9702/May Jun/12/2015/Q6

A single sheet of aluminium foil is folded twice to produce a stack of four sheets. The total thickness of the stack of sheets is measured to be (0.80 ± 0.02) mm. This measurement is made using a digital caliper with a zero error of (−0.20 ± 0.02) mm.

What is the percentage uncertainty in the calculated thickness of a single sheet?

A 1.0%                        B 2.0%                        C 4.0%                        D 6.7%

Solution:

Answer: C

This challenging question involves both random and systematic error.

The zero error (−0.20) can be removed but its uncertainty (± 0.02) must be added to the measurement uncertainty.

True value of total thickness = 0.80 + 0.20 = 1.0 mm  

Total uncertainty in measurement = ± (0.02 + 0.02) = ± 0.04

So the four sheets have a true thickness of (1.00 ± 0.04) mm.

A single sheet would have a thickness of (0.25 ± 0.01) mm.

Percentage error = (0.01 / 0.25) × 100% = 4%

Reference: PYQ - Oct/Nov 2015 Paper 12 Q6

9702/Oct Nov/11/2015/Q6

A light-meter measures the intensity I of the light incident on it. Theory suggests that I varies
inversely as the square of the distance d.

Which graph of the results supports this theory?

Solution:
Answer: C

Theory suggests that the intensity I varies as the inverse square of the distance d.

I α 1 / d²

I = k (1 / d²) = k / d²    where k is a constant

Consider the graph of I (on y-axis) against 1/d² (on x-axis).

Equation of a line: y = mx + x

Compare I = k (1 / d²) = k / d² with the equation y = mx + c:

y = I

x = 1 / d²

m = k

c = 0

So, if a graph of I against 1/d² is plotted, it should be a straight line with a positive gradient (the constant k is positive) and starting at the origin (y-intercept, c = 0 – that is, it intercepts the y-axis at y=0). [C is correct]

Options A and B both have the graph cutting an axis, which would not be the case for an inverse square law graph.

A graph of I against d² would be similar to a graph of y = 1 / x where y = I and x = d². Such a graph is a curve with a negative gradient and not touching the axes.

Type: graph y=1/x

at Google search to see to shape of the graph.

Reference: PYQ - Oct/Nov 2015 Paper 11 Q6

9702/Oct Nov/11/2016/Q6

Question 6

A cyclist pedals along a raised horizontal track. At the end of the track, he travels horizontally into

the air and onto a track that is vertically 2.0 m lower.

The cyclist travels a horizontal distance of 6.0 m in the air. Air resistance is negligible.

What is the horizontal velocity v of the cyclist at the end of the higher track?

A      6.3 ms^–1 

B      9.4 ms^–1 

C      9.9 ms^–1 

D      15 ms^–1

Solution:

There are important points to note in this question:

The horizontal velocity v is used to calculate the horizontal distance

The time to reach the maximum height is the time to travel the horizontal distance

At maximum height u = 0

Using H = ut + 1/2gt^2

2 = 0 + 1/2×9.81xt^2

(t=0.6395s)

Horizontal distance = horizontal velocity(v) x time(t)

6 = 0.6395v

V = 9.4ms^-2 (B is the correct option)

Ans: B

Reference PYQ - Oct/Nov 2016 Paper 11 Q6