TABEL HASIL PENGAMATANTabel 1. Massa jenis zat cair

No Jenis zat cair Massa (gram) Volume (ml)

1 Air |10,39±0,02| |10±1|

2 Gliserin |11,52±0,02| |10±1|

3 Minyak |8,06±0,02| |10±1|

Kegiatan 1. Pengaruh kedalaman terhadap tekanan hidrostatis

Jenis zat cair = Air

Tabel 2. Hubungan antara kedalaman zat cair dengan tekanan hidrostatis

No Kedalaman (cm) Perbedaan ketinggian zat cair pada pipa U (cm)

1 |1,70±0,05| 1. |1,50±0,05|

2. |1,60±0,05|

3. |1,70±0,05|

2 |2,80±0,05| 1. |2,40±0,05|

2. |2,80±0,05|

3. |2,10±0,05|

3 |4,00±0,05| 1. |3,80±0,05|

2. |3,70±0,05|

3. |3,60±0,05|

4 |4,90±0,05| 1. |4,90±0,05|

2. |5,00±0,05|

3. |4,50±0,05|

5 |5,70±0,05| 1. |5,80±0,05|

2. |6,30±0,05|

3. |5,80±0,05|

6 |7,80±0,05| 1. |7,60±0,05|

2. |7,50±0,05|

3. |7,70±0,05|

7 |8,70±0,05| 1. |8,00±0,05|

2. |8,10±0,05|

3. |8,00±0,05|

Kegiatan 2. Pengaruh massa jenis zat cair terhadap tekanan hidrostatik

Kedalaman = |4,00±0,05| cm

Tabel 3. Hubungan massa jenis zat cair dengan tekanan hidrostatis

No Massa Jenis Zat Cair Perbedaan Ketinggian zat cair

pada pipa U (cm)

1 1. |3,80 ±0,05|

2. | 3,70±0,05|

3. | 3,60±0,05|

2 1. | 3,90±0,05|

2. | 4,10±0,05|

3. |3,90 ±0,05|

3 1. |3,00 ±0,05|

2. | 3,20±0,05|

3. | 3,00±0,05|

ANALISIS DATA

I. Analisis perhitungan

1. Untuk Air

ρ=mv

=10,39gram

10cm3=1,039 g/cm3

Rambat ralat

ρ=mv−1

δρ=| δρδm|dm+|δρδv|dv

δρ=|δm v−1

δm |dm+|δmv−1

δv |dv δρ=v−1dm+mv−2dv

δρρ

= v−1

ρdm+mv

−2

ρdv

δρρ

= v−1

mv−1 dm+mv−2

mv−1 dv

δρρ

=∆mm

+∆vv

∆ ρ=|∆mm + ∆vv |ρ

∆ ρ=| 0,0210,39

+ 110|1,039g /cm3

∆ ρ=|0,0019249278+0,1|1,039 g /cm3

∆ ρ=0,1019249278×1,039 g/cm3

∆ ρ=0,1059 g/cm3

KR=∆ ρρ×100 %=0,1059

1,039×100 %=10,1925 %2 AB

DK = 100% - KR = 100% - 10,1925% = 89,8075%

ρ=|ρ±∆ ρ|=|1,0±0,1|g /cm3

2. Untuk Gliserin

ρ=mv

=11,52gram

10cm3=1,152 g/cm3

∆ ρ=|∆mm + ∆vv |ρ

∆ ρ=| 0,0211,52

+ 110|1,152g /cm3

∆ ρ=|0,0017361111+0,1|1,152 g/cm3

∆ ρ=0,1017361111×1,152 g/cm3

∆ ρ=0,1172 g/cm3

KR=∆ ρρ×100 %=0,1172

1,152×100 %=10,1736 %2 AB

DK = 100% - KR = 100% - 10,1736% = 89,8264%

ρ=|ρ±∆ ρ|=|1,2±0,1|g/cm3

3. Untuk Minyak

ρ=mv

=8,06 gram

10cm3=0,806g /cm3

∆ ρ=|∆mm + ∆vv |ρ

∆ ρ=|0,028,06

+ 110|0,806 g/cm3

∆ ρ=|0,0024813896+0,1|0,806 g/cm3

∆ ρ=0,1024813896×0,806g /cm3

∆ ρ=0,0826 g/cm3

KR=∆ ρρ×100 %=0,0826

0,806×100 %=10,2481 %2 AB

DK = 100% - KR = 100% - 10,2481% = 89,7519%

ρ=|ρ±∆ ρ|=|0,81±0,08|g /cm3

Kegiatan 1

1. Untuk kedalaman |1,70±0,05| cm

h=h1+h2+h3

3=1,50+1,60+1,70

3=4,80

3=1,6 cm

𝛿1 = |h – h1| = |1,6 – 1,50| = 0,1 cm𝛿2 = |h – h2| = |1,6 – 1,60| = 0 cm𝛿3 = |h – h3| = |1,6 – 1,70| = 0,1 cm𝛿maks = 𝛥h = 0,1cm

