Question

Two wires of same material of radius 'r' and '2r' respectively are welded together end to end. The combination is then used as a sonometer wire under tension 'T'. The joint is kept midway between the two bridges. The ratio of the number of loops formed in the wires such that the joint is a node is

• A. 1 : 5
• B. 1 : 2
• C. 1 : 4
• D. 1 : 3

Hint:

Linear mass density ($m$) scales with the square of the radius ($m \propto R^2$), meaning wave velocity drops linearly with a larger radius ($v \propto 1/R$). To keep a matching oscillation frequency, a lower velocity requires shorter loop lengths, which packs more loops into the same space ($p \propto R$). Since the radius doubles, the loop count must double too!

Answer 1

The Correct Option is B

Since the two segments are welded end to end and stretched between shared boundaries, the structural tension ($T$) and the operational vibration frequency ($n$) must remain identical across both wire components. Let the joint be a stationary node point separating $p_1$ loops in the first segment and $p_2$ loops in the second segment over equal half-lengths ($l$). The fundamental frequency relation for a wire of loop count $p$, length $l$, material density $\rho$, and cross-sectional radius $R$ expands as: $$n = \frac{p}{2l}\sqrt{\frac{T}{m}} = \frac{p}{2l}\sqrt{\frac{T}{\rho \cdot \pi R^2}} = \frac{p}{2lR}\sqrt{\frac{T}{\rho\pi}}$$ Equating frequencies for both joint sections ($n_1 = n_2$): $$\frac{p_1}{2l \cdot r}\sqrt{\frac{T}{\rho\pi}} = \frac{p_2}{2l \cdot (2r)}\sqrt{\frac{T}{\rho\pi}}$$ Canceling out the common length, tension, density, and radius variables from both sides simplifies the expression directly to: $$\frac{p_1}{r} = \frac{p_2}{2r} \implies \frac{p_1}{p_2} = \frac{1}{2}$$
Final Answer:
The ratio of the number of loops formed in the wires is 1 : 2, which corresponds to option (B).