Question

The force required to separate two glass plates of area \(10^{-2} \mathrm{~m}^{2}\) with a film of water \(0.05 \mathrm{~mm}\) thick between them is (surface tension of water is \(70 \times 10^{-3} \mathrm{~N} / \mathrm{m}\) ) (A) \(28 \mathrm{~N}\) (B) \(14 \mathrm{~N}\) (C) \(50 \mathrm{~N}\) (D) \(38 \mathrm{~N}\)

Short answer

The force required to separate the two glass plates with a water film between them can be calculated using the formula \(F = 2 \times T \times P\), where 'T' is the surface tension of water, and 'P' is the perimeter of the glass plates. Given the area of the glass plates \(A = 10^{-2}\mathrm{~m}^{2}\), the water film thickness \(t = 5\times10^{-5}\mathrm{~m}\), and the surface tension of water \(T = 70 \times 10^{-3}\mathrm{~N}/\mathrm{m}\), we can find the perimeter of the glass plates assuming a square shape. Then, substitute the values into the formula and calculate the force. The correct answer is (B) \(14 \mathrm{~N}\).

Step-by-step solution

Identify the given information

The given information in the problem is: - Area of glass plates: \(A = 10^{-2}\mathrm{~m}^{2}\) - Thickness of water film: \(t = 0.05\mathrm{~mm}\) or \(5\times10^{-5}\mathrm{~m}\) - Surface tension of water: \(T = 70 \times 10^{-3}\mathrm{~N}/\mathrm{m}\)

Convert thickness of water film to meters

Make sure that all quantities have the same units (meters in this case). So, we find the thickness in meters: \(t = 0.05\mathrm{~mm}\times\frac{1\mathrm{~m}}{1000\mathrm{~mm}} = 5\times10^{-5}\mathrm{~m}\)

Calculate the perimeter of the glass plates

Assume that the glass plates are square in shape. Let 's' be the length of one side. Then, \(A = s^2\). Thus, the length is: \(s = \sqrt{A} = \sqrt{10^{-2}\mathrm{~m}^{2}} = 10^{-1}\mathrm{~m}\) The perimeter (length of contact between the water film and the glass plate) is: \(P = 4s = 4\times10^{-1}\mathrm{~m}\)

Calculate the force required to separate the glass plates

The force required to separate the two glass plates can be found using the formula: \(F = 2 \times T \times P\) Here, we multiply by 2 because there are 2 interfaces between the glass plates and the water film (one between the top glass plate and the water film, and one between the bottom glass plate and the water film). \(F = 2 \times (70 \times 10^{-3}\mathrm{~N}/\mathrm{m}) \times (4\times10^{-1}\mathrm{~m})\)

Calculate the force

Plug in the given values and calculate the force: \(F = 2 \times (70 \times 10^{-3}\mathrm{~N}/\mathrm{m}) \times (4\times10^{-1}\mathrm{~m}) = 14\mathrm{~N}\)

Find the correct answer from the given options

The calculated force is \(14\mathrm{~N}\), which matches option (B). So, the force required to separate the two glass plates is \(14\mathrm{~N}\). The correct answer is (B) \(14 \mathrm{~N}\).

Key concepts

Force Calculation

Force calculation is an essential part of physics, especially when dealing with topics like surface tension. In problems like separating glass plates, the force is calculated using specific formulas involving the interplay of several variables. Here, the formula used is \[ F = 2 imes T imes P \], where:

• \( F \) is the force required to separate the glass plates,
• \( T \) is the surface tension of the liquid between the plates,
• \( P \) is the perimeter of the area in contact with the liquid.

The multiplication by 2 reflects that there are two interfaces here, indicating the presence of the liquid on both sides of the glass plate.Acquiring the right understanding of how surface tension and area interplay can lead you to correctly find the force required. In this problem, substituting the given values into the formula helps reach the solution accurately, providing the force required as 14 N.

Physics Problem Solving

Physics problem solving can appear daunting at first glance, but breaking it down into manageable steps makes it much easier. It's crucial to always start by identifying given information. This problem is a great example:

• Surface tension value for water: \(70 \times 10^{-3} \text{ N/m}\).
• Area of the glass plates: \(10^{-2} \text{ m}^2\).
• Film thickness: \(0.05 \text{ mm} \).

Once you've listed all known values, consider the formulas and principles applicable to the problem. Here, it's the formula for the force involving surface tension and perimeter. By practicing these systematic approaches, students improve their physics problem-solving skills, reduce errors, and develop a deeper understanding of concepts.

Units Conversion

Proper unit conversion is imperative for solving physics problems correctly. In this exercise, the thickness of the water film was initially given in millimeters, but calculations required the metric system. Thus, converting from millimeters to meters was vital:\[ 0.05\mathrm{~mm} \times \frac{1 \mathrm{~m}}{1000 \mathrm{~mm}} = 5 \times 10^{-5}\mathrm{~m} \]Every measurement should be expressed in consistent units, typically in meters in SI units, to ensure calculations are on the same scale. This prevents confusion and errors.When learning conversion:

• Always know your conversion factors.
• Implement a step-wise approach for accuracy.
• Double-check to confirm correct unit conversions.

By adhering to these techniques, students can navigate unit conversions with ease, making complex problem-solving more straightforward.

Perimeter Calculation

Perimeter calculation is fundamental in determining areas of contact in problems dealing with surface tension. Knowing if an object is square helps simplify calculations. For a square plate with given area \(A\):\[ s = \sqrt{A} \]From this, compute the perimeter \(P\) as:\[ P = 4s \]In our case, the side \( s \) is determined by:\[ s = \sqrt{10^{-2} \text{ m}^2} = 10^{-1} \text{ m} \]Subsequently, the perimeter turns out to be:\[ P = 4 \times 10^{-1} \mathrm{~m} \]Understanding the role of perimeter allows you to correctly determine its contribution to the surface tension force equation. Mastering these geometric fundamentals helps in a wider array of physics problems related to shapes and configurations.