Laporan praktikum fisika dasar 2 "dasar pengukuran dan ketidak pastian" untuk jurusan fisika khususnya yang berbahasa inggris.

1. 1
MEASUREMENT OF FUNDAMENTAL QUANTITIES AND
UNCERTAINTY
Frafti Rejeki S*), Karnila Sari, Lismawati, Mayang Segara
Fundamental Physics Laboratory Unit Physics Department
Universitas Negeri Makassar 2015
Abstact.
Key words : accuracy, measurement device, smallest value of scale, , uncertainty
RESEARCH QUESTIONS
1. How the uses of some basic devices measurement?
2.
EXPERIMENTAL RESULTS AND DATA ANALYSIS
EXPERIMENTAL RESULTS
1. Measurement of length
 NST Ruler =
batas ukur
jumlahskala
=
1 cm
10 scale
= 0.1 cm = 1 mm
∆x =
1
2
NST alat =
1
2
× NST =
1
2
× 1 mm = 0.5 mm
 NST vernier calipers
39 main scale = 20 skala nonius
1 skala nonius =
39
20
mm = 1.95 mm
NST jangka sorong = 2 mm – 1.95 mm = 0.05 mm
∆x =
1
1
NST alat =
1
1
× NST =
1
1
×0.05 mm = 0.05 mm
 NST micrometer =
nilai skala mendatar
jumlah skala putar
=
0,5 mm
50
= 0.01 mm
∆x =
1
2
NST alat =
1
2
× NST =
1
2
× 0.01 mm = 0.005 mm

2. 2
Table 1. Measurement Of Length
No.
Benda
Yang
Diukur
Besaran
yang
diukur
Measurement results (mm)
ruler Jangka Sorong Mikrometer Sekrup
1 Balok
kubus length
|18.5 ± 0.5| .|19.10 ± 0.05| |19.430 ± 0.005|
|18.5 ± 0.5| |19.10 ± 0.05| |19.410 ± 0.005|
|18.0 ± 0.5| |19.05 ± 0.05| |19.415 ± 0.005|
width
|18.0 ± 0.5| |19.05 ± 0.05| |18.383 ± 0.005|
|18.5 ± 0.5| |19.05 ± 0.05| |18.380 ± 0.005|
|18.0 ± 0.5| |19.05 ± 0.05| |18.390 ± 0.005|
high
|17.0 ± 0.5| |18.05 ± 0.05| |18.340 ± 0.005|
|17.5 ± 0.5| |18.10 ± 0.05| |18.335 ± 0.005|
|17.5 ± 0.5| |18.10 ± 0.05| |18.330 ± 0.005|
2 Ball
Diameter
|19.0 ± 0.5| |19.65 ± 0.05| |20.355 ± 0.005|
|19.0 ± 0.5| |19.60 ± 0.05| |20.335 ± 0.005|
|19.0 ± 0.5| |19.55 ± 0.05| |20.395 ± 0.005|
2. Mass measurement dengan Neraca ohauss 2610 gram
Nilai skala lengan 1 = 100 gr
Nilai skala lengan 2 = 10 gr
Nilai skala lengan 3 = 0.1 gr ⇨ Δx = ½ NST = ½ 0.1 gr = 0.05 gr
Massa beban gantung = 1000 gr
Table 2. Hasil pengukuran massa dengan Neraca Ohauss 2610 gram
Benda
Penun.
Lengan
1
Penun. Lengan
2
Penun.Lengan
3
Beban
Gantung
Massa Benda (g)
Balok
Kubus
0 |50.00 ± 0.05| |3.80 ± 0.05| 0 |53.80 ± 0.05|
0 |50.00 ± 0.05| |3.85 ± 0.05| 0 |53.85 ± 0.05|
0 |50.00 ± 0.05| |3.75 ± 0.05| 0 |53.75 ± 0.05|

3. 3
Bola
0 |30.00 ± 0.05| |2.40 ± 0.05| 0 |32.40 ± 0.05|
0 |30.00 ± 0.05| |2.60 ± 0.05| 0 |32.60 ± 0.05|
0 |30.00 ± 0.05| |2.50 ± 0.05| 0 |32.50 ± 0.05|
3. Pengukuran massa dengan neraca ohauss 311 gram
Nilai skala lengan 1 = 100 gr
Nilai skala lengan 2 = 10 gr
Nilai skala lengan 3 = 1 gram
Nilai skala lengan 4 = 0.01 gram ⇨ Δx = ½ NST = ½ × 0.01 gr = 0.005 gr
Table 3. mass measurement results with Neraca Ohauss 311 gram
Benda
Penun.
Lengan
1
Penun.
Lengan 2
Penun.Lengan
3
Penun.
Lengan 4
Massa Benda (g)
Balok
Kubus
0 |
50.000
±0.005
| |
3.000
±0.005
| |
0.880
±0.005
| |53.880 ± 0.005|
0 |
50.000
±0.005
| |
3.000
±0.005
| |
0.770
±0.005
| |53.770 ± 0.005|
0 |
50.000
±0.005
| |
3.000
±0.005
| |
0.880
±0.005
| |53.880 ± 0.005|
Bola
0 |
30.000
±0.005
| |
2.000
±0.005
| |
0.540
±0.005
| |32.540 ± 0.005|
0 |
30.000
±0.005
| |
2.000
±0.005
| |
0.550
±0.005
| |32.550 ± 0.005|
0 |
30.000
±0.005
| |
2.000
±0.005
| |
0.540
±0.005
| |32.540 ± 0.005|
4. Hasil pengukuran dengan neraca ohauss 310 gram
Nilai skala lengan 1 = 100 gr
Nilai skala lengan 2 = 10 gr
Nilai skala putar = 0.1 gr
Jumlah skala nonius = 10 skala
NST =
nilai skala putar
jumlah skala nonius
=
0.1
10
= 0.01 gr

