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Circular Motion question

2023 · 30 Jan · Shift 2 · Q70
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Circular Motion question

2023 · 30 Jan · Shift 2 · Q70

JEE MainPhysicsCircular MotionNumerical+4 / −1
A stone tied to 180 cm180 \mathrm{~cm}180 cm long string at its end is making 28 revolutions in horizontal circle in every minute. The magnitude of acceleration of stone is 1936xms−2\frac{1936}{x} ms^{-2}x1936​ms−2. The value of xxx‾\underline{\hspace{2cm}}​. (Take π=227\pi=\frac{22}{7}π=722​ )
Numerical answer
View written solutionFree

Correct answer: 125

  1. Given data

    • Length of string: 180 cm=1.8 m180\text{ cm} = 1.8\text{ m}180 cm=1.8 m
    • Number of revolutions: 282828 per minute
    • So frequency in revolutions per second is f=2860=715 s−1f = \frac{28}{60} = \frac{7}{15}\text{ s}^{-1}f=6028​=157​ s−1

  2. Angular speed ω=2πf=2⋅227⋅715=4415 rad/s\omega = 2\pi f = 2\cdot \frac{22}{7} \cdot \frac{7}{15} = \frac{44}{15}\text{ rad/s}ω=2πf=2⋅722​⋅157​=1544​ rad/s

  3. Centripetal acceleration Since the stone moves in a horizontal circle of radius equal to the string length, r=1.8=95 mr = 1.8 = \frac{9}{5}\text{ m}r=1.8=59​ m

    Magnitude of acceleration is a=ω2ra = \omega^2 ra=ω2r

    Substitute values: a=(4415)2⋅95a = \left(\frac{44}{15}\right)^2 \cdot \frac{9}{5}a=(1544​)2⋅59​

    a=1936225⋅95a = \frac{1936}{225} \cdot \frac{9}{5}a=2251936​⋅59​

    a=193625⋅5=1936125 m/s2a = \frac{1936}{25\cdot 5} = \frac{1936}{125}\text{ m/s}^2a=25⋅51936​=1251936​ m/s2

  4. Compare with given form Given, a=1936x m/s2a = \frac{1936}{x}\text{ m/s}^2a=x1936​ m/s2

    Therefore, x=125x = 125x=125

  5. Comparison with stored answer Stored correct answer = 125125125

    This matches our derived answer.

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