Physics - Fluid Question with Solution | TestHub

If $\mathrm{V}$ is the volume then we have $\begin{array}{rl}& \mathrm{V}=\frac{4}{3}\pi {\mathrm{r}}^3\\ & \mathrm{V}\propto {\mathrm{r}}^3\\ & \therefore \frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}={\left(\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}\right)}^3\\ & \frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}={\left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)}^{1/3}=(2{)}^{1/3}\\ & {\mathrm{W}}_1=8\pi {\mathrm{r}}_1^2\cdot \mathrm{T}\text{ and }{\mathrm{W}}_2=8\pi {\mathrm{r}}_2^2\mathrm{T}\\ & \therefore \frac{{\mathrm{W}}_2}{{\mathrm{W}}_1}={\left(\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}\right)}^2=(2{)}^{2/3}=(4{)}^{1/3}\\ & {\mathrm{W}}_2=(4{)}^{1/3}\mathrm{W}\end{array}$