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JEE Main 2020 (8 Jan, Shift 2) · Rolling Motion

Rolling Motion question from JEE Main 2020 (8 Jan, Shift 2)

Q62JEE Main 2020 · 8 JAN · SHIFT 2Easy

Rolling Motion

A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is:

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Answer: (c)

  1. Rolling sphere: K=12mv2(1+25)=710mv2K = \frac{1}{2}mv^2\left(1 + \frac{2}{5}\right) = \frac{7}{10}mv^2
  2. K=710×0.5×(0.05)2=0.35×2.5×10−3K = \frac{7}{10}\times 0.5\times(0.05)^2 = 0.35\times 2.5\times 10^{-3}
  3. K=8.75×10−4K = 8.75\times 10^{-4} J
  4. So the answer is (c) 8.75×10−48.75\times 10^{-4} J.
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