India tinkered again with their batting order by including all three openers in the playing XI in the first ODI against Australia at the Wankhede Stadium on Tuesday. In Pune, in the final game of the best-of-three T20I series against Sri Lanka last week that preceded the current ODI rubber, skipper Virat Kohli had batted as low as No. 6.
On Tuesday, all three in the running for the openers’ slots—Rohit Sharma, Shikhar Dhawan and KL Rahul— were included and batted in the same order.
This was already hinted by Kohli in the pre-game media conference on Monday.
Rohit was coming into action after being rested for the previous T20I series against Sri Lanka while Dhawan and Rahul had scored fifties in that series-decider in Pune on January 10 as openers.
There were some question marks over Dhawan going into the Pune game which he silenced with that knock of 54 in 36 balls.
The dilemma of who to leave out on Tuesday was solved by the easy way—by including all three. Rohit made only 10 and was consumed by Mitchell Starc who he had hit for two sweetly timed fours to the cover region earlier.
This brought together Dhawan and Rahul in the fifth over itself and the two put on a century stand worth 121 runs in 136 balls, while making 74 and 47 respectively.
It put India in a commanding position at 134 for one in the 28th over and a total in excess of 300 looked distinctly possible. But the Australian bowling attack, especially in pace, is far too superior to the Lankans’.
A mini collapse ensued in which India lost four wickets in 35 balls for the addition of 30 runs as the team slumped to 164 for 5.
However with Dhawan, who struck one six and nine fours in his 91-ball knock, and Rahul (47 in 61 balls) coming good, the team’s think tank is unlikely to leave one of them in the second game at Rajkot on Friday.