A spherical drop of liquid splits into 1000 identical spherical drops If $u _{ i }$ is the surfice energy of the original drop and $u_f$ is the total surface energy of the resulting drops, the (ignoring evaporation), $\frac{u_f}{u_i}=\left(\frac{10}{x}\right)$ Then value of $x$ is ___
To solve the problem, we first understand the relationships involved in surface energy with respect to a drop splitting. The surface energy of a sphere is given by \( u = \sigma \cdot 4\pi r^2 \), where \( \sigma \) is the surface tension and \( r \) is the radius of the sphere. Let the radius of the original drop be \( R \), so its surface energy is \( u_i = \sigma \cdot 4\pi R^2 \). After splitting, we have 1000 smaller drops, each with radius \( r \). The volume of each smaller drop is \( \frac{1}{1000} \)th of the original: \[ \frac{4}{3}\pi r^3 = \frac{1}{1000} \times \frac{4}{3}\pi R^3 \] Solving for \( r \), we find: \[ r^3 = \frac{R^3}{1000} \Rightarrow r = \frac{R}{10} \] The surface energy of each smaller drop becomes \( \sigma \cdot 4\pi r^2 = \sigma \cdot 4\pi \left(\frac{R}{10}\right)^2 = \sigma \cdot \frac{4\pi R^2}{100} \). The total surface energy of all 1000 drops is \( u_f = 1000 \times \sigma \cdot \frac{4\pi R^2}{100} = \sigma \cdot 40\pi R^2 \). So the ratio of the total surface energy of the final drops to the initial is: \[ \frac{u_f}{u_i} = \frac{\sigma \cdot 40\pi R^2}{\sigma \cdot 4\pi R^2} = 10 \] Comparing this with \( \frac{10}{x} \), we equate: \[ 10 = \frac{10}{x} \] Solving for \( x \), we find \( x = 1 \). Hence, the value of \( x \) is \( 1 \), fitting within the expected range (1,1).
