The length of the wire between two ends of sonometer is $100 \mathrm{~cm}$. What should be the positions of two bridges below the wire so that the three segments of the wire have their fundamental frequencies in the ratio $1: 3: 5$.

• A

  $\frac{1500}{23} \mathrm{~cm}, \frac{500}{23} \mathrm{~cm}$
• B

  $\frac{1500}{23} \mathrm{~cm}, \frac{300}{23} \mathrm{~cm}$
• C

  $\frac{300}{23} \mathrm{~cm}, \frac{1500}{23} \mathrm{~cm}$
• Correct answer
  D

  $\frac{1500}{23} \mathrm{~cm}, \frac{2000}{23} \mathrm{~cm}$

Solution

According to question Length of wire $=L=100 \mathrm{~cm}$ And ratio of fundamental frequencies $$ =1: 3: 5 $$
$$ \begin{aligned} L_1: L_2: L_3 & =\frac{1}{1}: \frac{1}{3}: \frac{1}{5} \\ & =15: 5: 3 \end{aligned} $$ Let $x=$ comman factor then $$ \begin{aligned} & 15 x+5 x+3 x=100 \\ & \Rightarrow \quad 2 x=100 \\ & \Rightarrow \quad x=\frac{100}{23} \\ & \therefore \quad L_1=15 \times \frac{100}{23} \\ & =\frac{1500}{23} \mathrm{~cm} \\ & \therefore \quad L_2=5 \times \frac{100}{23} \\ & =\frac{1500}{23} \mathrm{~cm} \\ & \end{aligned} $$ $$ \begin{aligned} \therefore \quad L_3 & =3 \times \frac{100}{23} \\ & =\frac{300}{23} \mathrm{~cm} \end{aligned} $$ This shows that bridge should be placed from $A$ at $\frac{1500}{23} \mathrm{~cm}$ and $\frac{2000}{23} \mathrm{~cm}$ respectively

Hence, the correct answer is option: 'D'.