Search any question & find its solution
Question:
Answered & Verified by Expert
A metal wire of linear mass density of $9.8 \mathrm{~g} / \mathrm{m}$ is stretched with a tension of $10 \mathrm{~kg}$-wt between two rigid supports 1 metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency $n$. The frequency $n$ of the alternating source is
Options:
Solution:
2403 Upvotes
Verified Answer
The correct answer is:
$50 \mathrm{~Hz}$
$50 \mathrm{~Hz}$
$n=\frac{1}{2 l} \sqrt{\frac{T}{\mu}} ; \quad I=1 m$
$\mathrm{T}=10 \mathrm{Kg}$ wt. $=10 \times 10=100 \mathrm{~N}$
$\mu=9.8 \mathrm{~g} / \mathrm{m}=9.8 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \quad \mathrm{n}=50 \mathrm{hz}$
$\mathrm{T}=10 \mathrm{Kg}$ wt. $=10 \times 10=100 \mathrm{~N}$
$\mu=9.8 \mathrm{~g} / \mathrm{m}=9.8 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \quad \mathrm{n}=50 \mathrm{hz}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.