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2022 HSC Physics Exam Paper Solutions

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Year 12 physics term, vce physics units 3 and 4.

The 2022 HSC Physics exam was held on 3 November. How did you go? Check your answers with the 2022 HSC Physics Exam Paper Solutions written by the Matrix Science Team!

Find the full 2022 HSC Physics Exam Paper Questions, here.

Section 1. Multiple Choice

Question 21

X has a greater surface temperature than Y (surface temperature is highest on the left of the diagram).

Y has a greater luminosity than X (luminosity increases upwards on the diagram)

X is performing fusion of hydrogen into helium in the core.

Y is fusing helium atoms into carbon.

Question 22

The continuous iron core of the transformer is a ferromagnetic material. This ensures the magnetic flux produced by the primary coil is transferred without leakage to the secondary coil. Induced EMF in the secondary is therefore the maximum possible amount.

The laminations of the core prevent large eddy currents from being induced in the transformer core. This improves efficiency by reducing energy dissipation.

Question 23

This question has multiple possible answers. Two examples are given below.

Cogwheel experiment

A beam of light is passed through the gaps between the cogs of a cogwheel. It reflects off a mirror placed some distance away and returns, passing through the gaps again and on to a screen.

If the wheel is rotated at such a speed that it moves half the distance between the cogs in the time it takes for the light to reflect and return, then the light will strike a cog upon its return and will be blocked from the screen.

The rotation speed of the wheel required to block light from the screen can be used to calculate the time of the light’s journey to and from the mirror. This can be combined with the distance to the mirror, \(d\), to find the speed of light, \(c = \frac{2d}{t}\).

Errors would arise from measurements of the cogwheel’s rotation speed, particularly if the beam width was narrower than the cogs, and a range of rotation speeds were able to block the light.

Cavity resonances

A microwave beam of a known frequency can be produced. The beam can be directed to the region between two mirrors, i.e. a cavity, such that it reflects back and forth between them.

The distance between the two mirrors and/or the frequency can be adjusted so as to form a standing wave.

The distance between the mirrors and the order of the standing wave can be determined and used to calculate the wavelength. This can be combined with the known frequency and used to calculate the wavelength, \(c = f\lambda\).

Errors would arise from measurements of the distance between the mirrors.

Question 24

The half-life of the substance is 16 hours.

Therefore the decay constant in \(\text{h}^{-1}\) can be calculated via \( \lambda = \frac{\text{ln } 2}{t_{1/2}} = 0.043 \text{ h}^{-1}\)

Using \( N = N_0e^{-\lambda t}\)

\(N = 8 \text{ }\mu \text{g}\) \(N_0 = 80 \text{ }\mu \text{g}\) \(\lambda = 0.043 \text{ h}^{-1}\)

Solve for \(t\) in hours: \(t = 53.15 \text{ hours}\)

Question 25

Calculate the kinetic energy at points 1 and 2 using \(K = \frac{1}{2}mv^2\)

\(K_1 = \frac{1}{2} \times 200 \times 5500^2 = 3.025 \times 10^9 \text{ J}\) \(K_2 = \frac{1}{2} \times 200 \times 2900^2 = 0.841 \times 10^9 \text{ J}\)

This gives the change in energy \(\Delta K = -2.18 \times 10^9 \text{ J} \approx -2.2 \times 10^9 \text{ J} \), which has the required magnitude.

We can equate the loss of kinetic energy to an increase in potential energy.

We know the rocket mass and the two positions relative to the planet’s centre.

Solving for the mass of the planet gives \(M = 8.5 \times 10^{23} \text{ kg} \)

Question 26

The work function \( \phi = 2.9 \text{ eV}\) equates to an energy of \(4.65 \times 10^{-19} \text{ J}\) that is lost by the electron as it escapes the metal.

The maximum kinetic energy of the photoelectron is: \(K = hf – \phi = 6.626 \times 10^{-34} \times 7.5 \times 10^{14}  – 4.65 \times 10^{-19} = 3.2 \times 10^{-20} \text{ J}\)

As the electron leaves the metal, it experiences a force downwards due to the electric field (it also experiences a gravitational force but this is so small we can ignore it).

This force will do negative work reducing the kinetic energy of the electron. Once the kinetic energy reduces to zero, the electron will stop. We can solve for the displacement required to achieve this:

\(W = F \times d = qE \times d = 1.602 \times 10^{-19} \times 5.2 \times d = 3.2 \times 10^{-20}\) \(d = 0.038 \text{ m}\)

Question 27

Newton considered light to be made of particles called corpuscles. Therefore the expected pattern would be two strips on the screen where the particles passed through the slits.

The angles at which the bright lines appear satisfy \(d \sin \theta = m \lambda \) In this case we are looking at the second order fringe, so \( m=2 \). \(d = 5.0 \times 10^{-5} \text{ m}\) \(\lambda = 655 \times 10^{-9} \text{ m}\) Hence \(\theta = 1.50^\circ \)

Using light with a shorter wavelength causes the angles between bright fringes to become smaller according to \(d \sin \theta = m \lambda \)

The bright lines would appear closer together on the screen.

Question 28

The energy released in each step depends on the mass defect associated with the reaction. Step Y involves a greater amount of mass converted to energy as calculated here:

Mass of reactants \(= 13.003 + 1.007 = 14.010 \text{ u}\)

Mass of products \(= 14.003 \text{ u}\)

Mass defect \( = 0.007 \text{ u}\)

At a rate of \( 931.5 \text{ MeV} / \text{c}^2 \), the energy released in Step Y is \( 6.5 \text{ MeV}\).

Question 29

We will describe the motion as seen from the ground.

The apple initially has zero vertical velocity. Its horizontal velocity at launch is  the sum of launch velocity to the East and the car’s velocity to the North.

The only force on the apple after it is launched is the gravitational force, acting vertically downwards. There are no forces in the horizontal plane, so the horizontal acceleration is zero and the apple will maintain its horizontal velocity.

The constant vertical acceleration due to gravity causes the velocity to increase downwards at a constant rate until the apple hits the ground.

Question 30

Einstein postulated that the speed of light should have the same value relative to all observers, regardless of which frame of reference they are in.

The observer inside the train sees the light travel a certain round trip distance from X to Y and back again. The observer outside the train sees the light travel distance as being longer due to the relative horizontal motion of the train.

Both observers agree on the speed of light, so they must disagree on the time interval during which the light was travelling. This is the basis for Einstein’s prediction of time dilation.

The observer inside the train measures the proper time for the journey, which is given as \(15 \text{ ns}\). Therefore the observer outside will measure a dilated time.

\(t_v = \frac{t_0}{\sqrt{1-v^2/c^2}}\) \(t_v = 53.6 \text{ ns}\)

Question 31

Before the Geiger-Marsden experiment, the main model of the atom was Thomson’s “plum pudding” model in which negatively charged electrons were embedded in a solid cloud of positively charged matter that made up the atom.

The Geiger-Marsden experiment involved firing alpha particles at a thin gold foil. Most of the alpha particles passed through the foil without any deflection, but some were deflected, with a few being deflected through very large angles. Rutherford concluded from these results that the positive charge in an atom was concentrated in a very small region at the centre which he referred to as the nucleus. The nucleus exerted repulsive forces on the positively charged alpha particles, and these forces had a greater magnitude the closer the alpha particle passed by.

Rutherford described the electrons as orbiting the nucleus in a ‘planetary’ fashion. However, this model could not account for the fixed wavelengths emitted by hydrogen atoms. Bohr proposed that the electrons orbited the nucleus in quantised orbits at fixed energies. One of his postulates was that electrons could transition between orbits as long as the exact energy difference was absorbed or emitted in the form of a photon with energy \(E = hf\).

