1.(JEE Main 2023 (Online) 30th January Evening Shift)

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$

Assertion A: The nuclear density of nuclides $_{5}^{10} B,_{3}^{6} Li,_{26}^{56} Fe,_{10}^{20} Ne$ and $_{83}^{209} Bi$ can be arranged as $ρ_{Bi}^{N} > ρ_{Fe}^{N} > ρ_{Ne}^{N} > ρ_{B}^{N} > ρ_{Li}^{N}$

Reason R: The radius $R$ of nucleus is related to its mass number $A$ as $R = R_{0} A^{1 / 3}$ , where $R_{0}$ is a constant.

In the light of the above statements, choose the correct answer from the options given below

A. $B o t h A$ and $R$ are true and $R$ is the correct explanation of $A$

B. Both $A$ and $R$ are true but $R$ is NOT the correct explanation of $A$

C. $A$ is false but $R$ is true

D. $A$ is true but $R$ is false

The Correct Answer is Option (C)

$R = R_{0} A^{\frac{1}{3}}$ , using this

$ρ = \frac{M}{\frac{4}{3} π R^{3}} = \frac{A m_{P}}{\frac{4}{3} π R_{0}^{3} A} = \frac{m_{P}}{\frac{4}{3} π R_{0}^{3}}$

$ρ$ is independent of mass number.

$∴$ A is false

2.(JEE Main 2023 (Online) 25th January Morning Shift)

The ratio of the density of oxygen nucleus ( $_{8}^{16} O$ ) and helium nucleus ( $_{2}^{4} He$ ) is

A. 4:1

B. 1:1

C. 2:1

D. 8:1

The Correct Answer is Option (B)

Nuclear density is independent of mass number

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

Also, $R = R_{0} A^{\frac{1}{3}}$

And $R^{3} = R_{0}^{3} A$

$\Rightarrow$ Nuclear density $= \frac{Au}{\frac{4}{3} π R_{0}^{3} A}$

Nuclear density $= \frac{3 u}{4 π R_{0}^{3}}$

$\Rightarrow$ Nuclear density is independent of $A$

3.(JEE Main 2023 (Online) 11th April Evening Shift)

A nucleus disintegrates into two nuclear parts, in such a way that ratio of their nuclear sizes is $1 : 2^{1 / 3}$ . Their respective speed have a ratio of $n : 1$ . The value of $n$ is __________.

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:

$\frac{m_{1}}{m_{2}} = {(\frac{1}{2^{1 / 3}})}^{3} = \frac{1}{2}$

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.

4.(JEE Main 2023 (Online) 25th January Evening Shift)

A nucleus disintegrates into two smaller parts, which have their velocities in the ratio 3 : 2. The ratio of their nuclear sizes will be ${(\frac{x}{3})}^{\frac{1}{3}}$ . The value of ' $x$ ' is :-

The Crrect Answer is

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\begin{aligned} \frac{v_{1}}{v_{2}} = \frac{3}{2} \\ m_{1} v_{1} = m_{2} v_{2} \Rightarrow \frac{m_{1}}{m_{2}} = \frac{2}{3} \end{aligned}

Since, Nuclear mass density is constant

\begin{aligned} \frac{m_{1}}{\frac{4}{3} π r_{1}^{3}} = \frac{m_{2}}{\frac{4}{3} π r_{2}^{3}} \\ {(\frac{r_{1}}{r_{2}})}^{3} = \frac{m_{1}}{m_{2}} \\ \frac{r_{1}}{r_{2}} = {(\frac{2}{3})}^{\frac{1}{3}} \\ So, x = 2 \end{aligned}

5.(JEE Main 2023 (Online) 24th January Morning Shift)

Assume that protons and neutrons have equal masses. Mass of a nucleon is $1.6 \times 10^{- 27}$ kg and radius of nucleus is $1.5 \times 10^{- 15} A^{1 / 3}$ m. The approximate ratio of the nuclear density and water density is $n \times 10^{13}$ . The value of $n$ is __________.

The Crrect Answer is

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

Volume $Volume = \frac{4 π}{3} r^{3}$

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

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

$\begin{aligned} = \frac{1.6 \times 3 \times 8 \times 10^{18}}{4 π \times 27} \\ = \frac{32}{9 π} \times 10^{17} \end{aligned}$

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

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

$= \frac{320}{9 π} \times 10^{13}$

$= 11.32 \times 10^{13}$

value of $n = 11$