NVAEasyJEE 2023Theorems of M.I (Parallel & Perpendicular Axis)

JEE Physics 2023 Question with Solution

Two identical solid spheres each of mass 2kg2 \, \text{kg} and radii 10cm10 \, \text{cm} are fixed at the ends of a light rod. The separation between the centers of the spheres is 40cm40 \, \text{cm}. The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is ×103kgm2\times 10^{-3} \, \text{kg}\cdot\text{m}^2:

Answer

Correct answer:176

Step-by-step solution

Using parallel axis theorem

Given: Each solid sphere has mass m=2kgm = 2 \, \text{kg} and radius r=10cm=0.1mr = 10 \, \text{cm} = 0.1 \, \text{m}. The separation between centers is 40cm40 \, \text{cm}, so the distance of each sphere from the midpoint is d=20cm=0.2md = 20 \, \text{cm} = 0.2 \, \text{m}.

Find: The moment of inertia of the system about an axis perpendicular to the rod through its midpoint.

Using parallel axis theorem,

Isys=(25mr2+md2)×2I_{\text{sys}} = \left(\frac{2}{5}mr^2 + md^2\right) \times 2

Substituting the values,

Isys=(25×2×0.01+2×0.04)×2I_{\text{sys}} = \left(\frac{2}{5} \times 2 \times 0.01 + 2 \times 0.04\right) \times 2 =(0.008+0.08)×2=0.088×2=0.176kgm2= (0.008 + 0.08) \times 2 = 0.088 \times 2 = 0.176 \, \text{kg}\cdot\text{m}^2

Therefore,

Isys=176×103kgm2I_{\text{sys}} = 176 \times 10^{-3} \, \text{kg}\cdot\text{m}^2

So the required numerical value is 176.

Common mistakes

  • Using d=0.4md = 0.4 \, \text{m} instead of d=0.2md = 0.2 \, \text{m}. The parallel axis theorem requires the distance from each sphere's center to the axis through the midpoint, not the full separation between the two spheres.

  • Ignoring the sphere's own moment of inertia 25mr2\frac{2}{5}mr^2. Each sphere is not a point mass, so both the self-rotation term and the shift term md2md^2 must be included.

  • Calculating the moment of inertia for only one sphere and forgetting to multiply by 22. The system contains two identical spheres placed symmetrically about the axis.

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