Australia elect to bowl against India in Rajkot ODI (Toss)

Rajkot: Australia captain Aaron Finch won the toss and elected to field in the second ODI against India at the Saurashtra Cricket Association Stadium on Friday.

Australia are going with the same playing XI which featured in Mumbai where they thrashed India by 10 wickets. “We are going to bowl first. It looks like a beautiful wicket. They controlled their length really well. Same team for us,” said Finch after winning the toss.

The hosts, on the other hand, have made two changes to line-up. Rishabh Pant and Shardul Thakur have been replaced by Manish Pandey and Navdeep Saini.

“We would’ve bowled first as well. But the good thing is that there’s not much dew. It looks really nice and hard. If we put enough runs on the board, we have a good chance of defending,” said India skipper Virat Kohli.

“You will be beaten comprehensively by sides like Australia at the international level. It’s about understanding what went wrong. We need to focus on the positives. We are going to be more brave than the last game, that’s for sure.”

Talking about the changes, Kohli said: “One forced change as Rishabh is out with a concussion. KL will don the gloves and Manish comes in for Rishabh. Also, Saini replaces Shardul.”

Playing XI: India: Rohit Sharma, Shikhar Dhawan, Virat Kohli(c), Lokesh Rahul(w), Shreyas Iyer, Manish Pandey, Ravindra Jadeja, Mohammed Shami, Navdeep Saini, Kuldeep Yadav, Jasprit Bumrah

Australia: David Warner, Aaron Finch(c), Marnus Labuschagne, Steven Smith, Ashton Turner, Alex Carey(w), Ashton Agar, Pat Cummins, Mitchell Starc, Kane Richardson, Adam Zampa

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