MCQ
A stretched wire of length $110 cm$ is divided into three segments whose frequencies are in ratio $1 : 2 : 3$. Their lengths must be
- A$20 cm ; 30 cm ; 60 cm$
- ✓$60 cm ; 30 cm ; 20 cm$
- C$60 cm ; 20 cm ; 30 cm$
- D$30 cm ; 60 cm ; 20 cm$
✓
Answer
Correct option: B.
$60 cm ; 30 cm ; 20 cm$
b
(b) ${l_1} + {l_2} + {l_3} = 110\,cm$ and ${n_1}{l_1} = {n_2}{l_2} = {n_3}{l_3}$
(b) ${l_1} + {l_2} + {l_3} = 110\,cm$ and ${n_1}{l_1} = {n_2}{l_2} = {n_3}{l_3}$
${n_1}:{n_2}:{n_3}::1:2:3$
$\because \frac{{{n_1}}}{{{n_2}}} = \frac{1}{2} = \frac{{{l_2}}}{{{l_1}}} \Rightarrow {l_2} = \frac{{{l_1}}}{2}$and $\frac{{{n_1}}}{{{n_3}}} = \frac{1}{3} = \frac{{{l_3}}}{{{l_1}}} \Rightarrow {l_3} = \frac{{{l_1}}}{3}$
$\therefore {l_1} + \frac{{{l_1}}}{2} + \frac{{{l_1}}}{3} = 110$ so ${l_1} = 60cm,{l_2} = 30cm,{l_3} = 20cm$
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