Hence the line spectrum of hydrogen corresponded to the allowed electron transitions to lower energies that gave visible light frequencies.

Rydberg’s equation, \(\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} – \frac{1}{n_i^2}\right)\) predicts the wavelengths emitted by hydrogen. The Balmer series, with \(n_f = 2\), gives the wavelengths shown in the diagram. For example, substituting in \(n_i = 3\) gives \(\lambda = 656 \text{ nm}\), corresponding to the rightmost (longest wavelength) line in the spectrum.

Question 32

The magnets lead to electromagnetic braking. As the conductive metal of the wheel passes between the magnets, eddy currents are induced in the metal due to Faraday’s Law. The induced magnetic flux interacts with the flux of the magnets to oppose the rotation of the flywheel, according to Lenz’s law, resulting in an opposing torque.

The magnets should be moved closer to the outer rim of the flywheel, this will produce a larger opposing torque, \(\tau = Fr\). The increased distance of the magnet from the centre of rotation increases \(r\).

The metal wheel is moving faster near the rim, so the rate of change of flux it experiences will increase, and so will the induced EMF (by Faraday’s Law) and the current. The opposing force will consequently increase.

Question 33

a) The projectile’s speed can be found by analysing its motion in a circle. \(v = \frac{2\pi r}{T} = 20.1 \text{ ms}^{-1}\)

If it is released when the direction is \(45^\circ\) to the horizontal, we can decompose this total velocity into components. The vertical component of the velocity is then \(20.1 \sin 45^\circ = 14.2 \text{ ms}^{-1}\) as required.

b) The projectile’s initial horizontal velocity is \(20.1 \sin 45^\circ = 14.2 \text{ ms}^{-1}\). To find the horizontal range we must find the time of flight. From release to landing: \(s_y = -1.2 \text{ m}\) \(a_y = -9.8 \text{ms}^{-2}\) \(u_y = 14.2 \text{ ms}^{-1}\)

Solving \(s = ut + \frac{1}{2}at^2\) for time and choosing the positive solution gives \( t = 2.98 \text{ s}\)

The horizontal range of the projectile is \(s_x = u_x t = 42.3 \text{ m}\).

Question 34

The force exerted on a particle in a magnetic field is \(F = qvB \sin \theta\). In the case of these particles, they all enter the field perpendicular to the field lines and so we can simplify this to \(F = qvB\).

Magnetic forces will act as a centripetal force in this situation, as they will exert force with constant magnitude that is always at right angles to the motion of the charged particle. The magnetic force can be equated to the centripetal force \(F_c = \frac{mv^2}{r}\) in order to derive an expression for the radius of the circular paths:

\(R = \frac{mv}{qB}\).

The three particles all enter the field with the same speed, and the field is uniform, so all three particles will have the same values of \(v\) and \(B\). Any differences between the particle trajectories will therefore be due to differences in their mass to charge ratio, \(m/q\), and the signs of the charges.

Particle Z is the only one to experience an initial force downwards. The Right Hand rule predicts that it must therefore be a negative charge.

Particle X initially deflects upwards, so by contrast it must be a positive charge. Both X and Z show the same path radius, so we can conclude that they have the same mass-to-charge ratio. Without more information we cannot determine whether their charges have the same magnitude.

Particle Y is also positive according to the Right Hand rule, but its radius of travel is approximately double that of particle X. This means its mass to charge ratio is double that of particle X. Again, without further information we cannot state definitively which of its properties, charge or mass, are different.

Question 35

If the hypothesis is true, then the force due to gravity between the capsule and the ISS will be equal to the centripetal force \(F_c = \frac{mv^2}{r}\) required to maintain the path shown.

Gravitational force: \(F_G = \frac{GMm}{r^2} = 8.40 \times 10^{-6} \text{ N}\)

Centripetal force required: \(F_c = \frac{mv^2}{r}= 3.26 \text{ N}\)

The gravitational force between the two objects is not sufficient to provide the necessary centripetal force. Therefore the hypothesis is incorrect and there must be additional forces present to achieve this motion.

Written by Matrix Science Team

© Matrix Education and www.matrix.edu.au, 2023. Unauthorised use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Matrix Education and www.matrix.edu.au with appropriate and specific direction to the original content.

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HSC Physics Module 7 Practice Questions with Solutions

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Test your exam readiness with these 10 Must Know Physics Module 7 questions

HSC Physics Module 7 The Nature of Light is considered by many students as the most interesting module in the Physics course. It explores the development of quantum mechanics and the “strangeness” of the Special Theory of Relativity.

Students find Special Relativity fascinating but difficult to comprehend.

In this article, we reveal:

• 11 important question types from the HSC Physics Module 7 Syllabus
• 10 Must Know questions to ace your next HSC Physics Module 7 Exam

What type of questions are commonly asked in HSC Physics Module 7?

11 commonly asked question types from the HSC Physics Module 7 Syllabus are listed below:

Source: NESA website

10 Must Know Questions for HSC Physics Module 7

Question 1 (5 marks).

A student conducts an experiment to investigate Malus’ law. Their light source is a laser that emits light polarised at an unknown angle. To conduct this experiment the student uses a detector with a light sensor and angle sensor.

The light sensor measures the intensity I of incident light. In front of the light sensor there is a polariser, which can be rotated or removed. The angle sensor measures the angle \theta of the polariser.

A schematic diagram of the apparatus is shown below, with the polariser currently in place and vertical at \theta = 0 \degree .

The student records the following data from conducting their experiment:

See Question 1 Solution

Question 2 (4 marks)

An experimental setup to demonstrate Young’s double slit experiment is shown below.

Monochromatic light is passed through a double slit and the interference pattern is projected onto a screen. The distance between the central bright fringe and the bright fringes adjacent to it is 10 \ cm .

See Question 2 Solution

Question 3 (3 marks)

The diagram below shows an AC voltage connected to the wires of an antenna. This is a common technique for producing radio waves.

See Question 3 Solution

Question 4 (7 marks)

A plot of some experimental blackbody radiation spectra at different temperatures is shown in the diagram below, together with a spectrum predicted by classical theory.

See Question 4 Solution

Question 5 (3 marks)

Spectroscopy can be used to determine a large amount of information about objects in the galaxy. The following spectrum of a star was recorded.

Explain what information about the star can be determined from this spectrum.

See Question 5 Solution.

Question 6 (6 marks)

Light is incident on a Zirconium surface in a vacuum. The graph below shows the variation of the stopping voltage V_{s} of the electrons emitted from the surface with the frequency f of the incident light.

See Question 6 Solution

Question 7 (3 marks)

The principle of relativity was proposed by Galileo Galilei in 1632.

See Question 7 Solution.

Question 8 (6 marks)

A spacecraft leaves Earth at a speed of 0.80 \ c as measured by an observer on Earth. It heads towards, and continues beyond, a distant planet in the star system Gliese 3325. The planet is 30 light years (ly) away from Earth as measured by an observer on Earth.

When the spacecraft leaves Earth, Emmet, one of the astronauts in the spacecraft, is 20 years old.

See Question 8 Solution

Question 9 (5 marks)

A deuteron (which consists of 1 proton and 1 neutron) is accelerated from rest to a speed of 0.8 \ c in a particle accelerator.

• Mass of a proton: 1.673 \times 10^{-27} \ kg
• Mass of a neutron: 1.675 \times 10^{-27} \ kg

See Question 9 Solution

Question 10 (6 marks)

Around 1850 a French scientist Leon Foucault performed an experiment to determine whether light travels faster in air or in water. His experimental setup is shown below.

In Foucault’s experiment a beam of light is reflected off mirror R to mirror M, back to mirror R, and then to position 1 on a screen, as shown in the left diagram.

