An alumminium rod and steel wire of same length and cross-section are attached end to end. Then compound wire is hung fron a rigid support and load is suspended from the free end. Y for steel is `((20)/(7))` times of aluminium. The ratio of increase in length of steel wire to the aluminium wire is

Options

A. `20 : 3`
B. `10 : 7`
C. `7 : 20`
D. `1 : 7`

Answer

C

Explanation

Here, `Y_(s)=(20)/(7)Y_(a)` or `(Y_(s))/(Y_(a))=(20)/(7)`
As `Y=(FL)/(A Delta L)` or `Delta L=(FL)/(AY)`
`Delta L prop(1)/(Y)` as both the wires are of same length.
`(Delta L_(s))/(Delta L_(a))=(Y_(a))/(Y_(s))=(1)/((Y_(s)//Y_(a)))=(1)/(20//7) =(7)/(20)`