Question

A stretched wire of length 110 cm is divided into three segments whose fundamental frequencies are in ratio 1 : 2 : 3. Their lengths must be

• A. 20 cm ; 30 cm ; 60 cm
• B. 60 cm ; 30 cm ; 20 cm
• C. 60 cm ; 20 cm ; 30 cm
• D. 30 cm ; 60 cm ; 20 cm

Answer 1

The correct option is B 60 cm ; 30 cm ; 20 cm

$l_1+l_2+l_3=110cmand{n}_1{l}_1=n_2{l}_2=n_3{l}_3$
$n_1$ : $n_2$ : $n_3$ :: 1 : 2 : 3
$\because \frac{n_1}{n_2}=\frac{1}{2}=\frac{l_2}{l_1}\Rightarrow l_2=\frac{l_1}{2}and\frac{n_1}{n_3}=\frac{1}{3}=\frac{l_3}{l_1}\Rightarrow l_3=\frac{l_1}{3}$
$\therefore l_1+\frac{i_1}{2}+\frac{l_1}{3}=110so{l}_1=60cm,l_2=30cm,l_3=20cm$