Mirror R is then spun clockwise at a high rate, as shown in the middle diagram. Due to the rotation of mirror R during the flight time of the beam of light, the returning beam is reflected to a displaced position on the screen, position 2.

In the third phase of the experiment a 3 m long tube of water is placed between mirrors M and R, in the path of the beam of light, as shown in the diagram on the right.

See Question 10 Solution

Solutions to HSC Physics Module 7 Practice Questions

Detailed, step-by-step solutions to the Module 5 Advanced Mechanics questions are provided below.

Marking practice exams is just as important as answering the questions

Question 1 Solution

• Direct the laser beam into the detector. Measure the laser intensity in the absence of the polariser to obtain the control variable I_{max} .
• Put the polariser in place and rotate it to specific angles, recording the measured laser intensity for each angle of the polariser. Use at least 5 different values of \theta .
• Analyse the data, specifically by plotting a graph of \frac{I}{I_{max}} vs cos^2 \theta .
• Malus’ Law is I = I_{max} cos^2 \theta , which will be verified if the graph produced in Step 3 exhibits a gradient of 1.

Given the laser intensity is I_{max} = 5.00 \ Wm^{-2} , the ratio \frac{I}{I_{max}} can be found from the intensity measurements recorded in the experiment.

Assuming Malus’ Law, \theta can be found as follows:

\begin{aligned} \dfrac{I}{I_{max}} &= \cos^2 \theta \\\\ \cos \theta &= \sqrt{\dfrac{I}{I_{max}}} \\\\ \therefore \theta &= \cos^{-1} \bigg(\sqrt{\dfrac{I}{I_{max}}}\bigg)  \end{aligned}

Calculating (\theta) for each measured intensity yields:

Step 1: Plot a graph of \dfrac{I}{I_{max}} \ vs \ \cos^2 \theta . Draw a line of best fit through your data.

Step 2: Calculate the gradient of the line of best fit.

The gradient of the line of best fit is calculated from m = \frac{rise}{run} :

\begin{aligned} m &= \dfrac{rise}{run} \\\\ &= \dfrac{0.6-0.4}{0.6-0.4} \\\\ &= \dfrac{0.2}{0.2} \\\\ &= 1 \end{aligned}

Therefore, y = x and hence \dfrac{I}{I_{max}} = \cos^2 \theta . Rearranging this equation, I = I_{max} \cos^2 \theta . This is Malus’ Law, which is thus verified by this experiment.

Back to Question 1

Question 2 Solution

For the bright fringe adjacent to the central bright fringe, m = 1 , in the relationship:

  d\sin \theta = m \lambda

For a small \theta , \sin \theta = \tan \theta= \frac{h}{L} :

\begin{aligned}\therefore \frac{dh}{L} &= \lambda \\\\ h &= \frac{ \lambda L}{d} \end{aligned}

The separation on the screen of the central bright fringe and the adjacent bright fringes can be increased from 10 \ cm to 30 \ cm by:

• Moving the screen to three times the initial distance, 3L .
• Increasing the wavelength of light to three times the initial wavelength, 3\lambda .
• Decreasing the slit spacing to one third of the initial spacing, \frac{1}{3}d .

Note that there are THREE possible changes that could be made, of which the question requires TWO, so choose two from the above list.

Young’s double slit experiment demonstrated the occurrence of interference effects with light, which produced the pattern of bright and dark fringes on the screen. The wave model of light was able to explain the interference effect, while the particle model could not provide an explanation. Hence, Young’s double slit experiment supported the wave model of light.

Back to Question 2

Question 3 Solution

Radio waves are electromagnetic waves. The AC powered antenna setup includes a dipole – the two separate metal rods. The application of AC causes charge to oscillate between the two rods of the dipole, resulting in changing electric and magnetic fields around the dipole antenna.

These changes in electric and magnetic field propagate sideways away from the antenna at the speed of light, as a transverse wave. This is an electromagnetic wave. The frequency of the wave is determined by the frequency of AC, so radio frequency AC produces radio waves.

Maxwell unified known, empirical laws of electricity and magnetism into the theory of electromagnetism, as expressed by Maxwell’s equations, which describe all electromagnetic phenomena. Maxwell’s equations can be combined to derive the wave equation, which predicts electromagnetic waves travel at a speed of v = 3 \times 10^8 \ ms^{-1} in a vacuum. This is the known speed of light.

The theory also predicts that electromagnetic waves are transverse. The agreement between the speed of the hypothesised electromagnetic waves and the known speed of light led Maxwell to propose that light must be an electromagnetic oscillation, i.e. an electromagnetic wave.

Back to Question 3

Question 4 Solution

When approaching questions about Planck’s explanation of black body radiation, students should be aware of the following context: Classical theory predicts that atoms can exist at any energy and oscillate at any frequency, and that there is no preferred frequency for emitted radiation. Hence all frequencies are emitted equally. Given c = fλ, hence λ = c/f, therefore shorter wavelengths should dominate at any temperature.

Planck proposed that the intensity and hence energy of radiation emitted by a black body could only be increased by fixed amounts proportional to the frequency of radiation, given by E = hf . This required atoms to have discrete (or quantised ) energy levels, and hence the atomic energy changes associated with radiation emission must be discrete changes. (1 mark)

At a given temperature one such discrete energy change would be most probable , hence occur most often, giving rise to the peak wavelength in the spectrum – a wavelength of light emitted more than any other and thus having the highest intensity . (1 mark)

Furthermore, atoms lack sufficiently large discrete energy changes to able to produce high frequency (short wavelength) radiation such as X-rays and gamma rays. Consequently, these wavelengths are not observed at any temperature. (1 mark)

Classical theory could not explain the photoelectric effect, while the concept of quantised light energy (photons with energy E = hf ) was able to fully explain the phenomenon. The maximum kinetic energy of the ejected electrons was found to be K_{max} = hf - \phi , where the photon energy hf must exceed the work function \phi of the metal, else electrons are not ejected.

Classical theory could also not explain the emission line spectra from gases in discharge tubes. The specific wavelengths of emission (while no other wavelengths were emitted) constitute direct evidence of discrete (i.e. quantised) energy levels in atoms.

Planck’s energy equation is E = hf . This can be re-written as E = \dfrac{hc}{\lambda} using the equation v = c = f \lambda .

By substituting the value for wavelength into this equation, we can calculate the energy of a quantum of red light:

\begin{aligned} E &= \dfrac{hc}{\lambda} \\\\ &= \dfrac{ 6.626 \times 10^{-34} \times 3 \times 10^8 }{ 700 \times 10^{-9} } \\\\ &= 2.84 \times 10^{-19} \ J \ (3 \ s.f.) \end{aligned}

Back to Question 4

Question 5 Solution

The surface temperature T of the star can be determined from the peak wavelength \lambda_{max} of the black body spectrum by using Wien’s Law: \lambda_{max} = \frac{2.898 \times 10^{-3}}{T} .

The chemical composition of the stellar atmosphere is determined from the presence of spectral lines. Laboratory spectra of atoms/molecules can be used to identify those elements/compounds in the spectrum of the star. The strengths (depths) of the absorption lines are proportional to the concentrations of the atoms/molecules, indicating how much of each chemical is in the star’s atmosphere.

Students could also discuss information gained from the broadening of spectral lines: The rotational velocity of the star can be determined from the extent of broadening of spectral lines (whole line) while density can be determined from the extent of pressure broadening (wings of line).

Back to Question 5

Question 6 Solution

The magnitude of the stopping voltage in volts is equal to the magnitude of the maximum kinetic energy in electron volts. Therefore, a graph of stopping voltage vs frequency can be interpreted as a graph of maximum kinetic energy vs frequency. Hence, the y-intercept of the graph represents the work function \phi , in electron volts.

