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NEET PHYSICSEasy

A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 s, what is the magnitude and direction of acceleration of the stone?

A

π24 m/s2\frac{\pi^2}{4} \text{ m/s}^2 and direction along the radius towards the center

B

π2 m/s2\pi^2 \text{ m/s}^2 and direction along the radius away from the center

C

π2 m/s2\pi^2 \text{ m/s}^2 and direction along the radius towards the center

D

π2 m/s2\pi^2 \text{ m/s}^2 and direction along the tangent to the circle

Step-by-Step Solution

  1. Identify Given Values: The problem input contains repeated digits due to formatting errors. Based on the options and standard physics problems (PMT 2005), the values are interpreted as: Radius (rr) = 1 m, Number of revolutions (nn) = 22, Time (tt) = 44 s.
  2. Calculate Angular Velocity (ω\omega): The angular velocity is given by ω=2πnt\omega = \frac{2\pi n}{t}. Substituting values: ω=2π(22)44=44π44=π rad/s\omega = \frac{2\pi (22)}{44} = \frac{44\pi}{44} = \pi \text{ rad/s} .
  3. Calculate Acceleration Magnitude: The acceleration in uniform circular motion is the centripetal acceleration, given by ac=ω2ra_c = \omega^2 r. ac=(π)2(1)=π2 m/s2a_c = (\pi)^2 (1) = \pi^2 \text{ m/s}^2.
  4. Determine Direction: By definition, centripetal acceleration is always directed along the radius towards the center of the circle .
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