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4 overs, 7 runs and 6 wickets… Australia will field Brahmastra against India in WTC final

London: Australian captain Pat Cummins on Tuesday confirmed that fast bowler Scott Boland will be part of the playing XI for the World Test Championship Final (WTC Final) against India, starting at The Oval. Boland will get a chance in place of Josh Hazlewood. At the same time, Michael Neser, who was included in the team in place of Hazlewood, will be out. Hazlewood has been ruled out of the match due to an ongoing left leg injury.

Boland can be a big threat

Scott Boland, 34, has played only 7 Tests for Australia, but his record is excellent. He made his Test debut towards the end of 2021 against Australia. In the fourth innings of that match played in Melbourne, Boland took 6 wickets for 7 runs in 4 overs. It included big wickets like Joe Root and Johnny Bairstow. He had completed 5 wickets in just 19 balls in that innings. He is the bowler to take 5 wickets at the lowest number of deliveries in Test history.

He took 28 wickets in his seven matches at an average of only 13.42. His average of 13.42 is the lowest of any Test cricketer to have bowled at least 1000 balls since the turn of the 20th century. This is the reason why he will be a big threat to Team India.

How will Australia’s attack be?

Australia will go in with three leading fast bowlers as well as one spinner. Along with captain Pat Cummins, Mitchell Starc and Scott Boland will handle the responsibility of fast bowling. Nathan Lyon will be the only spinner in the team. With this, Cameron Green will play as a fast bowling all-rounder. In a career spanning 20 Tests, Green has scored 941 runs and taken 23 wickets.

Why was Boland selected?

Regarding the inclusion of Boland in the playing eleven, captain Pat Cummins said – he is a different type of bowler, who hits good length and keeps the ball on the seam. Starc is a left-arm bowler and this will add variety to the attack. If Scott bowls in the right places, she will move.

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