By extrapolating the graph back to the y-axis, the work function of Zirconium is found to be 4 \ eV .

From the graph, a relationship between maximum kinetic energy and frequency of incident light can be obtained.

The gradient of the graph is equal to Planck’s constant h .

\begin{aligned} y &= mx + c \\ K_{max}& =gradient \times f - \phi \\ K_{max}& =hf - \phi \end{aligned}

From the photoelectric effect equation, we can infer:

• the maximum kinetic energy K_{max} of ejected photoelectrons is directly proportional to the frequency f of light.
• the energy absorbed by an electron is equal to the energy carried by a photon hf .

This supports the particle model of light, which stated that the energy of the photon is determined by its frequency by E = hf and there is a one-to-one interaction between an electron and a photon.

Back to Question 6

Question 7 Solution

Galileo’s principle of relativity states that the mechanical laws of physics are the same for every observer moving uniformly with constant speed in a straight line (i.e. for every observer in an inertial frame).

Galileo’s principle of relativity states that there is no physical way, from within an inertial frame, to differentiate between a body moving at a constant speed (one inertial frame of reference) and a body at rest (another inertial frame of reference). This means that constant motion cannot be detected in an inertial frame of reference.

However, acceleration can be detected in non-inertial frames of reference. This means that the principle of relativity holds true for any frames of reference that are not accelerating (i.e. inertial frames of reference).

Back to Question 7

Question 8 Solution

Using the formula, t = \frac{d}{v} , we can calculate the time taken for the journey to the planet.

t = \frac{30 \ \text{c years}}{0.80 \ c } = 37.5 \ years

1 light year can be written as 1 c year.

Using the formula for length contraction, we can calculate the distance between the Earth and the planet as measured by Emmet on the spacecraft:

\begin{aligned} l &= l_0 \times \sqrt{1-\frac{v^2}{c^2}} \\\\ &= 30 \times \sqrt{1-\frac{0.8c^2}{c^2}} \\\\ &= 30 \times \sqrt{1-0.64} \\\\ &= 18 \ light \ years \end{aligned}

Let the time taken for the signal to arrive at Earth be denoted by T . This means that the signal reaches Earth after travelling a distance of cT .

However, this distance is also equal to 18 light years (from part (b)) plus the distance travelled by the Earth in time T , i.e:

d = 18 \text{c years} + 0.80 cT

By rearranging this equation, we can calculate the time taken for the signal to arrive at earth:

\begin{aligned} cT &= 18c + 0.80 cT \\\\ \therefore 0.20 T &= 18 \ \text{c years} \\\\ T &= 90 \ years \end{aligned}

Back to Question 8

Question 9 Solution

\begin{aligned} E &= m_0 c^2 \\\\ &= (1.673 \times 10^{-27} + 1.675 \times 10^{-27})c^2 \\\\ &= 3.348 \times 10^{-27} \times (3 \times 10^8) ^2 \\\\ &= 3.0132 \times 10^{-10} \ J \end{aligned}

\begin{aligned} E_{0.8 \ c} &= m_v c^2 \\\\ &= \frac{ ( m_0 c^2 ) }{ \sqrt{ 1 - \frac{ v^2 }{ c^2 } } } \\\\ &= \frac{ ( 3.0132 \times 10^{-10} ) }{ \sqrt{ 1 - (0.8)^2 } } \\\\ &= 5.022 \times 10^{-10} \ J \end{aligned}

\begin{aligned} K &= m_{\nu} c^2 - m_0 c^2 \\ \Delta K&=qV \\ E_{0.8 \ c} - E_{rest}&= qV  \\ V&= \frac{E_{0.8c} - E_{rest}}{q} \\ &=\frac{5.022 \times 10^{-10} -3.0132 \times 10^{-10}}{1.602 \times 10^{-19}} \\ &= 1.254 \ GV \end{aligned}

Step 1: Calculate the total energy of the deuteron when it is travelling at 0.9 \ c .

Step 2: Calculate the work required to increase the energy of the deuteron from the previous value at v = 0.8 \ c to the new value at v = 0.9 \ c .

\begin{aligned} W &= E_{0.9 c} - E_{0.8 c} \\\\ &= 6.912 \times 10^{-10} - 5.022 \times 10^{-10} \\\\ &= 1.890 \times 10^{-10} \ J \end{aligned}

Back to Question 9

Question 10 Solution

Light travelling through only air reached the screen at position 2. Light travelling through air and the tube of water reached the screen at a more displaced position on the screen, position 3.

Foucault’s experimental results indicated that light travels slower in water than in air . The presence of the tube of water increased the flight time of the beam of light (by slowing the light down), which allowed the spinning mirror to rotate more and reflect the returning beam of light to a more displaced position on the screen.

At the time there were two major and competing models of light: Huygens’ wave model and Newton’s corpuscle model. For light entering a more dense medium:

• Huygens’ model predicted that waves of light will decrease in velocity.
• Newton’s model predicted that corpuscles of light will increase in velocity.

Hence Foucault’s result refuted Newton’s corpuscle model and supported Huygens’ wave model.

Back to Question 10

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State Ranker’s Guide to Year 12 HSC Physics Module 6 - Electromagnetism

Your complete guide to module 6 Electromagnetism in Year 12 HSC Physics!

Cory Aitchison

State Ranker & 99.95 ATAR

You’ll be hard pressed to go a day without encountering some application of electromagnetism in your life. Good luck turning on your lights, driving to school, or even just using your eyes. You’ll also be hard pressed to do well in your HSC Physics course without a strong understanding of Module 6 - Electromagnetism. The concepts in this module are crucial foundations which lead onto the key discoveries in Module 7 - The Nature of Light, and Module 8 - From the Universe to the Atom. Yet despite this importance, Module 6 is so often overlooked or underestimated, perhaps because it leans quite heavily on concepts from Year 11 Physics. My goal in this article is to help demystify Electromagnetism, and reveal the key themes and structures which will help you go on to ace this module in your assignments, tests, and the HSC exam.

My name is Cory and I graduated in 2018 with a 99.95 ATAR and two state ranks in Chemistry and Economics. I’ve gone on to complete a Bachelor of Science in physics at the University of Sydney, and am currently pursuing my Honours in quantum physics for 2022. I hope that my experiences can give some insight to help you do your best and ultimately achieve what you set out to achieve.

The Year 12 Syllabus

Before we get into electromagnetism, a quick side note. If you weren’t already aware, there’s this little document called the HSC Physics Syllabus . This treasure trove of information outlines exactly what you are expected to cover in each of the modules, and I highly recommend taking a look before setting out on your journey.

Each module is broken down into key subsections; in Module 6 there are four:

Charged Particles, Conductors and Electric and Magnetic Fields

The motor effect, electromagnetic induction.

• Applications of the Motor Effect

Within each section, there’s an Inquiry Question which guides the thinking, and individual dot points which highlight what it is you’ll be learning. Treat these dot points like a checklist to ensure you don’t miss anything! I would always try and structure my notes around these dot points to ensure I kept on track, and I would be using these dot points as revision nodes while preparing for any exam or assignment. I cannot emphasise this enough: these syllabus dot points are your best friend.

Key Concepts and Skills

It is crucial that you have a thorough understanding of each of those four subsections of the syllabus mentioned above. In this section, I’ll go through and try and demystify the key points from each. This is NOT a thorough explanation of all the ideas in each dot point! You will still need to do (quite) a bit of study by yourself to finish this module, but hopefully this section can give you a foundation to build upon.

Looking for a guide to answering short answers and long-response questions? Read Project Academy’s study guide for it!

