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Question: The work done in splitting a drop of water of 1mm radius into \({10^6}\) droplets is (surface tensio...

The work done in splitting a drop of water of 1mm radius into 106{10^6} droplets is (surface tension of water 72×10−3N/m72 \times {10^{ - 3}}N/m).
A. 5.98×10−5J5.98 \times {10^{ - 5}}J
B. 10.98×10−5J10.98 \times {10^{ - 5}}J
C. 16.95×10−5J16.95 \times {10^{ - 5}}J
D. 8.95×10−5J8.95 \times {10^{ - 5}}J

Explanation

Solution

One drop of water splits into many smaller droplets, so the volume remains constant.Assume the radius of the larger drop to be R and small drops to be r.Using the formula for volume of the sphere,i.e. V=43πR3V = \dfrac{4}{3}\pi {R^3} find the volume for both the larger and smaller drops.So, find out the radius of the new small droplet by equating the volume of the drop before splitting and total volume of drips after splitting.Then find out the increase in surface area ΔA\Delta A, and substitute into the formula W=T.ΔAW = T.\Delta A (W= work done, T=Surface Tension, ΔA\Delta A =Change in surface area)

Complete step by step answer:
One droplet splits into 106{10^6} smaller droplets, hence volume remains constant. So
43πR3=106×43πr3\dfrac{4}{3}\pi {R^3} = {10^6} \times \dfrac{4}{3}\pi {r^3}, where R=radius of big drop, and r=radius of small drop.
Substituting values (in SI Units) we get,
43π(0.001)3=106×43πr3\dfrac{4}{3}\pi {\left( {0.001} \right)^3} = {10^6} \times \dfrac{4}{3}\pi {r^3}
⇒r3=10−15\Rightarrow {r^3} = {10^{ - 15}}
⇒r=10−5cm\Rightarrow r = {10^{ - 5}}cm
As one big drop splits into 106{10^6} smaller droplets, there is an increase in surface area. We know that the surface area of the sphere is 4πr24\pi {r^2} .
So, increase in surface area is-
[4π×(r)2]n−[4π×(R)2]\left[ {4\pi \times {{\left( r \right)}^2}} \right]n - \left[ {4\pi \times {{\left( R \right)}^2}} \right] , {where n=number of smaller droplets}
⇒[4π×(10−5)2×106]−[4π×(10−3)2]\Rightarrow \left[ {4\pi \times {{\left( {{{10}^{ - 5}}} \right)}^2} \times {{10}^{^6}}} \right] - \left[ {4\pi \times {{\left( {{{10}^{ - 3}}} \right)}^2}} \right]
⇒[4π×(10−4)]−[4π×(10−6)]\Rightarrow \left[ {4\pi \times \left( {{{10}^{ - 4}}} \right)} \right] - \left[ {4\pi \times \left( {{{10}^{ - 6}}} \right)} \right]
⇒4π×10−6(100−1)\Rightarrow 4\pi \times {10^{ - 6}}\left( {100 - 1} \right)
⇒4π×10−6(99)\Rightarrow 4\pi \times {10^{ - 6}}\left( {99} \right)……eq1
Now using the formula, W=T.ΔAW = T.\Delta A and substituting values, we get:
W=72×4π×99×10−6×10−3W = 72 \times 4\pi \times 99 \times {10^{ - 6}} \times {10^{ - 3}}
∴W=8.95×10−5J\therefore W = 8.95 \times {10^{ - 5}}J

Hence, option D is correct.

Note: Sometimes students get confused about how to calculate the work done and consider the work done by a single drop and thus resulting into a wrong answer.So,always remember to multiply with the total number of drops that results into a work done.