Problem 980
Question
A beaker of radius \(15 \mathrm{~cm}\) is filled with liquid of surface tension \(0.075 \mathrm{~N} / \mathrm{m}\). Force across an imaginary diameter on the surface of liquid is (A) \(0.075 \mathrm{~N}\) (B) \(1.5 \times 10^{-2} \mathrm{~N}\) (C) \(0.225 \mathrm{~N}\) (D) \(2.25 \times 10^{-2} \mathrm{~N}\)
Step-by-Step Solution
Verified Answer
The force across the imaginary diameter on the surface of the liquid is 0.225 N, so the correct answer is (C) \(0.225 \mathrm{~N}\).
1Step 1: Identify the formula for force due to surface tension
We have surface tension (T) and radius of the beaker (R). We will use the formula for the force (F) exerted by the surface tension of the liquid acting across the imaginary diameter: \(F = 2 \cdot T \cdot L\), where L is the length along which the force is acting. Since we are considering an imaginary diameter, L is equal to the diameter of the beaker.
2Step 2: Calculate the diameter of the beaker
We are given the radius of the beaker, R = 15 cm. The diameter (D) is twice the value of R. So, the diameter is calculated as follows: \(D = 2 \cdot R = 2 \cdot 15 \mathrm{~cm} = 30 \mathrm{~cm}\). We'll convert the diameter to meters by dividing by 100: \(D = 0.30 \mathrm{~m}\).
3Step 3: Calculate the force due to surface tension
To find the force, we'll plug the values of surface tension and diameter into the formula we identified in step 1: \(F = 2 \cdot T \cdot L = 2 \cdot 0.075 \mathrm{~N/m} \cdot 0.30 \mathrm{~m} = 0.45 \mathrm{~N}\).
However, since the force is acting across an imaginary diameter, we need to divide the calculated force by 2: \(F = \frac{0.45 \mathrm{~N}}{2} = 0.225 \mathrm{~N}\).
The force across the imaginary diameter on the surface of the liquid is 0.225 N, so the correct answer is (C) \(0.225 \mathrm{~N}\).
Key Concepts
Force calculationBeaker physicsImaginary diameter
Force calculation
Calculating force due to surface tension involves utilizing the relationship between surface tension and the dimension over which it acts. Surface tension itself is the measure of how much force is exerted per unit length at the surface of a liquid. In mathematical terms, if you know the surface tension (T) and the length along which it acts (L), the formula for the force (F) is given by:
\[ F = 2 \cdot T \cdot L \]
\[ F = 2 \cdot T \cdot L \]
- The factor of 2 accounts for the surface tension acting along both edges of the length, since in a circular container like a beaker, surface tension affects both sides of the imaginary line spanning the diameter.
- It is important to ensure that the units are consistent, typically converting lengths to meters to match the units of surface tension (usually N/m).
Beaker physics
When dealing with problems involving liquids in a beaker, it's essential to understand the principles governing the behavior of fluids in containers. A beaker, often cylindrical, serves as a simple model to comprehend more complex fluid dynamics.
\[ D = 2 \cdot R \]
is fundamental. Also, converting measurements from centimeters to meters is vital for ensuring consistency with units of measurement.
Remember, proficiency in these basics is key to tackling more intricate physics challenges.
- Beakers have a round cross-section, making calculations involving diameters and radii straightforward with basic geometry rules.
- The beaker’s physical dimension, radius in this case (15 cm), is crucial as it directly pertains to determining critical measures like the diameter.
\[ D = 2 \cdot R \]
is fundamental. Also, converting measurements from centimeters to meters is vital for ensuring consistency with units of measurement.
Remember, proficiency in these basics is key to tackling more intricate physics challenges.
Imaginary diameter
The term "imaginary diameter" might sound mystical, but it is just a logical way to describe how we apply calculations over an assumed path or line, which, in real-world terms within the liquid, isn't physically marked.
- An imaginary diameter helps to conceptualize forces or actions distributed across the center of a beaker without needing an actual physical line.
- It is used to simplify the calculation of force and is an abstraction allowing us to apply mathematical formulas effectively.
Other exercises in this chapter
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