Answer:
emf will also be 10 times less as compared to when it has fallen [tex]40 \mathrm{m}[/tex]
Explanation:
We know, from faraday's law-
[tex]e m f=-N \frac{\Delta \Phi}{\Delta T}[/tex]
and [tex]\Phi=B . A[/tex]
So, as the height increases the velocity with which it will cross the ring will also increase. [tex](v=\sqrt{2 g h})[/tex]
Given
[tex]\mathrm{V} 1(\text { Speed at } 40 \mathrm{m})=2 \mathrm{x} \mathrm{V} 2(\text { speed at } 10 \mathrm{m})[/tex]
[tex]\sqrt{2 g h_{2}}=2 \times \sqrt{2 g h_{1}}=28.28 \mathrm{m} / \mathrm{s}[/tex]
Now, from [tex]40 \mathrm{cm}[/tex]
[tex]V_{3}=\sqrt{2 g h_{3}}=\sqrt{2 \times 10 \times 0.4}=2.82 \mathrm{m} / \mathrm{s}[/tex]
From equation a and b we see that velocity when dropped from [tex]40 \mathrm{m}[/tex] is 10 times greater when height is 40 [tex]\mathrm{cm}[/tex] so, emf will also be 10 times less as compared to when it has fallen [tex]40 \mathrm{m}[/tex]
