Let radius for the particle of mass M = OB = r and for the particle of mass m = OC = R
Let linear velocity for a particle of mass M = v1 and for the particle of mass m = v2
Let angular velocity for the particle having mass M = and for the particle having mass m=
Let Time period for the particle having mass M = T1 and for the particle having mass m = T2
\(T_1 = \frac{2\pi r}{v_1} \quad \text{and} \quad T_2 = \frac{2\pi R}{v_2}\)
Given: T1 = T2
\(⇒\)\(\frac{2\pi r}{v_1} = \frac{2\pi R}{v_2}\)
\(⇒\)\(\frac{r}{v_1} = \frac{R}{v_2}\)
\(⇒\)\(\frac{v_1}{r} = \frac{v_2}{R}\)
The above equation generated is the formula for angular velocity. hence:
\(⇒ \omega_1=\omega_2\)
\(⇒\frac{\omega_1}{\omega_2}=\frac{1}{1}\)
Therefore, the ratio of the angular velocity will be 1:1.
Therefore, the correct option is (C) : 1.
A string of length \( L \) is fixed at one end and carries a mass of \( M \) at the other end. The mass makes \( \frac{3}{\pi} \) rotations per second about the vertical axis passing through the end of the string as shown. The tension in the string is ________________ ML.
The current passing through the battery in the given circuit, is:
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :
A full wave rectifier circuit with diodes (\(D_1\)) and (\(D_2\)) is shown in the figure. If input supply voltage \(V_{in} = 220 \sin(100 \pi t)\) volt, then at \(t = 15\) msec: