IB Physics SL Paper 3 Question Bank


1.)  The circuit below can be used to measure the internal resistance of a cell

a.) Label the ammeter as A and the voltmeter as V in the circuit

The circle connected in series is the Ammeter and the circle connected in the parallel circuit is the Voltmeter. A voltmeter is placed in parallel with the voltage source to receive full voltage and must have a large resistance to limit its effect on the circuit. An ammeter is placed in series to get the full current flowing through a branch and must have a small resistance to limit its effect on the circuit.

b.) Upon performing an experiment, the graph for the current vs potential difference is shown below. Using the graph, calculate the best or closest estimate for the resistance of the cell.

To find the resistance of the cell, the gradient of the slope needs to be calculated. Therefore, to calculate the gradient of the line, two points that form a large triangle need to be taken. Therefore, for instance, 2 points on the line – (0.40, 1.25) and (2.00, 0.3) – were taken. To find the gradient of the slope, m = y2 – y1 / x2- x1 = (0.3 – 1.25) / (2.00 – 0.30) = 0.95 / 1.7 = 0.558 approximately, 0.56 A.

c.) The student forgets to take a zero error reading from the ammeter. Define zero error reading

When a circuit is not connected, the ammeter or voltmeter’s pointer is supposed to be at zero, Therefore, a zero error reading is when a non zero reading is displayed when it is supposed to show zero. It is essentially a calibration error.

d.) After the student has taken the zero error reading, she notices it to be a positive number. Explain the effect of a positive zero error reading on the internal resistance of the cell.

A positive number on the ammeter means that more electricity is being received by the battery than is being used, suggesting the regulator is not working efficiently. All the measurements would be higher and incorrect. Nevertheless, since the value of all of it increases, there will be no change in the internal resistance (the slope of the graph).

2.) This question is about a standing wave in a vibrating string

a.) Describe how a standing wave is formed when the string is fixed at both ends
The wave travels down the string, hits the end and reflects back. Therefore, the incident wave approaching the end and the reflected wave is interfered or superposed to create a standing wave.

b.) The string of a guitar vibrates at 350 Hz. The length of this guitar string is 0.69 m. Using this data, calculate the velocity of the wave in the string.
First the wavelength is calculated; λ = 2L → 2 x 0.69 = 1.38 m
Then, the velocity is calculated by multiplying it with the frequency:
v = λf → 1.38 x 350 = 483 m/s

3.) A student pours a can of a carbonated drink into a cylindrical cup after rigorously shaking the can. A large volume of foam is produced eventually filling the container. The graph below shows the time vs the height of the foam:

a.) Determine the time taken for the foam to drop to half of its original height

The initial height of the foam (0 seconds) was 37 cm. Therefore, half of its height will be 18.5 cm. To find the time taken, the data point of time when the height is 18.5 cm will be the answer. By checking the graph, the time taken is between 18 to 19 seconds.

b.) Determine the time taken for the foam to drop to quarter of its original height

Quarter of the foam height = ¾ x 37 = 27.75 cm
Since the height is reduced, it needs to be subtracted from 37 → 37 – 27.75 = 9.25 cm
Checking the data point, the time taken is 36 to 37 seconds.

SECTION B – Option D Astrophysics

1.) The surface temperature of the star Epsilon Indi is 4650 K
a.) Determine the peak wavelength of the emitted radiation by Epsilon Indi
The formula λ = b/T is used where b is the Wien’s displacement constant – 2.89 x 10-3
Therefore, substituting the temperature as well, we have: λ = 2.89 x 10-3/4650
λ = 621.5 nm = 622 nm

b.) The data shown below is about the Sun
Surface temperature – 5780 K
Luminosity – L0
Radius – R0
Mass – M0

