MEASUREMENT ERROR UNCERTAINTY PROBLEMS AND SOLUTIONS
Problems #1
Four students measure the same length of string and their results are as follows:l1 = 38.6 cm, l2 = 38.7 cm, l3 = 38.6 cm, l4 = 38.5 cm
(a) What is the average or mean measurement?
(b) What is the range of measured values?
(c) What is the mean and its absolute uncertainty for the length of string?
Answer:
(a) the average or mean measurement
lave = (l1 + l2 + l3 + l4)/nl = (38.6 cm + 38.7 cm + 38.6 cm + 38.5 cm)/4 = 38.6 cm
(b) the range of measured values
R = lmaks – lmin = 38.7 cm – 38.5 cm = 0.2 cm
(c) Repeated measurement of the same quantity can improve the overall acceptable value of that measurement. What is the mean and its absolute uncertainty for the length of string?
∆lave = R/n = 0.2 cm/4 = ±0.05 cm
but we have only 0.1 decimal place, hence dlave = ±0.1 cm
Therefore:
l ± ∆l = (38.6 ± 0.1) cm
Problems #2
What is the width and its absolute uncertainty of the object being measured in the sketch below?
Answer
w = w2 – w1 = 4.12 mm – 3.93 mm = 0.19 mm
Smallest division is ∆w = 0.01 mm
w ± ∆w = 0.19 mm ± 0.01 mm = (0.19 ± 0.01) mm
Problems #3
The analogue voltmeter below measures a voltage on a scale of zero to 5 volts. What is the measured voltage and what is the absolute uncertainty shown here?
w = w2 – w1 = 4.12 mm – 3.93 mm = 0.19 mm
Smallest division is ∆w = 0.01 mm
w ± ∆w = 0.19 mm ± 0.01 mm = (0.19 ± 0.01) mm
Problems #3
The analogue voltmeter below measures a voltage on a scale of zero to 5 volts. What is the measured voltage and what is the absolute uncertainty shown here?
Answer
Scale reads V = 3.18 V
We can discern one half the smallest division so ∆V = ±(0.5 x 0.1 V) = ±0.05V
V ± ∆V = (3.18 ± 0.05) mm
Problem #4
The digital stopwatch was started at a time t0 = 0 and then was used to measure ten swings of a simple pendulum to a time of t = 17.26 s.
If the time for ten swings of the pendulum is 17.26 s,
(a) what is the minimum absolute uncertainty in this measurement?
(b) What is the time (period T) of one complete pendulum swing and its absolute uncertainty?
Answer:
(a) the minimum absolute uncertainty in this measurement ∆(10T) = ±0.01 s
(b) the time (period T) of one complete pendulum swing and its absolute uncertainty
10T = (17.26 ± 0.01) s
T ± ∆T = (17.26 ± 0.01) s/10 = (1.726 ± 0.001) s
Problems #5
Given two masses, m1 = (100.0 ± 4.0) g dan m1 = (49.3 ± 0.3) g, what is their sum, m1 + m2 and what is their difference, m1 – m2 both expressed with uncertainties.
Answer
Sum: m1 + m2 = (149.3 ± 0.7) g
Difference: m1 – m2 = (50.7 ± 0.1) g
Problem #6
In an optical experiment a deflected ray of light is measured to be at an angle θ = (23 ± 1)0. Find the sine of this angle and then determine the minimum and maximum acceptable values of this experimental value. Next, express the experimental result in the form tan θ ± ∆tanθ.
Answer:
Sin 230 = 0.3907311 ≈ 0.39
For the maximum angle: sin (23 + 1)0 = sin 240 = 0.4067366 ≈ 0.41
For the minimum angle: sin (23 – 1)0 = sin 220 = 0.3746066 ≈ 0.37
tan θ ± ∆tan θ = (0.9 ± 0.2)
Problem #7
Density is the ratio of mass to volume, where ρ = m/V. What is the density of a material if m = 12.4 kg and V = 6.68 m3?. Determine the least uncertainty in the mass and in the volume, and then calculate the uncertainty in the density value.
Express the uncertainty in the density ρ as an absolute value ±∆ρ, as a fractional ratio ±∆ρ/ρ, and as a percentage ±∆ρ%.
Answer:
Density ρ = m/V = 12.4 kg/6.68 m3 = 1.8562874 kg/m3 ≈ 1.86 kg/m3 (to 3 SF)
Uncertainty in mass: ∆m = ±0.1 kg or (0.1 kg/12.4 kg) x 100% = 0.80645% ≈ 1%
Uncertainty in volume: ∆V = ±0.01 m3 or (0.01 kg/6.68 m3) x 100% = 0.1497006% ≈ 0.1%
Uncertainty in density is the sum of the uncertainty percentage of mass and volume, but the volume is one-tenth that of the mass, so we just keep the resultant uncertainty at 1%.
