MCQ
With the decrease of current in the primary coil from $2\,amperes$ to zero value in $0.01\,s$ the $emf$ generated in the secondary coil is $1000\,volts$. The mutual inductance of the two coils is......$H$
- A$1.25$
- B$2.50$
- ✓$5$
- D$10$
$\mathrm{e}=\frac{-\mathrm{MdI}}{\mathrm{dt}} \Rightarrow 1000=\mathrm{M}\left(\frac{2-0}{0.01}\right)$
(since current is reduced $dI$ $=-\text { ve }$)
$\Rightarrow \mathrm{M}=\frac{1000 \times 0.01}{2}=5.00 \mathrm{\,H}.$
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(Energy released per fission $= 200\,MeV, 1\,eV = 1.6 \times {10^{ - 19}}J)$

