WAEC SSCE - Physics - 1996 (Essay)
Tambaya 1 Rahoto
(a) Explain the following terms:
(i) viscosity;
(ii) terminal velocity.
(b)(i) Describe an experiment to determine the terminal velocity of a steel ball falling through a jar of glycerine..
(ii) State two precautions that should be taken to ensure accurate result.
(c) State two
(i) effects of viscosity
(ii) applications of viscosity.
(a)(i) Viscosity is the property of a fluid by which it opposes relative motion between its adjacent layers; it is a kind of internal friction in the fluid.
(ii) Terminal velocity is the constant (maximum) velocity attained by a body falling through a fluid when the net force on it is zero, i.e. when its weight is balanced by the upthrust plus the viscous drag.
(b)(i) Experiment (terminal velocity of a steel ball in glycerine):
- Fill a tall glass jar with glycerine and fix two rubber bands (marks) round it, well below the surface and a measured distance \(d\) apart.
- Gently release a small steel ball into the centre of the glycerine and let it fall.
- Using a stopwatch, time the ball \((t)\) as it travels between the two marks. In that lower region the velocity is constant.
- Terminal velocity \( v = \dfrac{d}{t} \). Repeat and average.
(ii) Two precautions: release the ball centrally so it falls far from the walls of the jar; take timings only in the lower region where the velocity has become steady, and avoid parallax when reading the marks.
(c)(i) Two effects of viscosity: it opposes/retards the motion of bodies moving through a fluid (drag); it reduces the rate of flow of the fluid and generates heat.
(ii) Two applications of viscosity: lubrication of machine parts by oils; use of viscous fluids in shock absorbers (dashpots) and hydraulic brakes.
Bayanin Amsa
(a)(i) Viscosity is the property of a fluid by which it opposes relative motion between its adjacent layers; it is a kind of internal friction in the fluid.
(ii) Terminal velocity is the constant (maximum) velocity attained by a body falling through a fluid when the net force on it is zero, i.e. when its weight is balanced by the upthrust plus the viscous drag.
(b)(i) Experiment (terminal velocity of a steel ball in glycerine):
- Fill a tall glass jar with glycerine and fix two rubber bands (marks) round it, well below the surface and a measured distance \(d\) apart.
- Gently release a small steel ball into the centre of the glycerine and let it fall.
- Using a stopwatch, time the ball \((t)\) as it travels between the two marks. In that lower region the velocity is constant.
- Terminal velocity \( v = \dfrac{d}{t} \). Repeat and average.
(ii) Two precautions: release the ball centrally so it falls far from the walls of the jar; take timings only in the lower region where the velocity has become steady, and avoid parallax when reading the marks.
(c)(i) Two effects of viscosity: it opposes/retards the motion of bodies moving through a fluid (drag); it reduces the rate of flow of the fluid and generates heat.
(ii) Two applications of viscosity: lubrication of machine parts by oils; use of viscous fluids in shock absorbers (dashpots) and hydraulic brakes.
Tambaya 2 Rahoto
(a) What is meant by dispersion of white light?
(b) State the colours in the spectrum of white light in ascending order of their wavelengths
(c) Which colour is deviated.
(i) least
(ii) most?
(d) Explain why white light is dispersed when it passes through a glass prism.
(e) Describe, with the aid of a labelled diagram, how a pure spectrum of white light can be produced on a screen.
(a) Dispersion of white light is the separation of white light into its constituent colours when it passes through a refracting medium such as a glass prism.
(b) In ascending order of wavelength:
Violet, Indigo, Blue, Green, Yellow, Orange, Red (VIBGYOR).
(c)
- The colour deviated least is red.
- The colour deviated most is violet.
(d) White light is made up of colours of different wavelengths. The refractive index of glass is different for the different colours. Consequently, the colours travel at different speeds in the prism and are refracted through different angles. Violet light is refracted most, while red light is refracted least; hence the colours spread out to form a spectrum.
(e) Production of a pure spectrum of white light
An illuminated narrow slit is placed at the principal focus of a converging lens, L1. The lens produces a parallel beam of white light which falls on a glass prism. The prism disperses the light into separate parallel beams of different colours. A second converging lens, L2, brings each coloured beam to a separate focus on a white screen placed in its focal plane. Thus, sharp non-overlapping images of the slit form a pure spectrum.