ρ = (1.86 kgm-3 ± 1%) (for a percentage of uncertainty)
Where 1% of the density is 0.0186 kgm-3 we can then write:
ρ = (1.86 kgm-3 ± 0.0186 kgm-3) = [1.86 ± (0.02/1.86)] kgm-3 for fractional uncertainty, and
ρ = (1.86 kgm-3 ± 0.0186 kgm-3) = [1.86 ± 0.02] kgm-3 for absolute uncertainty.
Problem #8
What is the uncertainty in the calculated area of a circle whose radius is determined to be r = (14.6 ± 0.5) cm?
Answer:
Area: A = πr2 where r = (14.6 ± 0.5) cm
Percentage of uncertainty: (0.5 cm/14.6 cm) x 100% = 3.42 %
The percentage is times 2 when squared = 2 x 3.42% = 6.849%
Area: A = π(14.6 cm)2 = 669.66189 cm2
Area uncertainty is about 7%, or in absolute terms, 6.849% x 699.6 cm2 = 45.86 cm2 = 46 cm2
Therefore area: A ± ∆A = (670 ± 46) cm2 ≈ (6.7 ± 0.5) x 102 cm2
Problem #9
An electrical resistor has a 2% tolerance and is marked R = 1800 Ω. What is the range of acceptable values that the resistor might have? An electrical current of I = (2.1 ± 0.1) mA flows through the resistor. What is the uncertainty in the calculated voltage across the resistor where the voltage is given as V = IR ?
Answer:
Resistance: 1800 Ω x 2% = ±36Ω with a range from 1764 Ω to 1836 Ω
Voltage: V = IR where ∆R% = 2% and ∆I% = (0.1 mA/2.1 mA) X 100% = 4.76%
The sum of the percentages is 2% + 4.76% = 6.76% ≈ 7%
and this is the uncertainty in the calculated voltage:
V = 7938 mV = 7.938 V at 6.76% x 7.938 V = 0.5366 V
Therefore: V ± ∆V = (7.9 ± 0.5) V
Problem #10
An accelerating object has an initial speed of u = (12.4 ± 0.1) ms-1 and a final speed of v = (28.8 ± 0.2) ms-1. The time interval for this change in speed is ∆t = (4.2 ± 0.1) s. Acceleration is defined as a = (v – u)/∆t. Calculate the acceleration and its uncertainty.
Answer:
a = (v – u)/∆t = (28.8 m/s – 12.4 m/s)/4.2 s = 3.90 m/s
Uncertainty in numerator: ∆u + ∆v = 0.1 m/s + 0.2 m/s = 0.3 m/s
As a percentage: (0.3 ms-1)/(28.8 m/s – 12.4 m/s) x 100% = 1.829%
Percentage in time: (0.1 s/4.2 s) x 100% = 2.380%
Percentage of uncertainty in acceleration is the sum: ∆a% = 1.829% + 2.380% = 4.209%
Therefore acceleration: 3.90 m/s ± 4% = (3.90 ± 0.16) m/s ≈ (3.9 ± 0.2) m/s
Scale reads V = 3.18 V
We can discern one half the smallest division so ∆V = ±(0.5 x 0.1 V) = ±0.05V
V ± ∆V = (3.18 ± 0.05) mm
Problem #4
The digital stopwatch was started at a time t0 = 0 and then was used to measure ten swings of a simple pendulum to a time of t = 17.26 s.
(a) what is the minimum absolute uncertainty in this measurement?
(b) What is the time (period T) of one complete pendulum swing and its absolute uncertainty?
Answer:
(a) the minimum absolute uncertainty in this measurement ∆(10T) = ±0.01 s
(b) the time (period T) of one complete pendulum swing and its absolute uncertainty
10T = (17.26 ± 0.01) s
T ± ∆T = (17.26 ± 0.01) s/10 = (1.726 ± 0.001) s
Problems #5
Given two masses, m1 = (100.0 ± 4.0) g dan m1 = (49.3 ± 0.3) g, what is their sum, m1 + m2 and what is their difference, m1 – m2 both expressed with uncertainties.
Answer
Sum: m1 + m2 = (149.3 ± 0.7) g
Difference: m1 – m2 = (50.7 ± 0.1) g
Problem #6
In an optical experiment a deflected ray of light is measured to be at an angle θ = (23 ± 1)0. Find the sine of this angle and then determine the minimum and maximum acceptable values of this experimental value. Next, express the experimental result in the form tan θ ± ∆tanθ.
