Figure shows the graph of elastic potential energy `(U)` stored versus extension, for a steel wire `(Y = 2xx10^(11) Pa)` of volume `200 c c`. If area of cross- section `A` and original length L, then

Options

A. `(a)A = 10^(-4) m^(2)`
B. `(b)A= 10 ^(-3) m^(2)`
C. `(c)L=1.5 m`
D. `(d)L = 2 m`

Answer

A, D

Explanation

`U =1/2 kx^(2)` . It is a parabola symmetric about
U- axis .
At `x = 0.2 mm`, `U =0.2 J` ( from the figure )
`:. 0.2 = 1/2 k(2xx10^(-4))^(2)`
`rArr k = (YA)/(L) `
`(A)/L = (K)/(Y) = (10^(7)) Nm^(-1)`
`k = (YA)/(L)`
`rArr (A)/(L) = (k)/(Y) = (10^(7))/(2xx10^(11)) = 5xx10^(-5)` ...(i)
`AL = "Volume" = 200xx10^(-6) m^(3)` ...(ii)
On solving Eqs. (i) and (ii) , We get
`A= 10^(-4) m^(2)` and `L = 2 m`