If \(0.8 \) Ampere current flows through metallic wire for 4 hours, the number of electrons flowing through the wire will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(7.18 \times 10^{22}\)
The charge carried by a steady current is the product of current and time, Q = It (Q in coulomb when I is in ampere and t in second). Each electron carries 1.6021 × 10−19 C, so the number of electrons is Q divided by this charge.
\( \displaystyle t = 4 \times 60 \times 60 = 14400\ \text{s} \)
\( \displaystyle Q = It = 0.8 \times 14400 = 11520\ \text{C} \)
\( \displaystyle N = \frac{Q}{e} = \frac{11520}{1.6021 \times 10^{-19}} = 7.19 \times 10^{22} \)
This is about 7.2 × 1022 electrons, which is option (2), printed as 7.18 × 1022.
Shortcut: 11520 C is about 0.12 F (11520/96500); 0.12 × 6 × 1023 = 7.2 × 1022.
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Intext Questions – “2.10 If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?”