The speed of a transverse wave travelling in a wire of length 50 cm, cross-sectional area 1 `mm^(2)` and mass 5 g is 80 `ms^(-1)`. The Young's modulus of the material of the wire is `4 xx 10^(11) Nm^(-2)`. The extension in the length of the wire is

Options

A. `8 xx 10^(-5)`m
B. `8 xx 10^(-4)` m
C. `16 xx 10^(-8)` m
D. `16 xx 10^(-4)`m

Answer

A

Explanation

Given , length of the wire , l = 50 cm
= 5` xx 10^(-2) ` m
Cross - sectional area of wire , A = 1 `mm^(2)`
` = 1 xx 10^(-6) m^(2)`
and mass of the wire, m = 5 g
speed of transverse wave, v = 80 m/s
Now, speed of transverse wave given as
`therefore " " v = sqrt((T)/(mu)) = sqrt((T)/(m) xx l )`
`rArr " " v = sqrt((T)/(m) xx l )`
`rArr " " v^(2) = (T)/( m ) xx l`
`rArr " " l = (v^(2) m)/(T ) " "` .... (i)
Now, Young.s modulus, Y = `((T)/(A))/((Delta l)/(l))`
or `" " Delta l = (T)/(AY). l " " ` .... (ii)
From Eqs (i) and (ii) , we get
` therefore " " Delta l = (v^(2) m)/(AY)`
Putting the given values, we get
`Delta l = ( 80^(2) xx 5 xx 10^(-3))/(1 xx 10^(-6) xx 4 xx 10^(11) ) m `
or ` " " Delta l = 8 xx 10^(-5) ` m
so, the extension in the length of the wire is
`Delta l = 8 xx 10^(-5)` m.