Question

Two wires, one made of capper and other of steel are joined end to end. (as show in figure). The area of cross-section of copper wire is twice that of steel wire, They ae placed under compressive force of magnitude F. Find the ratio of their lenghts such that change in lenght of both wire are same `(Y_(S)= 2 xx 10 ^(11) N//m^(2) and Y_(C) = 1.1 xx 10 ^(11) N//m^(2))`

A. 2.1
B. 1.1
C. 1.2
D. 2

Answer 1

Correct Answer - B
`Y_(s)= (FL_(s))/(A_(s)Deltal_(s)) and Y_(C)= (FL_(c))/(A_(c)Deltal_(c))`
`therefore " "(L_(c))/(L_(s))= (((Y_(c)A_(c)Deltal_(c)))/(F))/(((Y_(s)A_(s)Deltal_(s))/(F)))=((Y_(c))/(Y_(s)))(A_(C)/(A_(s)))((Deltal_(c))/(Deltal_(s)))`
` Here (A_(C))/(A_(S))= 2, (Deltal_(c))/(Deltal_(s))=1, (Y_(C))/(Y_(S))= (1.1)/(2)`
`therefore" " (L_(C))/(L_(s))=((1.1)/(2))(2)(1)=1.1`