Exams › NEET › Physics

A stone tied to a string of length 80 cm is whirled in a horizontal circle at constant speed, completing 14 revolutions in 25 s. What is the magnitude of the centripetal acceleration of the stone?

1. 850 cm/s²
2. 996 cm/s²
3. 720 cm/s²
4. 650 cm/s²

Correct answer: 996 cm/s²

Solution

The stone completes 14 revolutions in 25 s, so its frequency is 14/25 s^{-1}. Using $a_c = 4\pi^2 r f^2$ with $r = 80$ cm gives a value close to 996 cm/s².

Related NEET Physics questions

• A coin on a rotating turntable begins to slip when placed 4 cm from the center. If the turntable's angular speed is doubled, at what distance from the center will the coin just start to slip?
• The angular displacement of a rotating body is described by $\theta(t) = \omega_0 t + \frac{1}{2}\alpha t^2$, where $\omega_0 = 1\ \text{rad/s}$ and $\alpha = 1.5\ \text{rad/s}^2$. What is the angular velocity of the body at $t = 2\ \text{s}$?
• A coin rests on a gramophone record rotating at angular velocity \(\omega\). The coefficient of static friction is \(\mu\). The coin rotates without slipping if its distance \(r\) from the center satisfies:
• Two racing cars of masses \(m_1\) and \(m_2\) travel in circles of radii \(r_1\) and \(r_2\), respectively. Both complete one full revolution in the same time period \(t\). What is the ratio of the angular speed of the first car to the second car?
• A wheel starts from rest and undergoes uniform angular acceleration, reaching an angular velocity of 80 rad/s after 5 seconds. What is the total angular displacement of the wheel during this time?
• A particle moves in a circular path. If \(\omega\) is the angular velocity vector and \(\vec r\) is the position vector of the particle from the centre, which expression correctly gives the linear velocity \(\vec v\) of the particle?

⚔️ Practice NEET Physics free + battle 1v1 →