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JEE Main 20198 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 8 April
Shift: Morning Shift
Subject: Physics
Chapter: Waves and Sound

JEE Main 2019 · 8 April · Morning Shift · Physics · Waves and Sound

A wire of length 2L, is made by joining two wires A and B of same...

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 antinodes in wire

A

is

p

and that in

B

is

q

then ratio

p: q

is:

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

Options

A $3: 5$
B $4: 9$
C $1: 4$
D $1: 2$

Correct answer

D. $1: 2$

Step-by-step solution

For rod A,

{f}_{1} = \dfrac{{n}_{1}}{2l}\sqrt{\dfrac{T}{ρ{A}_{1}}}

For rod B,

{f}_{2} = \dfrac{{n}_{2}}{2l}\sqrt{\dfrac{T}{ρ{A}_{2}}}

Given

{f}_{1} = {f}_{2}

\dfrac{{n}_{1}}{2l}\sqrt{\dfrac{T}{ρ{A}_{1}}} = \dfrac{{n}_{2}}{2l}\sqrt{\dfrac{T}{ρ{A}_{2}}}

\dfrac{{n}_{1}}{2l}\sqrt{\dfrac{T}{ρ{A}_{1}}} = \dfrac{{n}_{2}}{2l}\sqrt{\dfrac{T}{ρ ( 4{A}_{1} )}}

\left[∵

\begin{array}{ll}{r}_{B} = 2{r}_{A} \\ {A}_{2} = 4{A}_{1}\end{array}

\right]

<br/>

{n}_{1} = \dfrac{{n}_{2}}{2}

<br/>

\dfrac{{n}_{1}}{{n}_{2}} = \dfrac{1}{2}

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