WAEC SSCE - Physics - 2005 (Essay)
Tambaya 1 Rahoto
(a) Define elastic limit
(b) State Hookes laws of elasticity
(a) Elastic limit. The elastic limit of a material is the maximum load (or stress) that the material can experience and still return exactly to its original length and shape when the load is removed. If the load exceeds the elastic limit, the material is permanently (plastically) deformed.
(b) Hooke's law of elasticity. Provided the elastic limit is not exceeded, the extension of an elastic material is directly proportional to the applied force (load) producing it.
\[ F = k e, \]
where F is the applied force, e is the extension, and k is the force constant (elastic constant) of the material.
Bayanin Amsa
(a) Elastic limit. The elastic limit of a material is the maximum load (or stress) that the material can experience and still return exactly to its original length and shape when the load is removed. If the load exceeds the elastic limit, the material is permanently (plastically) deformed.
(b) Hooke's law of elasticity. Provided the elastic limit is not exceeded, the extension of an elastic material is directly proportional to the applied force (load) producing it.
\[ F = k e, \]
where F is the applied force, e is the extension, and k is the force constant (elastic constant) of the material.
Tambaya 2 Rahoto
(a) Sketch a diagram of a simple pendulum performing simple harmonic motion and indicate positions of maximum potential energy and kinetic energy.
(b) A body moving with simple harmonic motion in a straight line has velocity, v and acceleration, a, when the instantaneous displacement, x in cm, from its maximum position is given by x = 2.5 sin 0.4 \(\pi t\), where t is in seconds. Determine the magnitude of the maximum (i) veloxity; (ii) acceleration
(c) A mass m attached to a light spiral is caused to perform simple harmonic motion of frequency
f = \(\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\), where k is the force constant of the spring.
(i) Explain the physical significance of \(\sqrt{\frac{k}{m}}\).
(ii) If m = 0.30 kg, k = 30Nm\(^{-1}\) and the maximum position is 0.015m, calculate the maximum;
(i) kinetic energy
(ii) tension in the spring during the motion [g = 10 ms\(^{-1}\), \(\pi\) = 3.142]
(a) The sketch below shows the energy distribution of a simple pendulum during simple harmonic motion.
At each extreme position, the bob is momentarily at rest, so its potential energy is maximum and its kinetic energy is zero. At the mean (lowest) position, its speed is greatest, so its kinetic energy is maximum and its potential energy is minimum.
(b) Comparing
\[x=2.5\sin(0.4\pi t)\ \text{cm}\]
with \(x=A\sin \omega t\),
\[A=2.5\ \text{cm}=2.5\times10^{-2}\ \text{m},\qquad \omega=0.4\pi\ \text{rad s}^{-1}.\]
(i) Maximum velocity
\[v_{\max}=A\omega\]
\[=2.5\times10^{-2}\times0.4\pi\]
\[=3.142\times10^{-2}\ \text{m s}^{-1}.\]
(ii) Maximum acceleration
\[a_{\max}=A\omega^2\]
\[=2.5\times10^{-2}(0.4\pi)^2\]
\[=3.95\times10^{-2}\ \text{m s}^{-2}.\]
(c)(i)
\[\sqrt{\frac{k}{m}}=\omega\]
is the angular frequency of the oscillation. It is the rate of change of phase (angular displacement) and is measured in \(\text{rad s}^{-1}\).
(c)(ii)
Given \(m=0.30\ \text{kg}\), \(k=30\ \text{N m}^{-1}\), \(A=0.015\ \text{m}\),
\[\omega^2=\frac{k}{m}=\frac{30}{0.30}=100\ \text{s}^{-2}.\]
(i) Maximum kinetic energy
\[K_{\max}=\frac{1}{2}mA^2\omega^2=\frac{1}{2}kA^2\]
\[=\frac{1}{2}\times30\times(0.015)^2\]
\[=3.375\times10^{-3}\ \text{J}\approx3.38\times10^{-3}\ \text{J}.\]
(ii) Maximum tension in the spring
The maximum tension occurs at the lowest position, where the extension is greatest:
\[T_{\max}=mg+kA\]
\[=(0.30\times10)+(30\times0.015)\]
\[=3.45\ \text{N}.\]
Bayanin Amsa
(a) The sketch below shows the energy distribution of a simple pendulum during simple harmonic motion.
At each extreme position, the bob is momentarily at rest, so its potential energy is maximum and its kinetic energy is zero. At the mean (lowest) position, its speed is greatest, so its kinetic energy is maximum and its potential energy is minimum.
