
A film of water is formed between two straight parallel wires of length 10 cm each separated by 0.5 cm. If their separation is increased by 1 mm while still maintaining their parallelism, how much work will have to be done on water? (Given that the surface tension of water $ = 7.2 \times {10^{ - 2}}{\text{ N/m}}$ )
A. $7.22 \times {10^{ - 6}}{\text{ J}}$
B. $1.44 \times {10^{ - 5}}{\text{ J}}$
C. $2.88 \times {10^{ - 5}}{\text{ J}}$
D. $5.76 \times {10^{ - 5}}{\text{ J}}$
Answer
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Hint: The above question is related to surface tension and surface energy of fluids. The total work done will be equivalent to the change in the surface energy of the water film. This surface energy is defined as the product of surface tension and the surface area of the object we are considering. Calculate the increase in the surface area when the separation is increased by 1 mm and then calculate its corresponding surface energy. This will give you the work to be done on water.
Complete answer:
Given values:
Length of the wires $ = 10{\text{ cm}}$ (length of the film)
Separation between the wires $ = 0.5{\text{ cm}}$ (initial breadth of the film)
Increased separation between the wires $ = 1{\text{ mm}}$ (increment in the breadth of the film)
Thus, considering both the inner and outer surfaces of the film,
The increase in the total surface area of the film $ = 2 \times 10 \times 0.1$
Therefore,
$\Delta A = 2{\text{ c}}{{\text{m}}^2} = 2 \times {10^{ - 4}}{\text{ }}{{\text{m}}^2}$ … (1)
Surface tension of water, $T = 7.2 \times {10^{ - 2}}{\text{ N/m}}$ … (2)
Now, we know that the work done is given by the change in the surface energy of the film and surface energy is given by:
Surface Energy $ = T\Delta A$
Hence, the work done $ = T\Delta A$
Substituting the values from equations (1) and (2) In the above equation,
Work done$ = 2 \times {10^{ - 4}} \times 7.2 \times {10^{ - 2}} = 14.4 \times {10^{ - 6}}{\text{ = 1}}{\text{.44}} \times {10^{ - 5}}{\text{ J}}$
Hence, the work done on the water film is ${\text{1}}{\text{.44}} \times {10^{ - 5}}{\text{ J}}$ respectively. Thus, the correct option is B.
Note: The above question is asking to calculate the work required to be done on water. When the separation between the wires is increased by 1 mm, the surface area of the film increases. As the surface area increases, the film will absorb some energy equivalent to the surface energy of the film. This energy is to be provided by working on water.
Complete answer:
Given values:
Length of the wires $ = 10{\text{ cm}}$ (length of the film)
Separation between the wires $ = 0.5{\text{ cm}}$ (initial breadth of the film)
Increased separation between the wires $ = 1{\text{ mm}}$ (increment in the breadth of the film)
Thus, considering both the inner and outer surfaces of the film,
The increase in the total surface area of the film $ = 2 \times 10 \times 0.1$
Therefore,
$\Delta A = 2{\text{ c}}{{\text{m}}^2} = 2 \times {10^{ - 4}}{\text{ }}{{\text{m}}^2}$ … (1)
Surface tension of water, $T = 7.2 \times {10^{ - 2}}{\text{ N/m}}$ … (2)
Now, we know that the work done is given by the change in the surface energy of the film and surface energy is given by:
Surface Energy $ = T\Delta A$
Hence, the work done $ = T\Delta A$
Substituting the values from equations (1) and (2) In the above equation,
Work done$ = 2 \times {10^{ - 4}} \times 7.2 \times {10^{ - 2}} = 14.4 \times {10^{ - 6}}{\text{ = 1}}{\text{.44}} \times {10^{ - 5}}{\text{ J}}$
Hence, the work done on the water film is ${\text{1}}{\text{.44}} \times {10^{ - 5}}{\text{ J}}$ respectively. Thus, the correct option is B.
Note: The above question is asking to calculate the work required to be done on water. When the separation between the wires is increased by 1 mm, the surface area of the film increases. As the surface area increases, the film will absorb some energy equivalent to the surface energy of the film. This energy is to be provided by working on water.
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