat energy is required to raise the temperature of 1 kg of copper by 1 K (or 1°C).
(b)(ii) Ratio of temperature rise
For each metal,
\[Q = mc\Delta\theta\]
Since the same quantity of heat is supplied to both metals:
\[m_Pc_P\Delta\theta_P = m_Qc_Q\Delta\theta_Q\]
Given:
\[c_P:c_Q = 3:1\]
\[m_P:m_Q = 1:2\]
Therefore,
\[\frac{\Delta\theta_P}{\Delta\theta_Q} = \frac{m_Qc_Q}{m_Pc_P}\]
\[\frac{\Delta\theta_P}{\Delta\theta_Q} = \frac{2\times 1}{1\times 3} = \frac{2}{3}\]
Hence,
\[\boxed{\Delta\theta_P:\Delta\theta_Q = 2:3}\]
(c)(i) Coefficient of thermal conductivity
The coeff
Bayanin Amsa
(a)(i) Dew point
Dew point is the temperature at which the water vapour present in air is just sufficient to saturate the air, so that condensation begins to occur.
(a)(ii) Why dew forms faster on metal than on rubber
Metal is a better conductor of heat than rubber. At night, the metal parts of the bicycle lose heat more rapidly and become colder faster than the rubber parts. Their temperature therefore falls below the dew point earlier, causing water vapour in the surrounding air to condense as dew on the metal surfaces. Rubber is a poor conductor of heat, so it cools more slowly and dew forms on it later.
(b)(i) Meaning of the specific heat capacity of copper being 400 J kg-1 K-1
It means that 400 J of heat energy is required to raise the temperature of 1 kg of copper by 1 K (or 1°C).
(b)(ii) Ratio of temperature rise
For each metal,
\[Q = mc\Delta\theta\]
Since the same quantity of heat is supplied to both metals:
\[m_Pc_P\Delta\theta_P = m_Qc_Q\Delta\theta_Q\]
Given:
\[c_P:c_Q = 3:1\]
\[m_P:m_Q = 1:2\]
Therefore,
\[\frac{\Delta\theta_P}{\Delta\theta_Q} = \frac{m_Qc_Q}{m_Pc_P}\]
\[\frac{\Delta\theta_P}{\Delta\theta_Q} = \frac{2\times 1}{1\times 3} = \frac{2}{3}\]
Hence,
\[\boxed{\Delta\theta_P:\Delta\theta_Q = 2:3}\]
(c)(i) Coefficient of thermal conductivity
The coeff
Tambaya 8 Rahoto
(a)(i) State Hooke's law. (ii) A spring has a length of 0.20 m when a mass of 0.30 kg hangs on it, and a length of 0.75 nm when a mass of 1.95 kg hangs on it. Calculate the: (i) force constant of the spring; (ii) length of the spring when it is unloaded. [g = 10m/s\(^2\)]
(b)(i) What is diffusion? (ii) State two factors that affect the rate of diffusion of a substance. (iii) State the exact relationship between the rate of diffusion of a gas and its density.
(c) A satellite of mass, m orbits the earth of mass. M with a velocity, v at a distance R from the centre of the earth. Derive the relationship between the period T, of orbit and R.
(a)(i) Hooke's law. Provided the elastic limit is not exceeded, the extension of an elastic material is directly proportional to the force (load) producing it. \(F = k e\).
(a)(ii) Spring calculation. (The second length is 0.75 m.)
Force when 0.30 kg hangs: \(F_1 = 0.30\times10 = 3\ \text{N}\), length \(L_1 = 0.20\ \text{m}\).
Force when 1.95 kg hangs: \(F_2 = 1.95\times10 = 19.5\ \text{N}\), length \(L_2 = 0.75\ \text{m}\).
