surface tension and capillarity in a water-filled capillary tube Given surface tension (sigma) = 0.0390 N/m; wetting angle (theta) = 29.0000 deg; specific weight (gamma) = 9,020 N/m^3; tube diameter (d) = 0.0045 m, determine the capillary rise (h) in m.

Start with the thinking

• The governing relation printed in this handbook section is Surface tension and capillarity (capillary rise).
• Everything except h is given, so isolate h symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A capillary tube experiment illustrates surface tension and capillarity in a small-diameter glass tube.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

2. Step 2 — Rearrange the relation so that h stands alone on the left-hand side.

3. Step 3 — List the givens: surface tension (sigma) = 0.0390 N/m, wetting angle (theta) = 29.0000 deg, specific weight (gamma) = 9,020 N/m^3, tube diameter (d) = 0.0045 m.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$h = 0.0034\ \text{m}$$

6. Step 6 — Check: returning h = 0.0034 m to

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0067 — kept a factor of two that cancels in the correct rearrangement.
• 0.0017 — dropped that same factor in the other direction.
• 0.0037 — rounded an intermediate value before the final step.

Reference: FE Handbook — Surface Tension and Capillarity

surface tension and capillarity of mercury in a glass tube Given wetting angle (theta) = 35.0000 deg; specific weight (gamma) = 9,070 N/m^3; tube diameter (d) = 0.0096 m; capillary rise (h) = 0.0030 m, determine the surface tension (sigma) in N/m.

Start with the thinking

• The governing relation printed in this handbook section is Surface tension and capillarity (capillary rise).
• Everything except sigma is given, so isolate sigma symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A capillary tube experiment illustrates surface tension and capillarity in a small-diameter glass tube.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

2. Step 2 — Rearrange the relation so that sigma stands alone on the left-hand side.

3. Step 3 — List the givens: wetting angle (theta) = 35.0000 deg, specific weight (gamma) = 9,070 N/m^3, tube diameter (d) = 0.0096 m, capillary rise (h) = 0.0030 m.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$\sigma = 0.0797\ \text{N/m}$$

6. Step 6 — Check: returning sigma = 0.0797 N/m to

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.1594 — kept a factor of two that cancels in the correct rearrangement.
• 0.0399 — dropped that same factor in the other direction.
• 0.0877 — rounded an intermediate value before the final step.

Reference: FE Handbook — Surface Tension and Capillarity

capillarity effects on a manometer reading due to surface tension Given surface tension (sigma) = 0.0650 N/m; wetting angle (theta) = 9.0000 deg; specific weight (gamma) = 9,380 N/m^3; capillary rise (h) = 0.0150 m, determine the tube diameter (d) in m.

Start with the thinking

• The governing relation printed in this handbook section is Surface tension and capillarity (capillary rise).
• Everything except d is given, so isolate d symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A capillary tube experiment illustrates surface tension and capillarity in a small-diameter glass tube.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

2. Step 2 — Rearrange the relation so that d stands alone on the left-hand side.

3. Step 3 — List the givens: surface tension (sigma) = 0.0650 N/m, wetting angle (theta) = 9.0000 deg, specific weight (gamma) = 9,380 N/m^3, capillary rise (h) = 0.0150 m.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$d = 0.0018\ \text{m}$$

6. Step 6 — Check: returning d = 0.0018 m to

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0037 — kept a factor of two that cancels in the correct rearrangement.
• 0.0009 — dropped that same factor in the other direction.
• 0.0020 — rounded an intermediate value before the final step.

Reference: FE Handbook — Surface Tension and Capillarity

surface tension and capillarity in a water-filled capillary tube Given surface tension (sigma) = 0.0470 N/m; wetting angle (theta) = 9.0000 deg; specific weight (gamma) = 9,580 N/m^3; tube diameter (d) = 0.0062 m, determine the capillary rise (h) in m.

