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Experimental Report 1 MEASUREMENT OF BASIC LENGTH Name: UHUST.COM T
Verification of the instructors AILIEUHUST.COM TAILI ID: Class: Group: I, EXPERIMENT T AL RESULTS AILIEUHUST .COM TAILIEUHUST.COM 1. Measurement Report:
Table 1: Metal hollow cylinder’s values
𝛥 = 0.02 a = 1 𝑀 = 35.10 𝑔 HUST.COM TAILIEUHUST.COM TAILIEU Trial D(mm) d(mm) h(mm) 1 43.74 35.20 7.96 2 43.76 35.22 7.98 3 43.76 35.20 7.96 4 43.74 35.24 7.98 5 43.78 35.20 7.98 TAILIEUHUST.COM TAILIEUHUST.COM Average value 43.76 35.21 7.97 UST.COM TAILIEUHUST.COM TAILIEUH
EUHUST.COM TAILIEUHUST.COM TAILIEUHUST.COM
Table 2: Steel ball’s diameter 𝛥 = 0.01 a = 0.5 ST .COM T Trial 1 AILIEUHUST 2 3 .COM T 4 5 A AILIEUHUST verage value .COM TAILIEU Parameter Db (mm) 10.04 10.25 10.05 10.04 10.03 10.04 M T
2. Calculation average values and uncertainties:
AILIEUHUST.COM TAILIEUHUST.COM TAILIEUHUST.
a.) For the Metal hollow cylinder 󰆽=𝐷
5=43.74 +43.76 +43.73 +43.74 +43.78
𝐷 1+ 𝐷2+ 𝐷3+ 𝐷4+ 𝐷5 5≈43.76( ) 𝑚𝑚
𝑑󰆽=𝑑1+ 𝑑2+ 𝑑3+ 𝑑 5 4 =+ 𝑑 35. 5
20 +35.22 + 35.20 +35.24 +35.20 5≈35.21(𝑚𝑚) ℎ󰆽=ℎ1+ ℎ2+ ℎ3+ ℎ 5 4 =+ 7. ℎ5 96 + 7.98 + 7.96 + 7.98 + 7.98 5≈ 7.97(𝑚𝑚)
EUHUST.COM TAILIEUHUST.COM TAILIEUHUST.COM T
𝛥𝐷 = 𝑠. 𝑑. = √(𝐷󰆽− 𝐷1)2+(𝐷 󰆽− 𝐷2)2+(𝐷 󰆽− 𝐷3)2+(𝐷 󰆽− 𝐷4)2+(𝐷 󰆽− 𝐷5)2 5
=√(43.74 −43.76)2+(43.76 −43.76)2+ 43 ( .76 −43.76)2+ 43 ( .74 −43.76)2+ 43 ( .78 −43.76)2 5
≈ 0.02 ⇒ 𝑆. 𝐷. = 𝑠. 𝑑 √5.=0.02 √5≈ 0.01 𝑚𝑚
ST.COM TAILIEUHUST.COM TAILIEUHUST.COM TAILIEU
𝛥ℎ = 𝑠. 𝑑. = √(𝑑󰆽− 𝑑1)2+ (𝑑󰆽− 𝑑2)2+ (𝑑󰆽− 𝑑3)2+ (𝑑󰆽− 𝑑4)2+ (𝑑󰆽− 𝑑5)2 5
=√(7.96 − 7.97)2+(7.98 − 7.97)2+(7.96 − 7.97)2+(7.98 − 7.97)2+(7.98 − 7.97)2 5
≈ 0.01 ⇒ 𝑆. 𝐷. = 𝑠. 𝑑 √5.=0.01 √5≈ 0.00𝑚𝑚
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𝛥𝑑 = 𝑠. 𝑑. = √(ℎ󰆽− ℎ1)2+ (ℎ󰆽− ℎ2)2+ (ℎ󰆽− ℎ3)2+ (ℎ󰆽− ℎ4)2+ (ℎ󰆽− ℎ5)2 5
