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

20 cm ; 30 cm ; 60 cm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

60 cm ; 30 cm ; 20 cm

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

60 cm ; 20 cm ; 30 cm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30 cm ; 60 cm ; 20 cm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 60 cm ; 30 cm ; 20 cm
$l_{1} + l_{2} + l_{3} = 110 c m a n d n_{1} l_{1} = n_{2} l_{2} = n_{3} l_{3}$
$n_{1} : n_{2} : n_{3} :: 1 : 2 : 3$
$∵ \frac{n_{1}}{n_{2}} = \frac{1}{2} = \frac{l_{2}}{l_{1}} \Rightarrow l_{2} = \frac{l_{1}}{2}$ and $\frac{n_{1}}{n_{3}} = \frac{1}{3} = \frac{l_{3}}{l_{1}} \Rightarrow l_{3} = \frac{l_{1}}{3}$
$∴ l_{1} + \frac{l_{1}}{2} + \frac{l_{1}}{3} = 110$ so $l_{1} = 60 c m, l_{2} = 30 c m, l_{3} = 20 c m$

Suggest Corrections

Similar questions

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

110 cm

3

1 : 2 : 3

An object is placed at 60 cm from the pole of a concave mirror, whose focal length is 20 cm. Find the distance at which the image is formed.

20 cm

O

30 cm

Join BYJU'S Learning Program