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Report 1 Measurement of basic constants Length, weight and time Hanoi University of Technology Instructor Class: BME-K54 Group: 1 Name: Phí Ngọc Tú I. EXPERIMETAL MOTIVATION Use the vernier caliper to measure the height , diameter of hollow cylinder and use vernier micrometer to measure diameter of steel ball. Then calculate the volume of cylinder and ball, the density of hollow cylinder by using some equation that learned. II. EXPERIMENTAL RESULTS 1. Table measurement data height (h), external and internal diameter( D and d, respectively) of metal hollow cylinder. Trial | D (mm) | Di- D(mm) | d (mm) | di- d (mm) | h (mm) | hi–h (mm) | 1 | 34,92 | 0,008 | 28.10 | 0,012 | 9,44 | 0,008 | 2 | 34,90 | -0,012 | 28,08 | -0,008 | 9,44 | 0,008 | 3 | 34,92 | 0,008 | 28,08 | -0,008 | 9,42 | -0,012 | 4 | 34,92 | 0,008 | 28,10 | 0,012 | 9,42 | -0,012 | 5 | 34,90 | -0,012 | 28,08 | -0,008 | 9,44 | 0,008 | Average | D=34,912 | | d=28,088 | | h=9,432 | | 2. Table measurement data diameter (Db) of the small ball. Trial | Db1(mm) | Db1 - Db1(mm) | Db2(mm) | Db2 - Db2(mm) | 1 | 11,83 | 0,000 | 7,51 | 0,004 | 2 | 11,84 | -0,010 | 7,52 | -0,006 | 3 | 11,83 | 0,000 | 7,52 | -0,006 | 4 | 11,82 | 0,010 | 7,51 | 0,004 | 5 | 11,83 | 0,000 | 7,51 | 0,004 | Average | D=11,830 | | D=7,514 | | 3. Calculate the volume and density of the metal hollow cylinder. Using the following equations. V = π4(D2- d2)h ρ = mV V = π4(34,9122 – 28,0882)9,432 = 3184,734 ρ = 26,523184,734 = 8,33 x10-3 (g/mm3) 4. Calculate the volume of the steel ball. Using the following equation. Vb = 16πDb3 Vb1 = 16π11,8303 = 866,868 (mm)3 Vb2 = 16π7,5143 = 222, 133 (mm)3 5. Calculate and comment the unsertanties of the volume and density of the metal hollow cylinder as well as the volume of the steel ball. Standard deviation of a single measurement Δx s.d. = Δx = i=1n(xi - x)2n + Metal hollow cylinder. The uncertainties of D, d and h ΔD = 0,0082+0,0122+0,0082+0,0082+0,01225 = 0,010 (mm) Δd = 0,0122+0,0082+0,0082+0,0122+0,00825 = 0,010 (mm) Δh = 0,0082+0,0082+0,0122+0,0122+0,00825 = 0,010 (mm) ΔV= (∂V∂DΔD)2+(∂V∂dΔd)2+ (∂V∂hΔh)2 = π2DhΔD2+π2dhΔd2+ π4D2- d2Δh2 = 7,45 (mm)3 Δρ = ∂ρ∂mΔm2+ ∂ρ∂VΔV2 = 1VΔm2+ -mV2ΔV2 = 0,02 x10-3 (g/mm3) + metal ball ΔD1 = 0,012+ 0,0125 =0,006 (mm) ΔD2 = 0,0042+ 0,0062+ 0,0062+ 0,0042+ 0,00425 = 0,005 (mm) ΔVb1 = ∂V1∂D1ΔD12 = 12πD12ΔD1 = 1,148 mm3 ΔVb2 = ∂V2∂D2ΔD22 = 12πD22ΔD2 = 0,666 mm3 6. The volume of the metal hollow cylinder V = V± ΔV = 3184,374 ±7,45 mm 3 The density of the metal hollow cylinder ρ=ρ±Δρ = 8,33 x10-3 (g/mm3) Volume of the ball one. V = V1± ΔV1 = 866,868 ±1,148 mm 3 Volume of the ball two V = V2± ΔV2 =222,133 ± 0,666 mm3