4. 4
Table 4. Hasil pengukuuran massa dengan Neraca Ohauss 310 gram
Benda
Penun.
Lengan
1
Penun.
Lengan 2
Penun. Skala
putar
Penun. Skala
Mendatar
Massa Benda (g)
Balok
Kubus
0 |
50.00
±0.01
| |
37.00
±0.01
|
|0.07 ± 0.01| |53.77 ± 0.01|
0 |
50.00
±0.01
| |
38.00
±0.01
|
|0.07 ± 0.01| |53.87 ± 0.01|
0 |
50.00
±0.01
| |
37.00
±0.01
| |0.02 ± 0.01| |53.72 ± 0.01|
Bola
0 |
30.00
±0.01
| |
24.00
±0.01
|
|0.05 ± 0.01| |32.45 ± 0.01|
0 |
30.00
±0.01
| |
24.00
±0.01
|
|0.01 ± 0.01| |32.41 ± 0.01|
0 |
30.00
±0.01
| |
24.00
±0.01
| |0.07 ± 0.01| |32.47 ± 0.01|
5. Pengukuran Waktu dan Suhu
NST thermometer =
𝐵𝑈
𝐽𝑆
=
10°𝐶
10
= 1°𝐶 ⇨ 𝛥𝑋 =
1
2
× 1°𝐶 = 0.5 °𝐶
Temperature mula-mula (To) = 30°C + 2°C = 32°C.
NST Stopwatch =
30
300
= 0.1 𝑆.
Table 5. Hasil pengukuran waktu dan suhu
No. Waktu (s) Temperatur (°𝐶) Perubahan Temperatur (°𝐶)
1 |60 ± 0.1| |34.0 ± 0.5| |2 ± 1|
2 |120 ± 0.1| |37.0 ± 0.5| |5 ± 1|
3 |180 ± 0.1| |40.0 ± 0.5| |8 ± 1|
4 |240 ± 0.1| |43.0 ± 0.5| |11 ± 1|
5 |300 ± 0.1| |46.0 ± 0.5| |14 ± 1|
6 |360 ± 0.1| |48.0 ± 0.5| |16 ± 1|
ANALISIS DATA
A. Pengukuran panjang
1. Balok Kubus
1.1.Mistar
Pengukuran panjang dengan menggunakan mistar

5. 5
Diketahui :
𝑥1=|18.5 ± 0.5| mm
𝑥2=|18.5 ± 0.5| mm
𝑥3=|18.0 ± 0.5| mm
𝑥̅ =
𝑥1 + 𝑥2 + 𝑥3
3
=
18.5 + 18.5 + 18.0
3
= 18.3 𝑚𝑚
𝛿1= | 𝑥1− 𝑥̅| = |18.5 − 18.3| mm = 0.2 𝑚𝑚
𝛿2= | 𝑥2− 𝑥̅| = |18.5 − 18.3| mm = 0.2 𝑚𝑚
𝛿3= | 𝑥3− 𝑥̅| = |18.0 − 18.3| mm = 0.3 𝑚𝑚
𝛿 𝑚𝑎𝑘𝑠 = 0.3 𝑚𝑚 ⇨ 𝛥𝑥 = 0.3 𝑚𝑚
KR =
∆𝑋
𝑋
× 100% =
0,3
18,3
× 100% = 0.01% (4 AP) sehingga
HP = [ X ± 𝛥𝑋 ] = [ 18.3 ± 0.3 ] 𝑚𝑚
Ketiga nilai X yaitu 𝑥1, 𝑥2,𝑑𝑎𝑛 𝑥3 , tercakup dalam interval [ 18.3 + 0.3 ] 𝑚𝑚 sampai
[ 18.3 – 0.3] mm.
Jika 𝛥𝑋 = 𝛿 𝑟𝑎𝑡𝑎−𝑟𝑎𝑡𝑎, maka :
𝛥𝑋 =
(0.2+0.2+0.7) 𝑚𝑚
3
= 0,3 mm
Jadi, {X} = [ X ± 𝛥𝑋 ] = [18.3 ± 0.3 ] mm
Pengukuran lebar dengan menggunakan mistar
Diketahui :
𝑥1=|18.0 ± 0.5| mm
𝑥2=|18.5 ± 0.5| mm
𝑥3=|18.0 ± 0.5| mm
𝑥̅ =
18.0+ 18.5+18.0
3
mm = 18.2 𝑚𝑚
𝛿1 = |18.0 − 18.2| mm = 0.2 𝑚𝑚
𝛿2 = |18.5 − 18.2| mm = 0.3 𝑚𝑚
𝛿3 = |18.0 − 18.2| mm = 0.2 𝑚𝑚
𝛿 𝑚𝑎𝑘𝑠 = 0.3 𝑚𝑚 ⇨ 𝛥𝑥 = 0.3 𝑚𝑚 ,
𝐾𝑅 =
∆𝑋
𝑋
× 100% =
0.3
18.2
𝑚𝑚 × 100% = 0.02 (4 𝐴𝑃) sehingga :

6. 6
HP = |18.20 ± 0.30| 𝑚𝑚
Ketiga nilai X yaitu 𝑥1, 𝑥2,𝑑𝑎𝑛 𝑥3 , tercakup dalam interval [ 18.2 + 0.3 ] 𝑚𝑚 sampai
[ 18.2 – 0.3 ] mm.
Jika 𝛥𝑋 = 𝛿 𝑟𝑎𝑡𝑎−𝑟𝑎𝑡𝑎, maka :
𝛥𝑋 =
(0.2+0.3+0.3) 𝑚𝑚
3
= 0.2 mm
Jadi, {X} = [ X ± 𝛥𝑋 ] = [18.2 ± 0.2 ] mm
Pengukuran tinggi dengan menggunakan mistar
Diketahui :
𝑥1=|17.0 ± 0.5| mm
𝑥2=|17.5 ± 0.5| mm
𝑥3=|17.5 ± 0.5| mm
𝑥̅ =
17.0+ 17.5+17.5
3
mm = 17.3 𝑚𝑚
𝛿1 = |17.0 − 17.3| mm = 0.3 𝑚𝑚
𝛿2 = |17.5 − 17.3| mm = 0.2 𝑚𝑚
𝛿3 = |17.5 − 17.3| mm = 0.2 𝑚𝑚
𝛿 𝑚𝑎𝑘𝑠 = 0.3 𝑚𝑚 ⇨ 𝛥𝑥 = 0.3 𝑚𝑚 ,
𝐾𝑅 =
0.3
17.3
× 100% = 0.01 (4 𝐴𝑃)sehingga :
HP = | 𝑋 ± ∆𝑋| = |17.30 ± 0.30| 𝑚𝑚
Ketiga nilai X yaitu 𝑥1, 𝑥2,𝑑𝑎𝑛 𝑥3, tercakup dalam interval [ 17,3 + 0,3 ] 𝑚𝑚 sampai [
17,3 - 0,3 ] mm.
Jika 𝛥𝑋 = 𝛿 𝑟𝑎𝑡𝑎−𝑟𝑎𝑡𝑎, maka :
𝛥𝑋 =
(0,3+0,2+0,2) 𝑚𝑚
3
= 0,2 mm
Jadi, {X} = [ X ± 𝛥𝑋 ] = [17.3 ± 0.2 ] mm
1.2. Jangka sorong

