Question

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

Answer: (B)

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