The total length of a sonometer wire fixed between two bridges is...

${l}_{1} : {l}_{2} : {l}_{3} = 6: 3: 2$

The frequancy in any ${n}^{th}$ mode for a segment is $f = \dfrac{nv}{2l} = constant$

no of loops ∝ length of the segment

so, no of loops are in the ratio 6:3:2

hence total loops =11

The string is divided in 60 cm, 30 cm, & 20 cm part such that for minimum frequency, the wavelength is maximum

$\dfrac{\lambda}{2} = \dfrac{1}{11}\text{x}110 = 10 cm$

$\text{f} = \dfrac{\text{V}}{\lambda} = \dfrac{1}{\lambda} \cdot \sqrt{\dfrac{\text{F}}{\mu}} = \dfrac{1}{\text{0.2}} \sqrt{\dfrac{4 0 0}{\text{0.01}}} = 1 0 0 0 Hz$