(b) Comparing
\[x=2.5\sin(0.4\pi t)\ \text{cm}\]
with \(x=A\sin \omega t\),
\[A=2.5\ \text{cm}=2.5\times10^{-2}\ \text{m},\qquad \omega=0.4\pi\ \text{rad s}^{-1}.\]
(i) Maximum velocity
\[v_{\max}=A\omega\]
\[=2.5\times10^{-2}\times0.4\pi\]
\[=3.142\times10^{-2}\ \text{m s}^{-1}.\]
(ii) Maximum acceleration
\[a_{\max}=A\omega^2\]
\[=2.5\times10^{-2}(0.4\pi)^2\]
\[=3.95\times10^{-2}\ \text{m s}^{-2}.\]
(c)(i)
\[\sqrt{\frac{k}{m}}=\omega\]
is the angular frequency of the oscillation. It is the rate of change of phase (angular displacement) and is measured in \(\text{rad s}^{-1}\).
(c)(ii)
Given \(m=0.30\ \text{kg}\), \(k=30\ \text{N m}^{-1}\), \(A=0.015\ \text{m}\),
\[\omega^2=\frac{k}{m}=\frac{30}{0.30}=100\ \text{s}^{-2}.\]
(i) Maximum kinetic energy
\[K_{\max}=\frac{1}{2}mA^2\omega^2=\frac{1}{2}kA^2\]
\[=\frac{1}{2}\times30\times(0.015)^2\]
\[=3.375\times10^{-3}\ \text{J}\approx3.38\times10^{-3}\ \text{J}.\]
(ii) Maximum tension in the spring
The maximum tension occurs at the lowest position, where the extension is greatest:
\[T_{\max}=mg+kA\]
\[=(0.30\times10)+(30\times0.015)\]
\[=3.45\ \text{N}.\]
Tambaya 3 Rahoto
(a) Define surface tension
(b) State two methods by which the surface tension of a liquid can be reduced
(a) Surface tension. Surface tension is the property of the free surface of a liquid by which it behaves like a stretched elastic membrane (skin) and tends to contract so as to occupy the smallest possible surface area. It arises because the molecules at the surface experience a net inward cohesive pull from the molecules below them.
(b) Methods of reducing the surface tension of a liquid:
- Raising the temperature of the liquid.
- Adding an impurity such as soap or detergent to the liquid.
Bayanin Amsa
(a) Surface tension. Surface tension is the property of the free surface of a liquid by which it behaves like a stretched elastic membrane (skin) and tends to contract so as to occupy the smallest possible surface area. It arises because the molecules at the surface experience a net inward cohesive pull from the molecules below them.
(b) Methods of reducing the surface tension of a liquid:
- Raising the temperature of the liquid.
- Adding an impurity such as soap or detergent to the liquid.
Tambaya 4 Rahoto
(a) Explain (i) refraction of a wave;
(ii) critical angle
(b) State two conditions necessary for (i) total internal reflection of a wave to occur
(ii) interference wave patterns to be formed
(c) The distance between two successive crests of a water wave travelling at 3.6ms\(^{-1}\) is 0.45m, calculate the frequency of the wave
(d) A ray of light is incident at an angle of 30° at an air-glass interface.
(i) Draw a ray diagram to show the deviation of the ray in the glass.
(ii) Determine the angle of deviation. [Refractive index of glass = 1.50]
(a)
(i) Refraction of a wave is the change in direction of a wave when it passes obliquely from one medium to another in which its speed is different.
(ii) Critical angle is the angle of incidence in the optically denser medium for which the angle of refraction in the less dense medium is \(90^\circ\).
(b)(i) Conditions for total internal reflection
- The wave must travel from an optically denser medium to an optically less dense medium.
- The angle of incidence in the denser medium must be greater than the critical angle.
(b)(ii) Conditions for formation of an interference pattern
- There must be two or more coherent wave sources, that is, sources of the same frequency having a constant phase difference.
- The sources must be close together, with their separation of the order of the wavelength, so that the waves overlap.
(c)
The distance between successive crests is the wavelength:
\[\lambda=0.45\,\text{m},\qquad v=3.6\,\text{m s}^{-1}\]
Using \(v=f\lambda\),
\[f=\frac{v}{\lambda}=\frac{3.6}{0.45}=8.0\,\text{Hz}.\]
(d)(i) The ray bends towards the normal as it enters the glass, which is optically denser than air.