Answer:
Sin 230 = 0.3907311 ≈ 0.39
For the maximum angle: sin (23 + 1)0 = sin 240 = 0.4067366 ≈ 0.41
For the minimum angle: sin (23 – 1)0 = sin 220 = 0.3746066 ≈ 0.37
tan θ ± ∆tan θ = (0.9 ± 0.2)
Problem #7
Density is the ratio of mass to volume, where ρ = m/V. What is the density of a material if m = 12.4 kg and V = 6.68 m3?. Determine the least uncertainty in the mass and in the volume, and then calculate the uncertainty in the density value.
Express the uncertainty in the density ρ as an absolute value ±∆ρ, as a fractional ratio ±∆ρ/ρ, and as a percentage ±∆ρ%.
Answer:
Density ρ = m/V = 12.4 kg/6.68 m3 = 1.8562874 kg/m3 ≈ 1.86 kg/m3 (to 3 SF)
Uncertainty in mass: ∆m = ±0.1 kg or (0.1 kg/12.4 kg) x 100% = 0.80645% ≈ 1%
Uncertainty in volume: ∆V = ±0.01 m3 or (0.01 kg/6.68 m3) x 100% = 0.1497006% ≈ 0.1%
Uncertainty in density is the sum of the uncertainty percentage of mass and volume, but the volume is one-tenth that of the mass, so we just keep the resultant uncertainty at 1%.
ρ = (1.86 kgm-3 ± 1%) (for a percentage of uncertainty)
Where 1% of the density is 0.0186 kgm-3 we can then write:
ρ = (1.86 kgm-3 ± 0.0186 kgm-3) = [1.86 ± (0.02/1.86)] kgm-3 for fractional uncertainty, and
ρ = (1.86 kgm-3 ± 0.0186 kgm-3) = [1.86 ± 0.02] kgm-3 for absolute uncertainty.
Problem #8
What is the uncertainty in the calculated area of a circle whose radius is determined to be r = (14.6 ± 0.5) cm?
Answer:
Area: A = πr2 where r = (14.6 ± 0.5) cm
Percentage of uncertainty: (0.5 cm/14.6 cm) x 100% = 3.42 %
The percentage is times 2 when squared = 2 x 3.42% = 6.849%
Area: A = π(14.6 cm)2 = 669.66189 cm2
Area uncertainty is about 7%, or in absolute terms, 6.849% x 699.6 cm2 = 45.86 cm2 = 46 cm2
Therefore area: A ± ∆A = (670 ± 46) cm2 ≈ (6.7 ± 0.5) x 102 cm2
Problem #9
An electrical resistor has a 2% tolerance and is marked R = 1800 Ω. What is the range of acceptable values that the resistor might have? An electrical current of I = (2.1 ± 0.1) mA flows through the resistor. What is the uncertainty in the calculated voltage across the resistor where the voltage is given as V = IR ?
Answer:
Resistance: 1800 Ω x 2% = ±36Ω with a range from 1764 Ω to 1836 Ω
Voltage: V = IR where ∆R% = 2% and ∆I% = (0.1 mA/2.1 mA) X 100% = 4.76%
The sum of the percentages is 2% + 4.76% = 6.76% ≈ 7%
and this is the uncertainty in the calculated voltage:
V = 7938 mV = 7.938 V at 6.76% x 7.938 V = 0.5366 V
Therefore: V ± ∆V = (7.9 ± 0.5) V
Problem #10
An accelerating object has an initial speed of u = (12.4 ± 0.1) ms-1 and a final speed of v = (28.8 ± 0.2) ms-1. The time interval for this change in speed is ∆t = (4.2 ± 0.1) s. Acceleration is defined as a = (v – u)/∆t. Calculate the acceleration and its uncertainty.
Answer:
a = (v – u)/∆t = (28.8 m/s – 12.4 m/s)/4.2 s = 3.90 m/s
Uncertainty in numerator: ∆u + ∆v = 0.1 m/s + 0.2 m/s = 0.3 m/s
As a percentage: (0.3 ms-1)/(28.8 m/s – 12.4 m/s) x 100% = 1.829%
Percentage in time: (0.1 s/4.2 s) x 100% = 2.380%
Percentage of uncertainty in acceleration is the sum: ∆a% = 1.829% + 2.380% = 4.209%
Therefore acceleration: 3.90 m/s ± 4% = (3.90 ± 0.16) m/s ≈ (3.9 ± 0.2) m/s
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