Question

$A,B$ and $C$ are $3$ wires of same material. The wave speed of $A$ is 1$0m/s$, then what is the wave speed of $B$?

• A. $10$
• B. $\frac{10}{\sqrt{2}}$
• C. $10\sqrt{2}$
• D. $none$

Answer 1

The correct option is B $\frac{10}{\sqrt{2}}$

Given,

Velocity =$v$ , tension = $T$ and linear density$=\mu$.

$v=\sqrt{\frac{T}{\mu }}$

$T=\mu v^2$

$v_A=10m{s}^{-1}$

Apply lami’s theorem on wire A, B and C

$\frac{T_A}{\sin {90}^o}=\frac{T_B}{\sin {150}^o}=\frac{T_C}{\sin {120}^o}$

$\frac{\mu v_A^2}{\sin {90}^o}=\frac{\mu v_B^2}{\sin {150}^o}=\frac{\mu v_C^2}{\sin {120}^o}$

$\frac{v_B^2}{\sin {150}^o}=\frac{v_A^2}{\sin {90}^o}$

$v_B=v_A\sqrt{\frac{\sin {150}^o}{\sin {90}^o}}=10\times \frac{1}{\sqrt{2}}$

Hence, velocity in B is $\frac{10}{\sqrt{2}}m{s}^{-1}$