IB Physics SL Theme B: The particulate nature of matter questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

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Question 1

Multiple choiceMedium1 mark

A star and the Sun are both modelled as black bodies. The star’s surface area is four times the Sun’s, and its surface temperature is half the Sun’s; both temperatures are absolute, measured in kelvin. What is the star’s luminosity in terms of LsunL_{\text{sun}}

Show answer and explanation

Answer: B. For a black body, L=σAT4L=\sigma AT^4 Comparing the star with the Sun cancels σ\sigma

LLsun=AAsun(TTsun)4=4(12)4=14.\frac{L}{L_{\text{sun}}} =\frac{A}{A_{\text{sun}}}\left(\frac{T}{T_{\text{sun}}}\right)^4 =4\left(\frac12\right)^4 =\frac14.

A: includes the temperature factor but omits the area ratio.
C: uses a square rather than the fourth power of the temperature ratio.
D: includes the area ratio but ignores the temperature dependence.
Remember: for black-body radiation, luminosity depends on the fourth power of absolute temperature.

Question 2

Paper 2Medium11 marks

An observer on Earth measures the apparent brightness of stars A and B shown in Figure 1. Assume that both stars emit uniformly in all directions and that absorption between each star and Earth is negligible.

Earth (observer)dA=2.0×1017 mdB=4.0×1017 mAbA=3.0×10−8 W m−2BbB=1.0×10−8 W m−2
(a)

State which star has the greater apparent brightness.

[1]
(b)

Determine the ratio LB/LAL_B/L_A

[3]
(c)

Calculate the luminosity of star A.

[2]
(d)

A third star, C, has twice the luminosity of star A and an apparent brightness of 1.2×10−8 W m−21.2\times10^{-8}\,\mathrm{W\,m^{-2}} Determine its distance from Earth.

[3]
(e)

Explain why the apparent brightness of a star decreases as its distance from Earth increases, if its luminosity is unchanged.

[2]

Question 3

Paper 2Medium8 marks

An ideal gas is in a vessel of uniform pressure. Figure 1 shows the gas and a surface SS of the vessel. The force shown is the resultant force exerted by the gas on SS

Use NA=6.02×1023 mol−1N_A=6.02\times10^{23}\ \mathrm{mol^{-1}} and R=8.31 J mol−1 K−1R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}} Give answers to three significant figures, except in (d), where give the temperature to the nearest degree Celsius.

surface S, A = 2.40 × 10−3 m2force exerted by gas, F = 240 NN = 2.50 × 1023 moleculesV = 1.00 × 10−2 m3
(a)

Determine the pressure of the gas.

[2]
(b)

Determine the amount of substance of the gas.

[2]
(c)

Using your answers to (a) and (b), determine the temperature of the gas in kelvin.

[3]
(d)

Express the temperature from (c) in degrees Celsius.

[1]

Question 4

Multiple choiceMedium1 mark

A resistor is tested by increasing the potential difference across it and recording the current. The measured values are plotted.

00.10.20.30.40.50.60.70.80.9100.511.522.533.544.555.56current, I / Apotential difference, V / VAB

Which row identifies the resistor’s behaviour and gives the ratio of power dissipated at B to that at A?

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Answer: C. At A, V/I=2.0/0.40=5.0 ΩV/I=2.0/0.40=5.0\ \Omega at B, V/I=5.0/0.80=6.25 ΩV/I=5.0/0.80=6.25\ \Omega Since the resistance is not constant, the behaviour is non-ohmic. Using P=IVP=IV PA=0.80 WP_A=0.80\ \mathrm{W} and PB=4.0 WP_B=4.0\ \mathrm{W} so PB/PA=5.0P_B/P_A=5.0

A: obtains the power ratio but treats the unequal V/IV/I values as equal, classifying the resistor as ohmic.
B: uses the current ratio, 0.80/0.40=2.00.80/0.40=2.0 as the power ratio and omits the change in potential difference.
D: assumes constant resistance and uses P∝I2P\propto I^2 giving a ratio of 4.04.0

Remember: compare V/IV/I to identify ohmic behaviour, and use P=IVP=IV for power at each operating point.

Question 5

Multiple choiceMedium1 mark

Two resistors of resistance RR and 2R2R are connected in series across an ideal source of potential difference VV They are then connected in parallel across an identical ideal source. What is the ratio of the current through the RR resistor in the series arrangement to the current through the 2R2R resistor in the parallel arrangement?

Show answer and explanation

Answer: B. In series, the total resistance is 3R3R so the current through RR is V/(3R)V/(3R) In parallel, the 2R2R resistor has the full potential difference VV so its current is V/(2R)V/(2R) The required ratio is [V/(3R)]/[V/(2R)]=2/3[V/(3R)]/[V/(2R)]=2/3 A: uses the current through the RR branch in parallel, V/RV/R as the denominator. C: assumes the currents are equal because the source potential differences are equal. D: omits the 2R2R resistor when finding the series resistance, giving a series current of V/RV/R Remember: series current depends on the total resistance, while each parallel branch has the full source potential difference.

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