India cricket team made no changes to their playing XI for their Champions Trophy semi-final clash against Australia. Rohit Sharma lost the toss ye again, the 14th time India have done so on Tuesday. The Men in Blue have been asked to bowl first in the contest in Dubai. This means Varun Chakravarthy retains his place in the playing XI and India would go ahead with 4 spinners.
India’s playing XI vs Australia
Rohit Sharma, Shubman Gill, Virat Kohli, Shreyas Iyer, KL Rahul, Axar Patel, Hardik Pandya, Ravindra Jadeja, Varun Chakravarthy, Kuldeep Yadav, Mohammed Shami
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After impressing with a five-wicket haul in just his 2nd ODI, India have decided to retain Varun Chakravarthy in their playing XI. Rohit has decided to go ahead with the same 4-spinner combo that they fielded in the game against New Zealand. The surface at the Dubai International Cricket stadium favours the spinners heaviily and it is evident from India’s previous contest against NZ.
9 out of 10 wickets were taken by spinners in India’s final group stage match against New Zealand. Chakravarthy, who was included in the Indian squad in the last-minute, surprised the Kiwi batters with his mystery component and ability to take the batters by surprise with his pace and variations. The Tamil Nadu spinner stunned New Zealand with figures of 5/42 in his very first Champions Trophy game.
Meanwhile, Harshit Rana continues to warm the bench and so do Rishabh Pant and Arshdeep Singh. The duo are yet to get a game in the ongoing Champions Trophy, with KL Rahul and Rana being preferred ahead of them. It is also pertinent to point out that Washington Sundar is also yet ot get a game in the tournament.
