[Australia], June 6 (ANI): Indian shuttler PV Sindhu was knocked out of the Australian Open after she faced a 21-19, 21-28 defeat against Thailand's Nitchaon Jindapol in the second round here on Thursday.
After securing an easy win over Indonesia's Choirunnisa, Sindhu struggled to win even a single set against Jindapol, she lost in straight sets.
In the opening set, Jindapol took an early 4-0 lead over Sindhu. However, Sindhu managed to bring the game at 5-5. The game proceeded in a similar manner and as the game was brought to 19-19 each. From there on, World number 21, Jindapol clinched two consecutive points to win the set.
Continuing a similar performance, Jindapol won the second set as well as she thrashed Sindhu by 21-18. With this defeat, Sindhu made a second-round exit from the Australian Open. (ANI)