Inquiry question: What happens to stationary and moving charged particles when they interact with an electric or magnetic field?

As the inquiry question tells us, this subsection is all about how charges interact with electric and magnetic fields. In particular, we want to be able to model and predict the behaviour of a particle, finding its acceleration, velocity, displacement, etc. Sound familiar? That’s because this module really is very similar to Module 5 - Advanced Mechanics! Just as we analysed the motion of projectiles in gravitational fields, we can use a lot of the same techniques and processes to analyse the motion of charges in electromagnetic fields.

Let’s consider two key cases:

1. Charges in electric fields

Let’s imagine an electron in an electric field. This electron will experience a force in the opposite direction of the field (because it is negatively charged!), with magnitude F = qE. Let’s assume a few things:

• The field is uniform, so that it is constant in direction and magnitude everywhere
• There are no other forces acting on the electron

Perhaps these assumptions are sounding familiar also - and they should! These are the exact same assumptions we made when discussing projectile motion in Module 5. Under these assumptions, we get a very specific type of motion for our electron: in one direction (along the field lines) we get acceleration following:

y = uyt + 0.5(at)2

in the other direction we have no acceleration and x = ux t. Note that we have assumed here that y is in the direction of the electric field, but we could just have easily made it x (or z, or whatever we want).

Just as in projectile motion, the charged particle will follow a perfect parabolic shape, and all your favourite SUVAT equations will apply! If you know how to do projectile motion, then you know how to figure out charges in electric fields!

You might be thinking - does assumption 2 actually hold? Surely we need to take into account other things, such as gravity?? Let’s look into this a bit more deeply. We know the acceleration due to gravity is g = 9.80 m.s2 regardless of the particle’s mass. On the other hand, the acceleration due to the electric field is qE/m. Using the charge and mass of an electron (check your formula sheet!) and a field of 1 N/C, we get 1.8x1011! Last time I checked, this is much larger than 9.80. So, it turns out the electric field is much, much stronger than gravity when dealing with such light particles - and so assumption 2 is pretty much valid.

2. Charges in magnetic fields

What about an electron in a magnetic field? This is where it gets really interesting. The force on the electron is F = qvB, but the magnitude is such that this acts perpendicular to both the velocity and the magnetic field lines. What other kinds of forces do we know that always act perpendicular to velocity? Centripetal forces! When we were considering circular motion in Module 5, we saw that we needed a constant radial force, perpendicular to the velocity, to cause an object to move in a circle. This is exactly what we have here.

Therefore, a charged particle, with an initial velocity perpendicular to a magnetic field, will undergo uniform circular motion in that field!

Inquiry question : Under what circumstances is a force produced on a current-carrying conductor in a magnetic field?

A large part of this module is getting us to understand how a DC motor and an AC motor operate. Before we can get to that, however, this subsection first takes us through how currents behave inside magnetic fields - which will be a crucial component of our motor.

The key idea is this: we know that a charged particle experiences a force when it moves with some velocity perpendicular to a magnetic field. Since currents are just movements of charges (positive charges, if we’re talking conventional current!), then it stands that we should also get a force. This force is known as the motor effect, and it causes current-carrying conductors to undergo accelerations in magnetic fields. Just as with a single charge, this force only arises when the current and field are not parallel.

To do well in this subsection, you need to understand how to calculate the forces on currents, especially taking into account parallel and perpendicular components, and also how to use your appropriate right hand rules to determine the direction of the force.

Inquiry question: How are electric and magnetic fields related?

No joke, this subsection is probably one of the most profound and intriguing aspects of the HSC Physics syllabus (at least to me!). Here we learn that electric and magnetic fields are fundamentally related - forming a combined “electromagnetism”. The core result of this is culminated through “Faraday’s Law” and Lenz’s Law - the tools that describe how changing magnetic fields can be used to create electromotive forces (emfs), electric fields, and currents.

There are so many consequences of this result. We get a brief taste of these implications by studying transformers: devices used to change the voltage of an electrical circuit, which are vital for efficient power transmission. More examples are explored in the final subsection.

To do well in this subsection, you need to be able to apply Faraday’s Law and Lenz’s Law to a variety of situations. Let’s consider Lenz’s Law in a bit more detail:

The law states: The current created by an induced emf is in the direction such that its magnetic field opposes the change in external magnetic field that created it.

To answer questions using this law, I like to follow three steps:

• Determine the change in external magnetic field. Is the field strength increasing or decreasing? Is it pointing into or out of the page?
• Determine the opposing change. If the field strength is decreasing, then the opposing change would be to increase it. If the field strength is increasing, then we want to oppose it by applying a field in the reverse direction.

• Determine the current direction that would create that opposing change. Typically we are dealing with solenoids, and so you would use your right hand rule to determine the appropriate current direction.

Like Project Academy’s style? Come enrol in our Physics course !

Application of the Motor Effect

Inquiry question: How has knowledge about the Motor Effect been applied to technological advances? We conclude this module by considering more applications of the Motor Effect, and of electromagnetic induction. As the name suggests, a key application is the electric motor.

This module includes a detailed understanding of simple DC motors, AC induction motors, and DC and AC generators. Note that the old syllabus also emphasised simple AC motors, but that is no longer the case.

The key operating mechanism of a simple DC motor can be summarised by the following process:

• Current through the coils of the motor creates a force due to the Motor Effect
• This force applies a torque, which causes rotational acceleration of the rotor
• As the rotor rotates, the angle between the force and the level arm changes, causing torque to decrease to a minimum when the coil is vertical
• To ensure that torque is maintained in the same direction, allowing a continuous rotation of the rotor, a split-ring commutator reverses the direction of current every half turn
• As the motor spins through the magnetic field, the coil experiences a change in flux. This causes a “back emf” to be induced via electromagnetic induction
• By Lenz’s Law, this back emf opposes the initial supply emf, so it reduced the current in the coils and therefore reduces the force and torque
• As the motor speeds up, this back emf increases until a point where it equals the supply. Then, there is no net current, force, or torque, and the motor has reached an equilibrium at its maximum speed

To do well in this module, you should be able to expand upon each of those dot points, linking the physical concepts to any relevant laws and supporting your answer with appropriate diagrams and calculations.

Sample Questions for Module 6 Electromagnetism

So, you think you understand electromagnetism? Let’s consider some example questions.

Explain the functioning of a DC motor with reference to each component. (6 marks)

In this question, we want to ensure that we reference all the key components of our DC motor. Although this is 6 marks, it is not necessarily deserving of a long-response-type answer. A typical marking scheme for this may award points based on the number of correctly described components.

Sample answer:

• FUNCTION: Transforms electrical potential energy into rotational kinetic energy through the Motor Effect (a current-carrying conductor in a magnetic field experiences a force perpendicular to the field and the current)
• STATOR: Non-rotating, consists of magnets fixed to a casting that provides the magnetic field; can be electromagnets or permanent magnets
• ARMATURE: A frame on which coils are wound (that carry the DC); Usually made from a ferromagnetic material such as a soft-iron core that enhances the magnetic flux threading and therefore the force/torque
• ROTOR: Consists of the armature and the coil
• SPLIT-RING COMMUTATOR: Reverses the direction of current during rotation to maintain constant direction of torque. Consists of spring-loaded brushes (graphite, act as a dry lubricant) against a two-piece conducting metal ring, each side connected to the coil; Changes direction every half-turn by alternating contact with the brushes (connected to the DC source)
• RADIAL MAGNETS: Can be used as part of the stator that maintain magnetic field lines always perpendicular to the plane of the coil, therefore maximising torque (torque = nβIAcos(θ))
• LAMINATIONS: The soft-iron core can be laminated with insulative layers to reduce the size of induced eddy currents, therefore minimising unwanted heating, and reduced efficiency through energy transformations

How is the direction of torque maximised as a DC motor’s coil rotates 360˚? (3 marks)

For this question, we really want to lean into the idea of CAUSE/LINK/EFFECT to justify the “how” verb in the question. The key understanding to get across is that the angle between the coil and the magnetic field affects the torque magnitude. A typical marking scheme would be (1) Identifies relationship between angle and torque; (2) Explain the conditions when torque is maximised; (3) Relates these conditions to the use of radial magnetic fields.