KR=∆hh×100 %=0,1

1,6×100 %=6,25 %3 AB

DK = 100% - 6,25% = 93,75%

h=|h±∆h|=|1,60±0,10|cm

Tekanan Hidrostatis

P= ρ gh

P=1,0 g /cm3×980cm /s2×1,60cm

P=1.568 g /cm s2 = 1,568 × 103 g/cm s2

δP=|δPδρ |dρ+|δPδh|dh

δP=|δρhδρ |dρ+|δρhδh |dh δP=hdρ+ ρ dh

δPP

= hPdρ+ ρ

Pdh

δPP

= hρhdρ+ ρ

ρhdh

δPP

= ∆ ρρ

+∆hh

∆ P=|∆ ρρ + ∆hh |P

∆ P=|0,11,0

+ 0,101,60|1,568×103g /cm s2

∆ P=|0,1+0,0625|1,568×103g /cm s2

∆ P=0,1625×1,568×103 g/cm s2

∆ P=0,2548×103 g/cm s2

KR=∆ PP×100 %=0,2548×103

1,568×103 ×100 %=16,25 %2 AB

DK = 100% - 16,25% = 83,75%

P=|P±∆ P|=|1,6±0,2|103g /cm s2

2. Untuk kedalaman |2,80±0,05| cm

h=h1+h2+h3

3=2,40+2,80+2,10

3=7,30

3=2,4333 cm

𝛿1 = |h – h1| = |2,4333 – 2,40| = 0,0333 cm𝛿2 = |h – h2| = |2,4333 – 2,80| = 0,3667 cm𝛿3 = |h – h3| = |2,4333 – 2,10| = 0,3333 cm

𝛿maks = 𝛥h = 0,3667 cm

KR=∆hh×100 %=0,3667

2,4333×100 %=15,0719 %2 AB

DK = 100% - 15,0719% = 84,9281%

h=|h±∆h|=|2,4±0,4|cm

Tekanan Hidrostatis

P= ρ gh

P=1,0 g /cm3×980cm /s2×cm

P=g /cm s2 = × 103 g/cm s2

∆ P=|∆ ρρ + ∆hh |P

∆ P=|0,11,0

+❑❑|×103g /cm s2

∆ P=¿

∆ P=××103g /cm s2

∆ P=×103 g/cm s2

KR=∆ PP×100 %=×103

×103 ×100 %=% AB

DK = 100% - % = %

P=|P±∆ P|=|±|103 g/cm s2

3. Untuk kedalaman |4,00±0,05| cm

h=h1+h2+h3

3=3,80+3,70+3,60

3=11,1

3=3,70cm

𝛿1 = |h – h1| = |3,70 – 3,80| = 0,1 cm𝛿2 = |h – h2| = |3,70 – 3,70| = 0 cm𝛿3 = |h – h3| = |3,70 – 3,60| = 0,1 cm𝛿maks = 𝛥h = 0,1cm

KR=∆hh×100 %= 0,1

3,70×100 %=2,7027 %3 AB

DK = 100% - 2,7027% = 97,2973%

h=|h±∆h|=|3,70±0,10|cm

Tekanan Hidrostatis

P= ρ gh

P=1,0 g /cm3×980cm /s2×cm

P=g /cm s2 = × 103 g/cm s2

∆ P=|∆ ρρ + ∆hh |P

∆ P=|0,11,0

+❑❑|×103g /cm s2

∆ P=¿

∆ P=××103g /cm s2

∆ P=×103 g/cm s2

KR=∆ PP×100 %=×103

×103×100 %=% AB

DK = 100% - % = %

P=|P±∆ P|=|±|103 g/cm s2

4. Untuk kedalaman |4,90±0,05| cm

h=h1+h2+h3

3=4,90+5,00+4,50

3=14,4

3=4,80cm

𝛿1 = |h – h1| = |4,80 – 4,90| = 0,1 cm𝛿2 = |h – h2| = |4,80 – 5,00| = 0,2 cm𝛿3 = |h – h3| = |4,80 – 4,50| = 0,3 cm𝛿maks = 𝛥h = 0,3cm