Force constant:
\[ k = \frac{F_2-F_1}{L_2-L_1} = \frac{19.5-3}{0.75-0.20} = \frac{16.5}{0.55} = 30\ \text{N/m} \]Unloaded (natural) length \(L_0\): using \(F_1 = k(L_1-L_0)\),
\[ 3 = 30(0.20 - L_0) \;\Rightarrow\; 0.20 - L_0 = 0.1 \;\Rightarrow\; L_0 = 0.10\ \text{m} \]Force constant \(= 30\ \text{N/m}\); natural length \(= 0.10\ \text{m}\).
(b)(i) Diffusion. Diffusion is the net movement of particles (molecules or ions) of a substance from a region of higher concentration to a region of lower concentration until they are evenly spread.
(b)(ii) Two factors affecting rate of diffusion. Temperature (higher temperature gives faster diffusion); the density or molar mass of the substance (lighter/less dense substances diffuse faster). (Also the concentration gradient.)
(b)(iii) Relationship with density. The rate of diffusion of a gas is inversely proportional to the square root of its density (Graham's law):
\[ \text{rate} \propto \frac{1}{\sqrt{\rho}} \](c) Period-radius relationship for a satellite. The gravitational pull provides the centripetal force:
\[ \frac{GMm}{R^{2}} = \frac{mv^{2}}{R} \;\Rightarrow\; v^{2} = \frac{GM}{R} \]The satellite covers the circumference \(2\pi R\) in one period, so \(v = \dfrac{2\pi R}{T}\). Substituting:
\[ \left(\frac{2\pi R}{T}\right)^{2} = \frac{GM}{R} \;\Rightarrow\; \frac{4\pi^{2}R^{2}}{T^{2}} = \frac{GM}{R} \] \[ \boxed{\,T^{2} = \frac{4\pi^{2}}{GM}\,R^{3}\,} \]Hence \(T^{2} \propto R^{3}\) (Kepler's third law).
Bayanin Amsa
(a)(i) Hooke's law. Provided the elastic limit is not exceeded, the extension of an elastic material is directly proportional to the force (load) producing it. \(F = k e\).
(a)(ii) Spring calculation. (The second length is 0.75 m.)
Force when 0.30 kg hangs: \(F_1 = 0.30\times10 = 3\ \text{N}\), length \(L_1 = 0.20\ \text{m}\).
Force when 1.95 kg hangs: \(F_2 = 1.95\times10 = 19.5\ \text{N}\), length \(L_2 = 0.75\ \text{m}\).
Force constant:
\[ k = \frac{F_2-F_1}{L_2-L_1} = \frac{19.5-3}{0.75-0.20} = \frac{16.5}{0.55} = 30\ \text{N/m} \]Unloaded (natural) length \(L_0\): using \(F_1 = k(L_1-L_0)\),
\[ 3 = 30(0.20 - L_0) \;\Rightarrow\; 0.20 - L_0 = 0.1 \;\Rightarrow\; L_0 = 0.10\ \text{m} \]Force constant \(= 30\ \text{N/m}\); natural length \(= 0.10\ \text{m}\).
(b)(i) Diffusion. Diffusion is the net movement of particles (molecules or ions) of a substance from a region of higher concentration to a region of lower concentration until they are evenly spread.
(b)(ii) Two factors affecting rate of diffusion. Temperature (higher temperature gives faster diffusion); the density or molar mass of the substance (lighter/less dense substances diffuse faster). (Also the concentration gradient.)
(b)(iii) Relationship with density. The rate of diffusion of a gas is inversely proportional to the square root of its density (Graham's law):
\[ \text{rate} \propto \frac{1}{\sqrt{\rho}} \](c) Period-radius relationship for a satellite. The gravitational pull provides the centripetal force:
\[ \frac{GMm}{R^{2}} = \frac{mv^{2}}{R} \;\Rightarrow\; v^{2} = \frac{GM}{R} \]The satellite covers the circumference \(2\pi R\) in one period, so \(v = \dfrac{2\pi R}{T}\). Substituting:
\[ \left(\frac{2\pi R}{T}\right)^{2} = \frac{GM}{R} \;\Rightarrow\; \frac{4\pi^{2}R^{2}}{T^{2}} = \frac{GM}{R} \] \[ \boxed{\,T^{2} = \frac{4\pi^{2}}{GM}\,R^{3}\,} \]Hence \(T^{2} \propto R^{3}\) (Kepler's third law).
