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Question: The diameter of one drop of water is 0.2 cm. The work done in breaking one drop into 1000 equal drop...

The diameter of one drop of water is 0.2 cm. The work done in breaking one drop into 1000 equal droplets will be:-
(Surface tension of water=7×10−2 N/m = 7 \times {10^{ - 2}}\,N/m)
(A) 7.9×10−6 J7.9 \times {10^{ - 6}}\,J
(B) 5.92×10−6 J5.92 \times {10^{ - 6}}\,J
(C) 2.92×10−6 J2.92 \times {10^{ - 6}}\,J
(D) 1.92×10−6 J1.92 \times {10^{ - 6}}\,J

Explanation

Solution

Since the volume of 1000 small droplets is the same as the volume of the large droplet. Using this, calculate the radius of each small droplet. Calculate the change in surface area after breaking the large droplet into 1000 small droplets. Use the formula for change in surface energy to determine the work done.

Formula used:
The formula for change in surface energy,
δU=TδA\delta U = T\delta A
Here, T is the surface tension and δA\delta A is the change in surface area.

Complete step by step answer:
We have given that the diameter of a large droplet is 0.2 cm. Therefore, the radius is obviously 0.1 cm.We can see that, on breaking the large droplet into 1000 small droplets, the volume of 1000 small droplets is the same as the volume of the large droplet. Therefore, we can write,
43πR3=1000(43πr3)\dfrac{4}{3}\pi {R^3} = 1000\left( {\dfrac{4}{3}\pi {r^3}} \right)
Here, R is the radius of a large droplet and r is the radius of each small droplet.
Simplifying the above equation, we get,
r=R10=0.110r = \dfrac{R}{{10}} = \dfrac{{0.1}}{{10}}
⇒r=0.01 cm\Rightarrow r = 0.01\,cm
Now, we can calculate the change in surface area of the droplet after breaking it into 1000 droplets as follows,
δA=(1000×4πr2)−4πR2\delta A = \left( {1000 \times 4\pi {r^2}} \right) - 4\pi {R^2}
We substitute r=R10r = \dfrac{R}{{10}} in the above equation.
δA=(1000×4π(R10)2)−4πR2\delta A = \left( {1000 \times 4\pi {{\left( {\dfrac{R}{{10}}} \right)}^2}} \right) - 4\pi {R^2}
⇒δA=(1000×4πR2100)−4πR2\Rightarrow \delta A = \left( {1000 \times \dfrac{{4\pi {R^2}}}{{100}}} \right) - 4\pi {R^2}
⇒δA=(10×4πR2)−4πR2\Rightarrow \delta A = \left( {10 \times 4\pi {R^2}} \right) - 4\pi {R^2}
⇒δA=36πR2\Rightarrow \delta A = 36\pi {R^2}
We have the formula for change in surface energy,
δU=TδA\delta U = T\delta A
Here, T is the surface tension.
We substitute 7×10−2 N/m7 \times {10^{ - 2}}\,N/m for T and 36πR236\pi {R^2} in the above equation.
δU=(7×10−2)(36πR2)\delta U = \left( {7 \times {{10}^{ - 2}}} \right)\left( {36\pi {R^2}} \right)
Substituting 0.1 cm for R in the above equation, we get,
δU=(7×10−2 N/m)×(36(3.14)(0.001 m )2)\delta U = \left( {7 \times {{10}^{ - 2}}\,N/m} \right) \times \left( {36\left( {3.14} \right){{\left( {0.001\,m\,} \right)}^2}} \right)
⇒δU=(7×10−2)×(1.13×10−4)\Rightarrow \delta U = \left( {7 \times {{10}^{ - 2}}} \right) \times \left( {1.13 \times {{10}^{ - 4}}} \right)
⇒δU=7.91×10−6 J\Rightarrow \delta U = 7.91 \times {10^{ - 6}}\,J
We know that change in surface energy is the work done. Therefore, the work done in breaking the droplet into 1000 small droplets is,W=7.91×10−6 JW = 7.91 \times {10^{ - 6}}\,J.

So, the correct answer is option (A).

Note: Students should note that the water droplet has only one surface that is outer surface unlike soup bubble has two surfaces: inner surface and outer surface. Therefore, don’t consider the twice of the surface area in case of water droplet while calculating the change in surface area. If the radius of the droplet is given in centimeters, you should convert it into meters.