7. 7
Pengukuran Panjang dengan menggunakan jangka sorong
Diketahui :
𝑥1=|19.10 ± 0.05| mm
𝑥2=|19.10 ± 0.05|mm
𝑥3=|19.05 ± 0.05|mm
𝑥̅ =
19.10+ 19.10+19.05
3
mm = 19.08 𝑚𝑚
𝛿1 = |19.10 − 19.08| mm = 0.02 𝑚𝑚
𝛿2 = |19.10 − 19.08| mm = 0.02 𝑚𝑚
𝛿3 = |19.05 − 19.08| mm = 0.03 𝑚𝑚
𝛿 𝑚𝑎𝑘𝑠 = 0.03 𝑚𝑚 ⇨ 𝛥𝑥 = 0.03 𝑚𝑚 ,
KR=
0.03
19.08
× 100% = 0.001% (4 𝐴𝑃), sehingga :
HP = |19.08 ± 0.03| 𝑚𝑚
Ketiga nilai X yaitu 𝑥1, 𝑥2,𝑑𝑎𝑛 𝑥3 , tercakup dalam interval [ 19,08 + 0,03 ] 𝑚𝑚 sampai
[ 19,08 - 0,03 ] mm.
Jika 𝛥𝑋 = 𝛿 𝑟𝑎𝑡𝑎−𝑟𝑎𝑡𝑎, maka :
𝛥𝑋 =
(0,02+0,02+0,03) 𝑚𝑚
3
= 0,02 mm
Jadi, {X} = [ X ± 𝛥𝑋 ] = [19,08 ± 0,02 ] mm
Pengukuran Lebar dengan menggunakan jangka sorong
Diketahui :
𝑥1=|19.05 ± 0.05| mm
𝑥2=|19.05 ± 0.05|mm
𝑥3=|19.05 ± 0.05|mm
𝑥̅ =
19,05+ 19,05+19,05
3
mm = 19.05 𝑚𝑚
𝛿1 = |19.05 − 19.05| mm = 0 𝑚𝑚
𝛿2 = |19.05 − 19.05| mm = 0 𝑚𝑚

8. 8
𝛿3 = |19.05 − 19.05| mm = 0 𝑚𝑚
𝛿 𝑚𝑎𝑘𝑠 = 0 𝑚𝑚 ⇨ 𝛥𝑥 =
1
2
𝑁𝑆𝑇 =
1
2
× 1 𝑚𝑚 = 0.5 𝑚𝑚
KR=
0.5
19.05
× 100% = 0.02 (4 AP), sehingga :
HP = |19.05 ± 0.50| 𝑚𝑚
Ketiga nilai X yaitu 𝑥1, 𝑥2,𝑑𝑎𝑛 𝑥3 , tercakup dalam interval [ 19,05 + 0,50 ] 𝑚𝑚 sampai
[ 19,05 - 0,50 ] mm.
Jika 𝛥𝑋 = 𝛿 𝑟𝑎𝑡𝑎−𝑟𝑎𝑡𝑎, maka :
𝛥𝑋 =
(0+0+0) 𝑚𝑚
3
= 0 mm
Jadi, {X} = [ X ± 𝛥𝑋 ] = [19,05 ± 0 ] mm
Pengukuran Tinggi dengan menggunakan jangka sorong
Diketahui :
𝑥1=|18.05 ± 0.05| mm
𝑥2=|18.10 ± 0.05|mm
𝑥3=|18.10 ± 0.05|mm
𝑥 =
18.05+ 18.10+18.10
3
mm = 18.08 𝑚𝑚
𝛿1 = |18.05 − 18.08| mm = 0.03 𝑚𝑚
𝛿2 = |18.10 − 18.08| mm = 0.02 𝑚𝑚
𝛿3 = |18.10 − 18.08| mm = 0.02 𝑚𝑚
𝛿 𝑚𝑎𝑘𝑠 = 0.03 𝑚𝑚 ⇨ 𝛥𝑥 = 0.03 𝑚𝑚
KR =
0.03
18.08
× 100% = 0.001 (4 𝐴𝑃) , sehingga :
HP = |18.08 ± 0.03 | 𝑚𝑚
Ketiga nilai X yaitu 𝑥1, 𝑥2,𝑑𝑎𝑛 𝑥3 , tercakup dalam interval [ 18,08 + 0,03 ] 𝑚𝑚 sampai
[ 18,08 - 0,03 ] mm.
Jika 𝛥𝑋 = 𝛿 𝑟𝑎𝑡𝑎−𝑟𝑎𝑡𝑎, maka :
𝛥𝑋 =
(0.03+0.02+0.02) 𝑚𝑚
3
= 0.02 mm

