Air India’s jumbo B747 plane is set to fly from Delhi to Wuhan on Friday to evacuate Indians from the Chinese city and the epicentre of the novel Coronavirus (2019-nCoV) that has resulted in over 200 deaths and infected another 10,000 globally.
The 423-seater B747 plane will depart from Delhi airport at 12.30 pm today and is expected to return by 2 am tomorrow.
“The B747 plane is all set to depart from Delhi at 12.30 pm. It came from Mumbai on Friday morning only,” said a senior airline official.
There will be five doctors from the Health Ministry and one paramedical staff on the plane. The doctors have reportedly been instructed to allow only non-infected people on the aircraft.
Earlier today, Air India Chairman and Managing Director Ashwani Lohani said no service will take place in the plane.
“Whatever food is there will be kept in seat pockets. As there will be no service, there will be no interaction (between cabin crew and passengers),” he said.
“Masks have been arranged for the crew and passengers. For our crew, we have also arranged complete protective gear,” he added.
The government has reached out to over 600 Indians out of the estimated 1,200 Indians residing in China’s Hubei province, the epicentre of the novel coronavirus outbreak, to ascertain their willingness to be brought back to India.
The death toll in China’s novel coronavirus outbreak has climbed to 213, with the number of confirmed cases totalling to 9,692. The Hubei province alone has reported 5,806 confirmed cases, including 204 deaths.
Meanwhile, the first confirmed case of the deadly virus was reported from India on Thursday. The affected person is a student who was studying at Wuhan University. He is currently admitted to a hospital in Kerala and under close observation.Back to latest news