Question

A wire having a linear density of 0.05 gm/cm is stretched between two rigid supports with a tension of $4.5\times {10}^7$ dynes. It is observed that the wire resonates at a frequency of 420 cycles/sec. The next higher frequency at which the same wire resonates is 490 cycles/sec. The length of wire is approximately

• A. 314 cm
• B. 254 cm
• C. 214 cm
• D. 354 cm

Answer 1

The correct option is C 214 cm

$\mu =0.05gm/cmT=4.5\times {10}^{7}dyneLet,\frac{nv}{2L}=420⟶\left(1\right)\&\frac{(n+1)v}{2L}=490⟶\left(2\right)$

Equation $\left(1\right)$-Equation $\left(2\right)$, we get

$\frac{(n+1)v}{2L}-\frac{nv}{2L}=490-420\frac{v}{2L}=70\frac{1}{2L}\sqrt{\frac{T}{\mu }}=70\frac{1}{2L}\sqrt{\frac{4.5\times {10}^7}{0.05}}=70\frac{1}{2L}\sqrt{9\times {10}^8}=70\frac{3\times {10}^4}{2\times 70}=L\therefore L=0.0214\times {10}^4=214cm$

Hence option (C) is correct