JEE Main 2014 — Waves and Sound Question with Solution

$l_1:l_2:l_3=6:3:2$

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

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

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

$\text{f}=\frac{\text{V}}{\lambda }=\frac{1}{\lambda }·\sqrt{\frac{\text{F}}{\mu }}=\frac{1}{\text{0.2}}\sqrt{\frac{400}{\text{0.01}}}=1000Hz$