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

Options

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

b

Explanation

Given `l_(1)+l_(2)+l_(3)=110` cm
and `n_(1)l_(1)=n_(2)l_(2)=n_(3)l_(3)` (`because` T and d are same)
`n_(1):n_(2):n_(3)::1 : 2: 3`
`because n_(1)/n_(2)=1/2=l_(2)/l_(1)rArr l_(2)=l_(1)/2`
and `because n_(1)/n_(3)=1/3=l_(3)/l_(1)rArr l_(3)=l_(1)/3`
`therefore l_(1) +l_(1)/2+l_(1)/3=110`
`rArr l_(1) = 60 ` cm
So, ` l_(1) = 60 ` cm, `l_(2) = 30` cm, ` l_(3) = 20` cm