Sample answer: 

• CAUSE: Torque = nβIAcos(θ), where theta is the angle between the plane of the coil and the magnetic field
• LINK: Torque is maximised when theta is 90 degrees (perpendicular)
• LINK: A radial magnetic field ensures that the plane is always perpendicular to the plane of the coil (except for at the vertical)
• EFFECT: Radial magnets maximise the torque as it rotates

Explain how induction is used in induction cooktops and electric braking. (4 marks)

Again, this question requires a CAUSE/LINK/EFFECT relationship to be developed for both applications. Given that it is worth 4 marks, we would take there to be 2 marks per application, with one mark awarded for showing understanding of its operation and another for explicitly describing the process of electromagnetic induction.

INDUCTION COOKTOPS:

• CAUSE: Eddy currents induced in a metal surface increase the temperature of the metal due to resistive heating, where collision of electrons with the metal lattice transfer some of their energy to heat energy
• LINK: In induction cooktops, a high voltage, high frequency AC input is supplied to an “induction coil” induces eddy currents in the ferrous material of a saucepan on top of a ceramic top plate by Faraday’s Law
• EFFECT: These currents cause the saucepan to heat up, with minimal loss of thermal energy, unlike in conventional gas/electric stoves that heat only via conduction (efficiency of 80% vs 43% for gas)

ELECTRIC BRAKING:

• CAUSE: A spinning metal disk attached to a rotating wheel passess through a magnetic field (controlled by electromagnets)
• LINK: By Faraday’s Law, this changing magnetic flux through the disk induces an emf, which by Lenz’s Law induces a current to oppose this change in flux
• EFFECT: The magnetic field created by these eddy currents opposes that of the electromagnets, therefore providing a resistive force as it repels when approaching an attracts when moving away, which acts to brake the wheel/vehicle (smoother than conventional braking with pads, and no friction reduces wear/maintenance)

Explain the production of back emf in a DC motor and relate it to the increased work required to operate a DC generator when a load is connected to the circuit. (5 marks)

There is quite a bit to unpack with this question, and we need to ensure that we address each point to get the marks. I would recommend structuring this kind of question by starting off by explaining how a back emf arises, then explaining DC generators, and finally linking the two.

CAUSE: When a DC motor’s rotor rotates with a magnetic field due to the motor effect, the coil experiences a changing flux.

LINK: By Faraday’s Law, a “back emf” is induced within the coil by ϵ = -(nΔϕ) / (Δt) , opposing the emf of the power supply (net emf = supply emf - back emf)

EFFECT: The rotor accelerates rotationally until back emf equals the input voltage, and therefore net voltage across the coil is zero (and no torque by 𝜏 = nβIAcosθ).

Link to DC Generators:

CAUSE: In a DC generator, an emf is similarly established in the coils due to Faraday’s Law and the change of magnetic flux in the rotor

LINK: If a circuit and load is connected to the coil, then a current is induced in the wire

LINK: By the motor effect, a current in a magnetic field experiences a force perpendicular to direction of flow

EFFECT: This force opposes the rotation of the rotor, therefore requiring increased torque (and work) to maintain the same rotation of the generator

Congratulations on reading all of this! I hope that some of these points have been useful in helping you understand some of the key points of Module 6 - Electromagnetism, or at least in identifying areas where you may need some work. I truly believe that this is a really exciting, interesting and ultimately important module of study - I am definitely still using a lot of these ideas today in my university work!

• Is HSC Physics hard?

Is an apple big? As cliche as it is, the answer really is relative! I know some students quickly grasp new physics concepts while others take a longer time to become accustomed to the way of thinking required to do HSC Physics. What I can say though is that it definitely is accessible - the content is not too mathy, nor too abstract, nor too hands-on, to make it unconquerable for any student.

• Is Year 11 content in the HSC Physics?

Yes! Although they won’t be testing you word-for-word what you had on your year 11 final exams, there is a strong overlap. Take a look at the formula sheet - quite a few of the formulas you learnt in year 11 are there. Moreover, the whole of Dynamics and Kinematics (Modules 1 and 2) were there to lay a foundation that is continued on in Module 5 - Advanced Mechanics. Module 4 - Electricity and Magnetism also continues on quite heavily within Module 5 - Electromagnetism. So, if you are looking to do well in HSC, it won’t hurt to brush up on your year 11 notes too :).

• Where can I find Year 12 Physics notes?

In your notebook hopefully! I believe that the best way to make notes is by actually making them yourself, not copying off your friend/teacher/enemy/cat. HOWEVER, it sometimes helps to have a foundation to base yours off, in which case there are fantastic resources such as your textbook or online - like at Project Academy’s website.

• What are the best Year 12 Physics textbooks?

Any of them! In terms of content, I have yet to find one that was significantly worse than any other. A few of them may have better questions, more worked examples, nicer illustrations, but I think whatever you have on hand will do the job well. Just make sure that it is from the new syllabus (2017 and later).

• Is Year 12 Physics similar to Year 11 Physics?

See question 2 for content. In terms of how to approach it, I think that Year 12 does place a greater emphasis on applications of understanding . Whereas a large part of Year 11 is getting you to be comfortable with the various concepts, mathematics, and language, the HSC will try and push you to apply what you learnt in entirely new settings. Memorising a bunch of answers from your notes won’t get you very far in the final exams. A good way to prepare then is to expose yourself to as many different kinds of questions and scenarios as possible, and think deeply about how each are different and how they all relate to the core HSC physics concepts.

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Physics Official 2023-2024 HSC Science (General) 12th Standard Board Exam Question Paper Solution

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General Instructions : The question paper is divided into four sections :

• Section A: Q. No. 1 contains Ten multiple choice type of questions carrying One mark each. Q. No. 2 contains Eight very short answer type of questions carrying One mark each.
• Section B: Q. No. 3 to Q. No. 14 contain Twelve short answers types of questions carrying Two marks each. (Attempt any Eight ).
• Section C: Q. No. 15 to Q. No. 26 contain Twelve short answer type of questions carrying Three marks each. (Attempt any Eight ).
• Section D: Q. No. 27 to Q. No. 31 contain Five long answer type of questions carrying Four marks each. (Attempt any Three ).
• Use of the log table is allowed. Use of calculator is not allowed
• Figures to the right indicate full marks.
• For multiple choice type questions, only the first attempt will be considered for evaluation.
• Physical Constants: (i) mass of electron m = 9.1 × 10 -31 kg (ii) ε 0 = 8.85 × 10 -12 `C^2/(Nm^2)` (iii) π = 3.142 (iv) charge on electron e = 1.6 × 10 -19 C (v) µ 0 = 4π × 10 -7 Wb/Am (vi) Planck's constant h = 6.63 × 10 -34 J.s. (vii) Speed of light c = 3 × 10 8 m/s (viii) g = 9.8 m/s 2 (ix) Rydberg's constant R H  = 1.097 × 10 7 m -1 (x) Stefan's constant σ = 5.67 × 10 -8 J m -2 s -1 K -4

The moment of inertia (MI) of a disc of radius R and mass M about its central axis is ______.

