The length of a sonometer wire AB is 176 cm. Where should the two...

Let the lengths of the three segments be $l_1$, $l_2$, and $l_3$.

The fundamental frequency of a stretched string is given by $f = \dfrac{v}{2l}$. Since the tension and mass per unit length are constant for the wire, $f \propto \dfrac{1}{l}$.

Given the ratio of fundamental frequencies is $f_1 : f_2 : f_3 = 1 : 2 : 3$.

Therefore, the ratio of their lengths is $l_1 : l_2 : l_3 = 1 : \dfrac{1}{2} : \dfrac{1}{3} = 6 : 3 : 2$.

Let $l_1 = 6x$, $l_2 = 3x$, and $l_3 = 2x$.

The total length of the wire is $l_1 + l_2 + l_3 = 176 \text{ cm}$.

$6x + 3x + 2x = 176 \Rightarrow 11x = 176 \Rightarrow x = 16 \text{ cm}$.

The lengths of the individual segments are:
$l_1 = 6 \times 16 = 96 \text{ cm}$
$l_2 = 3 \times 16 = 48 \text{ cm}$
$l_3 = 2 \times 16 = 32 \text{ cm}$

The positions of the two bridges from end A are $l_1$ and $l_1 + l_2$.

First bridge position $= 96 \text{ cm}$.

Second bridge position $= 96 + 48 = 144 \text{ cm}$.

Answer: 96 cm and 144 cm