JEE MAIN - Physics (2025 - 3rd April Morning Shift - No. 13)

$$ \text { Match the LIST-I with LIST-II } $$

List - I		List - II
A.	$$ { }_0^1 \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{54}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+2{ }_0^1 \mathrm{n} $$	I.	$$ \text { Chemical reaction } $$
B.	$$ 2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O} $$	II.	$$ \text { Fusion with +ve } \mathrm{Q} \text { value } $$
C.	$$ { }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \rightarrow{ }_2^3 \mathrm{He}+{ }_0^1 \mathrm{n} $$	III.	$$ \text { Fission } $$
D.	$$ { }_1^1 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} $$	IV.	$$ \text { Fusion with -ve } Q \text { value } $$

$$ \text { Choose the correct answer from the options given below: } $$

A-II, B-I, C-III, D-IV

A-III, B-I, C-II, D-IV

A-III, B-I, C-IV, D-II

A-II, B-I, C-IV, D-III

Here’s the correct matching:

$$_0^1\text{n}+_{92}^{235}\text{U}\to_{54}^{140}\text{Xe}+_{38}^{94}\text{Sr}+2\,_0^1\text{n}$$

–– This is nuclear fission ⇒ III.

$$2\text{H}_2+\text{O}_2\to2\text{H}_2\text{O}$$

–– This is a chemical reaction ⇒ I.

$$_1^2\text{H}+_1^2\text{H}\to_2^3\text{He}+_0^1\text{n}$$

–– This D–D fusion releases about 3.3 MeV ⇒ fusion with +ve Q ⇒ II.

$$_1^1\text{H}+_1^3\text{H}\to_1^2\text{H}+_1^2\text{H}$$

–– The mass of products exceeds reactants ⇒ requires energy ⇒ fusion with –ve Q ⇒ IV.

So the answer is

Option B: A–III, B–I, C–II, D–IV.