IPL 2020: MI wins toss, opts to bat first against SRH
4 October, 2020 | Rakshanda Afrin
Rohit Sharma led Mumbai Indians on Sunday won the toss and opted to bat first against SunRisers Hyderabad in the Indian Premier League match at the Sharjah Cricket Stadium.
Mumbai Indians on Sunday won the toss and opted to bat first against SunRisers Hyderabad (SRH) in the Indian Premier League (IPL) here at the Sharjah Cricket Stadium. SunRisers Hyderabad had to make one forced change as Bhuvneshwar Kumar got injured in the last game against Chennai Super Kings (CSK). David Warner-led SRH brought in Sandeep Sharma in place of Bhuvneshwar. Apart from this change, the SRH team made one another change to their lineup as they brought in Siddarth Kaul in place of Khaleel Ahmed.
Both Mumbai Indians and SunRisers Hyderabad have played four matches in the tournament so far. Both sides have won two out of their four matches. Mumbai Indians have defeated Kolkata Knight Riders and Kings XI Punjab while SRH outclassed Chennai Super Kings and Delhi Capitals.
Currently, Mumbai Indians is at the third position in the IPL 2020 standings while SRH is at the fourth. Whoever wins this contest, that side will go to the top of the table. If Mumbai Indians wins this contest, the side will go to the top of the table on the basis of net run rate.
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On the other hand, Mumbai Indians fielded the same playing XI from their last match against Kings XI Punjab.
Mumbai Indians’ playing XI: Rohit Sharma (c), Quinton de Kock, Suryakumar Yadav, Ishan Kishan, Kieron Pollard, Hardik Pandya, Krunal Pandya, James Pattinson, Rahul Chahar, Trent Boult, and Jasprit Bumrah. SunRisers Hyderabad playing XI: David Warner (c), Jonny Bairstow, Manish Pandey, Kane Williamson, Priyam Garg, Abhishek Sharma, Abdul Samad, Rashid Khan, Sandeep Sharma, T Natarajan, and Siddarth Kaul.
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