9. 9
Jadi, {X} = [ X ± 𝛥𝑋 ] = [18.08 ± 0.02 ] mm
1.3.Mikrometer sekrup
Pengukuran panjang dengan menggunakan mikrometer sekrup
Diketahui :
𝑥1=|19.430 ± 0.005| mm
𝑥2=|19.410 ± 0.005|mm
𝑥3=|19.415 ± 0.005|mm
𝑥 =
19.430+19.410+19.415
3
mm = 19.418 𝑚𝑚
𝛿1 = |19.430 − 19.418| mm = 0.012 𝑚𝑚
𝛿2 = |19.410 − 19.418| mm = 0.008 𝑚𝑚
𝛿3 = |19.415 − 19.418| mm = 0.003 𝑚𝑚
𝛿 𝑚𝑎𝑘𝑠 = 0.012 𝑚𝑚 ⇨ 𝛥𝑥 = 0.012 𝑚𝑚 , sehingga :
{X} = [ X ± 𝛥𝑋 ] = [ 19.418 ± 0.012 ] 𝑚𝑚
Ketiga nilai X yaitu 𝑥1, 𝑥2,𝑑𝑎𝑛 𝑥3 , tercakup dalam interval [ 19,418 +
0,012 ] 𝑚𝑚 sampai [ 19,418 - 0,012 ] mm.
Jika 𝛥𝑋 = 𝛿 𝑟𝑎𝑡𝑎−𝑟𝑎𝑡𝑎, maka :
𝛥𝑋 =
(0.012+0.008+0.003) 𝑚𝑚
3
= 0.007 mm
Jadi, {X} = [ X ± 𝛥𝑋 ] = [19.418 ± 0.007 ] mm
Pengukuran Lebar dengan menggunakan mikrometer sekrup
Diketahui :
𝑥1=|18.383 ± 0.005| mm
𝑥2=|18.380 ± 0.005|mm
𝑥3=|18.390 ± 0.005|mm
𝑥 =
18.383+18.380+18.390
3
mm = 18.384 𝑚𝑚
𝛿1 = |18.385 − 18.384| mm = 0.001 𝑚𝑚
𝛿2 = |18.380 − 18.384| mm = 0.004 𝑚𝑚

10. 10
𝛿3 = |18.390 − 18.384| mm = 0.006 𝑚𝑚
𝛿 𝑚𝑎𝑘𝑠 = 0.006 𝑚𝑚 ⇨ 𝛥𝑥 = 0,006 𝑚𝑚 ,
KR=
0,006
18,384
× 100% = 0,0003 (4 𝐴𝑃)sehingga :
HP = |18,38 ± 0,06 | 𝑚𝑚
Ketiga nilai X yaitu 𝑥1, 𝑥2,𝑑𝑎𝑛 𝑥3 , tercakup dalam interval [ 18,384 +
0,006 ] 𝑚𝑚 sampai [ 18,384 - 0,006 ] mm.
Jika 𝛥𝑋 = 𝛿 𝑟𝑎𝑡𝑎−𝑟𝑎𝑡𝑎, maka :
𝛥𝑋 =
(0,001+0,004+0,006) 𝑚𝑚
3
= 0,003 mm
Jadi, {X} = [ X ± 𝛥𝑋 ] = [18,384 ± 0,003 ] mm
Pengukuran Tinggi dengan menggunakan mikrometer sekrup
Diketahui :
𝑥1=|18,340 ± 0,005| mm
𝑥2=|18,335 ± 0,005|mm
𝑥3=|18,330 ± 0,005|mm
𝑥̅ =
18,340+18,335+18,330
3
mm = 18,335 𝑚𝑚
𝛿1 = |18,340 − 18,335| mm = 0,005 𝑚𝑚
𝛿2 = |18,335 − 18,335| mm = 0 𝑚𝑚
𝛿3 = |18,330 − 18,335| mm = 0,005 𝑚𝑚
𝛿 𝑚𝑎𝑘𝑠 = 0,005 𝑚𝑚 ⇨ 𝛥𝑥 = 0,005 𝑚𝑚 ,
KR=
0,005
18,335
× 100% = 0,0002 (4 𝐴𝑃) sehingga :
HP = |18,34 ± 0,05 | 𝑚𝑚
Ketiga nilai X yaitu 𝑥1, 𝑥2,𝑑𝑎𝑛 𝑥3 , tercakup dalam interval [ 18,335 +
0,005 ] 𝑚𝑚 sampai [ 18,335 - 0,005 ] mm.
Jika 𝛥𝑋 = 𝛿 𝑟𝑎𝑡𝑎−𝑟𝑎𝑡𝑎, maka :
𝛥𝑋 =
(0,005+0+0,005) 𝑚𝑚
3
= 0,003 mm

11. 11
Jadi, {X} = [ X ± 𝛥𝑋 ] = [18,335 ± 0,003 ] mm
Table 6.hasil pengukuran panjang, lebar, dan tinggi pada balok kubus
AlatUkur P (mm) L (mm) T (mm)
Mistar
|20,0 ± 0,5| |20,0 ± 0,5|
|20,0 ± 0,5|
Jangka Sorong
|20,10 ± 0,05| |20,14 ± 0,06| |20,12 ± 0,03|
Mikrometer Sekrup
|19,970
± 0,020|
|19,583
± 0,007|
|19,543
± 0,007|
Table 7. hasil pengukuran volume pada balok
Alat ukur Volume(𝒎𝒎 𝟑
)
mistar 8000 mm3
Jangka sorong
8144, 857 mm3
Micrometer
secrup
7642,730 mm3
2. BOLA
2.1.Mistar
Pengukuran diameter dengan menggunakan mistar
Diketahui : X1 = 16,0 mm
X2 = 16,0 mm
X3 = 16,0 mm
x̅ =
x1 + x2 + x3
3
=
16,0+16,0+16,0
3
= 16,0 mm
X = | 𝑥̅ ± ∆𝑥|
δ1 = |16,0 − 16,0| = 0 mm
δ2 = |16,0 − 16,0| = 0 mm
δ3 = |16,0 − 16,0| = 0 mm
Untuk ∆x = δmaks =
1
2
NST = 0,5 ,
KR=
0,5
16,0
× 100% = 0,03 (4 𝐴𝑃)maka :

12. 12
X = |16,00 ± 0,50 | mm dan rentang nilai diameter bola yaitu 15,5 mm –
16,5 mm
Pengukuran volumebola dengan menggunakan mistar
V =
4
3
𝜋𝑟3
=
4
3
3,14 (
1
2
. 𝑑)3
=
4
3
3,14 (8)3
= 2143,57 mm3
2.2.Jangka sorong
Pengukuran diameter dengan menggunakan jangka sorong
Diketahui : X1 = 14,60 mm
X2 = 14,40 mm
X3 = 14,65 mm
x̅ =
x1 + x2 + x3
3
=
14,60 + 14,40 + 14,65
3
= 14,55 mm
X = | 𝑥̅ ± ∆𝑥|
δ1 = |14,55 − 14,60| = 0,05 mm
δ2 = |14,55 − 14,40| = 0,15 mm
δ3 = |14,55 − 14,65| = 0,10 mm
Untuk ∆x = δmaks = 0,15 , maka :
X = |14,55 ± 0,15 | mm dan rentang nilai diameter bola yaitu 14,40 mm – 14,60
mm
Pengukuran volume dengan menggunakan jangka sorong
V =
4
3
𝜋𝑟3
=
4
3
3,14 (
1
2
. 𝑑)3
=
4
3
3,14 (7,275)3
= 1608,68 mm3
2.3.Mikrometer sekrup
Pengukuran diameter dengan menggunakan mikrometer sekrup
Diketahui : X1 = 15,880 mm

