Last Update on: January 19th, 2020 at 06:38 pm
Rohit Sharma, on Sunday, added yet another feather to his illustrious hat. The flamboyant batsman, who scored a couple in the second ball of India’s innings reached 9000-run mark in the ODI form of the game. By doing so, he became the third-fastest batsman to reach the feat. He surpassed the former skipper Sourav Ganguly in the elite list.
The opener has scored 8996 runs in 216 innings and scored the remaining 4 runs in the first couple of balls against Australia on Sunday at M Chinnaswamy Stadium. He clipped Pat Cummins towards square-leg for a brace to achieve the remarkable feat to his name.
Rohit Sharma took just 217 innings:
The batsman who was the fastest to reach 9000 runs in ODIs was India skipper Virat Kohli who achieved the feat in 194 runs and he is followed by South Africa batsman AB de Villiers (205 innings). He excelled the records of Ganguly (228 innings), Tendulkar (235 innings) and Lara (239 innings). Rohit took just 217 innings to amass 9000 ODI runs.
Rohit Sharma won the ICC ODI Cricketer of the Year award for the year 2019 earlier this week. The Indian opener produced a stellar display in the fifty-over format with the bat, slamming seven centuries throughout the year. Five of his centuries came at the 2019 World Cup, where he ended as the highest run-scorer.
Rohit had completed his first 2000 runs in the format in 82 innings, which was the third slowest by a batsman. But the Rohit Sharma’s fortunes turned around after he started to open the batting for India in 2013 and is now counted among the greatest white-ball cricketers in the sport.
India are taking on Australia in the three-match ODI series, which is the side’s last home series before it departs for New Zealand later this month. Australia registered a dominant 10-wicket victory in the first ODI in Mumbai, while India came back strongly in the second game, winning by 36 runs. Australia were restricted to 286/9 in the third ODI.