`(3"MR"^2)/2`

The dimensional formula of surface tension is ______.

[L -1 M 1 T -2 ]

[L 2 M 1 T -2 ]

[L 1 M 1 T -1 ]

[L 0 M 1 T -2 ]

Phase difference between a node and an adjacent antinode in a stationary wave is ______.

`(3pi)/4`rad

The work done in bringing a unit positive charge from infinity to a given point against the direction of electric field is known as ______.

electric flux

magnetic potential

electric potential

gravitational potential

To convert a moving coil galvanometer into an ammeter we need to connect a ______.

small resistance in parallel with it

large resistance in series with it

small resistance in series with it

large resistance in parallel with it

If the frequency of incident light falling on a photosensitive material is doubled, then the kinetic energy of the emitted photoelectron will be ______.

same as its initial value

two times its initial value

more than two times its initial value

less than two times its initial value

In a cyclic process, if ΔU = internal energy, W = work done, Q = Heat supplied then ______.

ΔU = Q

The current in a coil changes from 50A to 10A in 0.1 second. The self inductance of the coil is 20H. The induced e.m.f. in the coil is ______.

The velocity of bob of a second’s pendulum when it is 6 cm from its mean position and amplitude of 10 cm, is ______.

In biprism experiment, the distance of 20th bright band from the central bright band is 1.2 cm. Without changing the experimental set-up, the distance of 30 bright band from the central bright band will be ______.

 0.6 cm

1.2 cm 

Define centripetal force.

Why a detergent powder is mixed with water to wash clothes?

What is the value of resistance for an ideal voltmeter?

Write the formula for torque acting on rotating current carrying coil in terms of magnetic dipole moment, in vector form.

What is binding energy of a hydrogen atom?

What are surroundings in thermodynamics?

In a photoelectric experiment, the stopping potential is 1.5V. What is the maximum kinetic energy of a photoelectron?

Two capacitors of capacities 5 μF and 10 μF respectively are connected in series. Calculate the resultant capacity of the combination.

Explain the change in internal energy of a thermodynamic system (the gas) by heating it.

Explain the construction and propagation of a spherical wavefront using Huygens's principle.

Define Magnetization.

State its SI unit and dimensions.

Obtain the differential equation of linear simple harmonic motion.

A galvanometer has a resistance of 30Ω, and its full-scale deflection current is 20 microampere (ΩA). What resistance should be added to it to have a range of 0-10 volts?

Explain Biot Savart's Law.

What is a Light Emitting Diode?

Draw Light Emitting Diode circuit symbol.

An aircraft of wing span of 60 m flies horizontally in earth’s magnetic field of  6 x 10 -5 T at a speed of 500 m/s. Calculate the e.m.f. induced between the tips of the wings of the aircraft.

Derive an expression for maximum speed moving along a horizontal circular track.

A horizontal force of 0.5N is required to move a metal plate of area 10 -2 m 2 . with a velocity of 3×10 -2 m/s, when it rests on 0.5×10 -3 m thick layer of glycerin. Find the coefficient of viscosity of glycerin.

Two tuning forks having frequencies 320 Hz and 340 Hz are sounded together to produce sound waves. The velocity of sound in air is 340 m/s. Find the difference in wavelength of these waves. 

Calculate the change in angular momentum of the electron when it jumps from 3 rd orbit to 1 st orbit in hydrogen atom.

(Take h = 6.33 x 10 -34 Js)

A circular coil of wire is made up of 200 turns, each of radius 10 cm. If a current of 0.5A passes through it, what will be the Magnetic field at the centre of the coil?

Define photoelectric effect. 

Explain the experimental set-up of photoelectric effect.

Define α

Define β

Derive the relation between α and β.

Define the surface energy of the liquid.

Derive the relation between surface tension and surface energy per unit area.

What is an isothermal process?

Obtain an expression for the workdone by a gas in an isothermal process.

Derive an expression for the equation of stationary wave on a stretched string.

Explain the formulation of stationary waves by the analytical method. What are nodes and antinodes? Show that the distance between two successive nodes or antinodes is λ/2. 

Derive an expression for the impedance of an LCR circuit connected to an AC power supply.

Draw the phasor diagram for a series LRC circuit connected to an AC source.

Calculate the wavelength of the first two lines in the Balmer series of hydrogen atoms.

The radius of a circular track is 200 m. Find the angle of banking of the track, if the maximum speed at which a car can be driven safely along it is 25 m/sec.

Prove the Mayer's relation `C_p - C _v = R/J`

An alternating voltage is given by e = 8sin628.4t. Find

• peak value of e.m.f.
• frequency or e.m.f.
• instantaneous value of e.m.f. at time t = 10. ms

What is a transformer?

Explain transformer working with construction.

Derive the equation for a transformer.

Using the geometry of the double slit experiment, derive the expression for fringe width of interference bands.

Distinguish between an ammeter and a voltmeter.

The displacement of a particle performing simple harmonic motion is `1/3` rd of its amplitude. What fraction of total energy will be its kinetic energy?

Draw a neat labelled diagram of Ferry’s perfectly blackbody.

Compare the rms speed of hydrogen molecules at 227°C with the rms speed of oxygen molecules at 127°C. Given that molecular masses of hydrogen and oxygen are 2 and 32 respectively.

Derive an expression for energy stored in a capacitor.

A spherical metal ball of radius 15 cm carries a charge of 2μC. Calculate the electric field at a distance of 20 cm from the center of the sphere.

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HSC Physics 2nd Paper Question Solution 2023 | All Board

HSC Physics 2nd Paper Question Solution 2023. The exam for the HSC began on 17 August 2023. The Physics 2nd Paper exam was held on August 31, 2023. You’re looking for a solution to Question 2023 of the Second Physics Paper? Then you’ve arrived in the proper place. Today’s question was resolved by an expert physics teacher in this post. We have collected all board’s HSC Physics Solution for all education board.

The fundamental science of physics is the study of the physical nature of matter, its fundamental components, its movements and activities in space and time, as well as the relations between energy and force in the universe. The primary objective of physics is to explain what occurs in the universe. This paper will discuss the questions posed, as well as provide images and solutions to all Board Physics 2nd papers.

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Physics forms the foundation of many fundamental sciences. For instance, chemistry is grounded in quantum physics, the study of the microscopic structure of atoms and molecules. Physics is widely used in most engineering disciplines. For instance, architecture is largely based on physics, which is used to determine structural stability and acoustic properties. Heating, lighting and cooling systems are also based on physics. Geology is grounded in physics, which is based on the study of the non-living components of the Earth, such as atmospheres and lithospheres and hydrospheres. HSC Physics 2nd Paper Question Solution 2023. It is also used to study the Earth’s natural hazards, such as earthquakes, storms, volcanoes, floods, droughts, extreme heat, and more.

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M5-S7: Quantitative Analysis of Circular Motion

Solve problems, model and make quantitative predictions about objects executing uniform circular motion in a variety of situations, using the following relationships:.

• Tangential velocity v is defined as the total circumference of circular path divided by the total time taken to complete one revolution T , which is also known as the period of circular motion.

`v=(2pir)/T`

• Angular velocity w is defined as the total angle of one revolution, 2p, divided by the period, T , of circular motion. Its SI unit is radians per second (rad s -1 )

Alternatively, angular velocity is also defined as the angle completed by an object in circular motion in time t , where t does not necessarily need to be the period of circular motion.

 `omega=(Deltatheta)/t`

Figure shows the direction of tangential and angular velocity .  vector points perpendicular to the v vector, can be determined by curling the right hand into a fist form with thumb pointing outwards. Curled fingers represent direction of rotation while thumb points in the direction of angular velocity.

