Start free

Physics · Nuclear Physics

JEE Main 2024 — 9 April, Shift 1 — Question 41

The energy equivalent of 1 g of substance is:

  1. Option A:

    11.2×1024MeV11.2 \times 10^{24} \mathrm{MeV}

  2. Option B:

    5.6×1012MeV5.6 \times 10^{12} \mathrm{MeV}

  3. Option C:

    5.6 eV

  4. Option D:

    5.6×1026MeV5.6 \times 10^{26} \mathrm{MeV}

    Correct

Answer: D

Step-by-step solution

E=mC2\text{E}=\text{m}{{\text{C}}^{2}}

\begin{array}{*{35}{r}}\Rightarrow \text{E} & ~=\left( 1\times {{10}^{-3}} \right)\times {{\left( 3\times{{10}^{8}}\right)}^{2}}\text{ }\!\!~\!\!\text{ J} \\\Rightarrow \text{E} & ~=\left( {{10}^{-3}} \right)\left( 9\times {{10}^{16}} \right)\left( 6.241\times {{10}^{18}} \right)\text{eV}\\\text{E} & =56.169\times {{10}^{31}}\text{eV}\\\text{E} & ~\approx 5.6\times{{10}^{26}}\text{MeV}\\\end{array}

Answer key and solution verified before publishing.

Practise Nuclear Physics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2024
Paper
9 April, Shift 1
Subject
Physics
Chapter
Nuclear Physics
Topic
Mass Defect, Binding Energy and Q-Value of Nuclear Reaction

More Nuclear Physics questions from this paper

The energy equivalent of 1 g of substance is: | JEE Main 2024 PYQ with Solution · DhiX AI