
An oil drop of radius 1 cm is sprayed into 1000 small equal drops of same radius the surface tension of oil drop is $50\,dyne/cm$ then the work done is:
A. $18\pi \,ergs$
B. $180\pi \,ergs$
C. $1800\pi \,ergs$
D. $8000\pi \,ergs$
Answer
164.1k+ views
Hint: Surface tension is often a liquid's characteristic. All of the molecules in the liquid perceive force coming from all directions at once. Net force on such molecules will therefore be zero. Since there are no molecules above the molecules on the liquid surface, they will only be subject to force from below. They therefore possess some surface energy.
Formula used:
$W=T\times \Delta A$
W is the work done, T is the surface tension and $\Delta A$is the change in area
Complete answer:
Start with finding out the radius of each 1000 small oil drops.
Let the radius of each 1000 small oil drop be r.
Now from the question, we know that the volume of oil drop will remain the same. So, $\dfrac{4}{3}\pi {R^2} = \left( {\dfrac{4}{3}\pi {r^2}} \right)1000$
$R = 1$ (given in the question)
Therefore, $r = \dfrac{R}{{10}} = \dfrac{1}{{10}} = 0.1$
Now find the change in surface area of the oil drop. Let change in surface area be S.
$S = 4\pi {\left( {0.1} \right)^2}1000 - 4\pi {\left( 1 \right)^2}$
So,
$S = 4\pi \left( 9 \right)$
Work done = surface tension*change in surface area
$W = 50 \times 4\pi \left( 9 \right)$ (surface tension is 50 given in question)
$W = 1800\pi \,ergs$
Hence the correct answer is Option D.
Note: Here the radius of oil drop was given so it was easy to find the required answer. Also each 1000 oil drops has the same radius if it is different then the whole answer will change. Try to put all the values from the question in the formula of work done carefully. If any value is wrong every value in the formula will get affected.
Formula used:
$W=T\times \Delta A$
W is the work done, T is the surface tension and $\Delta A$is the change in area
Complete answer:
Start with finding out the radius of each 1000 small oil drops.
Let the radius of each 1000 small oil drop be r.
Now from the question, we know that the volume of oil drop will remain the same. So, $\dfrac{4}{3}\pi {R^2} = \left( {\dfrac{4}{3}\pi {r^2}} \right)1000$
$R = 1$ (given in the question)
Therefore, $r = \dfrac{R}{{10}} = \dfrac{1}{{10}} = 0.1$
Now find the change in surface area of the oil drop. Let change in surface area be S.
$S = 4\pi {\left( {0.1} \right)^2}1000 - 4\pi {\left( 1 \right)^2}$
So,
$S = 4\pi \left( 9 \right)$
Work done = surface tension*change in surface area
$W = 50 \times 4\pi \left( 9 \right)$ (surface tension is 50 given in question)
$W = 1800\pi \,ergs$
Hence the correct answer is Option D.
Note: Here the radius of oil drop was given so it was easy to find the required answer. Also each 1000 oil drops has the same radius if it is different then the whole answer will change. Try to put all the values from the question in the formula of work done carefully. If any value is wrong every value in the formula will get affected.
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