(a) d1 + d2 / 2
(b)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEghzE7gj-4INkJsoLrlP04773hzVD6otS8XFq7iw13sZk3y4xrgQtxst-DgzpByv4IQmudUMnIOTt59aZ4grJlb-6XXjHxSQrWo_A9HYtR5acFfIw-tKJz7ZvOeH_WAgEtiwhs6O5fEP7g/s1600/2.jpg)
(c)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj08S32heZaE4f7fAo9IBMexneN01d6AJuL3YmaLH_1sEOphsS_KYnKaICv9uj2UCT8bfRSXWoFSkseep7kKPDIiBUrVFimHWwLn2wz7pIGibA_BAUZDgi7bdJxsYviQ6SKUgNE03H_ql0/s1600/4.jpg)
(d)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjn50u31m5umMPCgCqul-EFExQ0gz78oG7Kt2wC6eE212a2ePguG0FWFW7QpNbFSGGl38fXAYnkGqRRY13-_XFEhuIQ5qVMrITaW4LKOS7e6QXZDJhl0qzRp5zqb1XUW51pQe_pnZv8yRA/s1600/3.jpg)
Ans: (b)
Explanation: E = Stress / Strain = WL / ΔL A
For a tapering rod,
A = π d1 d2 / 4
ΔL = 4WL / π d1 d2 E
For a uniform cross section with diameter d,
A = π d2 / 4
ΔL = 4WL / π d2 E
The bars have the same length l and are subjected to the same axial pull W. Then for the same extension ΔL
4WL / π d1 d2 E = 4WL / π d2 E
1/ d1d2 = 1/ d2
so d =
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEghzE7gj-4INkJsoLrlP04773hzVD6otS8XFq7iw13sZk3y4xrgQtxst-DgzpByv4IQmudUMnIOTt59aZ4grJlb-6XXjHxSQrWo_A9HYtR5acFfIw-tKJz7ZvOeH_WAgEtiwhs6O5fEP7g/s1600/2.jpg)