In the question number 36, at which distance a weight may be hung along the rod, in order to produce equal strains in both the wires?

In the question number 36, at which distance a weight may be hung alon

1.	In the question number 36, at which distance a weight may be hung along the rod, in order to produce equal strains in both the wires?
Answer» `(4)/(3)m`from steel wire `(4)/(3)m`from brass wire 1 m from steel wire `(1)/(4)m`from brass wire Solution :As Strain `= ("Srtress")/("Young's Modulus")` LTBR `therefore` Strainin stell wire `= (T_(S))/(A_(S)Y_(S))` For equal strain in both the wire, `(T_(S))/(A_(S)Y_(S)) = (T_(B))/(A_(B)Y_(B))` `therefore (T_(S))/(T_(B))=(A_(S))/(A_(B)) (Y_(S))/(Y_(B)) = (0.1 cm^(2))/(0.2 cm^(2)) xx (2xx 10^(11)N m^(-2))/(1 xx 10^(11) N m^(-2))= 1 .......(i)` For therotational equlibrium of the rod, `T_(S)x = T_(B) (2-x)` `(2-x)/(x) = (T_(S))/(T_(B))= 1""["Using (i)"]` `2 - x = x or x=1m `