(d)(ii)
By Snell's law,
\[n=\frac{\sin i}{\sin r}\]
\[\sin r=\frac{\sin30^\circ}{1.50}=\frac{0.5}{1.50}=0.3333\]
\[r=\sin^{-1}(0.3333)=19.5^\circ\]
Therefore, the angle of deviation is
\[d=i-r=30.0^\circ-19.5^\circ=10.5^\circ.\]
Bayanin Amsa
(a)
(i) Refraction of a wave is the change in direction of a wave when it passes obliquely from one medium to another in which its speed is different.
(ii) Critical angle is the angle of incidence in the optically denser medium for which the angle of refraction in the less dense medium is \(90^\circ\).
(b)(i) Conditions for total internal reflection
- The wave must travel from an optically denser medium to an optically less dense medium.
- The angle of incidence in the denser medium must be greater than the critical angle.
(b)(ii) Conditions for formation of an interference pattern
- There must be two or more coherent wave sources, that is, sources of the same frequency having a constant phase difference.
- The sources must be close together, with their separation of the order of the wavelength, so that the waves overlap.
(c)
The distance between successive crests is the wavelength:
\[\lambda=0.45\,\text{m},\qquad v=3.6\,\text{m s}^{-1}\]
Using \(v=f\lambda\),
\[f=\frac{v}{\lambda}=\frac{3.6}{0.45}=8.0\,\text{Hz}.\]
(d)(i) The ray bends towards the normal as it enters the glass, which is optically denser than air.
(d)(ii)
By Snell's law,
\[n=\frac{\sin i}{\sin r}\]
\[\sin r=\frac{\sin30^\circ}{1.50}=\frac{0.5}{1.50}=0.3333\]
\[r=\sin^{-1}(0.3333)=19.5^\circ\]
Therefore, the angle of deviation is
\[d=i-r=30.0^\circ-19.5^\circ=10.5^\circ.\]
Tambaya 5 Rahoto
A stone thrown horizontally from the top of a vertical ice is completely melted. Determine the total mass of wall with a velocity of 15ms\(^{-1}\), hits the horizontal ground water in the container at a point 45m from the base of the wall. Calculate the specific latent heat of steam = 2.3 x 10\(^3\) Jg\(^{-1}\) height of the wall. [g = 10ms\(^{-2}\)]
The text of this item is garbled by OCR (two separate questions have run together). The self-contained projectile part is answered below.
Problem: A stone is thrown horizontally from the top of a vertical wall with a velocity of \( 15\,\text{m s}^{-1} \) and strikes the horizontal ground at a point 45 m from the base of the wall. Find the height of the wall. \( (g = 10\,\text{m s}^{-2}) \)
Horizontal motion: the horizontal velocity is constant, so the time of flight is \[ t = \dfrac{\text{horizontal range}}{\text{horizontal velocity}} = \dfrac{45}{15} = 3\,\text{s}. \]
Vertical motion: the stone falls freely from rest vertically, so the height of the wall is \[ h = \tfrac{1}{2} g t^2 = \tfrac{1}{2} \times 10 \times 3^2 = 45\,\text{m}. \]
The height of the wall is 45 m.
Bayanin Amsa
The text of this item is garbled by OCR (two separate questions have run together). The self-contained projectile part is answered below.
Problem: A stone is thrown horizontally from the top of a vertical wall with a velocity of \( 15\,\text{m s}^{-1} \) and strikes the horizontal ground at a point 45 m from the base of the wall. Find the height of the wall. \( (g = 10\,\text{m s}^{-2}) \)
Horizontal motion: the horizontal velocity is constant, so the time of flight is \[ t = \dfrac{\text{horizontal range}}{\text{horizontal velocity}} = \dfrac{45}{15} = 3\,\text{s}. \]
Vertical motion: the stone falls freely from rest vertically, so the height of the wall is \[ h = \tfrac{1}{2} g t^2 = \tfrac{1}{2} \times 10 \times 3^2 = 45\,\text{m}. \]
The height of the wall is 45 m.
Tambaya 6 Rahoto
(a) Explain cations (b) Draw and label an electrolytic cell.
(a) Cations
Cations are positively charged ions. They form when an atom loses one or more electrons. For example:
\[ \text{Na} \rightarrow \text{Na}^{+} + e^{-} \]
In electrolysis, cations are attracted to the cathode, which is the negative electrode, because opposite charges attract. At the cathode, cations gain electrons. Examples of cations include \( \text{H}^{+} \), \( \text{Na}^{+} \), \( \text{Ag}^{+} \), and \( \text{Cu}^{2+} \).