Epsilon Indi has a radius of 0.731 R0. Determine the luminosity of the star
L/L0 = (0.731 R0 / R0)2 x (4650 / 5780)4
L/L0 = 0.53436 x 0.41889
L = 0.22383L0 → 0.224 L0

c.) Epsilon Indi is a star in the main sequence. Determine the mass of the star
The relationship between luminosity and mass is L = M3.5
Therefore, M = L1/3.5 M0
Substituting the luminosity we have, M = 0.2241/3.5 M0
M = 0.652 M0

d.) Outline the stages the star, Epsilon Indi undergoes once it leaves main sequence till its final stable state
A star stays in the main sequence until all of the hydrogen has fused to form helium. The helium core now starts to contract further and reactions begin to occur in a shell around the core. Once the core is hot enough for the helium to fuse to form carbon, the outer layers begin to expand, cool and shine less brightly. The expanding star is now called a Red Giant. The helium core runs out, and the outer layers drift away from the core as a gaseous shell, this gas that surrounds the core is called a Planetary Nebula. The remaining core is now in its final stages. When the core becomes a White Dwarf the star eventually cools and dims. When it stops shining, it becomes a dead star and is called a Black Dwarf.

2.) Shown below is a Hertzsprung-Russell Diagram. The following data is available for a star X with respect to the Solar mass (M0) and the Solar Radius (R0):

Mx = 4.7 M0
Rx = 3.4 R0

a.) By how much is the luminosity of the star X greater than the luminosity of the sun
Lx = 4.73.5 M0 since the relationship between luminosity and mass is L = M3.5
Therefore, 4.73.5 = 225 L0

b.) The parallax angle of star X from Earth is 0.174 arc second. Outline how the parallax angle of a star can be measured
The position of a star is recorded in a 6 month gap (Month 1 and Month 7). The parallax angle is calculated by seeing how much the star has moved relative to the distant stars present in the background.

c.) Star X eventually becomes a White Dwarf.
i.) State the star type for A, B and C
A: White Dwarfs
B: Main sequence
C: Red Giant

ii.) State what prevents Star X from collapsing further after becoming a white dwarf
An electron degeneracy. Electron degeneracy pressure will halt the gravitational collapse of a star if its mass is below the Chandrasekhar limit (1.44 solar masses) which it is since we’re talking about the Sun here. This is the pressure that prevents a white dwarf star from collapsing.

3.) Light from a galaxy far away observed on Earth shows a redshift of 0.13.
a.) Define redshift
Redshift is when the wavelength of the light is stretched, so the light is seen as ‘shifted’ towards the red part of the spectrum. It means that the received wavelength is longer than the emitted wavelength.

b.) Determine the distance to this galaxy using the Hubble constant assuming that it is H0 = 67 km/s Mpc
First the velocity needs to be determined using the formula v = zc where z is the redshift and c is the speed of light. Therefore, substituting the respective values, we have: v = 0.13 x 3.0 x 105 = 3.9 x 104 km/s. To find the distance, the formula d = v / H0 can be used. Putting in the values, we have, d = 3.9 x 104 / 67 = 582 Mpc

c.) Suggest how the cosmic microwave background radiation provides strong evidence for the existence of the Big Bang model.
The CMB radiation is essentially a black body spectrum that is highly isotropic across the universe. The wavelength of this CMB radiation matched the predicted wavelength of the Big Bang had it increased by expansion.

4.) The question is with regards to planets and stars
a.) 4 of the outer planets in the solar system are Jupiter, Saturn, Uranus and Neptune
i.) List these planets in order of increasing distance from the Sun
Jupiter, Saturn, Uranus, Neptune

ii.) List these planets in order of increasing mass
Uranus, Neptune, Saturn, Jupiter

b.) Define apparent brightness and luminosity
Apparent brightness is a measure of the amount of light received by Earth from a star or other object. Luminosity is an absolute measure of the radiant power emitted by a light-emitting object.

c.) The apparent brightness of a Star C is 3.4 x 10-10 W/m2. The luminosity of the Sun is 3.9 x 1026 W. Determine the distance of C from Earth
The formula used here is d = (L / 4πb)½
Substituting the values we have, d = (3.9 x 1026 x 104 / 4π x 3.4 x 10-10)1/2
d = 3.02 x 1019 m