The Correct Answer is Option (C)

$R=R_0{A}^{\frac{1}{3}}$, using this

$\rho =\frac{M}{\frac{4}{3}\pi R^3}=\frac{A{m}_P}{\frac{4}{3}\pi R_0^{3}A}=\frac{m_P}{\frac{4}{3}\pi R_0^3}$

$\rho$is independent of mass number.

$\therefore$ A is false

The Correct Answer is Option (B)

Nuclear density is independent of mass number

As nuclear density $=\frac{\mathrm{Au}}{\frac{4}{3}\pi {\mathrm{R}}^3}$

Also, $\mathrm{R}={\mathrm{R}}_0{\mathrm{A}}^{\frac{1}{3}}$

And $R^3=R_0^{3}A$

$\Rightarrow$ Nuclear density $=\frac{\mathrm{Au}}{\frac{4}{3}\pi {\mathrm{R}}_0^3\mathrm{A}}$

Nuclear density $=\frac{3\mathrm{u}}{4\pi {\mathrm{R}}_0^3}$

$\Rightarrow$ Nuclear density is independent of $\mathrm{A}$

The Crrect Answer is

Let the masses of the two nuclear parts be $m_1$ and $m_2$, and their respective speeds be $v_1$ and $v_2$. According to the problem, the ratio of their nuclear sizes is $1:2^{1/3}$. Since the nuclear size is proportional to the cube root of the mass, we can write:

$\frac{m_1}{m_2}={\left(\frac{1}{2^{1/3}}\right)}^3=\frac{1}{2}$

Now, according to the conservation of linear momentum, the momentum before disintegration is equal to the momentum after disintegration:

$m_1{v}_1=m_2{v}_2$

From the problem statement, the ratio of their respective speeds is $n:1$, so we can write:

$v_1=n\cdot v_2$

Substitute the expression for $v_1$ into the momentum conservation equation:

$m_1(n\cdot v_2)=m_2{v}_2$

We know the mass ratio, so substitute that into the equation:

$\frac{1}{2}{m}_2(n\cdot v_2)=m_2{v}_2$

Divide both sides by $m_2{v}_2$:

$\frac{1}{2}n=1$

Now, solve for $n$:

$n=2$

Thus, the value of $n$ is 2.

The Crrect Answer is

$\begin{array}{rl}& \frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\frac{3}{2}\\ \\ & {\mathrm{m}}_1{\mathrm{v}}_1={\mathrm{m}}_2{\mathrm{v}}_2\Rightarrow \frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}=\frac{2}{3}\end{array}$

Since, Nuclear mass density is constant

$\begin{array}{rl}& \frac{{\mathrm{m}}_1}{\frac{4}{3}\pi {\mathrm{r}}_1^3}=\frac{{\mathrm{m}}_2}{\frac{4}{3}\pi {\mathrm{r}}_2^3}\\ \\ & {\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^3=\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}\\ \\ & \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}={\left(\frac{2}{3}\right)}^{\frac{1}{3}}\\ \\ & \text{ So, }\mathrm{x}=2\end{array}$

The Crrect Answer is

Radius $=1.5\times {10}^{-15}{A}^{1/3}$

Volume$\text{ Volume }=\frac{4\pi }{3}{r}^3$

Mass of nucleus $=(1.6\times {10}^{-27})\mathrm{A}\mathrm{kg}$

Density of nucleus $\text{ Density of nucleus }=\frac{1.6\times {10}^{-27}\times A}{\frac{4}{3}\times \pi \times {(1.5\times {10}^{-15}{A}^{\frac{1}{3}})}^3}$

$\begin{array}{rl}& =\frac{1.6\times 3\times 8\times {10}^{18}}{4\pi \times 27}\\ \\ & =\frac{32}{9\pi }\times {10}^{17}\end{array}$

Density of water $=1000\mathrm{kg}/{\mathrm{m}}^3$

Density of nucleus Density of water $\frac{\text{ Density of nucleus }}{\text{ Density of water }}=\frac{\frac{32}{9\pi }\times {10}^{17}}{1000}$

$=\frac{320}{9\pi }\times {10}^{13}$

$=11.32\times {10}^{13}$

value of $n=11$

The Correct Answer is Option (C)

