Physics · Nuclear Physics

JEE Main 2024 — 31 January, Shift 2 — Question 57

A nucleus has mass number $A_{1}$ and volume $V_{1}$ . Another nucleus has mass number $A_{2}$ and volume $V_{2}$ . If relation between mass number is $A_{2}=4 A_{1}$ , then $\frac{V_{2}}{V_{1}}=$ $\_\_\_\_\_\_\_$ .

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

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

\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

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Exam: JEE Main 2024
Paper: 31 January, Shift 2
Subject: Physics
Chapter: Nuclear Physics
Topic: Introduction to Nucleus and its constituents

The mass number of nucleus having radius equal to half of the radius of nucleus with mass number 192 is:

Answer: 4

Step-by-step solution

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