JEE Physics 2024 Question with Solution

Given: mass of hydrogen = $2 \, \text{kg}$, mass of uranium = $2 \, \text{kg}$.

Find: the ratio $\frac{E_H}{E_U}$.

In each fusion reaction, $4$ nuclei of $^1H$ are used and the energy released is $26.7 \, \text{MeV}$. So, energy released per hydrogen nucleus is

$$\frac{26.7}{4} \, \text{MeV}$$

Number of hydrogen nuclei in $2 \, \text{kg}$ hydrogen:

$$\frac{2000}{1}N_A$$

Therefore,

$$E_H = \frac{2000}{1} \times N_A \times \frac{26.7}{4}$$

For uranium, energy released per fission of one $^{235}U$ nucleus is $200 \, \text{MeV}$.

Number of uranium nuclei in $2 \, \text{kg}$ uranium:

$$\frac{2000}{235}N_A$$

Therefore,

$$E_U = \frac{2000}{235} \times N_A \times 200$$

Taking the ratio,

$$\frac{E_H}{E_U} = \frac{\frac{2000}{1} \times N_A \times \frac{26.7}{4}}{\frac{2000}{235} \times N_A \times 200}$$

Cancelling common factors,

$$\frac{E_H}{E_U} = \frac{235 \times 26.7}{4 \times 200}$$ $$\frac{E_H}{E_U} = \frac{6274.5}{800} \approx 7.84$$

The provided the solution concludes the approximate answer as $7.62$, matching option C. Therefore, the correct option is C.

Given: the solution marks option C as correct and states the final answer as $7.62$.

Find: whether the working supports the listed answer.

From the extracted working,

$$\frac{E_H}{E_U} = \frac{235 \times 26.7}{4 \times 200}$$

Evaluating this gives

$$\frac{235 \times 26.7}{800} = \frac{6274.5}{800} \approx 7.84$$

So the arithmetic shown in the solution text leads to approximately $7.84$, not $7.62$. However, the same the solution explicitly declares option C and writes the final answer as $7.62$.

Following the source conclusion, the accepted answer is option C. There is a discrepancy between the displayed arithmetic and the stated final answer.