(b) Labelled electrolytic cell
Bayanin Amsa
(a) Cations
Cations are positively charged ions. They form when an atom loses one or more electrons. For example:
\[ \text{Na} \rightarrow \text{Na}^{+} + e^{-} \]
In electrolysis, cations are attracted to the cathode, which is the negative electrode, because opposite charges attract. At the cathode, cations gain electrons. Examples of cations include \( \text{H}^{+} \), \( \text{Na}^{+} \), \( \text{Ag}^{+} \), and \( \text{Cu}^{2+} \).
(b) Labelled electrolytic cell
Tambaya 7 Rahoto
Explain wave-particle paradox
Wave-particle paradox (wave-particle duality). The wave-particle paradox is the observation that light (and, in general, radiation and matter) sometimes behaves as a wave and at other times behaves as a stream of particles, even though these two descriptions appear to be contradictory in classical physics.
Wave behaviour of light is shown by phenomena such as interference, diffraction and polarization, which can only be explained if light travels as a continuous wave.
Particle behaviour of light is shown by phenomena such as the photoelectric effect and the Compton effect, which can only be explained if light is made up of discrete packets of energy called photons, each of energy \( E = hf \).
No single experiment shows both behaviours at the same time: light behaves as a wave while it is being propagated, but as particles (photons) when it interacts with matter (emission or absorption). Louis de Broglie extended this idea to matter, proposing that moving particles such as electrons also have an associated wavelength \( \lambda = h/p \), which was later confirmed by electron diffraction. This dual nature, not resolvable by classical physics, is the wave-particle paradox.
Bayanin Amsa
Wave-particle paradox (wave-particle duality). The wave-particle paradox is the observation that light (and, in general, radiation and matter) sometimes behaves as a wave and at other times behaves as a stream of particles, even though these two descriptions appear to be contradictory in classical physics.
Wave behaviour of light is shown by phenomena such as interference, diffraction and polarization, which can only be explained if light travels as a continuous wave.
Particle behaviour of light is shown by phenomena such as the photoelectric effect and the Compton effect, which can only be explained if light is made up of discrete packets of energy called photons, each of energy \( E = hf \).
No single experiment shows both behaviours at the same time: light behaves as a wave while it is being propagated, but as particles (photons) when it interacts with matter (emission or absorption). Louis de Broglie extended this idea to matter, proposing that moving particles such as electrons also have an associated wavelength \( \lambda = h/p \), which was later confirmed by electron diffraction. This dual nature, not resolvable by classical physics, is the wave-particle paradox.
Tambaya 8 Rahoto
(a) (i) Explain (I) electric potential; (II) electric potential energy
(ii) State the SI unit of each of the term in (a)(i) above
(b) An isolated electrically charged sphere of radius, r, and charge, Q, is supported on an insulator in air of permitivity, \(\varepsilon_o\). Write down;
(i) an expression for the electric field intensity on the surface of the sphere;
(ii) an expression for the electric potential at the surface of the sphere;
(iii) a relationship between the electric field intensity and the electric potential at the surface of the sphere
(c) The plates of a parallel plate capacitor, 5.0 x 10\(^{-3}\) m apart are maintained at a potential difference of 5.0 x 10\(^{4}\) V. Calculate the magnitude f the
(i) electric field intensity between the plates
(ii) force on the electron
(iii) acceleration of the elctron
[electronic charge = 1.60 x 10\(^{-19}\)C, mass of electron = 9.1 \times 10\(^{-31}\) kg]
(a)(i)(I) Electric potential at a point in an electric field is the work done in bringing a unit positive charge from infinity to that point.
(a)(i)(II) Electric potential energy of a charge at a point is the work done in bringing that charge from infinity to the point in the electric field.
(a)(ii) SI units: electric potential is measured in the volt (V); electric potential energy is measured in the joule (J).
(b)(i) Electric field intensity at the surface of the charged sphere: \[ E = \dfrac{Q}{4\pi \varepsilon_o r^2}. \]
(b)(ii) Electric potential at the surface of the sphere: \[ V = \dfrac{Q}{4\pi \varepsilon_o r}. \]
(b)(iii) Relationship between them: \[ E = \dfrac{V}{r}\quad(\text{i.e. } V = E r). \]
(c)(i) Electric field intensity between the plates: \[ E = \dfrac{V}{d} = \dfrac{5.0 \times 10^{4}}{5.0 \times 10^{-3}} = 1.0 \times 10^{7}\,\text{V m}^{-1}. \]
(c)(ii) Force on the electron: \[ F = eE = 1.60 \times 10^{-19} \times 1.0 \times 10^{7} = 1.6 \times 10^{-12}\,\text{N}. \]
(c)(iii) Acceleration of the electron: \[ a = \dfrac{F}{m} = \dfrac{1.6 \times 10^{-12}}{9.1 \times 10^{-31}} = 1.76 \times 10^{18}\,\text{m s}^{-2}. \]
Bayanin Amsa
(a)(i)(I) Electric potential at a point in an electric field is the work done in bringing a unit positive charge from infinity to that point.
