**Online Experiment 1: Error and data Analysis **

*Lab Group: 13 *

Phys 2125- 11149

__The instruction pages:__

__OBJECTIVE:__

To study the concept of significant figures, and to calculate errors in measurements.

__EQUIPMENT:__

- Pencil/pen and ruler
- Sheets of graph paper
- Calculator / computer with some graphing software
__DIAGRAM:__

__THEORY AND EQUATIONS:__

__ANSWERS TO QUESTIONS:__

**Q1)** In an experiment to measure the value of ^{π} , the following results are obtained:

3.12 3.10 3.19 3.09 3.10

- Calculate the Mean Value. ___
**12**_____ - Calculate the percent Error. ___
**64%**___ - Calculate the True Standard Deviation. ___
**0406**_ _

**Q2) **One of the equations of one-dimensional motion is: X = Xo + V T, where X is the position of the object in meters at time T (in seconds), Xo is the initial position of the object in meters (i.e. at time = 0.0), and V is the average velocity during this time in meters per second. Since this is a linear equation, the plot between X (on the y-axis) and T (in the x-axis) should be a straight line, with the slope being the value of the velocity V.

In an experiment, the following data has been obtained for position of an object at different times.

Time (seconds) T 0 1 2 3 4 5

Position (meters) X 0 2.7 3.6 6.4 7.3 9.4

- Determine the equation (write as y = mx + b) for the best-fit line from this data, by using the method of least squares. (6 points)

Equation: ___**y=1.86x+0.35**_________________________

- Draw a graph of this data by using the scale: on X-axis 5 squares = 1.00 s, on Y-axis 5 squares = 1.00 m. On the graph, clearly label the axes, plot the data points, and draw the straight line representing the equation that you found in part ‘a’. (Note: this best-fit line will not pass through all the data points). (3+3 points)

- Use the slope of the line to find the velocity. (2 points) Velocity: ___
**82m/s**________

__E-01 Error and data Analysis__

- A metal cylinder is measured to have a length of 5.00 cm and a diameter of 1.26 cm. Compute the volume of the cylinder in cm
^{3}. Use pi =3.14159. Write your final answer with the correct number of significant figures.

h=5.0 cm d=1.26cm

pi=3.1415

Volume of Cylinder: pi*r^2*h r=d/2

so V= pi*(d/2)^2*h ----> V= (pi*d^2*h)/4 ----> V= (3.14159*1.26^2*5)/4 ----> **V=6.2 cm^3 **

- In an experiment to determine the value of π, a sphere is measured to have an average value of 6.37 cm for its diameter and an average value of 20.11 cm for its circumference. What is the experimental value of π to the correct number of significant figures? d=6.37 cm

C=20.11 cm

C=pi*d ----> pi=C/d ---> pi =20.11/6.37 ----> **pi= 3.16 **

- If the theoretical value of a quantity is 3.142, and the calculated value is 3.16, what is the fractional error?

Theoretical Value: V*th*= 3.142

Calculated Value: V*c*=3.16

Fractional Error= (V*c*-V*th*)/V*th*----> (3.16-3.142)/3.142 ----> **Fractional Error= 0.0057**

- In an experiment to measure the acceleration due to gravity g, two independent equally reliable measurements gave 9.67 m/s
^{2}and 9.88 m/s^{2}. Determine the percent difference of the measurements.

X1 = 9.67 m/s^{2}

X2= 9.88 m/s^{2}

Percent Difference: [(X2-X1 )/((X2+X1 )/2)]*100----> [(9.88-9.67)/((9.88+9.67)/2)] **Percent Difference= 2.15% **

- Calculate, to the correct number of significant figures: 4 + 15.1 + 17.24 + 5.004

**58.7**

- In an experiment to measure the value of pi, the following results are obtained for pi:

3.14 3.11 3.20 3.06 3.08

- Calculate the Mean Value.---> (3.14+ 3.11+ 3.20+ 3.06+ 3.08)/5---->
**Mean Value= 3.12** - Calculate the Average Deviation from the Mean---->
**03** - Calculate the: True Standard Deviation---->
**041** - Calculate the: Standard Deviation of the Mean.----->
**045** - If the correct value is 3.14159, calculate the: Percent Error---->
**127%**

- In an experiment to measure the density of iron, the following data was obtained. Book value is 7.86 g/cm
^{3}. One of the two sets of data is more precise, while one is more accurate. Which of these two sets has the greater precision and which has the greater accuracy - 7.72, 7.74. 7.73. 7.75, 7.74 b. 7.86, 7.90, 7.78, 7.93, 7.83

Standard Value is 7.86. So we can see data in set B are more close K standard value than set A. Hence **Set B is more accurate **

- What determines the number of significant figures used in reporting measured values?
**The accuracy of the calculations**b. The accuracy of the equipment

- A rectangle is measured to be 6.4 ± 0.2 cm by 8.3 ± 0.2 cm.
- Calculate its area in square cm. width= 6.4 cm ± 2 cm length= 8.3 cm ± 0.2cm

Area of Rectangle----> A= ab----> 8.3cm * 6.4 cm----> **53.12cm2 **

- Calculate the uncertainty in its area. lnA= ln(ab) lnA= lna + lnb

Change in area= A[(△a/a)+ (△b/b)]

----> 53.12cm2*[(0.2/8.3)+(0.2/6.4)]----> **2.94 cm2 **

- A rectangle is measured to be 6.4 ± 0.2 cm by 8.3 ± 0.2 cm.
- Calculate its perimeter in cm

Perimeter: z= 2(a+b)----> 2(8.3+6.4)----> **29.4cm **

- Calculate the uncertainty in its perimeter.

△*p*= ± (△L+△b)----> 0.2 cm ± 0.2cm---> **±0.4cm **

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