(A) \(8.3 \times 10^{-3}\)
(B) \(4.75 \times 10^{-3}\)
(C) \(1.86 \times 10^{-2}\)
(D) \(9.5 \times 10^{-3}\)
Explanation (Complete Step by Step Calculation)
Masses are:
\[
m_1 = 0.4\,\text{kg}, \quad m_2 = 0.35\,\text{kg}
\]
Radius of pulley:
\[
R = 2\,\text{cm} = 0.02\,\text{m}
\]
Distance fallen by heavier mass:
\[
s = 81\,\text{cm} = 0.81\,\text{m}
\]
Time taken:
\[
t = 9\,\text{s}
\]
Using equation of motion \( s = \frac{1}{2}at^2 \):
\[
0.81 = \frac{1}{2} a (9)^2
\]
\[
0.81 = 40.5a
\]
\[
a = 0.02\,\text{m/s}^2
\]
For the two masses, equations of motion are:
\[
m_1 g - T_1 = m_1 a
\]
\[
T_2 - m_2 g = m_2 a
\]
Torque equation for the pulley:
\[
(T_1 - T_2)R = I \alpha
\]
Angular acceleration:
\[
\alpha = \frac{a}{R}
\]
Substituting tensions and simplifying:
\[
(m_1 - m_2)g - (m_1 + m_2)a = \frac{Ia}{R^2}
\]
Substitute numerical values:
\[
(0.4 - 0.35)(9.8) - (0.75)(0.02) = \frac{I(0.02)}{(0.02)^2}
\]
\[
0.49 - 0.015 = \frac{I(0.02)}{0.0004}
\]
\[
0.475 = 50I
\]
\[
I = 9.5 \times 10^{-3}\,\text{kg·m}^2
\]
Thus, the rotational inertia of the pulley is \(9.5 \times 10^{-3}\,\text{kg·m}^2\).