(a)(i)(II) Electric potential energy of a charge at a point is the work done in bringing that charge from infinity to the point in the electric field.
(a)(ii) SI units: electric potential is measured in the volt (V); electric potential energy is measured in the joule (J).
(b)(i) Electric field intensity at the surface of the charged sphere: \[ E = \dfrac{Q}{4\pi \varepsilon_o r^2}. \]
(b)(ii) Electric potential at the surface of the sphere: \[ V = \dfrac{Q}{4\pi \varepsilon_o r}. \]
(b)(iii) Relationship between them: \[ E = \dfrac{V}{r}\quad(\text{i.e. } V = E r). \]
(c)(i) Electric field intensity between the plates: \[ E = \dfrac{V}{d} = \dfrac{5.0 \times 10^{4}}{5.0 \times 10^{-3}} = 1.0 \times 10^{7}\,\text{V m}^{-1}. \]
(c)(ii) Force on the electron: \[ F = eE = 1.60 \times 10^{-19} \times 1.0 \times 10^{7} = 1.6 \times 10^{-12}\,\text{N}. \]
(c)(iii) Acceleration of the electron: \[ a = \dfrac{F}{m} = \dfrac{1.6 \times 10^{-12}}{9.1 \times 10^{-31}} = 1.76 \times 10^{18}\,\text{m s}^{-2}. \]
Tambaya 9 Rahoto
(a) State two;
(i) differences between nuclear fusion and nuclear fission;
(ii) peaceful uses of atomic energy
(b)(i) Explain chain reaction
(ii) State (I) one condition necessary for chain reaction to occur.
(II) two components in a nuclear reactor used to control chain reaction.
(c)(i) A nuclear reaction is given \(^2_1H + ^3_1H = ^4_0n\) + energy
What type of nuclear reaction is it?
(ii) The isotope of a nuclide has a half life of 5.40 x 10\(^3\) s, Calculate its decay constant.
(a)(i) Two differences between nuclear fusion and nuclear fission:
| Nuclear fusion | Nuclear fission |
|---|---|
| Two light nuclei join to form a heavier nucleus. | A heavy nucleus splits into two lighter nuclei. |
| Requires extremely high temperature and pressure to occur. | Can be started at ordinary temperature by bombardment with a neutron. |
(a)(ii) Two peaceful uses of atomic energy:
- Generation of electricity in nuclear power stations.
- Production of radioisotopes used in medicine (radiotherapy) and in agriculture/industry (tracers).
(b)(i) Chain reaction: a self-sustaining nuclear reaction in which the neutrons released by the fission of one nucleus go on to cause the fission of further nuclei, each of which releases still more neutrons, so that the process continues on its own.
(b)(ii)(I) A necessary condition: there must be at least a critical mass of the fissile material present.
(b)(ii)(II) Two control components in a nuclear reactor: control rods (of boron or cadmium) that absorb excess neutrons, and a moderator (such as graphite or heavy water) that slows down the neutrons.
(c)(i) The reaction \( {}^{2}_{1}\text{H} + {}^{3}_{1}\text{H} \rightarrow {}^{4}_{2}\text{He} + {}^{1}_{0}n + \text{energy} \) is a nuclear fusion reaction (light nuclei combining).
(c)(ii) Decay constant: \[ \lambda = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{5.40 \times 10^{3}} = 1.28 \times 10^{-4}\,\text{s}^{-1}. \]
Bayanin Amsa
(a)(i) Two differences between nuclear fusion and nuclear fission:
| Nuclear fusion | Nuclear fission |
|---|---|
| Two light nuclei join to form a heavier nucleus. | A heavy nucleus splits into two lighter nuclei. |
| Requires extremely high temperature and pressure to occur. | Can be started at ordinary temperature by bombardment with a neutron. |
(a)(ii) Two peaceful uses of atomic energy:
- Generation of electricity in nuclear power stations.
- Production of radioisotopes used in medicine (radiotherapy) and in agriculture/industry (tracers).
