\(R\) : Radius of bigger drop
\(r :\) Radius of smaller drop
Volume will remain same
\(\frac{4}{3} \pi R ^3=1000 \times \frac{4}{3} \pi r^3\)
\(R =10 r\)
\(u _{ i }= T \cdot 4 \pi R ^2\)
\(u _{ f }= T .4 \pi r ^2 \times 1000\)
\(\frac{ u _{ f }}{ u _{ i }}=\frac{1000 r ^2}{ R ^2}\)
\(\frac{ u _{ f }}{ u _{ i }}=\frac{10}{1}\)
So, \(x =1\)