
On this day (July 20) in 2017, Harmanpreet Kaur played a majestic knock against Australia to help India qualify for the final of the ICC Women’s World Cup. Kaur’s merciless hitting saw her carve out an unbeaten 171 off 115 balls with the help of 20 fours and seven sixes. The Indian bowlers complemented her batting effort to knock out the defending champions.
India got off to a terrible start after captain Mithali Raj won the toss and opted to bat first. They lost opener Smriti Mandana in the very first over, and their innings progressed at a snail’s pace. Kaur arrived at the crease when her team was struggling at 35 for 2 in 9.2 overs. She took time to settle and eventually unleashed her fury on the Aussie bowlers.
The right-hander reached her first fifty of the innings in 64 balls by making brief partnerships with skipper Mithali and all-rounder Deepti Sharma. Kaur didn’t hesitate to switch gears, and her acceleration saw her second fifty coming in another 26 balls and her third in just 17 balls.
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Rain robs Kaur’s double hundred
Had the match not been reduced to 42 overs due to rain, Harmanpreet Kaur could have scored her maiden ODI double-century. However, her 171* is recorded as the third-highest score in the World Cup and fifth-highest overall. On the back of her third ODI hundred, India reached a massive total of 281/4.
Australia, in reply, struggled to make a strong start as they were reduced to 3-21 in 7.2 overs. However, their three middle-order batters, Ellyse Perry, Elyse Villani, and Alex Blackwell, brought them back into the game. Blackwell threatened with a 56-ball 90, but Deepti dismissed the Aussie all-rounder to wrap up the innings to 245/10 in 40.1 overs.
India thus advanced to the final for the first time since 2005. However, the Women in Blue suffered a heartbreaking loss to England by 9 runs in the summit clash at Lord’s.