=√(35.20 −35.21)2+(35.22 −35.21)2+ 35 ( .20 −35.21)2+ 35 ( .24 −35.21)2+ 35 ( .20 −35.21)2 5
ST.COM TAILIEUHUST.COM TAILIEUHUST.COM TAILIEU
≈ 0.02 ⇒ 𝑆. 𝐷. = 𝑠. 𝑑 √5.=0.02 √5≈ 0.01𝑚𝑚
⇨ 𝑉󰆽=𝜋 4(𝐷󰆽2− 𝑑󰆽2).ℎ󰆽=3,14 4(43.762−35.212)× 7.97 ≈ 4.22 ×103(𝑚𝑚3) = 4.22 ×10−6(𝑚3)
⇨ 𝜌 =𝑚𝑉󰆽=35.14.022×103≈ 8.32 × 10−3𝑔/𝑚𝑚3= 8.32 × 103𝑘𝑔/𝑚3 b.) For the steel ball M T 󰆽𝑏=𝐷𝑏1+ 𝐷𝑏2 AILIEUHUST +𝐷𝑏3+𝐷 5 𝑏 = 4 .COM+ 10. 𝐷𝑏 04 T 5+10.05 +10.05 +10.04 +10 AILIEUHUST .03 .COM TAILIEUHUST. 𝐷 5 ≈10.04(𝑚𝑚)
𝑠. 𝑑. = 𝛥𝐷𝑏=√(𝐷 󰆽𝑏− 𝐷𝑏1)2+ (𝐷 󰆽𝑏− 𝐷𝑏2)2+ (𝐷 󰆽𝑏− 𝐷𝑏3)2+ (𝐷 󰆽𝑏− 𝐷𝑏4)2+ (𝐷 󰆽𝑏− 𝐷𝑏5)2 5
=√(10.04 −10.04)2+(10.05 −10.04)2+ 10 ( .05 −10.04)2+ 10 ( .04 −10.04)2+ 10 ( .03 −10.04)2 5
EUHUST.COM TAILIEUHUST.COM TAILIEUHUST.COM T ≈ 0.01(mm)
⇨ 𝑉󰆽𝑏=1 6. 𝜋. 𝐷󰆽𝑏3=1 6× 3.14 ×10.043≈ 0.53 × 103(𝑚𝑚3) = 0.53 × 10−6𝑚3
3. Calculation the uncertainties of volume and density:
a/ For metal hollow cylinder: with 𝑀 = 35.10 𝑔  = 𝜋 V 2 – d2 ).h 4 (D
ST.COM TAILIEUHUST.COM TAILIEUHUST.COM TAILIEU
⇨ ∆V = V √(𝛥𝜋 )2+ (∆(𝐷2𝐷−2𝑑−2𝑑)2)2+ (∆ℎ ℎ)2
= V√(∆𝜋 )2+ ( 2 √(∆𝐷𝐷 )2+ (∆𝑑  )2×1 𝐷2−𝑑2)2+ (∆ℎℎ )2
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= 4.22 ×103×√(0,013,14)2+ ( 2 √(0,0423.76)2+ (0,07.197)2×1 43.762−7.972)2+ ( 0,0325. 1)2
≈13.65 (mm 3) = 0.01 × 10 −6 (m 3)
ST.COM TAILIEUHUST.COM TAILIEUHUST.COM TAILIEU
Then V = (4.22 ± 0.01) × 10 −6 (m 3) Hence:
V = (4.22± 0.01) × 10 −6 (m 3) M T
𝑊𝑒 ℎ𝑎𝑣𝑒: 𝜌 = 𝑚𝑉󰆽⇒𝛥𝜌 AILIEUHUST = 𝜌√(𝛥𝑉 .COM𝑉)2=
T (8.27 × 10−3)√(0.014. × 22 10 × AILIEUHUST 3103)2≈ 0.01 .COM × 10 T −3𝑔/𝑚𝑚3 AILIEUHUST. = 0.01 × 103𝑘𝑔/𝑚3 Hence
𝜌 = (8.27 ± 0.01) × 103𝑘𝑔/𝑚3 b/ For steel ball:
EUHUST.COM TAILIEUHUST.COM TAILIEUHUST.COM T 𝑉󰆽𝑏=1 󰆽
6. 𝜋. 𝐷󰆽3⇒ 𝛥𝑉𝑏= 𝑉󰆽𝑏√(𝛥𝜋𝜋)2+ (3 × 𝛥𝐷𝑏𝑏)2= (0.53 × 103)√(0.01 3.14)2+(3 × 0.01 𝑏 10.04)2 𝐷
= 2.31𝑚𝑚3≈ 0.00 × 10−6𝑚3 Hence
𝑉𝑏= (0.53 ± 0.00) × 10−6 𝑚3
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