(b)(i) Chain reaction: a self-sustaining nuclear reaction in which the neutrons released by the fission of one nucleus go on to cause the fission of further nuclei, each of which releases still more neutrons, so that the process continues on its own.
(b)(ii)(I) A necessary condition: there must be at least a critical mass of the fissile material present.
(b)(ii)(II) Two control components in a nuclear reactor: control rods (of boron or cadmium) that absorb excess neutrons, and a moderator (such as graphite or heavy water) that slows down the neutrons.
(c)(i) The reaction \( {}^{2}_{1}\text{H} + {}^{3}_{1}\text{H} \rightarrow {}^{4}_{2}\text{He} + {}^{1}_{0}n + \text{energy} \) is a nuclear fusion reaction (light nuclei combining).
(c)(ii) Decay constant: \[ \lambda = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{5.40 \times 10^{3}} = 1.28 \times 10^{-4}\,\text{s}^{-1}. \]
Tambaya 10 Rahoto
(a) Define Young modulus of elasticity;
(b) A spiral spring extends from a length of 10.0 cm to 10.01 cm when a force of 20 N is applied on it. Calculate the force constant of the spring.
(a) Young modulus of elasticity is the ratio of the tensile stress to the tensile strain of a material, provided the elastic limit is not exceeded.
\[ E = \dfrac{\text{tensile stress}}{\text{tensile strain}} = \dfrac{F/A}{e/l} = \dfrac{F l}{A e}. \]
(b) The force constant is \( k = \dfrac{F}{e} \), where e is the extension.
Extension: \[ e = 10.01 - 10.00 = 0.01\,\text{cm} = 1.0 \times 10^{-4}\,\text{m}. \]
\[ k = \dfrac{F}{e} = \dfrac{20}{1.0 \times 10^{-4}} = 2.0 \times 10^{5}\,\text{N m}^{-1}. \]
The force constant of the spring is \( 2.0 \times 10^{5}\,\text{N m}^{-1} \).
Bayanin Amsa
(a) Young modulus of elasticity is the ratio of the tensile stress to the tensile strain of a material, provided the elastic limit is not exceeded.
\[ E = \dfrac{\text{tensile stress}}{\text{tensile strain}} = \dfrac{F/A}{e/l} = \dfrac{F l}{A e}. \]
(b) The force constant is \( k = \dfrac{F}{e} \), where e is the extension.
Extension: \[ e = 10.01 - 10.00 = 0.01\,\text{cm} = 1.0 \times 10^{-4}\,\text{m}. \]
\[ k = \dfrac{F}{e} = \dfrac{20}{1.0 \times 10^{-4}} = 2.0 \times 10^{5}\,\text{N m}^{-1}. \]
The force constant of the spring is \( 2.0 \times 10^{5}\,\text{N m}^{-1} \).
Tambaya 11 Rahoto
A ball is projected specific latent heat of ice = 3.4 x 10\(^{2}\) Jg\(^{-1}\) horizontally from a height of 20m above the ground specific heat capacity of water = 4.2 Jg\(^{-1}\) K\(^{-1}\) with an initial velocity of 0.4ms\(^{-1}\). Calculate the horizontal distance moved by the ball before hitting the ground. [g = 10ms\(^{-1}\)]
The stem contains OCR noise (latent-heat constants merged in). The self-contained projectile part is answered.
Problem: A ball is projected horizontally from a height of 20 m above the ground with an initial velocity of \( 0.4\,\text{m s}^{-1} \). Find the horizontal distance moved before it hits the ground. \( (g = 10\,\text{m s}^{-2}) \)
Time of flight (vertical fall from rest): \[ h = \tfrac{1}{2} g t^2 \Rightarrow t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2 \times 20}{10}} = \sqrt{4} = 2\,\text{s}. \]
Horizontal distance (constant horizontal velocity): \[ x = u \times t = 0.4 \times 2 = 0.8\,\text{m}. \]
The horizontal distance moved is 0.8 m.
Bayanin Amsa
The stem contains OCR noise (latent-heat constants merged in). The self-contained projectile part is answered.
Problem: A ball is projected horizontally from a height of 20 m above the ground with an initial velocity of \( 0.4\,\text{m s}^{-1} \). Find the horizontal distance moved before it hits the ground. \( (g = 10\,\text{m s}^{-2}) \)
Time of flight (vertical fall from rest): \[ h = \tfrac{1}{2} g t^2 \Rightarrow t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2 \times 20}{10}} = \sqrt{4} = 2\,\text{s}. \]
Horizontal distance (constant horizontal velocity): \[ x = u \times t = 0.4 \times 2 = 0.8\,\text{m}. \]
The horizontal distance moved is 0.8 m.