Linear momentum is conserved

so, $p_{M^{\prime }/3}=p_{2M^{\prime }/3}$

so, $\frac{{\lambda }_{M^{\prime }/3}}{{\lambda }_{2M^{\prime }/3}}=\frac{1}{1}$

The Correct Answer is Option (C)

$\therefore$$R=R_0{A}^{\frac{1}{3}}$

$\Rightarrow \frac{R_1}{R_2}={\left(\frac{A_1}{A_2}\right)}^{\frac{1}{3}}={\left(\frac{4}{3}\right)}^{\frac{1}{3}}$

$\therefore$ Density ratio, $\frac{{\rho }_1}{{\rho }_2}=\frac{A_1/V_1}{A_2/V_2}$

$=\left(\frac{A_1}{A_2}\right)\times {\left(\frac{R_2}{R_1}\right)}^3$

$=1:1$

The Correct Answer is Option (B)

We know that

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

$\Rightarrow \underset{y}{\underbrace{\ln \left(\frac{R}{R_0}\right)}}=\underset{m}{\underbrace{\frac{1}{3}}}\underset{x}{\underbrace{\ln (A)}}$

$\Rightarrow$ Straight line

The Correct Answer is Option ()

$\lambda =\frac{h}{p}$

both the particles will move with momentum same in magnitude & opposite in direction.

So De-Broglie wavelength of both will be same i.e. ratio 1 : 1

The Correct Answer is Option (C)

$R=(1.3\times {10}^{-15})A^{\frac{1}{3}}$

We know, $m=pV$

$\Rightarrow$ $p=\frac{m}{V}$

$\Rightarrow$ $p=\frac{m_{p}A}{\frac{4}{3}\pi R^3}$

$p=\frac{m_{p}A}{\frac{4}{3}\pi \times {(1.3\times {10}^{-15})}^{3}A}$

$p\approx {10}^{17}kg/m$

The correct answer is option (A)

Densities of nucleus happens to be constant, irrespective of mass number.

The correct answer is option (C)

The two nuclei have velocity in ratio 8 : 27. By conservation of momentum, we have

$m_1{v}_1=m_2{v}_2\Rightarrow \frac{v_1}{v_2}=\frac{m_2}{m_1}\Rightarrow \frac{m_2}{m_1}=\frac{8}{27}$

Now, since $m=\rho \frac{4}{3}\pi r^3$

Therefore, $\frac{m_2}{m_1}=\frac{\rho \frac{4}{3}\pi r_2^3}{\rho \frac{4}{3}\pi r_1^3}\Rightarrow \frac{m_2}{m_1}={\left(\frac{r_2}{r_1}\right)}^3\Rightarrow {\left(\frac{r_2}{r_1}\right)}^3=\frac{8}{27}$

$\Rightarrow \frac{r_2}{r_1}=\frac{2}{3}$

Thus, ratio of radii of nuclei $r_1:r_2=3:2$.

The correct answer is option (B)

KEY CONCEPT : $R=R_0{\left(A\right)}^{1/3}$

Here A = Mass number

$\therefore$ $\frac{R_1}{R_2}={\left(\frac{A_1}{A_2}\right)}^{1/3}$

$={\left(\frac{27}{125}\right)}^{1/3}=\frac{3}{5}$

$R_2=\frac{5}{3}\times 3.6=6$ fermi

The correct answer is option (B)

From conservation of momentum $m_1{v}_1=m_2{v}_2$

$\Rightarrow \left(\frac{m_1}{m_2}\right)=\left(\frac{v_2}{v_1}\right)$ given $\frac{v_1}{v{}_2}=2$

$\Rightarrow \frac{m_1}{m_2}=\frac{1}{2}$

$\Rightarrow \frac{r_1^3}{r_2^3}=\frac{1}{2}$

$\Rightarrow \left(\frac{r_1}{r_2}\right)={\left(\frac{1}{2}\right)}^{1/3}$

The correct answer is option (C)

The wavelength of spectrum is given by

$\frac{1}{\lambda }=R{z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

where $R=\frac{1.097\times {10}^7}{1+\frac{m}{M}}$$m=$ mass of electron

$M=$ mass of nucleus.

For different $M,R$ is different and therefore $\lambda$ is different