Tambaya 9 Rahoto
State three observable phenomena where a particle behaves like waves. State the scientific principle underlying the operation of fibre optics.
(b) Explain each of the following terms as used in fibre optics: (i) core; (ii) cladding
Three observable phenomena in which particles behave like waves.
- Electron diffraction: a beam of electrons passed through a thin crystal (or graphite film) produces diffraction rings, just as X-rays do (Davisson-Germer experiment).
- Interference of electrons: electrons passed through a double slit build up an interference pattern of bright and dark fringes.
- The electron microscope: electrons are focused and form images because they behave like short-wavelength waves; neutron diffraction by crystals is a further example.
Scientific principle underlying fibre optics. Fibre optics works on the principle of total internal reflection of light. When light inside the denser core strikes the core-cladding boundary at an angle greater than the critical angle, it is totally reflected back into the core, so light is guided along the fibre with very little loss even when the fibre bends.
(b) Terms used in fibre optics.
- (i) Core: the central, thin strand of very pure glass (or plastic) of higher refractive index along which the light signal actually travels by repeated total internal reflection.
- (ii) Cladding: the outer layer of glass of lower refractive index surrounding the core. Because it is optically less dense than the core, it ensures that light striking the boundary is totally internally reflected and kept inside the core.
Bayanin Amsa
Three observable phenomena in which particles behave like waves.
- Electron diffraction: a beam of electrons passed through a thin crystal (or graphite film) produces diffraction rings, just as X-rays do (Davisson-Germer experiment).
- Interference of electrons: electrons passed through a double slit build up an interference pattern of bright and dark fringes.
- The electron microscope: electrons are focused and form images because they behave like short-wavelength waves; neutron diffraction by crystals is a further example.
Scientific principle underlying fibre optics. Fibre optics works on the principle of total internal reflection of light. When light inside the denser core strikes the core-cladding boundary at an angle greater than the critical angle, it is totally reflected back into the core, so light is guided along the fibre with very little loss even when the fibre bends.
(b) Terms used in fibre optics.
- (i) Core: the central, thin strand of very pure glass (or plastic) of higher refractive index along which the light signal actually travels by repeated total internal reflection.
- (ii) Cladding: the outer layer of glass of lower refractive index surrounding the core. Because it is optically less dense than the core, it ensures that light striking the boundary is totally internally reflected and kept inside the core.
Tambaya 10 Rahoto
(a) Name two artificial satellites.
(b) A geostationary satellite moves in an orbit of radius 6300 km. Calculate the speed with which it moves in the orbit. π = \(_{22}{7}\)
(a) Two artificial satellites. A communication satellite and a weather (meteorological) satellite. (Other acceptable examples: a navigation/GPS satellite, a spy/reconnaissance satellite, the International Space Station.)
(b) Speed of the geostationary satellite. A geostationary satellite has the same period as the earth's rotation, so its period is
\[ T = 24\ \text{hours} = 24\times 60\times 60 = 86400\ \text{s} \]Radius of orbit \(r = 6300\ \text{km} = 6.3\times10^{6}\ \text{m}\).
The satellite moves once round the circular orbit (circumference \(2\pi r\)) in time T, so its orbital speed is
\[ v = \frac{2\pi r}{T} = \frac{2\times\frac{22}{7}\times 6.3\times10^{6}}{86400} \] \[ v = \frac{39\,600\,000}{86400} \approx 458\ \text{m s}^{-1} \]The satellite moves in its orbit with a speed of about 458 m s\(^{-1}\).
Bayanin Amsa
(a) Two artificial satellites. A communication satellite and a weather (meteorological) satellite. (Other acceptable examples: a navigation/GPS satellite, a spy/reconnaissance satellite, the International Space Station.)
