A uniform circular disc of radius 50 cm at rest is free to tum about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s⁻². Its net acceleration in ms⁻² at the end of 2.0 s is approximately:

8.0 

Torque : \(\tau = I \alpha\)

\(\alpha\ = 2rad/s^{-2}\)

Tangential acceleration 

a = r\(\alpha\)

a_t = \(\frac{1}{2}\) × 2 = 1ms^-2

a_t = 1ms^-2

v = u + at

   = 0 + 2

v = 2m/s

Radial acceleration

a_r = \(\frac{v^2}{r}\) = \(\frac{4}{0.4}\) = 8ms^-2

Net acceleration 

a = \(\sqrt{a_r^2 + a_t^2}\)

   = \(\sqrt{64+ 1}\)

   = 8ms^-2 (approx)