KR=∆hh×100 %= 0,3

4,80×100 %=6,25 %3 AB

DK = 100% - 6,25% = 93,75%

h=|h±∆h|=|4,80±0,30|cm

Tekanan Hidrostatis

P= ρ gh

P=1,0 g /cm3×980cm /s2×cm

P=g /cm s2 = × 103 g/cm s2

∆ P=|∆ ρρ + ∆hh |P

∆ P=|0,11,0

+❑❑|×103g /cm s2

∆ P=¿

∆ P=××103g /cm s2

∆ P=×103 g/cm s2

KR=∆ PP×100 %=×103

×103 ×100 %=% AB

DK = 100% - % = %

P=|P±∆ P|=|±|103 g/cm s2

5. Untuk kedalaman |5,70±0,05| cm

h=h1+h2+h3

3=5,80+6,30+5,80

3=17,9

3=5,9667cm

𝛿1 = |h – h1| = |5,9667 – 5,80| = 0,1667 cm𝛿2 = |h – h2| = |5,9667 – 6,30| = 0,3333 cm𝛿3 = |h – h3| = |5,9667 – 5,80| = 0,1667 cm𝛿maks = 𝛥h = 0,3333cm

KR=∆hh×100 %=0,3333

5,9667×100 %=5,5860 %3 AB

DK = 100% - 5,5860% = 94,414%

h=|h±∆h|=|5,97±0,33|cm

Tekanan Hidrostatis

P= ρ gh

P=1,0 g /cm3×980cm /s2×cm

P=g /cm s2 = × 103 g/cm s2

∆ P=|∆ ρρ + ∆hh |P

∆ P=|0,11,0

+❑❑|×103g /cm s2

∆ P=¿

∆ P=××103g /cm s2

∆ P=×103 g/cm s2

KR=∆ PP×100 %=×103

×103 ×100 %=% AB

DK = 100% - % = %

P=|P±∆ P|=|±|103 g/cm s2

6. Untuk kedalaman |7,80±0,05| cm

h=h1+h2+h3

3=7,60+7,50+7,70

3=22,8

3=7,60cm

𝛿1 = |h – h1| = |7,60 – 7,60| = 0 cm𝛿2 = |h – h2| = |7,60 – 7,50| = 0,1 cm𝛿3 = |h – h3| = |7,60 – 7,70| = 0,1 cm𝛿maks = 𝛥h = 0,1cm

KR=∆hh×100 %= 0,1

7,60×100 %=1,3158 %3 AB

DK = 100% - 1,3158% = 98,6842%

h=|h±∆h|=|7,60±0,10|cm

Tekanan Hidrostatis

P= ρ gh

P=1,0 g /cm3×980cm /s2×cm

P=g /cm s2 = × 103 g/cm s2

∆ P=|∆ ρρ + ∆hh |P

∆ P=|0,11,0

+❑❑|×103g /cm s2

∆ P=¿

∆ P=××103g /cm s2

∆ P=×103 g/cm s2

KR=∆ PP×100 %=×103

×103×100 %=% AB

DK = 100% - % = %

P=|P±∆ P|=|±|103 g/cm s2

7. Untuk kedalaman |8,70±0,05| cm

h=h1+h2+h3

3=8,00+8,10+8,00

3=24,1

3=8,0333cm

𝛿1 = |h – h1| = |8,0333 – 8,00| = 0,0333 cm𝛿2 = |h – h2| = |8,0333 – 8,10| = 0,0667 cm𝛿3 = |h – h3| = |8,0333 – 8,00| = 0,0333 cm𝛿maks = 𝛥h = 0,0667 cm

KR=∆hh×100 %=0,0667

8,0333×100 %=0,8303 %3 AB

DK = 100% - 0,8303% = 99,1697%

h=|h±∆h|=|8,03±0,07|cm

Tekanan Hidrostatis

P= ρ gh

P=1,0 g /cm3×980cm /s2×cm

P=g /cm s2 = × 103 g/cm s2

∆ P=|∆ ρρ + ∆hh |P

∆ P=|0,11,0

+❑❑|×103g /cm s2

∆ P=¿

∆ P=××103g /cm s2

∆ P=×103 g/cm s2

KR=∆ PP×100 %=×103

×103×100 %=% AB

DK = 100% - % = %

P=|P±∆ P|=|±|103 g/cm s2

Kegiatan 2

Kedalaman = |4,00±0,05| cm

1. Untuk massa jenis air |1,0±0,1|g /cm3

h=h1+h2+h3

3=3,80+3,70+3,60

3=11,10

3=3,70 cm

𝛿1 = |h – h1| = |3,70 – 3,80| = 0,1 cm𝛿2 = |h – h2| = |3,70 – 3,70| = 0 cm𝛿3 = |h – h3| = |3,70 – 3,60| = 0,1 cm𝛿maks = 𝛥h = 0,1 cm

