Types_of_measurement_&_Errors_measurement_grad_10_lesson_2_2.pdf

Types of measurement & Errors measurement LESSON 2 Points of Direct measurement Indirect measurement comparison Number of 1 measuring tool is used More than 1 measuring measuring tool is used tools Number of One measurement More than 1 measuring...

Types of measurement & Errors measurement LESSON 2 Points of Direct measurement Indirect measurement comparison Number of 1 measuring tool is used More than 1 measuring measuring tool is used tools Number of One measurement More than 1 measuring process measurement process processes Mathematical No mathematical A mathematical relations relations is applied relation applied to find quantity Number of 1 measurement More than 1 measurement errors error measurement error  Errors in measurements  Reasons of measurement error : 1.Choosing improper tool. 2.A defect in the measuring tool. 3.Wrong procedure due to unexperienced persons. 4.Enviromental conditions ( humidity – temperature – air current ).  Estimating error of measurement 1. Estimating error in direct measurement Absolute error (∆x ) Relative error (r): It is the difference It is the ratio between between the real the absolute error (∆X) (actual) value to the real value (𝑋0 ). (𝑋0 )and the measured value (X). r= ∆𝑥 𝑥0 ∆X = |𝑋0 -X| Q. A student measured the length of a pencil. It was found to be 9.9 cm, meanwhile its actual length is 10 cm. Another student measured the classroom length. It was found to be 9.13 m, meanwhile its actual length is 9.11 m. Estimate both the absolute error and the relative error in each case. Answer: Case (1): ∆𝑥 = 𝑥0 -x = 10-9.9 =0.1 ∆𝑥 0.1 r= = =0.01 𝑥0 10 Case (2): ∆𝑥 = 𝑥0 − 𝑥 = 9.11-9.13 = 0.02 ∆𝑥 0.02 r= = =2.195*10−3 𝑥0 9.11  2 Estimating error in case of indirect measurement Mathematical Example How to calculate operation error Summation Measuring the volume of two The absolute error = The amounts of a liquid. V = V1+ V2 absolute error in first measurement + The absolute error in second Subtraction Finding the volume of a coin by measurement subtracting the volume of water ∆𝑥 = ∆𝑥1 + ∆𝑥2 before dropping the coin into the = 𝑥01 −𝑥1 + 𝑥02 −𝑥2 measuring cylinder from that after dropping it *relative error: 𝑣𝑐𝑜𝑖𝑛 =𝑣2 −𝑣1 ∆𝑥 r= 𝑥0 Multiplication Finding the area of a The relative error = The rectangle by measuring its relative error in first length and its width then measurement + The multiplying them. relative error in second measurement r=𝑟1 +𝑟2 ∆𝑥1 ∆𝑥2 + 𝑥01 𝑥02 Division Finding the density of a liquid by measuring its mass and its volume then dividing them. The absolute error(∆𝑥): ∆𝑥 = 𝑟 ∗ 𝑥0 Calculate each of the relative error and the absolute error when measuring the area of a rectangle (A) that has length (6±0.1) m and width (5±0.2) m. Answer: ∆𝑥1 = 0.1 ∆𝑥2 = 0.2 A(𝑥0 )=length × width=5×6=30 𝑐𝑚2. To find the absolute error in the area: ∆𝑥1 ∆𝑥2 0.1 0.2 r=𝑟1 +𝑟2 = + = + =0.0567. 𝑥01 𝑥02 6 5 absolute error: ∆𝑥 = 𝑟 ∗ 𝑥0 ∆𝑥 =0.0567*30=1.701 𝑚2. To Calculate the relative error: ∆𝑥 1.7 Relative Error= = ≈0.05667 𝑥0 30 If X = 20±04, calculate the following: (a) 2 x (b)𝑥 2 Answer: (a)2𝑋 The value of 2𝑋is: 2𝑋=2×20=40 To find the absolute error in 2𝑋, we multiply the absolute error of 𝑋 by 2: Δ(2𝑋)=2×0.4=0.8 Thus, the result is: 2𝑋=40±0.8 (b)𝑥 2 𝑥 2 =202 =400 To calculate the absolute error in 𝑥 2 , we use the formula for the propagation of uncertainty for multiplication: 2 2∆𝑥 0.8 ∆𝑥 2 =𝑥. =400⋅ =400*0.04=16 𝑥 20 𝑥 2 =400±16 Thank you