13. 13
X2 = 15,600 mm
X3 = 15,870 mm
x̅ =
x1 + x2 + x3
3
=
15,880 + 15,600 + 15,870
3
= 15,783 mm
X = | 𝑥̅ ± ∆𝑥|
δ1 = |15,783 − 15,880| = 0,097 mm
δ2 = |15,783 − 15,600| = 0,183 mm
δ3 = |15,783 − 15,870| = 0,087 mm
Untuk ∆x = δmaks = 0,183 , maka :
X = |15,783 ± 0,183 | mm dan rentang nilai diameter bola yaitu 15,600
mm – 15,966 mm
Pengukuran volume dengan menggunakan mikrometer sekrup
V =
4
3
𝜋𝑟3
=
4
3
3,14 (
1
2
. 𝑑)3
=
4
3
3,14 (7,89)3
= 2056,360 mm3
Table 8. hasil pengukuran diameter pada bola
No Alat Ukur Diameter (Mm)
1.
Mistar |16,0 ± 0,5|
2.
Jangka Sorong |14,55 ± 0,15|
3.
Micrometer Sekrup |15,783 ± 0,183|
Table 9. hasil pengukuran volume pada bola
no Alat ukur Volume (𝒎𝒎 𝟑)
1
Mistar 2143,57
2
Jangka Sorong 1608,68
3
Micrometer Sekrup 2056,360

14. 14
B. Pengukuran massa
1. Balok
pengukuran massa dengan menggunakan neraca ohauss 2610 gram
Diketahui :
m1 = 62,40 gram
m2 = 62,20 gram
m3 = 62,40 gram
𝑚̅ =
m1+m2+m3
3
=
62,40+62,20+62,40
3
= 62,33 gram
m = |𝑚̅ ± ∆𝑚|
𝛿1 = |𝑚̅ − 𝑚1| = |62,33 – 62,40 | gram = 0,07 gram
𝛿2 = |𝑚̅ − 𝑚2| = |62,33 – 62,20 | gram = 0,13 gram
𝛿3 = |𝑚̅ − 𝑚3| = |62,33 – 62,40 | gram = 0,07 gram
untuk ∆𝑚 = 𝛿max= 0,13 gram, maka :
m = |62,33 ± 0,13 | gram ,dan rentang nilai massa balok yaitu 62,20 gram –
62,46 gram .
pengukuran massa dengan menggunakan neraca ohauss 311 gram
Diketahui :
m1 = 62,410 gram
m2 = 62,420 gram
m3 = 62,380 gram
𝑚̅ =
m1+m2+m3
3
=
62 ,410+62,420 +62,380
3
= 62,403 gram
m = |𝑚̅ ± ∆𝑚|
𝛿1 = |𝑚̅ − 𝑚1| = |62,403 – 62,410 | gram = 0,007 gram
𝛿2 = |𝑚̅ − 𝑚2| = |62,403 – 62,420 | gram = 0,017 gram
𝛿3 = |𝑚̅ − 𝑚3| = |62,403 – 62,380 | gram = 0,023 gram
untuk ∆𝑚 = 𝛿max= 0,023 gram, maka :
m = |62,403 ± 0,023 | gram ,dan rentang nilai massa balok yaitu 62,380 gram –
62,426 gram .
pengukuran massa dengan menggunakan neraca ohauss 310 gram
Diketahui : m1 = 62,24 gram
m2 = 62,20 gram
m3 = 62,23 gram
𝑚̅ =
m1+m2+m3
3
=
62,24+62,20+62,23
3
= 62,22 gram
m = |𝑚̅ ± ∆𝑚|

15. 15
𝛿1 = |𝑚̅ − 𝑚1| = |62,22 – 62,24 | gram = 0,02 gram
𝛿2 = |𝑚̅ − 𝑚2| = |62,22 – 62,20 | gram = 0,02 gram
𝛿3 = |𝑚̅ − 𝑚3| = |62,22 – 62,23 | gram = 0,01 gram
untuk ∆𝑚 = 𝛿max= 0,02 gram, maka :
m = |62,22 ± 0,02 | gram ,dan rentang nilai massa balok yaitu 62,20 gram
– 62,24 gram .
2. Bola
pengukuran massa dengan menggunakan neraca ohauss 2610 gram
Diketahui :
m1 = 5,90 gram
m2 = 5,90 gr
m3 = 5,80 gram
𝑚̅ =
m1+m2+m3
3
=
5,90+5,90+5,80
3
= 5,86 gram
m = |𝑚̅ ± ∆𝑚|
𝛿1 = |𝑚̅ − 𝑚1| = |5,86 – 5,90 | gram = 0,04 gram
𝛿2 = |𝑚̅ − 𝑚2| = |5,86 – 5,90 | gram = 0,04 gram
𝛿3 = |𝑚̅ − 𝑚3| = |5,86 – 5,80 | gram = 0,06 gram
untuk ∆𝑚 = 𝛿max= 0,06 gram, maka
m = |5,86 ± 0,06 | gram ,dan rentang nilai massa balok yaitu 5,80 gram – 5,92
gram .
pengukuran massa dengan menggunakan neraca ohauss 311 gram
Diketahui :
m1 = 5,810 gram
m2 = 5,800 gram
m3 = 5,800 gram
𝑚̅ =
m1+m2+m3
3
=
5,810+5,800+5,800
3
= 5,803 gram
m = |𝑚̅ ± ∆𝑚|
𝛿1 = |𝑚̅ − 𝑚1| = |5,803 – 5,810 | gram = 0,007 gram
𝛿2 = |𝑚̅ − 𝑚2| = |5,803 – 5,800 | gram = 0,003 gram
𝛿3 = |𝑚̅ − 𝑚3| = |5,803 – 5,800 | gram = 0,003 gram
untuk ∆𝑚 = 𝛿max= 0,007 gram, maka :
m = |5,803 ± 0,007 | gram ,dan rentang nilai massa balok yaitu 5,796 gram –
5,810 gram .

