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JEE Main 2023 (10 Apr, Shift 1) · Conservation of Energy

Conservation of Energy question from JEE Main 2023 (10 Apr, Shift 1)

Q19JEE Main 2023 · 10 APR · SHIFT 1Medium

Conservation of Energy

A closed circular tube of average radius 15 cm, whose inner walls are rough, is kept in vertical plane. A block of mass 1 kg just fits inside the tube. The speed of block is 22 m/s, when it is introduced at the top of tube. After completing five oscillations, the block stops at the bottom region of tube. The work done by the tube on the block is __________ J. (Given g = 10 m/s2{}^{2}).

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Numerical answer type. Work it out, then open the answer.

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Answer: 245

  1. Work-energy theorem: Wtube+Wgravity=ΔKW_{tube} + W_{gravity} = \Delta K.
  2. Block goes from top to bottom: Wgravity=mg(2r)=1×10×0.3=3 JW_{gravity} = mg(2r) = 1 \times 10 \times 0.3 = 3\text{ J}.
  3. ΔK=0−12×1×222=−242 J\Delta K = 0 - \frac{1}{2} \times 1 \times 22^2 = -242\text{ J}
  4. Wtube=−242−3=−245 JW_{tube} = -242 - 3 = -245\text{ J}, i.e. magnitude 245 J245\text{ J} (negative work by friction).
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