A wire of length `2L`, is made by joining two wires `A` and `B` of same length but different radii `r` and `2r` and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antionodes in wire `A` is P and that in `B` is `q` then the ratio `p:q` is:

Options

A. `3:5`
B. `1:4`
C. `4:9`
D. `1:2`

Answer

D

Explanation

Frequency will be same in both
`(pc_(1))/(2l)=(qc_(2))/(2l) rArr pc_(1) =qc_(2)`
Tension will be same in both but mass per unit length will be 4 times in 2 (as radius is twice) `therefore mu_(2) = 4mu_(1)` `therefore Psqrt((T)/(mu_(1))) = q sqrt((T) /(mu_(2)))` `therefore (p)/(q)= sqrt(mu_(1))/(mu_(2)) = (1)/(2)`