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Question
AnswersA liquid drop of diameter \[D\] breaks up into \[27\] drops. Find the resultant change in energy. Surface tension of the liquid is \[T\].
a. \[ 2\pi T{D^2}\]
b. \[ \pi T{D^2}\]
c. \[ \dfrac{{\pi T{D^2}}}{2}\]
d. \[ 4\pi T{D^2}\]
Answer
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Hint: First we have to find the total volume of large drop, so we have to construct the volume of small drop and volume of \[27\] drops is multiplied. Then we have to equate that and by doing some simplification we get the required answer.
Formula Used:
Change in energy \[ = \]\[(\]Change in Area \[ \times \] Surface Tension \[)\]
Volume of a sphere is \[ = \dfrac{4}{3}\pi {r^3}\], here \[r\] is the radius of the given sphere.
Area of a sphere is \[ = 4\pi {r^2}\].
Complete step by step answer:
A liquid drop is like a sphere.
So, the volume of the drop will be equal to the volume of the sphere.
Now we have to use the formula for volume of a sphere is \[ = \dfrac{4}{3}\pi {r^3}\], here \[r\] is the radius of the given sphere.
Let's say, the radius of each small drop is\[d\]unit.
The diameter of the large drop is \[ = D\] .
Radius of the large drop is \[ = \dfrac{D}{2}\] .
Volume of the large drop is \[ = \dfrac{4}{3}\pi {\left( {\dfrac{D}{2}} \right)^3}\] .
Volume of each small drop is \[ = \dfrac{4}{3}\pi {d^3}\].
Total volume of \[27\] drops will be\[ = \dfrac{4}{3}\pi {d^3} \times 27 = 9 \times 4\pi {d^3} = 36\pi {d^3}\] .
So, the total volume of \[27\]drops will be equal to the total volume of the large drop.
So, we derive the following equation:
\[ \Rightarrow 36\pi {d^3} = \dfrac{4}{3}\pi {\left( {\dfrac{D}{2}} \right)^3}\]
By simplifying the cubic form we get:
\[ \Rightarrow 36\pi {d^3} = \dfrac{4}{3}\pi \left( {\dfrac{{{D^3}}}{8}} \right)\]
Calculate the denominator parts in R.H.S:
\[ \Rightarrow 36\pi {d^3} = \pi \left( {\dfrac{{{D^3}}}{6}} \right)\]
Cancel out the \[\pi \]from both sides:
\[ \Rightarrow 36{d^3} = \left( {\dfrac{{{D^3}}}{6}} \right)\]
Divide R.H.S by \[36\]we get:
\[ \Rightarrow {d^3} = \left( {\dfrac{{{D^3}}}{6}} \right) \times \dfrac{1}{{36}}\]
Simplifying the above form:
\[ \Rightarrow {d^3} = \left( {\dfrac{{{D^3}}}{{{6^3}}}} \right)\]
Applying the rule of indices:
\[ \Rightarrow d = \left( {\dfrac{D}{6}} \right)\]
Initial area of the large liquid was \[ = 4\pi {\left( {\dfrac{D}{2}} \right)^2} = \pi {D^2}\] .
After division in \[27\] drops, the total area of all these drops becomes \[ = 27 \times 4\pi {d^2}\] .
But \[d = \left( {\dfrac{D}{6}} \right)\].
The changed area will be \[ = 27 \times 4\pi {\left( {\dfrac{D}{6}} \right)^2} = 3\pi {D^2}\] .
Change in total area will be \[ = (3\pi {D^2} - \pi {D^2}) = 2\pi {D^2}\] .
Resultant change in energy will be \[ = \]( Change in area \[ \times \] Surface Tension of the liquid) .
So, change in energy \[ = (2\pi {D^2} \times T) = 2\pi T{D^2}\].
Hence, the correct answer is option (A).
Note: A surface tension is the tendency of minimums possible and the liquid surfaces to shrink into it. So, change in energy is calculated alongside the change in area of liquid.
Formula Used:
Change in energy \[ = \]\[(\]Change in Area \[ \times \] Surface Tension \[)\]
Volume of a sphere is \[ = \dfrac{4}{3}\pi {r^3}\], here \[r\] is the radius of the given sphere.
Area of a sphere is \[ = 4\pi {r^2}\].
Complete step by step answer:
A liquid drop is like a sphere.
So, the volume of the drop will be equal to the volume of the sphere.
Now we have to use the formula for volume of a sphere is \[ = \dfrac{4}{3}\pi {r^3}\], here \[r\] is the radius of the given sphere.
Let's say, the radius of each small drop is\[d\]unit.
The diameter of the large drop is \[ = D\] .
Radius of the large drop is \[ = \dfrac{D}{2}\] .
Volume of the large drop is \[ = \dfrac{4}{3}\pi {\left( {\dfrac{D}{2}} \right)^3}\] .
Volume of each small drop is \[ = \dfrac{4}{3}\pi {d^3}\].
Total volume of \[27\] drops will be\[ = \dfrac{4}{3}\pi {d^3} \times 27 = 9 \times 4\pi {d^3} = 36\pi {d^3}\] .
So, the total volume of \[27\]drops will be equal to the total volume of the large drop.
So, we derive the following equation:
\[ \Rightarrow 36\pi {d^3} = \dfrac{4}{3}\pi {\left( {\dfrac{D}{2}} \right)^3}\]
By simplifying the cubic form we get:
\[ \Rightarrow 36\pi {d^3} = \dfrac{4}{3}\pi \left( {\dfrac{{{D^3}}}{8}} \right)\]
Calculate the denominator parts in R.H.S:
\[ \Rightarrow 36\pi {d^3} = \pi \left( {\dfrac{{{D^3}}}{6}} \right)\]
Cancel out the \[\pi \]from both sides:
\[ \Rightarrow 36{d^3} = \left( {\dfrac{{{D^3}}}{6}} \right)\]
Divide R.H.S by \[36\]we get:
\[ \Rightarrow {d^3} = \left( {\dfrac{{{D^3}}}{6}} \right) \times \dfrac{1}{{36}}\]
Simplifying the above form:
\[ \Rightarrow {d^3} = \left( {\dfrac{{{D^3}}}{{{6^3}}}} \right)\]
Applying the rule of indices:
\[ \Rightarrow d = \left( {\dfrac{D}{6}} \right)\]
Initial area of the large liquid was \[ = 4\pi {\left( {\dfrac{D}{2}} \right)^2} = \pi {D^2}\] .
After division in \[27\] drops, the total area of all these drops becomes \[ = 27 \times 4\pi {d^2}\] .
But \[d = \left( {\dfrac{D}{6}} \right)\].
The changed area will be \[ = 27 \times 4\pi {\left( {\dfrac{D}{6}} \right)^2} = 3\pi {D^2}\] .
Change in total area will be \[ = (3\pi {D^2} - \pi {D^2}) = 2\pi {D^2}\] .
Resultant change in energy will be \[ = \]( Change in area \[ \times \] Surface Tension of the liquid) .
So, change in energy \[ = (2\pi {D^2} \times T) = 2\pi T{D^2}\].
Hence, the correct answer is option (A).
Note: A surface tension is the tendency of minimums possible and the liquid surfaces to shrink into it. So, change in energy is calculated alongside the change in area of liquid.
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