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Question

The system of 2 rods shown in the figure vibrates at the same frequency and forms a standing wave. The ratio of the number of antinodes in the two rods if the radius of rod B is twice the radius of A is:


  1. 1

  2. 2

  3. 3

  4. 4

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Solution

The correct option is B

2


Step 1: Given Data

Let the equal frequency of the two rods be f1=f2

Let the antinode of rod A be m.

Let the antinode of rod B be n.

The radius of the rod A is Ar

Radius of rod B is Br=2Ar

Let the area of rod A be AA.

Area of rod B is BA=4AA(Ar2)

Let the length of the rods be l.

Let the time period be T.

Let the density be ρ.

Step 2: Formula Used

A=πr2

A is the area

The area of the rod A is AA

Area of rod B is BA

f1=n2lTρAA,f2=m2lTρBA

Step 3: Calculate the Ratio

f1=n2lTρAAf2=m2lTρ4AAGiven,f1=f2f1f2=nm×2mn=2

Hence, the correct option is B.


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