Q114 / 150 · MHT CET 2026 (15 April Shift 1) · Physics
A particle performing U.C.M. of radius $\dfrac{\pi}{2}$ m makes 'x' revolutions in time t. Its tangential velocity is
- A$\dfrac{\pi x}{t}$
- Correct answerB$\dfrac{\pi^2 x}{t}$
- C$\dfrac{\pi}{xt}$
- D$\dfrac{xt}{\pi}$
Solution
The radius of the circular path is $r = \dfrac{\pi}{2}$ m.
The particle makes $x$ revolutions in time $t$, so its frequency is $n = \dfrac{x}{t}$.
The angular velocity is given by $\omega = 2\pi n = \dfrac{2\pi x}{t}$.
The tangential velocity is $v = r\omega$.
Substituting the values, we get $v = \left(\dfrac{\pi}{2}\right) \left(\dfrac{2\pi x}{t}\right) = \dfrac{\pi^2 x}{t}$.
Answer: $\dfrac{\pi^2 x}{t}$
The particle makes $x$ revolutions in time $t$, so its frequency is $n = \dfrac{x}{t}$.
The angular velocity is given by $\omega = 2\pi n = \dfrac{2\pi x}{t}$.
The tangential velocity is $v = r\omega$.
Substituting the values, we get $v = \left(\dfrac{\pi}{2}\right) \left(\dfrac{2\pi x}{t}\right) = \dfrac{\pi^2 x}{t}$.
Answer: $\dfrac{\pi^2 x}{t}$
Hence, the correct answer is option: 'B'.