A mild steel bar is in three parts, each 200mm long as shown in fig. The diameters of the parts are indicated on the diagram. The bar is subjected to axial pull of (P) N. If $E=2 \times 10^5$ MPa and the elongations in the three parts AB,BC, and CD are $\Delta 1$,$\Delta 2$,and $\Delta 3$ respectively, then the ratio of the greatest elongation to the least elongation will be :

i) 9

ii) 4

iii) 3

iv) 2

For equilibrium, along axial axis,

$\sum F_x=0$

a) Portion AB

Elongation is given by,

$\Delta 1=\frac{P*l_{AB}}{A_{AB}*E_{AB}}$

$\Delta 1=\frac{P*200}{\frac{\pi}{4}*20^2*2*10^5}=3.183*10^{-6} P$

b)Portion BC

Elongation is $\Delta 2=\frac{P*l_{BC}}{A_{BC}*E_{BC}}$

$\Delta 2=\frac{P*200}{\frac{\pi}{4}*10^2*2*10^5}=1.273*10^{-5} P$

c)Portion CD:

Elongation is $\Delta 3=\frac{P*200}{\frac{\pi}{4}*30^2*2*10^5}=1.414*10^{-6} P$

The ratio of greatest elongation to the least elongation is $\frac{\Delta 2}{\Delta 3}$

$=\frac{1.273*10^{-5} P}{1.414*10^{-6} P}$

= 9