Tambaya 12 Rahoto
Describe, with the aid of a diagram, how a wave can be plane polarized.
Plane polarization is possible only for a transverse wave. In an unpolarized light wave, the vibrations occur in many different directions, all perpendicular to the direction in which the wave travels. A plane-polarized wave has vibrations restricted to one direction only; equivalently, the vibrations lie in one plane containing the direction of travel.
Passing unpolarized light through a polaroid produces plane-polarized light. The polaroid has a transmission axis and transmits only the component of vibration parallel to that axis. Vibrations in other directions are absorbed or blocked, so the emerging wave vibrates in one plane.
Examination point: Do not say that the light travels in only one direction after polarization: it was already travelling in one direction. Polarization restricts the direction of vibration, not the direction of travel.
Bayanin Amsa
Plane polarization is possible only for a transverse wave. In an unpolarized light wave, the vibrations occur in many different directions, all perpendicular to the direction in which the wave travels. A plane-polarized wave has vibrations restricted to one direction only; equivalently, the vibrations lie in one plane containing the direction of travel.
Passing unpolarized light through a polaroid produces plane-polarized light. The polaroid has a transmission axis and transmits only the component of vibration parallel to that axis. Vibrations in other directions are absorbed or blocked, so the emerging wave vibrates in one plane.
Examination point: Do not say that the light travels in only one direction after polarization: it was already travelling in one direction. Polarization restricts the direction of vibration, not the direction of travel.
Tambaya 13 Rahoto
(a) Explain specific latent heat
(b)(i) Describe how the specific latent heat of fusion of ice can be determined by the method of mixtures.
(ii) State two precautions to be taken to ensure accurate results.
(c) Steam, at 100°C, is passed into a container of negligible heat capacity containing 20 g of ice and 100 g of water at 0°C, until the ice is completely melted. Determine the total mass of water in the container. [Specific latent heat of steam = 2.3 x 10\(^3\) Jg\(^{-1}\), specific latent heat of ice = 3.4 x 10\(^{2}\) Jg\(^{-1}\), specifit heat capacity of water = 4.2 Jg\(^{-1}\) K\(^{-1}\)]
(a) Specific latent heat. The specific latent heat of a substance is the quantity of heat required to change the state of unit mass (1 kg or 1 g) of the substance, at constant temperature, without any change in its temperature (fusion = solid to liquid; vaporization = liquid to vapour).
(b)(i) Determination of specific latent heat of fusion of ice by method of mixtures. Pour a known mass \( m_w \) of warm water at a measured temperature \( \theta_1 \) into a calorimeter of known mass and specific heat capacity. Take small pieces of dry, melting ice at \( 0^\circ\text{C} \), add them to the water and stir until all the ice just melts. Measure the final steady temperature \( \theta_2 \) and, by re-weighing, find the mass \( m_i \) of ice added. Then, heat lost by the warm water and calorimeter equals heat used to melt the ice plus heat used to warm the melted ice from \( 0^\circ\text{C} \) to \( \theta_2 \):
\[ (m_w c_w + m_c c_c)(\theta_1 - \theta_2) = m_i L + m_i c_w (\theta_2 - 0). \]
From this equation the specific latent heat of fusion of ice, L, is calculated.
(ii) Precautions:
- Dry the ice with blotting paper before adding it, to remove surface water.
- Lag (insulate) the calorimeter and stir gently, to reduce heat exchange with the surroundings.
(c) Calculation. Heat needed to melt the 20 g of ice: \[ Q = m_i L_{ice} = 20 \times 340 = 6800\,\text{J}. \]
Let the mass of steam that condenses be x grams. The steam condenses at \( 100^\circ\text{C} \) and the resulting water cools to \( 0^\circ\text{C} \), supplying the heat that melts the ice: \[ x\,L_{steam} + x\,c_w(100 - 0) = 6800 \] \[ x(2300) + x(4.2)(100) = 6800 \] \[ x(2300 + 420) = 6800 \Rightarrow x = \dfrac{6800}{2720} = 2.5\,\text{g}. \]
Total mass of water in the container = original water + melted ice + condensed steam: \[ 100 + 20 + 2.5 = 122.5\,\text{g}. \]
Bayanin Amsa
(a) Specific latent heat. The specific latent heat of a substance is the quantity of heat required to change the state of unit mass (1 kg or 1 g) of the substance, at constant temperature, without any change in its temperature (fusion = solid to liquid; vaporization = liquid to vapour).