Bayanin Amsa
(a) Dispersion of white light is the separation of white light into its constituent colours when it passes through a refracting medium such as a glass prism.
(b) In ascending order of wavelength:
Violet, Indigo, Blue, Green, Yellow, Orange, Red (VIBGYOR).
(c)
- The colour deviated least is red.
- The colour deviated most is violet.
(d) White light is made up of colours of different wavelengths. The refractive index of glass is different for the different colours. Consequently, the colours travel at different speeds in the prism and are refracted through different angles. Violet light is refracted most, while red light is refracted least; hence the colours spread out to form a spectrum.
(e) Production of a pure spectrum of white light
An illuminated narrow slit is placed at the principal focus of a converging lens, L1. The lens produces a parallel beam of white light which falls on a glass prism. The prism disperses the light into separate parallel beams of different colours. A second converging lens, L2, brings each coloured beam to a separate focus on a white screen placed in its focal plane. Thus, sharp non-overlapping images of the slit form a pure spectrum.
Tambaya 3 Rahoto
(a) Define the apparent cubic expansivity of a liquid
(b)(i) Describe with the aid of a labelled diagram, an experiment to determine the apparent cubic expansivity of a liquid.
(ii) State two precuations that should be taken to ensure accurate results.
(c) A density glass bottle contains 44.25g of a liquid at 0°C and 42.02g at 50°C. Calculate the real cubic expansivity of the liquid. (Linear expansivity of glass = 1.0 x 10-5K\(^{-1}\))
(a) The apparent cubic expansivity of a liquid is the apparent increase in volume per unit original volume of the liquid per unit rise in temperature. It is the expansion observed when the expansion of the containing vessel is not allowed for.
(b)(i) Determination of apparent cubic expansivity
The apparatus is arranged as shown below.
A clean, dry density bottle fitted with a stopper having a fine capillary hole is weighed empty. Let its mass be \(M_1\). It is filled completely with the liquid at the initial temperature \(T_1\), the stopper is inserted, and the outside is wiped dry. The mass of the bottle and liquid is found as \(M_2\).
The bottle is suspended in a water bath, with the stopper and capillary orifice above the water level. A thermometer is placed in the bath. The water is heated slowly to a final temperature \(T_2\). As the liquid expands, some of it escapes through the capillary hole. Heating is continued until no further liquid escapes at \(T_2\).
The bottle is removed, allowed to cool, wiped dry and weighed again. Let its final mass be \(M_3\).
Thus,
Therefore, the apparent cubic expansivity is
(ii) Precautions
- Heat the water bath slowly and stir it so that the liquid and the bath are at a uniform temperature.
- Read the thermometer at eye level to avoid parallax error.
- Keep the capillary orifice above the water level and wipe the bottle completely dry before each weighing.
(c) Calculation of the real cubic expansivity
For a temperature rise of \(50\,\mathrm{K}\):
The cubic expansivity of the glass bottle is
Since \(\gamma_r=\gamma_a+\gamma_g\),
Bayanin Amsa
(a) The apparent cubic expansivity of a liquid is the apparent increase in volume per unit original volume of the liquid per unit rise in temperature. It is the expansion observed when the expansion of the containing vessel is not allowed for.
(b)(i) Determination of apparent cubic expansivity
The apparatus is arranged as shown below.
A clean, dry density bottle fitted with a stopper having a fine capillary hole is weighed empty. Let its mass be \(M_1\). It is filled completely with the liquid at the initial temperature \(T_1\), the stopper is inserted, and the outside is wiped dry. The mass of the bottle and liquid is found as \(M_2\).
The bottle is suspended in a water bath, with the stopper and capillary orifice above the water level. A thermometer is placed in the bath. The water is heated slowly to a final temperature \(T_2\). As the liquid expands, some of it escapes through the capillary hole. Heating is continued until no further liquid escapes at \(T_2\).
The bottle is removed, allowed to cool, wiped dry and weighed again. Let its final mass be \(M_3\).
Thus,
Therefore, the apparent cubic expansivity is
(ii) Precautions
- Heat the water bath slowly and stir it so that the liquid and the bath are at a uniform temperature.
