A brass of length `2 m` and cross-sectional area `2.0 cm^(2)` is attached end to end to a steel rod of length and cross-sectional area `1.0 cm^(2)`. The compound rod is subjected to equal and opposite pulls of magnitude `5 xx 10^(4) N` at its ends. If the elongations of the two rods are equal, the length of the steel rod `(L)` is
{`Y_(Brass) = 1.0 xx 10^(11) N//m^(2) ` and `Y_(steel) = 2.0 xx 10^(11) N//m^(2)`

Options

A. 1.5 m
B. 1.8 m
C. 1 m
D. 2 m

Answer

D

Explanation

(d) `therefore`Since the elongation of the two rods are equal
`(Deltal)_(b)=(Deltal)_(s)`

or `((Fl)/(AY))_(b)=((Fl)/(AY))_(s)" "(F_(b)=F_(s))`
or `((l)/(AY))_(b)=((l)/(AY))_(s)`
`therefore` The length of the steel rod
`l_(s)=((A_(s)Y_(s))/(A_(b)Y_(b)))l_(b)=((1.0xx2.0xx10^(11))/(2.0xx1.0xx10^(11)))(2m)=2m`