Sonika Vaid Advances to the Top 24 in ‘American Idol’

Sonika Vaid Advances to the Top 24 in ‘American Idol’

Massachusetts-based Indian American singer Sonika Vaid has secured a place in the Top 24 of the farewell season of “American Idol.” “Sonika! Welcome to the top 24!” exclaimed “American Idol” judge Jennifer Lopez on the Feb. 4 episode of the show.

The 51 remaining singers gave one final performance for the judges during the last night of Hollywood Week following which the Top 24 were chosen.

Twenty-year-old Vaid rendered a cover version of the classic pop number, “I Surrender” by Celine Dion, during the last night of Hollywood Week. Though Vaid has garnered praise for the originality in her voice, the judges felt she still needed to work on her performance skill.

“You gave a great performance yesterday vocally,” Lopez said. “But we worry in that there’s just something missing in your performance. There’s something that needs to come alive, and that’s what we argued over.”

“So after much debate, we decided that we would like you to be part of the Top 24, and maybe work on that part of things because of that amazing voice you have.”

“Oh, my God, I’m so excited, I can’t believe this!” said an ecstatic Vaid on the show. The competition continues next week, which will see the remaining singers perform in groups of 12. Six former “Idols” will perform duets with them.

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