Start with the thinking

• The governing relation printed in this handbook section is Surface tension and capillarity (capillary rise).
• Everything except h is given, so isolate h symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A capillary tube experiment illustrates surface tension and capillarity in a small-diameter glass tube.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

2. Step 2 — Rearrange the relation so that h stands alone on the left-hand side.

3. Step 3 — List the givens: surface tension (sigma) = 0.0470 N/m, wetting angle (theta) = 9.0000 deg, specific weight (gamma) = 9,580 N/m^3, tube diameter (d) = 0.0062 m.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$h = 0.0031\ \text{m}$$

6. Step 6 — Check: returning h = 0.0031 m to

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0063 — kept a factor of two that cancels in the correct rearrangement.
• 0.0016 — dropped that same factor in the other direction.
• 0.0034 — rounded an intermediate value before the final step.

Reference: FE Handbook — Surface Tension and Capillarity

surface tension and capillarity of mercury in a glass tube Given wetting angle (theta) = 15.0000 deg; specific weight (gamma) = 9,300 N/m^3; tube diameter (d) = 0.0025 m; capillary rise (h) = 0.0540 m, determine the surface tension (sigma) in N/m.

Start with the thinking

• The governing relation printed in this handbook section is Surface tension and capillarity (capillary rise).
• Everything except sigma is given, so isolate sigma symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A capillary tube experiment illustrates surface tension and capillarity in a small-diameter glass tube.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

2. Step 2 — Rearrange the relation so that sigma stands alone on the left-hand side.

3. Step 3 — List the givens: wetting angle (theta) = 15.0000 deg, specific weight (gamma) = 9,300 N/m^3, tube diameter (d) = 0.0025 m, capillary rise (h) = 0.0540 m.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$\sigma = 0.3249\ \text{N/m}$$

6. Step 6 — Check: returning sigma = 0.3249 N/m to

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.6499 — kept a factor of two that cancels in the correct rearrangement.
• 0.1625 — dropped that same factor in the other direction.
• 0.3574 — rounded an intermediate value before the final step.

Reference: FE Handbook — Surface Tension and Capillarity

capillarity effects on a manometer reading due to surface tension Given surface tension (sigma) = 0.0710 N/m; wetting angle (theta) = 33.0000 deg; specific weight (gamma) = 9,270 N/m^3; capillary rise (h) = 0.0465 m, determine the tube diameter (d) in m.

Start with the thinking

• The governing relation printed in this handbook section is Surface tension and capillarity (capillary rise).
• Everything except d is given, so isolate d symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A capillary tube experiment illustrates surface tension and capillarity in a small-diameter glass tube.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

2. Step 2 — Rearrange the relation so that d stands alone on the left-hand side.

3. Step 3 — List the givens: surface tension (sigma) = 0.0710 N/m, wetting angle (theta) = 33.0000 deg, specific weight (gamma) = 9,270 N/m^3, capillary rise (h) = 0.0465 m.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$d = 0.0006\ \text{m}$$

6. Step 6 — Check: returning d = 0.0006 m to

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0011 — kept a factor of two that cancels in the correct rearrangement.
• 0.0003 — dropped that same factor in the other direction.
• 0.0006 — rounded an intermediate value before the final step.

Reference: FE Handbook — Surface Tension and Capillarity

surface tension and capillarity in a water-filled capillary tube Given surface tension (sigma) = 0.0560 N/m; wetting angle (theta) = 7.0000 deg; specific weight (gamma) = 9,610 N/m^3; tube diameter (d) = 0.0041 m, determine the capillary rise (h) in m.

Start with the thinking

• The governing relation printed in this handbook section is Surface tension and capillarity (capillary rise).
• Everything except h is given, so isolate h symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A capillary tube experiment illustrates surface tension and capillarity in a small-diameter glass tube.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

2. Step 2 — Rearrange the relation so that h stands alone on the left-hand side.

3. Step 3 — List the givens: surface tension (sigma) = 0.0560 N/m, wetting angle (theta) = 7.0000 deg, specific weight (gamma) = 9,610 N/m^3, tube diameter (d) = 0.0041 m.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$h = 0.0056\ \text{m}$$

6. Step 6 — Check: returning h = 0.0056 m to

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0113 — kept a factor of two that cancels in the correct rearrangement.
• 0.0028 — dropped that same factor in the other direction.
• 0.0062 — rounded an intermediate value before the final step.