(b) Speed of the geostationary satellite. A geostationary satellite has the same period as the earth's rotation, so its period is
\[ T = 24\ \text{hours} = 24\times 60\times 60 = 86400\ \text{s} \]Radius of orbit \(r = 6300\ \text{km} = 6.3\times10^{6}\ \text{m}\).
The satellite moves once round the circular orbit (circumference \(2\pi r\)) in time T, so its orbital speed is
\[ v = \frac{2\pi r}{T} = \frac{2\times\frac{22}{7}\times 6.3\times10^{6}}{86400} \] \[ v = \frac{39\,600\,000}{86400} \approx 458\ \text{m s}^{-1} \]The satellite moves in its orbit with a speed of about 458 m s\(^{-1}\).
Tambaya 11 Rahoto
(a)(i) What is meant by the root-mean-square value of an alternating current? (ii) Define impedance of an alternating current circuit.
(b) An electrical device rated 120 V, 60 W is opened on a 240 V, 50Hz mains supply. The circuit has a capacitor connected in series with ihe electrical device and the supply. Calculate the capacitance of the capacitor. [π=3.142].
(c)(i) Define the capacitance of a capacitor.
(ii)
The circuit diagram above illustrates two capacitors of capacitance C\(_1\) and C\(_2\) connected in series across a 2V source.
(i)Obtain an expression for the total capacitance in terms of C\(_2\). 2 mm 5 n (ii) Calculate the potential difference across each capacitor.
(a)(i) Root-mean-square (r.m.s.) value of an alternating current
The r.m.s. value of an alternating current is the value of a steady direct current that would produce the same heating effect, at the same rate, in the same resistor.
For a sinusoidal current:
\[ I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}} \]
(a)(ii) Impedance
Impedance is the total opposition offered by an a.c. circuit to the flow of alternating current. It includes resistance and reactance due to capacitors and inductors.
\[ Z=\frac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}} \]
Its SI unit is the ohm, \(\Omega\).
(b) Capacitance required
The device must operate at its rated values, so the current in the series circuit is:
\[ I=\frac{P}{V}=\frac{60}{120}=0.50\ \text{A} \]
The voltage across the device is \(V_R=120\ \text{V}\), whereas the supply voltage is \(240\ \text{V}\). The capacitor voltage and the device voltage are \(90^\circ\) out of phase, so they must be added using Pythagoras, not by ordinary addition:
\[ V^2=V_R^2+V_C^2 \]
\[ V_C=\sqrt{240^2-120^2}=\sqrt{43200}=207.8\ \text{V} \]
The capacitive reactance is therefore:
\[ X_C=\frac{V_C}{I}=\frac{207.8}{0.50}=415.7\ \Omega \]
For a capacitor:
\[ X_C=\frac{1}{2\pi fC} \]
Therefore:
\[ C=\frac{1}{2\pi fX_C} \]
\[ C=\frac{1}{2(3.142)(50)(415.7)}=7.66\times10^{-6}\ \text{F} \]
\[ \boxed{C\approx7.7\ \mu\text{F}} \]
The expression \(C=\dfrac{X_C}{2\pi f}\) is incorrect: capacitance is inversely proportional to capacitive reactance.
(c)(i) Capacitance
The capacitance of a capacitor is the charge stored on either plate per unit potential difference across the capacitor:
\[ C=\frac{Q}{V} \]
Its SI unit is the farad (F).
(c)(ii) Capacitors in series
The diagram gives:
\[ C_1=\frac{5}{2}C_2 \]
For capacitors in series:
\[ \frac{1}{C_T}=\frac{1}{C_1}+\frac{1}{C_2} \]
\[ C_T=\frac{C_1C_2}{C_1+C_2} \]
Substitute \(C_1=\dfrac{5}{2}C_2\):
\[ C_T=\frac{\left(\frac{5}{2}C_2\right)C_2}{\frac{5}{2}C_2+C_2} \]
\[ C_T=\frac{\frac{5}{2}C_2^2}{\frac{7}{2}C_2}=\boxed{\frac{5}{7}C_2} \]
Capacitors in series carry the same charge. Since \(V=\dfrac{Q}{C}\), the smaller capacitance has the larger potential difference. Thus:
\[ \frac{V_1}{V_2}=\frac{C_2}{C_1}=\frac{C_2}{\frac{5}{2}C_2}=\frac{2}{5} \]
Also:
\[ V_1+V_2=2\ \text{V} \]
The voltage is divided in the ratio \(2:5\), giving:
\[ \boxed{V_1=\frac{4}{7}\ \text{V}} \]
\[ \boxed{V_2=\frac{10}{7}\ \text{V}} \]
Examination point: In a series capacitor circuit, charge is the same on each capacitor, but the potential difference is inversely proportional to capacitance.
