A round steel bar of overall length $40 c m$ consists of two equal portions of $20 c m$ each having diameters of $10 c m$

and $8 c m$ respectively. If the rod is subjected to a tensile load of $10 t o n n e s$ , the elongation will

- 1
  
  $\frac{1}{10 π} (\frac{1}{25} + \frac{1}{16}) c m$
- 2
  
  $\frac{2}{10 π} (\frac{1}{25} + \frac{1}{16}) c m$
- 3
  
  $\frac{3}{10 π} (\frac{1}{25} + \frac{1}{16}) c m$
- 4
  
  $\frac{4}{10 π} (\frac{1}{25} + \frac{1}{16}) c m$

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