Two identical solid spheres each of mass $2 \mathrm{~kg}$ and radii $10 \mathrm{~cm}$ are fixed at the ends of a light rod. The separation between the centres of the spheres is $40 \mathrm{~cm}$. The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is __________ $\times 10^{-3} \mathrm{~kg}~\mathrm{m}^{2}$

<p>The problem requires calculating the moment of inertia of the system consisting of two identical solid spheres fixed at the ends of a light rod. We need to find the moment of inertia about an axis perpendicular to the rod and passing through its midpoint.</p> <p>First, let’s identify the moment of inertia of each solid sphere about its own center, which is given by the formula:</p> <p>$I_{\text{sphere}} = \frac{2}{5} m r^2$</p> <p>Where:</p> <ul> <li>$m$ is the mass of the sphere = $2 \mathrm{~kg}$</li> <li>$r$ is the radius of the sphere = $0.1 \mathrm{~m}$</li> </ul> <p>Substituting the values:</p> <p>$$I_{\text{sphere}} = \frac{2}{5} \times 2 \mathrm{~kg} \times (0.1 \mathrm{~m})^2 = \frac{4}{5} \times 0.01 \mathrm{~kg}~\mathrm{m}^2 = 0.008 \mathrm{~kg}~\mathrm{m}^2$$</p> <p>Now, we need the moment of inertia of the two spheres about the axis passing through the midpoint of the rod. This requires using the parallel axis theorem, which states:</p> <p>$I_{\text{total}} = I_{\text{sphere}} + m d^2$</p> <p>Where:</p> <ul> <li>$d$ is the distance from the center of the sphere to the axis through the rod’s midpoint = $0.2 \mathrm{~m}$</li> </ul> <p>Calculating the additional inertia due to the parallel axis theorem for one sphere:</p> <p>$$I_{\text{parallel}} = m d^2 = 2 \mathrm{~kg} \times (0.2 \mathrm{~m})^2 = 2 \mathrm{~kg} \times 0.04 \mathrm{~m}^2 = 0.08 \mathrm{~kg}~\mathrm{m}^2$$</p> <p>The total moment of inertia for one sphere about the midpoint of the rod is:</p> <p>$$I_{\text{one sphere, total}} = I_{\text{sphere}} + I_{\text{parallel}} = 0.008 \mathrm{~kg}~\mathrm{m}^2 + 0.08 \mathrm{~kg}~\mathrm{m}^2 = 0.088 \mathrm{~kg}~\mathrm{m}^2$$</p> <p>Since there are two identical spheres, the total moment of inertia of the system is:</p> <p>$$I_{\text{system}} = 2 \times 0.088 \mathrm{~kg}~\mathrm{m}^2 = 0.176 \mathrm{~kg}~\mathrm{m}^2$$</p> <p>Converting the result to the given form:</p> <p>$0.176 \mathrm{~kg}~\mathrm{m}^2 = 176 \times 10^{-3} \mathrm{~kg}~\mathrm{m}^2$</p> <p>So, the moment of inertia of the system about the given axis is:</p> <p>$176 \times 10^{-3} \mathrm{~kg}~\mathrm{m}^2$</p>