Substituting  into centripetal force equation (`F_c=(mv^2)/r`) gives `F_c=mromega^2` .

  Derivation:

Practice Question 1 (NESA Sample Question) 

A 15-gram metal ball bearing on a string is swung around a pole in a circle of radius 0.8 m. The plane of the circular path is horizontal. The angular velocity of the motion is `4pi   rad s^-1`.

What is the magnitude of the centripetal force required to maintain the motion of the ball?

Practice Question 2

A particle is moving around in a circle of radius 1.5 m with a constant speed of 2 ms -1 . Calculate its

Practice Question 3 (HSC 2013)

The diagram shows a futuristic space station designed to simulate gravity in a weightless environment.

If the space station has a diameter of 550 m, calculate the rotational speed needed to simulate 1g of gravitational acceleration.

Practice Question 4

A 15 kg weight is attached to a rope and spun in uniform circular motion with a 0.5 m radius as shown.

If the rope makes an angle of 30º with the horizontal during circular motion, calculate the angular velocity of the weight.

Practice Question 1

`F_c=mromega^2`

Convert grams into SI unit (kg): 

`F_c=(15/1000)(0.8)(4pi)^2` 

`F_c=1.9` N

(a) We will use `a_c=v^2/r` to find centripetal acceleration:

(b) We will use `omega=v/r` to find angular velocity:

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20 Practice Questions for Year 11 Physics Module 3: Waves and Thermodynamics

On the hunt for practice questions for Year 11 Physics Module 3: Waves and Thermodynamics? Well, luckily we’re here to help you out! Just in case you missed it, we broke down Module 3 in this article here!

We’ve structured the questions for Physics Module 3: Waves and Thermodynamics by breaking them up into five parts (based on each inquiry question). So today we are going to have a look at:

Wave Properties

Wave behaviour, sound waves, ray model of light, thermodynamics.

Let’s get started on these Physics Module 1: Waves and Thermodynamics practice questions!

Wave Properties Wave Behaviour Sound Waves Ray Model of Light Thermodynamics

If two waves have the same frequency, which other wave property must also be the same?   Solve problems and/or make predictions by modelling and applying the following relationships to a variety of situations

Distinguish the similarities and differences between transverse matter waves and transverse electromagnetic waves.   Conduct practical investigations to explain and analyse the differences between: transverse and longitudinal waves (ACSPH068) mechanical and electromagnetic waves (ACSPH070, ACSPH074)

A wave has a wavelength of 0.30 m and a period of 5.0 x 10-4s. What is the velocity and frequency of the wave?   S olve problems and/or make predictions by modelling and applying the following relationships to a variety of situations

The amount of energy carried by a transverse matter wave is dependent on which wave characteristic? Explain your reasoning.   C onduct a practical investigation involving the creation of mechanical waves in a variety of situations in order to explain: the role of the medium in the propagation of mechanical waves the transfer of energy involved in the propagation of mechanical waves (ACSPH067, ACSPH070)

When a sound wave travelled from air to a container of Krypton gas, the speed of the wave decreased from 340 ms-1 to 220 ms-1, If the frequency of the sound wave in air was 700 Hz, what is the frequency of the wave in the container?   Explain the behaviour of waves in a variety of situations by investigating the phenomena of: reflection refraction diffraction wave superposition (ACSPH071, ACSPH072)

Define resonance and provide an example to help explain your answer.   C onduct an investigation to explore resonance in mechanical systems and the relationships between: driving frequency natural frequency of the oscillating system amplitude of motion transfer/transformation of energy within the system (ACSPH073)

How would classical musicians use beats to tune their instruments?   Analyse qualitatively and quantitatively the relationships of the wave nature of sound to explain: beats the Doppler effect

A bystander notices that when stationary, the frequency of a police siren is 1800Hz. What would the frequency of the siren be if a bystander approaches the police car at 2ms-1, and the police car approaches the bystander at 20ms-1?   Analyse qualitatively and quantitatively the relationships of the wave nature of sound to explain: beats the Doppler effect

Explain which wave properties will create beats.   Analyse qualitatively and quantitatively the relationships of the wave nature of sound to explain: beats the Doppler effect

Question 10

A person is standing between two cliffs. He is 800m from the nearest cliff and yells. If the velocity of sound in air is 340 ms-1, and he hears the second echo from the cliff behind him 5.0s after he hears the first echo. What is the distance between the two cliffs?   C onduct investigations to analyse the reflection, diffraction, resonance and superposition of sound waves (ACSPH071)

Question 11

What is the relationship between the harmonic of a standing wave in a string and the frequency of the standing wave?   Investigate and model the behaviour of standing waves on strings and/or in pipes to relate quantitatively the fundamental and harmonic frequencies of the waves that are produced to the physical characteristics (eg length, mass, tension, wave velocity) of the medium (ACSPH072)

Question 12

Two people are whispering to each other, although they are whispering at the same volume, one has a much higher pitch than the other. Someone else stands just outside and was able to hear one person talk, who was this person? Justify your answer.   C onduct a practical investigation to relate the pitch and loudness of a sound to its wave characteristics

Question 13

Explain differences in image formed by a concave and convex lens. Use diagrams to strengthen your answer.   C onduct a practical investigation to analyse the formation of images in mirrors and lenses via reflection and refraction using the ray model of light (ACSPH075)

Question 14

Create an expression for the critical angle of a substance and the relative refractive index of the boundary involved in the interaction using Snell’s law.   Solve problems or make quantitative predictions in a variety of situations by applying the following relationships to: for the refractive index of medium ?, Snell’s Law

Question 15

What is total internal reflection?   P redict quantitatively, using Snell’s Law, the refraction and total internal reflection of light in a variety of situations

Question 16

How does the inverse square law for electromagnetic radiation complicate our ability to study our Universe?   C onduct an investigation to demonstrate the relationship between inverse square law, the intensity of light and the transfer of energy (ACSPH077)

Question 17

Find the final temperature and total heat transfer if 200g of ice is placed in a room with a temperature of 25ºC. The specific heat capacity of water is 4184 Jkg-1K-1 and the latent heat of fusion is 334000 Jkg-1.   A pply the following relationships to solve problems and make quantitative predictions in a variety of situations: ?=??Δ? where c is the specific heat capacity of a substance

Question 18

Two metal squares of the same material have a surface area of 3m2 and 6m2. The two squares are heated evenly in the same environment and have the same thickness. Predict the rate at which each square loses energy as it becomes cooler. Justify your answer.   A pply the following relationships to solve problems and make quantitative predictions in a variety of situations: ?=??Δ? where c is the specific heat capacity of a substance

Question 19

A house has 200 m2 worth of windows that are 2.5 mm thick. The glass has a thermal conduction constant of 0.75 Wm-1K-1. If the outside temperature is -5ºC and the internal temperature is 27ºC, how much energy is transferred via the windows per hour?   A pply the following relationships to solve problems and make quantitative predictions in a variety of situations: ?=??Δ? where c is the specific heat capacity of a substance

Question 20

What is the relationship between the latent heat and water going through the different states? Explain your reasoning.   C onduct an investigation to analyse qualitatively and quantitatively the latent heat involved in a change of state

And that’s all for our 20 practice questions for Physics Module 3: Waves and Thermodynamics. Good luck!  

Looking for some extra help with Physics Module 3: Waves and Thermodynamics?

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To find out more and get started with an inspirational HSC Physics tutor and mentor,  get in touch today or give us a ring on 1300 267 888!

Ivy Zhang is a tutor at Art of Smart and an undergraduate student at the University of Sydney. While studying Physics and Maths in order to become a high school teacher, Ivy enjoys running and reading about modern history and the universe.

• Topics: ⚡️ Physics , ✍️ Learn

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