A circular plate of radius $4{\text{ cm}}$ and weight $W$ is made to rest on the surface of the water. If a minimum pull of $W+F$ is required to clear the plate off the water surface, then find F. Given surface tension of water ${S_W}{\text{ = 0}}{\text{.072 N}}{{\text{m}}^{ - 1}}$.
Answer
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Hint: in order to solve this question first of all we need to convert the radius from centimetre to the meter and then we have to find the circumference of the plate. After then by applying the formula we will find the force (F). (F) is the part of the minimum pull required to clear the plate off the water surface.
Formula used:
$F{\text{ }} = {\text{ }}{S_W}{\text{ }} \times {\text{ }}Circumference{\text{ }}of{\text{ }}the{\text{ }}plate$
Here, $F$ denotes the force due to surface tension and ${S_W}{\text{ }}$denotes the surface tension of water.
$Circumference{\text{ }}of{\text{ }}the{\text{ }}plate{\text{ = 2}}\pi r$
This is the formula to find the circumference of the plate where $r$ denotes the radius of the plate and value of $\pi {\text{ = 3}}{\text{.14}}$.
Complete step by step answer:
Changing the radius from centimetre to meter
$1cm{\text{ = 0}}{\text{.01}}m$
Here we are given radius is equal to 4cm
$4cm{\text{ = 0}}{\text{.04}}m$
Now we will find the circumference of the plate
$\text{Circumference of the plate} = 2 \pi r$
$ \Rightarrow \text{Circumference of the plate} =2 \times \pi \times 0.04 \\ $
$ \Rightarrow \text{Circumference of the plate} =2 \times 3.14 \times 0.04 \\ $
$ \Rightarrow \text{Circumference of the plate} =0.25{\text{ m}}$
Here we are given that the downward force acting on the plate of W+F is required to clear the plate off the water surface. Now we will have to find force (F) caused by the surface tension according to the question. Formula of applied force by surface tension
$F{\text{ }} = {\text{ }}{S_W}{\text{ }} \times \text{Circumference of the plate} \\ $
It is given that ${S_W}{\text{ = 0}}{\text{.072 N}}{{\text{m}}^{ - 1}}$
And we have found,
$\text{Circumference of the plate} = 0{\text{.25 m}}$
Now applying formula
$F{\text{ }} = {\text{ 0}}{\text{.072 N}}{{\text{m}}^{ - 1}}{\text{ }} \times\text{Circumference of the plate} \\ $
$ \Rightarrow F{\text{ }} ={\text{ 0}}{\text{.072 N}}{{\text{m}}^{ - 1}}{\text{ }} \times {\text{ 0}}{\text{.25 m}} \\ $
$ \therefore F{\text{ }} ={\text{ 0}}{\text{.018 N}}$
Therefore, the force applied is equal to ${\text{ 0}}{\text{.018 N}}$.
Note: Most of the students make the mistake by not converting the radius from centimetre to the meter this small mistake can make the whole answer incorrect another common mistake is to forgot to write the unit along the question unit should be there on each step as it avoids the confusion for example in above question we found out that final answer was in newton and in the above step we get to know how meter unit cut down and how answer was in newton.
Formula used:
$F{\text{ }} = {\text{ }}{S_W}{\text{ }} \times {\text{ }}Circumference{\text{ }}of{\text{ }}the{\text{ }}plate$
Here, $F$ denotes the force due to surface tension and ${S_W}{\text{ }}$denotes the surface tension of water.
$Circumference{\text{ }}of{\text{ }}the{\text{ }}plate{\text{ = 2}}\pi r$
This is the formula to find the circumference of the plate where $r$ denotes the radius of the plate and value of $\pi {\text{ = 3}}{\text{.14}}$.
Complete step by step answer:
Changing the radius from centimetre to meter
$1cm{\text{ = 0}}{\text{.01}}m$
Here we are given radius is equal to 4cm
$4cm{\text{ = 0}}{\text{.04}}m$
Now we will find the circumference of the plate
$\text{Circumference of the plate} = 2 \pi r$
$ \Rightarrow \text{Circumference of the plate} =2 \times \pi \times 0.04 \\ $
$ \Rightarrow \text{Circumference of the plate} =2 \times 3.14 \times 0.04 \\ $
$ \Rightarrow \text{Circumference of the plate} =0.25{\text{ m}}$
Here we are given that the downward force acting on the plate of W+F is required to clear the plate off the water surface. Now we will have to find force (F) caused by the surface tension according to the question. Formula of applied force by surface tension
$F{\text{ }} = {\text{ }}{S_W}{\text{ }} \times \text{Circumference of the plate} \\ $
It is given that ${S_W}{\text{ = 0}}{\text{.072 N}}{{\text{m}}^{ - 1}}$
And we have found,
$\text{Circumference of the plate} = 0{\text{.25 m}}$
Now applying formula
$F{\text{ }} = {\text{ 0}}{\text{.072 N}}{{\text{m}}^{ - 1}}{\text{ }} \times\text{Circumference of the plate} \\ $
$ \Rightarrow F{\text{ }} ={\text{ 0}}{\text{.072 N}}{{\text{m}}^{ - 1}}{\text{ }} \times {\text{ 0}}{\text{.25 m}} \\ $
$ \therefore F{\text{ }} ={\text{ 0}}{\text{.018 N}}$
Therefore, the force applied is equal to ${\text{ 0}}{\text{.018 N}}$.
Note: Most of the students make the mistake by not converting the radius from centimetre to the meter this small mistake can make the whole answer incorrect another common mistake is to forgot to write the unit along the question unit should be there on each step as it avoids the confusion for example in above question we found out that final answer was in newton and in the above step we get to know how meter unit cut down and how answer was in newton.
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