A wire of length \(2\,L\) is made by joining two wires \(A\) and \(B\) of same lengths but different radii \(r\) and \(2r\) and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire \(A\) is \(p\) and that in \(B\) is \(q\) then the ratio \(p : q\) is

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Answer & Solution

Correct Answer

(B) \(1 : 2\)

Detailed explanation

Let mass per unit length of wires are \(\mu_{1}\) and \(\mu_{2}\) respectively.

\(\because\) Materials are same, so density \(\rho\) is same.

\(\mu_{1}=\frac{\rho \pi r^{2} L}{L}=\mu \text { and } \mu_{2}=\frac{\rho 4 \pi r^{2} L}{L}=4 \mu\)

Tension in both are same \(=\mathrm{T},\) let speed of wave in wires are \(V_{1}\) and \(V_{2}\)

\(V_{1}=\frac{V_{1}}{2 L}=\frac{V}{2 L} \& f_{02}=\frac{V_{2}}{2 L}=\frac{V}{4 L}\)

Frequency at which both resonate is \(L.C.M.\) of both frequencies i.e. \(\frac{\mathrm{V}}{2 \mathrm{L}}\)

Hence number of loops in wires are \(1\) and \(2\) respectively

So, ratio of number of antinodes is \(1:2\)

Same subject

From JEE Mains

List \(I\)	List \(II\)
\((1)\) \(700\, nm\) to \(1\,mm\)	\((i)\) Vibration of atoms and molecules
\((2)\) \(1\,nm\) to \(400\, nm\)	\((ii)\) Inner shell electrons in atoms moving from one energy level to a lower level
\((3)\) \( < 10^{-3}\,nm\)	\((iii)\) Radioactive decay of the nucleus
\((4)\) \(1\,mm\) to \(0.1\,m\)	\((iv)\) Magnetron valve