KR=∆hh×100 %= 0,1

3,70×100 %=2,7027 %3 AB

DK = 100% - 2,7027% = 97,2973%

h=|h±∆h|=|3,70±0,10|cm

Tekanan Hidrostatis

P= ρ gh

P=1,0 g /cm3×980cm /s2×cm

P=g /cm s2 = × 103 g/cm s2

∆ P=|∆ ρρ + ∆hh |P

∆ P=|0,11,0

+❑❑|×103g /cm s2

∆ P=¿

∆ P=××103g /cm s2

∆ P=×103 g/cm s2

KR=∆ PP×100 %=×103

×103 ×100 %=% AB

DK = 100% - % = %

P=|P±∆ P|=|±|103 g/cm s2

2. Untuk massa jenis gliserin |1,2±0,1|g /cm3

h=h1+h2+h3

3=3,90+4,10+3,90

3=11,9

3=3,9667 cm

𝛿1 = |h – h1| = |3,9667 – 3,90| = 0,0667 cm𝛿2 = |h – h2| = |3,9667 – 4,10| = 0,1333 cm𝛿3 = |h – h3| = |3,9667 – 3,90| = 0,0667 cm

𝛿maks = 𝛥h = 0,1333 cm

KR=∆hh×100 %=0,1333

3,9667×100 %=3,3605 %3 AB

DK = 100% - 3,3605% = 96,6395%

h=|h±∆h|=|3,97±0,13|cm

Tekanan Hidrostatis

P= ρ gh

P=1,0 g /cm3×980cm /s2×cm

P=g /cm s2 = × 103 g/cm s2

∆ P=|∆ ρρ + ∆hh |P

∆ P=|0,11,0

+❑❑|×103g /cm s2

∆ P=¿

∆ P=××103g /cm s2

∆ P=×103 g/cm s2

KR=∆ PP×100 %=×103

×103 ×100 %=% AB

DK = 100% - % = %

P=|P±∆ P|=|±|103 g/cm s2

3. Untuk massa jenis minyak |0,81±0,08|g/cm3

h=h1+h2+h3

3=3,00+3,20+3,00

3=9,2

3=3,0667 cm

𝛿1 = |h – h1| = |3,0667 – 3,00| = 0,0667 cm𝛿2 = |h – h2| = |3,0667 – 3,20| = 0,1333 cm𝛿3 = |h – h3| = |3,0667 – 3,00| = 0,0667 cm𝛿maks = 𝛥h = 0,1333 cm

KR=∆hh×100 %=0,1333

3,0667×100 %=4,3467 % 3 AB

DK = 100% - 4,3467% = 95,6533%

h=|h±∆h|=|3,07±0,13|cm

Tekanan Hidrostatis

P= ρ gh

P=1,0 g /cm3×980cm /s2×cm

P=g /cm s2 = × 103 g/cm s2

∆ P=|∆ ρρ + ∆hh |P

∆ P=|0,11,0

+❑❑|×103g /cm s2

∆ P=¿

∆ P=××103g /cm s2

∆ P=×103 g/cm s2

KR=∆ PP×100 %=×103

×103×100 %=% AB

DK = 100% - % = %

P=|P±∆ P|=|±|103 g/cm s2

II. Tabel Perbandingan

Kegiatan 1

Tabel 2.1. Hubungan antara kedalaman dengan tekanan hidrostatis

Kedalaman (cm) Tekanan Hidrostatis (g/cms2)

|1,70±0,05|

|2,80±0,05|

|4,00±0,05|

|4,90±0,05|

|5,70±0,05|

|7,80±0,05|

|8,70±0,05|

Kegiatan 2

Tabel 3.1 . Hubungan antara massa zenis zat cair dengan tekanan

hidrostatis

Massa Jenis Zat Cair (g/cm3) Tekanan Hidrostatis (g/cms2)

Air |1,0±0,1|

Gliserin |1,2±0,1|

Minyak |0,81±0,08|

III. Analisis Grafik

Kegiatan 1