Question

Two wires A and B made of the same material have equal lengths. If the volumes of wires A and B are in the ratio \(1:8\) and the tensions applied to the wires A and B are in the ratio \(1:2\), then the ratio of the speeds of transverse waves in wires A and B is:

• A. \(1:1\)
• B. \(2:1\)
• C. \(4:1\)
• D. \(8:1\)

Hint:

For wires of the same material and equal length, volume ratio directly gives the ratio of linear mass densities.

Answer 1

The Correct Option is A

Concept: The speed of a transverse wave in a stretched string is \[ v=\sqrt{\frac{T}{\mu}} \] where \[ T=\text{tension} \] \[ \mu=\text{mass per unit length} \] For the same material, \[ \mu=\rho A \] and since lengths are equal, \[ A\propto \text{Volume} \]

Step 1: Find the ratio of linear mass densities. Since volumes are in the ratio \[ 1:8 \] and lengths are equal, \[ \mu_A:\mu_B=1:8 \]

Step 2: Apply wave speed relation. \[ \frac{v_A}{v_B} = \sqrt{ \frac{T_A/\mu_A} {T_B/\mu_B} } \] Substituting, \[ = \sqrt{ \frac{1/1} {2/8} } \] \[ = \sqrt4 \] \[ =2 \] Therefore, \[ v_A:v_B=2:1 \]

Step 3: Final answer. \[ \boxed{2:1} \]