Reference: FE Handbook — Surface Tension and Capillarity

surface tension and capillarity of mercury in a glass tube Given wetting angle (theta) = 34.0000 deg; specific weight (gamma) = 9,560 N/m^3; tube diameter (d) = 0.0025 m; capillary rise (h) = 0.0510 m, determine the surface tension (sigma) in N/m.

Start with the thinking

• The governing relation printed in this handbook section is Surface tension and capillarity (capillary rise).
• Everything except sigma is given, so isolate sigma symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A capillary tube experiment illustrates surface tension and capillarity in a small-diameter glass tube.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

2. Step 2 — Rearrange the relation so that sigma stands alone on the left-hand side.

3. Step 3 — List the givens: wetting angle (theta) = 34.0000 deg, specific weight (gamma) = 9,560 N/m^3, tube diameter (d) = 0.0025 m, capillary rise (h) = 0.0510 m.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$\sigma = 0.3676\ \text{N/m}$$

6. Step 6 — Check: returning sigma = 0.3676 N/m to

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.7351 — kept a factor of two that cancels in the correct rearrangement.
• 0.1838 — dropped that same factor in the other direction.
• 0.4043 — rounded an intermediate value before the final step.

Reference: FE Handbook — Surface Tension and Capillarity

capillarity effects on a manometer reading due to surface tension Given surface tension (sigma) = 0.0730 N/m; wetting angle (theta) = 36.0000 deg; specific weight (gamma) = 9,310 N/m^3; capillary rise (h) = 0.0260 m, determine the tube diameter (d) in m.

Start with the thinking

• The governing relation printed in this handbook section is Surface tension and capillarity (capillary rise).
• Everything except d is given, so isolate d symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A capillary tube experiment illustrates surface tension and capillarity in a small-diameter glass tube.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

2. Step 2 — Rearrange the relation so that d stands alone on the left-hand side.

3. Step 3 — List the givens: surface tension (sigma) = 0.0730 N/m, wetting angle (theta) = 36.0000 deg, specific weight (gamma) = 9,310 N/m^3, capillary rise (h) = 0.0260 m.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$d = 0.0010\ \text{m}$$

6. Step 6 — Check: returning d = 0.0010 m to

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0020 — kept a factor of two that cancels in the correct rearrangement.
• 0.0005 — dropped that same factor in the other direction.
• 0.0011 — rounded an intermediate value before the final step.

Reference: FE Handbook — Surface Tension and Capillarity

surface tension and capillarity in a water-filled capillary tube Given surface tension (sigma) = 0.0790 N/m; wetting angle (theta) = 23.0000 deg; specific weight (gamma) = 9,030 N/m^3; tube diameter (d) = 0.0037 m, determine the capillary rise (h) in m.

Start with the thinking

• The governing relation printed in this handbook section is Surface tension and capillarity (capillary rise).
• Everything except h is given, so isolate h symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• A capillary tube experiment illustrates surface tension and capillarity in a small-diameter glass tube.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

2. Step 2 — Rearrange the relation so that h stands alone on the left-hand side.

3. Step 3 — List the givens: surface tension (sigma) = 0.0790 N/m, wetting angle (theta) = 23.0000 deg, specific weight (gamma) = 9,030 N/m^3, tube diameter (d) = 0.0037 m.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$h = 0.0087\ \text{m}$$

6. Step 6 — Check: returning h = 0.0087 m to

   $$h = \dfrac{4 \sigma \cos\theta}{\gamma d}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0174 — kept a factor of two that cancels in the correct rearrangement.
• 0.0044 — dropped that same factor in the other direction.
• 0.0096 — rounded an intermediate value before the final step.

Reference: FE Handbook — Surface Tension and Capillarity