- Read the thermometer at eye level to avoid parallax error.
- Keep the capillary orifice above the water level and wipe the bottle completely dry before each weighing.
(c) Calculation of the real cubic expansivity
For a temperature rise of \(50\,\mathrm{K}\):
The cubic expansivity of the glass bottle is
Since \(\gamma_r=\gamma_a+\gamma_g\),
Tambaya 4 Rahoto
(a) Explain what is meant by the statement: The capacitance of a parallel-plate capacitor is 2\(\mu\)F
(b) State: (i) three factors on which its capacitance depends
(ii) three uses of capacitors.
(c) Derive a formula for the energy W stored in a charged capacitor of capacitance C carrying a charge Q on either plate.,
(d) Two parallel-plate capacitors of capacitances 2\(\mu\)F and 3\(\mu\)F are connected in parallel and the combination is connected to a 50V d.c. source. Draw the circuit diagram of the arrangement and determine the:
(i) charge on either plate of each capacitor
(ii) potential difference across each capacitor
(iii) energy of the combinad capacitors.
(a) Meaning of \(2\,\mu\text{F}\)
Capacitance is defined by \(C=\dfrac{Q}{V}\). A capacitance of \(2\,\mu\text{F}\) means that the capacitor stores \(2\,\mu\text{C}\) of charge on each plate for every \(1\,\text{V}\) potential difference across its plates:
\[C=2\,\mu\text{F}=2\times10^{-6}\,\text{F}\]
Thus, when \(V=1\,\text{V}\), \(Q=CV=2\times10^{-6}\,\text{C}=2\,\mu\text{C}\). The two plates carry equal charges of opposite sign.
(b)(i) Factors affecting the capacitance of a parallel-plate capacitor
- The area of overlap of the plates: larger area gives greater capacitance.
- The separation between the plates: greater separation gives smaller capacitance.
- The permittivity of the dielectric between the plates: a dielectric with higher permittivity gives greater capacitance.
For parallel plates, this is summarised by:
\[C=\frac{\varepsilon A}{d}\]
(b)(ii) Uses of capacitors
- Storing charge or energy, for example in a camera flash.
- Smoothing the output from a rectifier in a power supply.
- Separating a.c. signals from d.c. components in electronic circuits.
(c) Derivation of the energy stored
During charging, the potential difference is not constant: it rises from \(0\) to its final value \(V\). When the charge already on the capacitor is \(q\),
\[V=\frac{q}{C}\]
The work done in bringing a further small charge \(\mathrm{d}q\) onto the capacitor is:
\[\mathrm{d}W=V\,\mathrm{d}q=\frac{q}{C}\,\mathrm{d}q\]
Therefore, the total energy stored when the final charge is \(Q\) is:
\[ W=\int_0^Q\frac{q}{C}\,\mathrm{d}q =\frac{1}{C}\left[\frac{q^2}{2}\right]_0^Q =\frac{Q^2}{2C} \]
Since \(Q=CV\), equivalent forms are:
\[ \boxed{W=\frac{Q^2}{2C}=\frac{1}{2}QV=\frac{1}{2}CV^2} \]
The use of \(\tfrac12 QV\), rather than \(QV\), is important because the p.d. increases gradually while the capacitor is charged.
(d) Capacitors connected in parallel to a \(50\,\text{V}\) d.c. supply
In a parallel connection, each capacitor is connected directly across the source. Therefore, the potential difference across each is \(50\,\text{V}\).
(i) Charge on either plate of each capacitor
For the \(2\,\mu\text{F}\) capacitor:
\[ Q_1=C_1V=(2\times10^{-6})(50) =1.0\times10^{-4}\,\text{C} =100\,\mu\text{C} \]
For the \(3\,\mu\text{F}\) capacitor:
\[ Q_2=C_2V=(3\times10^{-6})(50) =1.5\times10^{-4}\,\text{C} =150\,\mu\text{C} \]
Each capacitor has these charge magnitudes on its plates: one plate is positive and the other has an equal negative charge.
