Bengaluru, Jan 19: India vice-captain Rohit Sharma became the third fastest batsman to reach 9,000 ODI runs during the series deciding third and final game against Australia here on Sunday.

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Rohit needed four runs to get to the landmark and the senior India opener got to it in the final ball of the first over of India's innings to sit behind skipper Virat Kohli and South Africa's AB de Villiers.

Kohli achieved the feat in 194 innings and he is followed by de Villiers (205 innings).

Rohit (216 innings) beat batting greats Sourav Ganguly (228 innings), Sachin Tendulkar (235 innings) and Brian Lara 239 innings) to the feat.

The series is locked 1-1 and here the Aussies rode Steve Smith's 131 to post 286/9 in 50 overs after winning the toss and electing to bat first.

India, in reply, opener with K.L. Rahul and Rohit after Shikhar Dhawan had to be taken for an x-ray. Dhawan was taken off the field after he hurt his left shoulder in the series-deciding tie after he dived to save an Aaron Finch shot in the cover region before hurting his left shoulder in the fifth over.

About the Author

Saurabh Sharma
Saurabh Sharma is the Editorial Head of Cricketnmore Hindi and a passionate cricket journalist with over 14 years of experience in sports media. He began his journalism career with Navbharat Times, part of the Times of India Group, before moving to television media with Sadhna News. In 2014, he joined Cricketnmore and currently serves as the editor of the platform.
Known for his deep understanding of cricket statistics and unique storytelling approach, Saurabh specializes in cricket news, match analysis, records, and feature stories. Along with editorial responsibilities, he also works as a show producer for popular cricket video series such as Cricket Tales, Cricket Flashback, and Cricket Trivia. Read More
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