16. 16
pengukuran massa dengan menggunakan neraca ohauss 310 gram
Diketahui :
m1 = 5,87 gram
m2 = 5,74 gram
m3 = 5,87 gram
𝑚̅ =
m1+m2+m3
3
=
5,87+5,74+5,87
3
= 5,82 gram
m = |𝑚̅ ± ∆𝑚|
𝛿1 = |𝑚̅ − 𝑚1| = |5,82 – 5,87 | gram = 0,05 gram
𝛿2 = |𝑚̅ − 𝑚2| = |5,82 – 5,74 | gram = 0,08 gram
𝛿3 = |𝑚̅ − 𝑚3| = |5,82 – 5,87 | gram = 0,05 gram
untuk ∆𝑚 = 𝛿max= 0,08 gram, maka
m = |5,82 ± 0,08 | gram ,dan rentang nilai massa balok yaitu 5,74 gram – 5,90
gram .
C. Rambat Ralat
1) Rambat ralat volume
a) Balok
rambat ralat volume untuk mistar
Diketahui : P = |20,0 ± 0,5 |mm
L = |20,0 ± 0,5 |mm
T = |20,0 ± 0,5 |mm
V = 8000 mm3
V = P × L× T
 ∆𝑉 =
𝛿𝑉
𝛿𝑝
dp+
𝛿𝑉
𝛿𝑙
dl +
𝛿𝑉
𝛿𝑡
dt
𝛿𝑣 =
𝛿 𝑝 𝑙 𝑡
𝛿 𝑝
𝑑𝑝 +
𝛿 𝑝 𝑙 𝑡
𝛿 𝑙
𝑑𝑙 +
𝛿 𝑝 𝑙 𝑡
𝛿 𝑡
𝑑𝑡
𝛿𝑣 = 𝑙 𝑡 𝑑𝑝 + 𝑝 𝑡 𝑑𝑙 + 𝑝 𝑙 𝑑𝑡
𝛿𝑣
𝑣
=
𝑙 𝑡
𝑣
𝑑𝑝 +
𝑝 𝑡
𝑣
𝑑𝑙 +
𝑝 𝑙 𝑑𝑡
𝑣
𝛿𝑣
𝑣
=
𝑙 𝑡
𝑝 𝑙 𝑡
𝑑𝑝 +
𝑝 𝑡
𝑝 𝑙 𝑡
𝑑𝑙 +
𝑝 𝑙
𝑝 𝑙 𝑡
𝑑𝑡
𝛿𝑣
𝑣
=
𝑑𝑝
𝑝
+
𝑑𝑙
𝑙
+
𝑑𝑡
𝑡
∆𝑣
𝑣
=
∆𝑝
𝑝
+
∆𝑙
𝑙
+
∆𝑡
𝑡
∆𝑣 = |
∆𝑝
𝑝
+
∆𝑙
𝑙
+
∆𝑡
𝑡
| 𝑣
∆𝑣 = |
0,5
20,0
+
0,5
20,0
+
0,5
20,0
|8000
∆𝑣 = |
1,5
20,0
| 8000

17. 17
∆v = 600 mm3
Untuk V = 8000 mm3 dan ∆v = 600 mm3 , maka :
V= |8000 ± 600| mm3
 KR =
∆v
𝑣
× 100 %
=
600
8000
× 100 %
= 7,5 % → 2 AP
 DK = 100% - KR
= 100% - 7,5 %
= 92,5 %
rambat ralat volume untuk jangka sorong
Diketahui : P = |20,10 ± 0,05 |mm
L = |20,14 ± 0,06 |mm
T = |20,12 ± 0,03 |mm
V = 8144,875 mm3
V = P × L× V
∆𝑣 = |
∆𝑝
𝑝
+
∆𝑙
𝑙
+
∆𝑡
𝑡
| 𝑣
∆𝑣 = |
0,05
20,10
+
0,06
20,14
+
0,03
20,12
|8144,857
∆𝑣 = |0,005| 8144,857
∆v = 32,579 mm3
Untuk V = 8144,857 mm3 dan ∆v = 32,579 mm3 , maka :
V= |8144,00 ± 32,57| mm3
 KR =
∆v
𝑣
× 100 %
=
32,579
8144,857
× 100 %
= 0,3 % → 4 AP
 DK = 100% - KR
= 100% - 0,3 %
= 99,7 %
rambat ralat volume untuk micrometer sekrup
Diketahui : P = |19,970 ± 0,020 |mm
L = |19,583 ± 0,007|mm

18. 18
T = |19,543 ± 0,007 |mm
V = 7642,730 mm3
V = P × L× V
∆𝑣 = |
∆𝑝
𝑝
+
∆𝑙
𝑙
+
∆𝑡
𝑡
| 𝑣
∆𝑣 = |
0,020
19,970
+
0,007
19,583
+
0,007
19,543
|7642,730
∆𝑣 = |0,0016| 7642,730
∆v = 12,228 mm3
Untuk V = 7642,730 mm3 dan ∆v = 12,228 mm3 , maka :
V= |7642,00 ± 12,22| mm3
 KR =
∆v
𝑣
× 100 %
=
12,228
7642,730
× 100 %
= 0,1 % → 4 AP
 DK = 100% - KR
= 100% - 0,1 %
= 99,9 %
b) Bola
rambat ralat volume untuk mistar
Diketahui : D = |16,0 ± 0,5 |mm
R = |8,00 ± 0,25 |mm
V = 2143,57 mm3
V=
4
3
𝜋r3
 ∆𝑣 = r3
∆𝑣 = |
4𝜋𝑟2
4
3
𝜋𝑟3
|
∆𝑣
𝑣
= |3
∆𝑟
𝑟
|
∆𝑣 = |3
∆𝑟
𝑟
| 𝑣
∆𝑣 = |3
0,25
8,00
| 2143,57
∆𝑣 = |0,09| 2143,57
∆𝑣 = 192,92 mm3

