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 $L_{\text{sun}}$

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

\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

Multiple choiceMedium1 mark

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

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, $V/I=2.0/0.40=5.0\ \Omega$ at B, $V/I=5.0/0.80=6.25\ \Omega$ Since the resistance is not constant, the behaviour is non-ohmic. Using $P=IV$ $P_A=0.80\ \mathrm{W}$ and $P_B=4.0\ \mathrm{W}$ so $P_B/P_A=5.0$

A: obtains the power ratio but treats the unequal $V/I$ values as equal, classifying the resistor as ohmic.
B: uses the current ratio, $0.80/0.40=2.0$ as the power ratio and omits the change in potential difference.
D: assumes constant resistance and uses $P\propto I^2$ giving a ratio of $4.0$

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

Question 3

Multiple choiceMedium1 mark

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

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

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