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Chapter 8: Problem 37

The surface tension of a liquid is \(5 \mathrm{~N} / \mathrm{m}\). If a film is held on a ring of area \(0.02 \mathrm{~m}^{2}\), its total surface energy is about (A) \(5 \times 10^{-2} \mathrm{~J}\) (B) \(2.5 \times 10^{-2} \mathrm{~J}\) (C) \(2 \times 10^{-1} \mathrm{~J}\) (D) \(3 \times 10^{-1} \mathrm{~J}\)

Short Answer

Expert verified
The total surface energy of the film held on a ring of area \(0.02 m^2\) with a surface tension of \(5 N/m\) is \(0.2 J\), which corresponds to option (C) \(2 \times 10^{-1} J\).

Step by step solution

01

1. Identify the given variables

We are given the surface tension as \(5 N/m\) and the area of the ring as \(0.02 m^2\).
02

2. Recall surface energy formula

The formula for surface energy (E) is given by the product of surface tension (T) and the area (A): \(E = T \times A\).
03

3. Substitute the values and find surface energy

Now we can substitute the given values of surface tension \(T = 5 N/m\) and area \(A = 0.02 m^2\) into the formula: \[E = 5 \times 0.02\]
04

4. Calculate the product

On calculating the product, we get: \[E = 0.1 J\] Now, it's important to remember that the film has two sides, so we actually have twice this surface energy for the whole system.
05

5. Multiply by 2 to account for both sides of the film

We multiply the calculated surface energy by 2 to account for both sides of the film: \[E_{total} = 2 \times 0.1\]
06

6. Calculate the total surface energy

On calculating the total surface energy, we get: \[E_{total} = 0.2 J\] This corresponds to the answer (C) \(2 \times 10^{-1} J\).

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Most popular questions from this chapter

A uniform rod of length \(L\) has a mass per unit length \(\lambda\) and area of cross-section \(A .\) The elongation in the rod is \(l\) due to its own weight if it is suspended from the ceiling of a room. The Young's modulus of the rod is (A) \(\frac{2 \lambda g L^{2}}{A l}\) (B) \(\frac{\lambda g L^{2}}{2 A l}\) (C) \(\frac{2 \lambda g L}{A l}\) (D) \(\frac{\lambda g l^{2}}{A L}\)

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The weight of a body in air is \(100 \mathrm{~N}\). Its weight in water, if it displaces \(400 \mathrm{cc}\) of water is (A) \(90 \mathrm{~N}\) (B) \(94 \mathrm{~N}\) (C) \(98 \mathrm{~N}\) (D) None of these

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