Physics · Nuclear Physics

JEE Main 2024 — 1 February, Shift 1 — Question 51

The radius of a nucleus of mass number 64 is 4.8 fermi. Then the mass number of another nucleus having radius of 4 fermi is $\frac{1000}{x}$ , where x is $\qquad$ .

Answer: 27

Numerical answer — enter this value.

Step-by-step solution

R={{R}_{0}}{{A}_{1/3}}

$\begin{array}{*{35}{r}}{} & {{\text{R}}^{3}}\propto \text{ }\!\!~\!\!\text{ A} \\{} & {{\left( \frac{4.8}{4} \right)}^{3}}=\frac{64}{\text{ }\!\!~\!\!\text{ A}} \\{} & ~=\frac{64}{\text{ }\!\!~\!\!\text{ A}}={{(1.2)}^{3}} \\{} & \frac{64}{\text{ }\!\!~\!\!\text{ A}}=1.44\times 1.2 \\{} & \text{ }\!\!~\!\!\text{ A}=\frac{64}{1.44\times 1.2}=\frac{1000}{\text{x}} \\{} & \text{x}=\frac{144\times 12}{64}=27 \\\end{array}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 1
Subject: Physics
Chapter: Nuclear Physics
Topic: Introduction to Nucleus and its constituents