Bayanin Amsa
(a)(i) Root-mean-square (r.m.s.) value of an alternating current
The r.m.s. value of an alternating current is the value of a steady direct current that would produce the same heating effect, at the same rate, in the same resistor.
For a sinusoidal current:
\[ I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}} \]
(a)(ii) Impedance
Impedance is the total opposition offered by an a.c. circuit to the flow of alternating current. It includes resistance and reactance due to capacitors and inductors.
\[ Z=\frac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}} \]
Its SI unit is the ohm, \(\Omega\).
(b) Capacitance required
The device must operate at its rated values, so the current in the series circuit is:
\[ I=\frac{P}{V}=\frac{60}{120}=0.50\ \text{A} \]
The voltage across the device is \(V_R=120\ \text{V}\), whereas the supply voltage is \(240\ \text{V}\). The capacitor voltage and the device voltage are \(90^\circ\) out of phase, so they must be added using Pythagoras, not by ordinary addition:
\[ V^2=V_R^2+V_C^2 \]
\[ V_C=\sqrt{240^2-120^2}=\sqrt{43200}=207.8\ \text{V} \]
The capacitive reactance is therefore:
\[ X_C=\frac{V_C}{I}=\frac{207.8}{0.50}=415.7\ \Omega \]
For a capacitor:
\[ X_C=\frac{1}{2\pi fC} \]
Therefore:
\[ C=\frac{1}{2\pi fX_C} \]
\[ C=\frac{1}{2(3.142)(50)(415.7)}=7.66\times10^{-6}\ \text{F} \]
\[ \boxed{C\approx7.7\ \mu\text{F}} \]
The expression \(C=\dfrac{X_C}{2\pi f}\) is incorrect: capacitance is inversely proportional to capacitive reactance.
(c)(i) Capacitance
The capacitance of a capacitor is the charge stored on either plate per unit potential difference across the capacitor:
\[ C=\frac{Q}{V} \]
Its SI unit is the farad (F).
(c)(ii) Capacitors in series
The diagram gives:
\[ C_1=\frac{5}{2}C_2 \]
For capacitors in series:
\[ \frac{1}{C_T}=\frac{1}{C_1}+\frac{1}{C_2} \]
\[ C_T=\frac{C_1C_2}{C_1+C_2} \]
Substitute \(C_1=\dfrac{5}{2}C_2\):
\[ C_T=\frac{\left(\frac{5}{2}C_2\right)C_2}{\frac{5}{2}C_2+C_2} \]
\[ C_T=\frac{\frac{5}{2}C_2^2}{\frac{7}{2}C_2}=\boxed{\frac{5}{7}C_2} \]
Capacitors in series carry the same charge. Since \(V=\dfrac{Q}{C}\), the smaller capacitance has the larger potential difference. Thus:
\[ \frac{V_1}{V_2}=\frac{C_2}{C_1}=\frac{C_2}{\frac{5}{2}C_2}=\frac{2}{5} \]
Also:
\[ V_1+V_2=2\ \text{V} \]
The voltage is divided in the ratio \(2:5\), giving:
\[ \boxed{V_1=\frac{4}{7}\ \text{V}} \]
\[ \boxed{V_2=\frac{10}{7}\ \text{V}} \]
Examination point: In a series capacitor circuit, charge is the same on each capacitor, but the potential difference is inversely proportional to capacitance.
