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JEE Main 2023 (6 Apr, Shift 1) · Work by a Variable Force

Work by a Variable Force question from JEE Main 2023 (6 Apr, Shift 1)

Q8JEE Main 2023 · 6 APR · SHIFT 1Medium

Work by a Variable Force

A particle of mass 10 g10\text{ g} moves in a straight line with retardation 2x2x, where xx is the displacement in SI units. Its loss of kinetic energy for above displacement is (10x)−n\left(\frac{10}{x}\right)^{-n} J. The value of nn will be __________

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

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Answer: 2

  1. Retarding force F=m(2x)=0.01×2x=0.02xF = m(2x) = 0.01 \times 2x = 0.02x N.
  2. Loss of KE =∫0x0.02x dx=0.01x2=(x10)2= \int_0^x 0.02x\,dx = 0.01x^2 = \left(\frac{x}{10}\right)^2
  3. (x10)2=(10x)−2\left(\frac{x}{10}\right)^2 = \left(\frac{10}{x}\right)^{-2}, so n=2n = 2.
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