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JEE Main 2023 (30 Jan, Shift 2) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from JEE Main 2023 (30 Jan, Shift 2)

Q18JEE Main 2023 · 30 JAN · SHIFT 2Easy

Centripetal Acceleration & Force

A stone tied to 180 cm180\text{ cm} long string at its end is making 28 revolutions in horizontal circle in every minute. The magnitude of acceleration of stone is 1936x m s−2\frac{1936}{x}\text{ m s}^{-2}. The value of xx ________. (Take π=227\pi = \frac{22}{7})

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

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Answer: 125

  1. r=1.8 mr = 1.8\text{ m}, n=2860=715 Hzn = \frac{28}{60} = \frac{7}{15}\text{ Hz}.
  2. ω=2πn=2×227×715=4415 rad/s\omega = 2\pi n = 2 \times \frac{22}{7} \times \frac{7}{15} = \frac{44}{15}\text{ rad/s}
  3. a=ω2r=1936225×1.8=1936125 m/s2a = \omega^2 r = \frac{1936}{225} \times 1.8 = \frac{1936}{125}\text{ m/s}^2
  4. So x=125x = 125.
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