When India face Australia to keep the series alive in the 2nd ODI at Rajkot on Friday, Rohit Sharma needs just 46 runs to beat the likes of Sourav Ganguly, Sachin Tendulkar and Brian Lara and become the third fastest to 9000 ODIs runs. Rohit has 8954 runs in 215 innings so far. If he gets the required runs on Friday against Australia then he would comfortably beat Ganguly, who had taken 228 innings, Tendulkar (235 innings) and Brian Lara (239 innings).
The record for the fastest to 9000 ODI runs stands with India skipper Virat Kohli, who achieved the feat in just 194 innings followed by South Africa’s AB de Villiers in 205 innings.
Rohit had a fantastic 2019. He was awarded ICC ODI Player of the Year for scoring 7 centuries – a record 5 in World Cup alone – in ODIs last year. Rohit was also the highest run-scorer in ODIs in 2019.
Some more Rohit Sharma records to look forward to in the 2nd ODI against Australia
1 more century: Rohit needs 1 more century to become the player with 4th most centuries in ODIs. He currently has 28 ODI centuries, which is joint 4th most along with Sanath Jayasuriya. That century will also make Rohit the 4th highest century-getter for India across formats. The list is currently lead by Sachin Tendulkar with 100 centuries, followed by Virat Kohli with 70.
Hit-man or 6-man? Rohit Sharma needs 7 sixes to become the 1st player to hit 100 sixes against Australia across formats.