The ratios of length areas of cross-section and Young's modulii of steel of that of brass wires shown in the figure are a,b and c respectively . The ratio of increase in the lengths of brass of that of steel wires is [Assume that the masses of steel and brass wires are negligible]

Options

A. `(4a)/(7bc)`
B. `(7bc)/(4a)`
C. `(4bc)/(7a)`
D. `(7a)/(4bc)`

Answer

C

Explanation

Let the ratio of length be a ratio of area be b
Ratio of Young .s modulus be c
we know that ,
`Y = (F //A)/(Delta L//L)`
`:. Delta L = (FL)/(AY)`
`(DeltaL_(s))/(DeltaL_(B))= ((F_(S))/(F_(B))) ((L_(S))/(L_(B))) ((A_(B))/(A_(S)))((Y_(B))/(Y_(S)))`
`= (7)/(4) xx(a) ((1)/(b)) xx ((1)/(c))`
`(Delta L_(S))/(DeltaL_(B))= (7a)/(4bc)`
`(Delta L_(B))/(Delta_(S)) = (4bc)/(7a)`