A steel wire of negligible mass, length `2I`, cross sectional area 'A' and Young's modulus 'Y' is held between two rigid walls. A small body of mass 'm' is suspended from its middle point 'O' as shown. The vertical descent of the middle point 'O' of the wire is

Options

A. `l((mg)/(AY))^(1//3)`
B. `l((2mg)/(AY))^(1//3)`
C. `l ((4mg)/(AY))^(1//3)`
D. `l ((6mg)/(AY))^(1//3)`

Answer

A

Explanation

`2F sin theta = mg`
`F = (mg)/(2 sin theta) = (mgl)/(2x) [ therefore theta "is small"]`
`Deltal = (sqrt(l^(2) + x^(2)) - l)`
`= l [(1 + (x^(2))/(l^(2)))^(1//2) - 1 ]`
`Deltal = [1 + (1)/(2) (x^(2))/(l^(2)) - 1] [ :' x lt lt l]`
`:. (Deltal)/(l) = (x^(2))/(2l^(2))`
Now `(F)/(A) = Y (Deltal)/(l)`
`F = AY (Deltal)/(l)`
`(mgl)/(2x) = AY (x^(2)/(2l^(2)))`
`x^(3) = (mgl^(3))/(AY)`
`:. x = l ((mg)/(AY))^(1//3)`