MCQ
Current in the circuit will be


- A$\frac{5}{{40}}A$
- ✓$\frac{5}{{50}}A$
- C$\frac{5}{{10}}A$
- D$\frac{5}{{20}}A$

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(Consider the fusion reaction as $4{ }_1^1 \mathrm{H}+2 \mathrm{e}^{-} \rightarrow{ }_2^4 \mathrm{He}+2 \mathrm{v}+6 \gamma+26.7 \mathrm{MeV}$, energy released in the fission reaction of ${ }^{235} \mathrm{U}$ is $200 \mathrm{MeV}$ per fission nucleus and $\mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23}$ )

