- Home
- Standard 11
- Physics
14.Waves and Sound
hard
A wire having a linear mass density $9.0 \times 10^{-4} \;{kg} / {m}$ is stretched between two rigid supports with a tension of $900\; {N}$. The wire resonates at a frequency of $500\;{Hz}$. The next higher frequency at which the same wire resonates is $550\; {Hz}$. The length of the wire is $...... {m}$
A
$50$
B
$100$
C
$10$
D
$2$
(JEE MAIN-2021)
Solution
$\mu=9.0 \times 10^{-4}\, \frac{{kg}}{{m}}$
${T}=900\, {N}$
${V}=\sqrt{\frac{{T}}{\mu}}=\sqrt{\frac{900}{9 \times 10^{-4}}}=1000 \,{m} / {s}$
${f}_{1}=500 \,{Hz}$
${f}=550$
$\frac{{nV}}{2 \ell}=500 \ldots . \text { (i) }$
$\frac{({n}+1) {V}}{2 \ell}=500 \ldots(ii)$
$(ii)\, (i)$ $\frac{{V}}{2 \ell}=50$
$\ell=\frac{1000}{2 \times 50}=10$
Standard 11
Physics
Similar Questions
medium