19. 19
Untuk V = 2143,57 mm3 dan ∆v = 192,92 mm3 , maka :
V= |2143,57 ± 192,92| mm3
 KR =
∆v
𝑣
× 100 %
=
192,92
2143,57
× 100 %
= 8 % → 2 AP
 DK = 100% - KR
= 100% - 8 %
= 92 %
rambat ralat volume untuk jangka sorong
Diketahui : D = |14,55 ± 0,15 |mm
R = |7,275 ± 0,075 |mm
V = 1608,68 mm3
V=
4
3
𝜋r3 ,
∆𝑣 = |3
∆𝑟
𝑟
| 𝑣
∆𝑣 = |3
0,075
7,275
| 1608,68
∆𝑣 = |0,030| 1608,68
∆𝑣 = 48,260 mm3
Untuk V = 1608,68 mm3 dan ∆v = 48,260 mm3 , maka :
V= |1608,68 ± 48,26| mm3
 KR =
∆v
𝑣
× 100 %
=
48,26
1608,68
× 100 %
= 2 % → 2 AP
 DK = 100% - KR
= 100% - %
= 98 %
rambat ralat volume untuk micrometer sekrup
Diketahui : D = |15,783 ± 0,183 |mm
R = |7,891 ± 0,091 |mm
V = 2056,360 mm3

20. 20
V=
4
3
𝜋r3
∆𝑣 = |3
∆𝑟
𝑟
| 𝑣
∆𝑣 = |3
0,091
7,891
| 2056,360
∆𝑣 = |0,034| 2056,360
∆𝑣 = 69,916 mm3
Untuk V = 2056,360 mm3 dan ∆v = 69,916 mm3 , maka :
V= |2056,360± 69,916| mm3
 KR =
∆v
𝑣
× 100 %
=
69,916
2056,360
× 100 %
= 3 % → 3 AP
 DK = 100% - KR
= 100% - 3 %
= 97 %
2) Rambat ralat massa jenis
1. Balok
Rambat ralat massa jenis dengan neraca ohauss 310 gram dan mistar
Diketahui : M = |62,22 ± 0,02| gram
V = |8000 ± 600| mm3
𝜌 =
𝑚
𝑣
= 𝑚𝑣−1
𝜌 =
62,22 𝑔𝑟𝑎𝑚
8000 𝑚𝑚3 = 0,007
𝑔𝑟𝑎𝑚
𝑚𝑚3
 ∆𝜌 = 𝛿𝜌 =
𝛿𝜌
𝛿𝑚
𝑑𝑚 +
𝛿𝜌
𝛿𝑣−1 𝑑𝑣
𝛿𝜌 =
𝛿𝑚𝑣−1
𝛿𝑚
𝑑𝑚 +
𝛿𝑚𝑣−1
𝛿𝑣−1 𝑑𝑣
𝛿𝜌 = 𝑣−1
𝑑𝑚 + 𝑚𝑣−2
𝑑𝑣
𝛿𝜌
𝜌
=
𝑣−1
𝜌
𝑑𝑚 +
𝑚𝑣−2
𝜌
𝑑𝑣
𝛿𝜌
𝜌
=
𝑣−1
𝑚𝑣−1 𝑑𝑚 +
𝑚𝑣−2
𝑚𝑣−1 𝑑𝑣
𝛿𝜌
𝜌
=
𝑑𝑚
𝑚
+
𝑑𝑣
𝑣

21. 21
∆𝜌
𝜌
=
∆𝑚
𝑚
+
∆𝑣
𝑣
∆𝜌 = |
∆𝑚
𝑚
+
∆𝑣
𝑣
| 𝜌
∆𝜌 = |
0,02
62,22
+
600
8000
| 0,007
∆𝜌 = |0,0753|0,007
∆𝜌 = 0,0005
𝑔𝑟𝑎𝑚
𝑚𝑚3
Untuk 𝜌 =0,007
𝑔𝑟𝑎𝑚
𝑚𝑚3 dan ∆ρ = 0,0005
𝑔𝑟𝑎𝑚
𝑚𝑚3 , maka :
𝜌 = |0,0070 ± 0,0005|
𝑔𝑟𝑎𝑚
𝑚𝑚3
 KR =
∆𝜌
𝜌
× 100%
=
0,0005
0,007
× 100%
= 7,1% → 2AP
 DK = 100% - KR
= 100% - 7,1%
= 92,9%
Rambat ralat massa jenis dengan neraca ohauss 310 gram dan jangka sorong
Diketahui : M = |62,22 ± 0,02| gram
V = |8144,857± 32,579| mm3
𝜌 =
𝑚
𝑣
= 𝑚𝑣−1
𝜌 =
62,22 𝑔𝑟𝑎𝑚
8144,857 𝑚𝑚3 = 0,007
𝑔𝑟𝑎𝑚
𝑚𝑚3
∆𝜌 = |
∆𝑚
𝑚
+
∆𝑣
𝑣
| 𝜌
∆𝜌 = |
0,02
62,22
+
32,579
8144 ,857
| 0,007
∆𝜌 = |0,0033|0,007
∆𝜌 = 0,00002
𝑔𝑟𝑎𝑚
𝑚𝑚3
Untuk 𝜌 =0,00700
𝑔𝑟𝑎𝑚
𝑚𝑚3 dan ∆ρ = 0,00002
𝑔𝑟𝑎𝑚
𝑚𝑚3 , maka :
𝜌 = |0,00700 ± 0,00002|
𝑔𝑟𝑎𝑚
𝑚𝑚3

