Question

A heavy nucleus splits into two fragment nuclei whose surface area are in the ratio $9:25$. Ratio of speeds of the two fragments is

• A. $27:16$
• B. $64:27$
• C. $81:16$
• D. $125:27$

Answer 1

The correct option is D $125:27$

$\text{Given-* Ratio of surface areas of two nuclei}$

$=9:25$

$\Rightarrow$ $\text{Ratio of surface area considering them as spherical}=\frac{4\pi r_1^2}{4\pi r_2^2}=\frac{9}{25}$

$\begin{array}{c}\text{Where}{r}_1\text{and}{r}_2\text{are radius of}\\ \Rightarrow \frac{r_1^2}{r_2^2}=\frac{9}{25}\Rightarrow \frac{r_1}{r_2}=\frac{3}{5}\\ \text{Now, let the density of the nuclei be (p)}\\ \Rightarrow \text{Ratio of their masses}\to \\ \frac{M_1}{M_2}=\frac{\rho \cdot \frac{4}{3}\pi r_1^3}{\rho \cdot \frac{4}{3}\pi r_2^3}=\frac{27}{125}\end{array}$

$\begin{array}{c}\text{* Now by conservation of momentum -}\\ \begin{array}{cc}{M}_1{v}_1=M_2{v}_2& v_1=\text{speed of nuclei}1\\ & v_2=\text{speed of nuclei}2\\ \Rightarrow & \frac{v_1}{v_2}=\frac{M_2}{M_1}=\frac{125}{27}\\ \Rightarrow \text{Ratio of their velocity is}125:27& \end{array}\\ \text{Hence, opt}\left(d\right)\text{is correct.}\end{array}$