Australia batsman Steven Smith during his half-century knock on Friday. (BCCI)
Mohali: India will come out searching for ways to dislodge Steven Smith on the third morning at the PCA Stadium, following the batsman's resolute innings of 58 on Friday that helped Australia add 122 runs from the time Phillip Hughes was dismissed. India need three wickets to bowl Australia out; the tourists are 273 for 7 in 104 overs.
Smith, recalled for the tour of India as the Australian selectors found promise in his batting against spin, finished not out on 58 in his first Test since January 2011. He walked in at No. 5 after Hughes' exit for 2 and looked on as Ed Cowan (86), Brad Haddin (21) and Moises Henriques (0) got out. Smith shared stands of 47 and 46 with Cowan and Haddin, and then put on 22 for the eighth wicket with Mitchell Starc.
Having batted sensibly and fluently against pace and spin, Smith is the wicket India will chase on Saturday morning. His innings contained seven fours and one six, and Smith has outshone the likes of Hughes and Haddin with his application.
India may have inched ahead on day two, but if Smith extends his stand with Starc (20*) and gets to three figures, then Australia will have a platform from which they can look to put pressure on India.
India's most successful bowler on Friday was left-arm spinner Ravindra Jadeja, who removed David Warner and Michael Clarke in successive deliveries and then added Peter Siddle for a duck before stumps. Ishant Sharma got among the wickets by removing Haddin and Henriques in the same over late in the day, and will look to chip away if given the ball early on Saturday.
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