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One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to total initial surface energy is
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Verified Answer
The correct answer is:
$1: 10$
Let radius of big drop $=R$
Radius of small drop $=r$
Number of small drops, $n=1000$
In this process, the volume remains same
$V_f=V_i$
$\frac{4}{3} \pi R^3=n\left(\frac{4}{3} \pi r^3\right)$
$R^3=1000 r^3 \Rightarrow R=10 r$ ...(i)
Initial surface area,
$A_1=n 4 \pi r^2$
Final surface area, $A_2=4 \pi R^2$
Now, increase in surface area,
$\Delta A=A_2-A_1=4 \pi\left[R^2-n r^2\right]$
We know that, surface energy,
$\mathrm{SE}=\mathrm{TA}$ [where, $T=$ surface tension of liquid]
According to question,
$\frac{\mathrm{SE}_f}{\mathrm{SE}_i}=\frac{\mathrm{TA}_2}{\mathrm{TA}_1}=\frac{4 \pi R^2}{n\left(4 \pi r^2\right)}=\frac{R^2}{1000 r^2}$ [From Eq. (i)]
$=\frac{(10 r)^2}{1000 r^2}=\frac{100 r^2}{1000 r^2}=\frac{1}{10}$ or $1: 10$
Radius of small drop $=r$
Number of small drops, $n=1000$
In this process, the volume remains same
$V_f=V_i$
$\frac{4}{3} \pi R^3=n\left(\frac{4}{3} \pi r^3\right)$
$R^3=1000 r^3 \Rightarrow R=10 r$ ...(i)
Initial surface area,
$A_1=n 4 \pi r^2$
Final surface area, $A_2=4 \pi R^2$
Now, increase in surface area,
$\Delta A=A_2-A_1=4 \pi\left[R^2-n r^2\right]$
We know that, surface energy,
$\mathrm{SE}=\mathrm{TA}$ [where, $T=$ surface tension of liquid]
According to question,
$\frac{\mathrm{SE}_f}{\mathrm{SE}_i}=\frac{\mathrm{TA}_2}{\mathrm{TA}_1}=\frac{4 \pi R^2}{n\left(4 \pi r^2\right)}=\frac{R^2}{1000 r^2}$ [From Eq. (i)]
$=\frac{(10 r)^2}{1000 r^2}=\frac{100 r^2}{1000 r^2}=\frac{1}{10}$ or $1: 10$
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