(ii) Potential difference across each capacitor
\[ \boxed{V_1=V_2=50\,\text{V}} \]
(iii) Energy stored by the combined capacitors
For capacitors in parallel, capacitances add:
\[ C_{\text{total}}=2\,\mu\text{F}+3\,\mu\text{F}=5\,\mu\text{F} =5\times10^{-6}\,\text{F} \]
\[ W=\frac12C_{\text{total}}V^2 =\frac12(5\times10^{-6})(50)^2 =6.25\times10^{-3}\,\text{J} \]
\[ \boxed{W=6.25\times10^{-3}\,\text{J}} \]
Examination reminder: In parallel, the potential difference is the same across every capacitor, while the charge on each capacitor is found separately using \(Q=CV\).
Bayanin Amsa
(a) Meaning of \(2\,\mu\text{F}\)
Capacitance is defined by \(C=\dfrac{Q}{V}\). A capacitance of \(2\,\mu\text{F}\) means that the capacitor stores \(2\,\mu\text{C}\) of charge on each plate for every \(1\,\text{V}\) potential difference across its plates:
\[C=2\,\mu\text{F}=2\times10^{-6}\,\text{F}\]
Thus, when \(V=1\,\text{V}\), \(Q=CV=2\times10^{-6}\,\text{C}=2\,\mu\text{C}\). The two plates carry equal charges of opposite sign.
(b)(i) Factors affecting the capacitance of a parallel-plate capacitor
- The area of overlap of the plates: larger area gives greater capacitance.
- The separation between the plates: greater separation gives smaller capacitance.
- The permittivity of the dielectric between the plates: a dielectric with higher permittivity gives greater capacitance.
For parallel plates, this is summarised by:
\[C=\frac{\varepsilon A}{d}\]
(b)(ii) Uses of capacitors
- Storing charge or energy, for example in a camera flash.
- Smoothing the output from a rectifier in a power supply.
- Separating a.c. signals from d.c. components in electronic circuits.
(c) Derivation of the energy stored
During charging, the potential difference is not constant: it rises from \(0\) to its final value \(V\). When the charge already on the capacitor is \(q\),
\[V=\frac{q}{C}\]
The work done in bringing a further small charge \(\mathrm{d}q\) onto the capacitor is:
\[\mathrm{d}W=V\,\mathrm{d}q=\frac{q}{C}\,\mathrm{d}q\]
Therefore, the total energy stored when the final charge is \(Q\) is:
\[ W=\int_0^Q\frac{q}{C}\,\mathrm{d}q =\frac{1}{C}\left[\frac{q^2}{2}\right]_0^Q =\frac{Q^2}{2C} \]
Since \(Q=CV\), equivalent forms are:
\[ \boxed{W=\frac{Q^2}{2C}=\frac{1}{2}QV=\frac{1}{2}CV^2} \]
The use of \(\tfrac12 QV\), rather than \(QV\), is important because the p.d. increases gradually while the capacitor is charged.
(d) Capacitors connected in parallel to a \(50\,\text{V}\) d.c. supply
In a parallel connection, each capacitor is connected directly across the source. Therefore, the potential difference across each is \(50\,\text{V}\).
(i) Charge on either plate of each capacitor
For the \(2\,\mu\text{F}\) capacitor:
\[ Q_1=C_1V=(2\times10^{-6})(50) =1.0\times10^{-4}\,\text{C} =100\,\mu\text{C} \]
For the \(3\,\mu\text{F}\) capacitor:
\[ Q_2=C_2V=(3\times10^{-6})(50) =1.5\times10^{-4}\,\text{C} =150\,\mu\text{C} \]
Each capacitor has these charge magnitudes on its plates: one plate is positive and the other has an equal negative charge.
(ii) Potential difference across each capacitor
\[ \boxed{V_1=V_2=50\,\text{V}} \]
(iii) Energy stored by the combined capacitors
For capacitors in parallel, capacitances add:
\[ C_{\text{total}}=2\,\mu\text{F}+3\,\mu\text{F}=5\,\mu\text{F} =5\times10^{-6}\,\text{F} \]
\[ W=\frac12C_{\text{total}}V^2 =\frac12(5\times10^{-6})(50)^2 =6.25\times10^{-3}\,\text{J} \]
\[ \boxed{W=6.25\times10^{-3}\,\text{J}} \]
Examination reminder: In parallel, the potential difference is the same across every capacitor, while the charge on each capacitor is found separately using \(Q=CV\).
