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Question 20N.3.SL.TZ0.1

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Date November 2020 Marks available [Maximum mark: 9] Reference code 20N.3.SL.TZ0.1
Level SL Paper 3 Time zone TZ0
Command term Comment, Identify, Show that, Sketch, State, Suggest Question number 1 Adapted from N/A
1.
[Maximum mark: 9]
20N.3.SL.TZ0.1

A spherical soap bubble is made of a thin film of soapy water. The bubble has an internal air pressure Pi and is formed in air of constant pressure Po. The theoretical prediction for the variation of Pi-Po is given by the equation

(Pi-Po)=4gR

where γ is a constant for the thin film and R is the radius of the bubble.

Data for Pi-Po and R  were collected under controlled conditions and plotted as a graph showing the variation of Pi-Po with 1R.

(a)

Suggest whether the data are consistent with the theoretical prediction.

[2]

Markscheme

«theory suggests» Pi-Po is proportional to 1R 


graph/line of best fit is straight/linear «so yes»
OR
graph/line of best fit passes through the origin «so yes»

 

MP1: Accept ‘linear’

MP2 do not award if there is any contradiction
eg: graph not proportional, does not pass through origin.

Examiners report

Many students obtained full marks here although a significant number did not acknowledge that the graph was through the origin and lost a mark.

(b(i))

Show that the value of γis about 0.03.

[2]

Markscheme

gradient = «4γ» =0.10
OR
use of equation with coordinates of a point 


γ=0.025 

 

MP1 allow gradients in range 0.098 to 0.102

MP2 allow a range 0.024 to 0.026 for γ

 

Examiners report

Very well answered either by obtaining the gradient or replacing with the coordinates of a point.

(b(ii))

Identify the fundamental units of γ.

[1]

Markscheme

kgs-2

 

Accept kgs2

 

Examiners report

Although the question was specifically about the fundamental units, several candidates lost the mark by answering Pa m.

(b(iii))

In order to find the uncertainty for γ, a maximum gradient line would be drawn. On the graph, sketch the maximum gradient line for the data.

[1]

Markscheme

straight line, gradient greater than line of best fit, and within the error bars

Examiners report

Almost all candidates were able to draw the correct maximum gradient line.

(b(iv))

The percentage uncertainty for γ is 15%. State γ, with its absolute uncertainty.

[2]

Markscheme

«15% of 0.025» = 0.00375
OR
«15% of 0.030» = 0.0045

rounds uncertainty to 1sf
±0.004
OR
±0.005

 

Allow ECF from (b)(i)
Award
[2] marks for a bald correct answer

Examiners report

Well answered. A significant number did not round the uncertainty to match the value of gamma.

(b(v))

The expected value of γ is 0.027. Comment on your result.

[1]

Markscheme

Experimental value matches this/correct, as expected value within the range
OR
experimental value does not match/incorrect, as it is not within range