JEE Main 2026 · 4 Apr · Shift 1 · Physics Q29

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PhysicsSingle correct+4 / -1Question 29 of 75

Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass numbers 3, 4 and 10 are 5.6 MeV, 7.4 MeV and 6.1 MeV, respectively, then in the process, $\Delta Mc^2$ = ________ MeV.

6.9

7.9

2.2

4.3

Answer and worked solution

Official answer: C

Binding energies.

Before: $2(3\times5.6)+4\times7.4=63.2$ MeV.
After: $10\times6.1=61.0$ MeV.

Difference. The binding energy falls by 2.2 MeV (the reaction absorbs energy): $|\Delta M|c^2=2.2$ MeV.

(C)

Topics:Mass defect and binding energy Nuclear fission and fusion

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JEE Main 2026 · 4 Apr · Shift 1, Physics question 29