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Question: A mercury drop of radius \(1cm\) is sprayed into \({10^6}\) drops of equal size. The energy expended...

A mercury drop of radius 1cm1cm is sprayed into 106{10^6} drops of equal size. The energy expended in joule is (surface tension of mercury is 460×10−3Nm−1460 \times {10^{ - 3}}N{m^{ - 1}}
A.0.0570.057
B.5.75.7
C.5.7×10−45.7 \times {10^{ - 4}}
D.5.7×10−65.7 \times {10^{ - 6}}

Explanation

Solution

Recall the concept of surface tension. Surface tension is defined as the ability of a liquid surface, by which the liquids can shrink into a minimum surface area that they can. The liquid molecules act as a stretched elastic membrane and resist effect caused by any external force. This can be observed in any liquid drop with spherical shape.

Complete answer:
Step I:
Let the radius of big drop of mercury be =R = R
The radius of small drop when it is sprayed be =r = r
The total volume of the drop remains conserved after it is sprayed. Therefore it can be written that
Step II:
⇒43πR3=n×43πr3\Rightarrow \dfrac{4}{3}\pi {R^3} = n \times \dfrac{4}{3}\pi {r^3}
Where ‘R’ is the radius of the mercury drops before spraying into drops and ‘r’ is the radius after splitting in drops and ‘n’ is the number of drops into which the mercury drop is broken.
⇒R=n13.r\Rightarrow R = {n^{\dfrac{1}{3}}}.r---(i)
Step III:
Given that the radius of drop is 1cm1cm, substituting the value and solving
r=102×1r = {10^2} \times 1
r=0.01cmr = 0.01cm
Step IV:
Energy required will be equal to the force of surface tension and the change in surface area of the drop. Therefore,
⇒E=T×ΔA\Rightarrow E = T \times \Delta A
⇒E=T×4π(R2−nr2)\Rightarrow E = T \times 4\pi ({R^2} - n{r^2})
Where ‘T’ is the surface tension
‘E’ is the energy required
Step V:
Substitute the value of r2{r^2} from equation (i),
⇒E=460×10−3×4×3.14[(R)2−n(Rn13)2]\Rightarrow E = 460 \times {10^{ - 3}} \times 4 \times 3.14[{(R)^2} - n{(\dfrac{R}{{{n^{\dfrac{1}{3}}}}})^2}]
⇒E=460×10−3×4×3.14×[(R)2−n1−23(R)2]\Rightarrow E = 460 \times {10^{ - 3}} \times 4 \times 3.14 \times [{(R)^2} - {n^{1 - \dfrac{2}{3}}}{(R)^2}]
⇒E=460×4×3.14×10−3(R)2[1−n13]\Rightarrow E = 460 \times 4 \times 3.14 \times {10^{ - 3}}{(R)^2}[1 - {n^{\dfrac{1}{3}}}]
⇒E=460×4×3.14×10−3×1×10−4[1−1063]\Rightarrow E = 460 \times 4 \times 3.14 \times {10^{ - 3}} \times 1 \times {10^{ - 4}}[1 - {10^{\dfrac{6}{3}}}]
⇒E=−(460×4×3.14×10−7×99)\Rightarrow E = - (460 \times 4 \times 3.14 \times {10^{ - 7}} \times 99)
⇒E=571982.4107\Rightarrow E = \dfrac{{571982.4}}{{{{10}^7}}}
⇒E=0.057J\Rightarrow E = 0.057J
Therefore the energy expanded will be 0.057J0.057J

⇒\Rightarrow Option A is the right answer.

Note:
It is to be noted that the term energy means the ability to do work. When the mercury drop is broken in parts, energy will be required. It is to be remembered that energy is a property and it can not be created or destroyed. It can be transferred from one form to another.