Virat Kohli, the India captain, fell to leg-spinner Adam Zampa yet again, marking the fourth time he has perished to the Australian in ODIs.
That makes Zampa the Australian, alongside Jhye Richardson, to have dismissed Kohli the most number of times in ODIs. Off-spinners Suraj Randiv of Sri Lanka and Graeme Swann of England have also taken out Kohli four times. Only three men – New Zealand’s Tim Southee and Sri Lanka’s Thisara Perera (five each) and West Indies fast bowler Ravi Rampaul (six times) have accounted for Kohli’s wicket more times in ODIs.
Kohli latched on to a short ball from Zampa, on the first ball of the 32nd over in the first ODI at Mumbai’s Wankhede Stadium on January 14, and laced a pull over backward square leg for six. Zampa corrected his length next ball, but tempted Kohli to go after him by beautifully flighting the ball up. The line was straight, around middle and off, and the length was really full, meaning that Kohli couldn’t fully put his wrists to use and was cramped for room. He ended up drilling the ball straight back towards the bowler, who, with a fraction of a second to react, put his hands in front of him and grabbed a reverse-cupped stunner.
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Kohli v Zampa was a contest many had their eyes on, after the tough time the leg-spinner gave the Indian captain the last time Australia toured India, in early 2019. Zampa had largely stifled Kohli with leg-side lines on that occasion, and picked up his wicket once in the T20I series and twice in the ODI series, both of which Australia won.
Kohli’s wicket on Tuesday constituted a part of a batting collapse, as India went from 134-1 to 164-5, exposing the much weaker middle order to Australia’s potent attack. Kohli had walked in at No.4, at the fall of KL Rahul’s wicket, only the second time he has batted from that position since October 2015.
