The 2nd Test match between india and australia has now reached an exciting stage. Playing in the last innings is usually very difficult on indian pitches, so whoever scores the most runs in the first innings in a Test match wins. But it was expected that australia would get such a thing in this Test. The reason was that the Australian team, who batted first, was bowled out for 263 runs in the first innings. After this, the front-line players of the indian team lost the game by a few runs.
Star player virat kohli was bowled out for 44 runs, and at one stage the indian team was 139 for 7. At one stage, australia was expected to take a lead of over 100 runs in the first innings, but Ashwin and Axar patel combined to rescue india from the slump. Ashwin was out on 37 runs and Axar patel top-scored with 74 runs.
As a result, the indian team was bowled out for 262 runs. With this, the Australian team took a lead of only 1 run. After this, the Australian team is playing in the 2nd innings. In this situation, we can now see the maximum score scored in the last innings in Delhi. The highest score in the 4th innings in delhi was 276 by west indies in 1986 and india in 2011.
Similarly, in the 36 Test matches played in delhi, the target has been successfully chased in the last innings 11 times. Due to this, australia should be bowled out within 200 runs in the 2nd innings. 150 or more will be difficult for india to win.
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