MHT CET · Physics · Waves and Sound

A stretched wire of length 260 cm is set into vibrations. It is divided into three segments whose frequencies are in the ratio $2 : 3 : 4$ . Their lengths must be

A 80 cm, 60 cm, 120 cm
B 120 cm, 80 cm, 60 cm
C 60 cm, 80 cm, 120 cm
D 120 cm, 60 cm, 80 cm

Verified Solution

Answer & Solution

Correct Answer

(B) 120 cm, 80 cm, 60 cm

Step-by-step Solution

Detailed explanation

The frequency produced by a stretched wire is given by

f = \frac{p}{2 l} \sqrt{\frac{T}{m}}

… (i)
Where, p = number of loops formed in vibration,

l

= length of wire,
T = tension in wire,
and m = mass per unit length of wire
So, form Eq (i), we get

f \propto \frac{1}{l}

Here, the ratio of frequencies of three segments is

f_{1} : f_{2} : f_{3} = 2 : 3 : 4

\Rightarrow l_{1} : l_{2} : l_{3} = \frac{1}{2} : \frac{1}{3} : \frac{1}{4}

= 6 : 4 : 3

The length of wire, L = 260 cm, so

l_{1} = \frac{6}{13} \times 260 = 120

l_{2} = \frac{4}{13} \times 260 = 80

a n d l_{3} = \frac{3}{13} \times 260 = 60

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A stretched wire of length 260 cm is set into vibrations. It is divided into three segments whose frequencies are in the ratio 2:3:4 . Their lengths must be

Step-by-step Solution

A stretched wire of length 260 cm is set into vibrations. It is divided into three segments whose frequencies are in the ratio $2 : 3 : 4$ . Their lengths must be