22. 22
 KR =
∆𝜌
𝜌
× 100%
=
0,00002
0,00700
× 100%
= 0,2% → 4 AP
 DK = 100% - KR
= 100% - 0,2%
= 99,8%
Rambat ralat massa jenis dengan neraca ohauss 310 gram dan micrometer
sekrup
Diketahui : M = |62,22 ± 0,02| gram
V= |7642,730± 12,228| mm3
𝜌 =
𝑚
𝑣
= 𝑚𝑣−1
𝜌 =
62,22 𝑔𝑟𝑎𝑚
7642,730𝑚𝑚3 = 0,008
𝑔𝑟𝑎𝑚
𝑚𝑚3
∆𝜌 = |
∆𝑚
𝑚
+
∆𝑣
𝑣
| 𝜌
∆𝜌 = |
0,02
62,22
+
12,228
7642 ,730
| 0,008
∆𝜌 = |0,0013|0,008
∆𝜌 = 0,00001
𝑔𝑟𝑎𝑚
𝑚𝑚3
Untuk 𝜌 =0,00800
𝑔𝑟𝑎𝑚
𝑚𝑚3 dan ∆ρ = 0,00001
𝑔𝑟𝑎𝑚
𝑚𝑚3 , maka :
𝜌 = |0,00800 ± 0,00001|
𝑔𝑟𝑎𝑚
𝑚𝑚3
 KR =
∆𝜌
𝜌
× 100%
=
0,00001
0,00800
× 100%
= 0,1% → 4 AP
 DK = 100% - KR
= 100% - 0,1%
= 99,9%
2. Bola
Rambat ralat massa jenis dengan neraca ohauss 310 gram dan mistar
Diketahui : M = |5,82 ± 0,08| gram
V= |2143,57 ± 192,92| mm3

23. 23
𝜌 =
𝑚
𝑣
= 𝑚𝑣−1
𝜌 =
5,82 𝑔𝑟𝑎𝑚
2143,57 𝑚𝑚3 = 0,002
𝑔𝑟𝑎𝑚
𝑚𝑚3
 ∆𝜌 = 𝛿𝜌 =
𝛿𝜌
𝛿𝑚
𝑑𝑚 +
𝛿𝜌
𝛿𝑣−1 𝑑𝑣
𝛿𝜌 =
𝛿𝑚𝑣−1
𝛿𝑚
𝑑𝑚 +
𝛿𝑚𝑣−1
𝛿𝑣−1 𝑑𝑣
𝛿𝜌 = 𝑣−1
𝑑𝑚 + 𝑚𝑣−2
𝑑𝑣
𝛿𝜌
𝜌
=
𝑣−1
𝜌
𝑑𝑚 +
𝑚𝑣−2
𝜌
𝑑𝑣
𝛿𝜌
𝜌
=
𝑣−1
𝑚𝑣−1 𝑑𝑚 +
𝑚𝑣−2
𝑚𝑣−1 𝑑𝑣
𝛿𝜌
𝜌
=
𝑑𝑚
𝑚
+
𝑑𝑣
𝑣
∆𝜌
𝜌
=
∆𝑚
𝑚
+
∆𝑣
𝑣
∆𝜌 = |
∆𝑚
𝑚
+
∆𝑣
𝑣
| 𝜌
∆𝜌 = |
0,08
5,82
+
192,92
2143 ,57
| 0,002
∆𝜌 = |0,001|0,002
∆𝜌 = 0,000002
𝑔𝑟𝑎𝑚
𝑚𝑚3
Untuk 𝜌 =0,002000
𝑔𝑟𝑎𝑚
𝑚𝑚3 dan ∆ρ = 0,000002
𝑔𝑟𝑎𝑚
𝑚𝑚3 , maka :
𝜌 = |0,002000 ± 0,000002| 𝑔𝑟𝑎𝑚
𝑚𝑚3
 KR =
∆𝜌
𝜌
× 100%
=
0,000002
0,002000
× 100%
= 0,1 % → 2AP
 DK = 100% - KR
= 100% - 0,1%
= 99,9 %
Rambat ralat massa jenis dengan neraca ohauss 310 gram dan jangka sorong
Diketahui : M = |5,82 ± 0,08| gram

24. 24
V= |1608,68 ± 48,26| mm3
𝜌 =
𝑚
𝑣
= 𝑚𝑣−1
𝜌 =
5,82 𝑔𝑟𝑎𝑚
1608,68 𝑚𝑚3 = 0,003
𝑔𝑟𝑎𝑚
𝑚𝑚3
∆𝜌 = |
∆𝑚
𝑚
+
∆𝑣
𝑣
| 𝜌
∆𝜌 = |
0,08
5,82
+
48,26
1608 ,68
| 0,003
∆𝜌 = |0,0004|0,003
∆𝜌 = 0,000001
𝑔𝑟𝑎𝑚
𝑚𝑚3
Untuk 𝜌 =0,003000
𝑔𝑟𝑎𝑚
𝑚𝑚3 dan ∆ρ = 0,000001
𝑔𝑟𝑎𝑚
𝑚𝑚3 , maka :
𝜌 = |0,003000 ± 0,000001|
𝑔𝑟𝑎𝑚
𝑚𝑚3
 KR =
∆𝜌
𝜌
× 100%
=
0,000001
0,003000
× 100%
= 0,03 % → 2AP
 DK = 100% - KR
= 100% - 0,03%
= 99,97 %
Rambat ralat massa jenis dengan neraca ohauss 310 gram dan micrometer
sekrup
Diketahui : M = |5,82 ± 0,08| gram
V= |2056,360± 69,916| mm3
𝜌 =
𝑚
𝑣
= 𝑚𝑣−1
𝜌 =
5,82 𝑔𝑟𝑎𝑚
2056,360 𝑚𝑚3 = 0,002
𝑔𝑟𝑎𝑚
𝑚𝑚3
∆𝜌 = |
∆𝑚
𝑚
+
∆𝑣
𝑣
| 𝜌
∆𝜌 = |
0,08
5,82
+
69,916
2056 ,360
| 0,002
∆𝜌 = |0,0476|0,002
∆𝜌 = 0,000095
𝑔𝑟𝑎𝑚
𝑚𝑚3

25. 25
Untuk 𝜌 =0,002000
𝑔𝑟𝑎𝑚
𝑚𝑚3 dan ∆ρ = 0,000095
𝑔𝑟𝑎𝑚
𝑚𝑚3 , maka :
𝜌 = |0,002000 ± 0,000095|
𝑔𝑟𝑎𝑚
𝑚𝑚3
 KR =
∆𝜌
𝜌
× 100%
=
0,00009
0,00200
× 100%
= 4,5 % → 2AP
 DK = 100% - KR
= 100% - 0,03%
= 95,5 %
 PF = 𝜌 = |0,002000 ± 0,000095|
𝑔𝑟𝑎𝑚
𝑚𝑚3