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India took a 2-1 lead in the series after winning the Oval Test against England. In the fourth match of the five-match series, Team India won by 157 runs even after trailing by 99 runs in the first innings. The performance of a player of the Indian team in this match was the best. With batting and bowling, Shardul Thakur showed a game that turned the tide of the match.
The Indian team scored 191 runs in the first innings after losing the toss in the Oval Test. In reply, England’s team scored 290 runs and took a 99-run lead. Team India scored 466 runs in the second innings on the strength of opener Rohit Sharma’s century and gave the host a target of 368 runs. In the second innings, England’s team could only score 210 runs on the last day and Team India took the lead in the series with victory.
Strong performance by Shardul:
In the first innings, Shardul played a blistering innings of 57 runs off 36 balls after the Indian team’s top order batsmen flopped. This innings, played with the help of 7 fours and 3 sixes, took the Indian team to the score of 191 runs. In the second innings, he once again showed the power of the bat and scored 60 runs in 72 balls with the help of 7 fours and 1 six.
The most important wicket in bowling:
Shardul did it only when India needed a wicket in both the innings. He bowled Ollie Pope, who was becoming difficult for India by scoring a half-century in the first innings. Due to the fall of this wicket, England’s innings was shattered. In the second innings, he got India the first wicket on the fifth day.
By dismissing Rory Burns, who had scored 50 runs, Shardul broke the 100-wicket partnership of England’s first wicket. The second wicket he got in the series against India was the highest run scorer of the English captain Joe Root. This wicket turned the match in India’s direction.
