An iron circular disc is rotated about its axis with a constant...

The linear speed of a particle on the rim of the disc is given by $v = R \omega$, where $R$ is the radius and $\omega$ is the angular velocity.

Since the angular velocity $\omega$ is maintained constant, the fractional change in the linear speed is equal to the fractional change in the radius.

$\dfrac{\Delta v}{v} = \dfrac{\Delta R}{R}$

Due to thermal expansion, the change in radius is given by $\Delta R = R \alpha \Delta T$, which implies $\dfrac{\Delta R}{R} = \alpha \Delta T$.

Therefore, the percentage change in linear speed is:

$\dfrac{\Delta v}{v} \times 100 = \alpha \Delta T \times 100$

Substituting the given values, \alpha = 1.2 \times 10^{-5} ^{{}^{{}^{\circ}}}\textC^{-1} and $\Delta T = 50^{{}^{{}^{\circ}}}\text{C} - 20^{{}^{{}^{\circ}}}\text{C} = 30^{{}^{{}^{\circ}}}\text{C}$:

Percentage change $= 1.2 \times 10^{-5} \times 30 \times 100 = 0.036$