A round steel bar of overall length 40 cm consists of two equal portions of 20 cm each having diameters of 10 cm and 8 cm respectively. Take E as 2 × 106 kg/cm2. If the rod is subjected to a tensile load of 10 tonnes, the elongation in cm will be given by

• A. \(\frac{3}{{10\pi }}\left[ {\frac{1}{{20}} + \frac{1}{{15}}} \right]\)
• B. \(\frac{8}{{10\pi }}\left[ {\frac{1}{{10}} + \frac{1}{{15}}} \right]\)
• C. \(\frac{2}{{10\pi }}\left[ {\frac{1}{{25}} + \frac{1}{{30}}} \right]\)
• D. \(\frac{1}{{10\pi }}\left[ {\frac{1}{{25}} + \frac{1}{{16}}} \right]\)