Kinematics

JEE Main level, numerical answer, medium

291 of 308

Question 291 of 308

A stone tied to 180cm180cm long string at its end is making 2828 revolutions in horizontal circle in every minute. The magnitude of acceleration of stone is 1936xms−2\dfrac{1936}{x}m{s}^{-2}. The value of xx ______.
[Take π=227\pi =\dfrac{22}{7}]

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Answer: 125

Angular velocity, ω=28×2π60rads−1\omega =\dfrac{28\times 2\pi }{60}rad{s}^{-1}.
Required centripetal acceleration is given by,
a=ω2Ra={\omega }^{2}R
=(5660×227)2×1.8={\left(\dfrac{56}{60}\times \dfrac{22}{7}\right)}^{2}\times 1.8
=(44)2225×1.8=1936125=\dfrac{(44{)}^{2}}{225}\times 1.8 =\dfrac{1936}{125}
Hence, required x=125x=125.

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