(b)(i) Determination of specific latent heat of fusion of ice by method of mixtures. Pour a known mass \( m_w \) of warm water at a measured temperature \( \theta_1 \) into a calorimeter of known mass and specific heat capacity. Take small pieces of dry, melting ice at \( 0^\circ\text{C} \), add them to the water and stir until all the ice just melts. Measure the final steady temperature \( \theta_2 \) and, by re-weighing, find the mass \( m_i \) of ice added. Then, heat lost by the warm water and calorimeter equals heat used to melt the ice plus heat used to warm the melted ice from \( 0^\circ\text{C} \) to \( \theta_2 \):
\[ (m_w c_w + m_c c_c)(\theta_1 - \theta_2) = m_i L + m_i c_w (\theta_2 - 0). \]
From this equation the specific latent heat of fusion of ice, L, is calculated.
(ii) Precautions:
- Dry the ice with blotting paper before adding it, to remove surface water.
- Lag (insulate) the calorimeter and stir gently, to reduce heat exchange with the surroundings.
(c) Calculation. Heat needed to melt the 20 g of ice: \[ Q = m_i L_{ice} = 20 \times 340 = 6800\,\text{J}. \]
Let the mass of steam that condenses be x grams. The steam condenses at \( 100^\circ\text{C} \) and the resulting water cools to \( 0^\circ\text{C} \), supplying the heat that melts the ice: \[ x\,L_{steam} + x\,c_w(100 - 0) = 6800 \] \[ x(2300) + x(4.2)(100) = 6800 \] \[ x(2300 + 420) = 6800 \Rightarrow x = \dfrac{6800}{2720} = 2.5\,\text{g}. \]
Total mass of water in the container = original water + melted ice + condensed steam: \[ 100 + 20 + 2.5 = 122.5\,\text{g}. \]
Tambaya 14 Rahoto
State (a) the principle upon which the lightening in fluorescent tubes operate;
(b) two factors on which the colour of light from a fluorescent tube depends.
(a) Principle of operation of a fluorescent tube. When an electric discharge is passed through mercury vapour at low pressure inside the tube, the mercury atoms are excited and emit ultraviolet radiation. This ultraviolet radiation strikes the fluorescent powder (phosphor) coated on the inside wall of the tube, causing it to fluoresce, that is, to absorb the ultraviolet and re-emit the energy as visible light.
(b) Factors on which the colour of the light depends:
- The nature (chemical composition) of the fluorescent powder (phosphor) coating the inside of the tube.
- The type of gas or vapour used in the tube (and its pressure).
Bayanin Amsa
(a) Principle of operation of a fluorescent tube. When an electric discharge is passed through mercury vapour at low pressure inside the tube, the mercury atoms are excited and emit ultraviolet radiation. This ultraviolet radiation strikes the fluorescent powder (phosphor) coated on the inside wall of the tube, causing it to fluoresce, that is, to absorb the ultraviolet and re-emit the energy as visible light.
(b) Factors on which the colour of the light depends:
- The nature (chemical composition) of the fluorescent powder (phosphor) coating the inside of the tube.
- The type of gas or vapour used in the tube (and its pressure).
Tambaya 15 Rahoto
Using the kinetic theory of matter, explain the definite structure of solids.
Definite structure of solids (kinetic theory). According to the kinetic theory of matter, the molecules of a solid are very closely packed together in a regular, orderly arrangement (a lattice) and are held in position by strong intermolecular forces of attraction.
Because these forces are strong, the molecules cannot move about freely from place to place; they can only vibrate to and fro about fixed mean positions. Since each molecule keeps to a fixed position within the ordered lattice, the solid maintains a fixed arrangement of its particles.
This fixed, orderly packing gives a solid both a definite shape and a definite volume, that is, a definite structure.
Bayanin Amsa
Definite structure of solids (kinetic theory). According to the kinetic theory of matter, the molecules of a solid are very closely packed together in a regular, orderly arrangement (a lattice) and are held in position by strong intermolecular forces of attraction.
Because these forces are strong, the molecules cannot move about freely from place to place; they can only vibrate to and fro about fixed mean positions. Since each molecule keeps to a fixed position within the ordered lattice, the solid maintains a fixed arrangement of its particles.
This fixed, orderly packing gives a solid both a definite shape and a definite volume, that is, a definite structure.
