The lengths of two wires of the same material are in the ratio 1:2, their tensions are in the ratio 1:2 and their diameters are in the ratio 1:3. The ratio of the notes they emit, when sounded together by the same source, is

\sqrt{2}

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\sqrt{3}

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\sqrt[2]{3}

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\sqrt[3]{2}

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Solution

The correct option is D $\sqrt[3]{2}$
The velocity of a wave is given by,

$v = \sqrt{T / μ}$

Now, v = $2 f L$

Hence, $f = \sqrt{T / μ} / 2 L$

Now, $μ = m / L = ρ A$

Hence,

$f_{1} / f_{2} = \sqrt{T_{1} / A_{1}} \times L_{2} / (\sqrt{T_{2} A_{2}} \times L_{1})$

Substituting the values,

$f_{1} / f_{2} = 2^{1 / 3}$

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The lengths of two wires of the same material are in the ratio 1:2, their tensions are in the ratio 1:2 and their diameters are in the ratio 1:3. The ratio of the notes they emit, when sounded together by the same source, is

The correct option is D 3√2The velocity of a wave is given by,v=√T/μNow, v = 2fLHence, f=√T/μ/2LNow, μ=m/L=ρAHence, f1/f2=√T1/A1×L2/(√T2A2×L1)Substituting the values,f1/f2=21/3