Question

a stretched wire of length 110 cm is divided into three segments whose frequency are in ratio 1:2:3 their length must be

Answer 1

Answer:

The length of the wire is divided into three segments 60 cm, 30 cm and 20 cm.

Explanation:

Given that,

Length l = 110 cm

A stretched wire is divided into three segments whose frequency are in 1:2:3.

We know that,

The frequency of the wave is inversely proportional to the wavelength.

$f\propto\dfrac{1}{l}$

$f_{1}:f_{2}:f_{3}=1:2:3$

$l_{1}:l_{2}:l_{3}=\dfrac{1}{1}:\dfrac{1}{2}:\dfrac{1}{3}$

$l_{1}:l_{2}:l_{3}=6:3:2$

The length of first frequency is

$l_{1}=\dfrac{6}{11}\times110$

$l_{1}=60$

The length of first frequency is

$l_{2}=\dfrac{3}{11}\times110$

$l_{2}=30$

The length of first frequency is

$l_{3}=\dfrac{2}{11}\times110$

$l_{3}=20$

Hence, The length of the wire is divided into three segments 60 cm, 30 cm and 20 cm.