Correct answer option is (B)

$\begin{array}{rl}& {\mathrm{R}}_1=\frac{{\mathrm{R}}_2}{2}\\ & {\mathrm{R}}_0{\left({\mathrm{A}}_1\right)}^{1/3}=\frac{{\mathrm{R}}_0}{2}{\left({\mathrm{A}}_2\right)}^{1/3}\\ & {\mathrm{A}}_1=\frac{1}{8}{\mathrm{A}}_2\\ & {\mathrm{A}}_1=\frac{192}{8}=24\end{array}$

Correct answer is 27

According to the empirical formula relating the radius of a nucleus ($R$) with its mass number ($A$), we know that the radius of a nucleus is proportional to the cube root of its mass number. This relationship is given as:

$R=R_0{A}^{1/3}$

where $R_0$ is a constant with an approximate value of 1.2 fermis.

Given that for a nucleus with a mass number 64 has a radius of 4.8 fermis, we can write:

$4.8\text{ fermi}=R_0\times {64}^{1/3}$

Now another nucleus has a radius of 4 fermis:

$4\text{ fermi}=R_0\times A^{\prime 1/3}$

Where $A^{\prime }$ is the mass number of the other nucleus.

Let's solve for $R_0$ from the first equation:

$R_0=\frac{4.8\text{ fermi}}{{64}^{1/3}}$

Now we're going to find the mass number $A^{\prime }$ using the second equation and substituting $R_0$ from the above:

$4\text{ fermi}=\left(\frac{4.8\text{ fermi}}{{64}^{1/3}}\right)\times A^{\prime 1/3}$

Now, we want to find $A^{\prime }$ in terms of $x$ as given by the equation in the question:

$A^{\prime }=\frac{1000}{x}$

Substitute $A^{\prime }$ in the equation above, we get:

$4=\left(\frac{4.8}{{64}^{1/3}}\right)\times {\left(\frac{1000}{x}\right)}^{1/3}$

Let's solve for $x$:

$(4{)}^3={\left(\frac{4.8}{{64}^{1/3}}\right)}^3\times \frac{1000}{x}$

$4^3\times x={\left(\frac{4.8}{{64}^{1/3}}\right)}^3\times 1000$

$x=\left(\frac{{\left(\frac{4.8}{{64}^{1/3}}\right)}^3\times 1000}{4^3}\right)$

Now calculate the values:

$x=\left(\frac{{\left(\frac{4.8}{4}\right)}^3\times 1000}{64}\right)$

$x=\left(\frac{{1.2}^3\times 1000}{64}\right)$

$x=\left(\frac{1.728\times 1000}{64}\right)$

$x=\left(\frac{1728}{64}\right)$

$x=27$

Therefore, the value of $x$ is 27.

Correct answer is 4

For a nucleus

Volume: $\mathrm{V}=\frac{4}{3}\pi {\mathrm{R}}^3$

$\begin{array}{rl}& \mathrm{R}={\mathrm{R}}_0(\mathrm{A}{)}^{1/3}\\ & \mathrm{V}=\frac{4}{3}\pi {\mathrm{R}}_0^3\mathrm{A}\\ & \Rightarrow \frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}=\frac{{\mathrm{A}}_2}{{\mathrm{A}}_1}=4\end{array}$