A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:

Correct Answer: $\frac{7}{57}$

Step I: Let the density of the material be $\rho$. The mass of the cut-out smaller sphere of radius $R$ is $m = \rho \cdot \frac{4}{3}\pi R^3$.
Step II: The original larger sphere has radius $2R$. Its mass is $M = \rho \cdot \frac{4}{3}\pi (2R)^3 = 8 \left(\rho \frac{4}{3}\pi R^3\right) = 8m$.
Step III: Calculate the MOI of the original large sphere about the Y-axis (which passes through its center): $I_{orig} = \frac{2}{5} M (2R)^2 = \frac{2}{5} (8m) (4R^2) = \frac{64}{5} mR^2$.
Step IV: The smaller sphere has radius $R$ and its center is located at distance $R$ from the Y-axis. Calculate its MOI about the Y-axis using the parallel axis theorem: $I_{small} = I_{cm} + m(R)^2 = \frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2$.
Step V: The MOI of the remaining (rest) part is $I_{rest} = I_{orig} - I_{small} = \frac{64}{5}mR^2 - \frac{7}{5}mR^2 = \frac{57}{5}mR^2$.
Step VI: The ratio of the MOI of the smaller sphere to the MOI of the rest part is $\frac{I_{small}}{I_{rest}} = \frac{\frac{7}{5}mR^2}{\frac{57}{5}mR^2} = \frac{7}{57}$.

सही उत्तर: $\frac{7}{57}$

Step I: माना सामग्री का घनत्व $\rho$ है। R त्रिज्या के कटे हुए छोटे गोले का द्रव्यमान $m = \rho \cdot \frac{4}{3}\pi R^3$ है।
Step II: मूल बड़े गोले की त्रिज्या 2R है। इसका द्रव्यमान $M = \rho \cdot \frac{4}{3}\pi (2R)^3 = 8 \left(\rho \frac{4}{3}\pi R^3\right) = 8m$ है।
Step III: Y-अक्ष (जो इसके केंद्र से होकर गुजरता है) के परितः मूल बड़े गोले के MOI की गणना करें: $I_{orig} = \frac{2}{5} M (2R)^2 = \frac{2}{5} (8m) (4R^2) = \frac{64}{5} mR^2$।
Step IV: छोटे गोले की त्रिज्या R है और इसका केंद्र Y-अक्ष से R दूरी पर स्थित है। समानांतर अक्ष प्रमेय का उपयोग करके Y-अक्ष के परितः इसके MOI की गणना करें: $I_{small} = I_{cm} + m(R)^2 = \frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2$।
Step V: शेष भाग का MOI है $I_{rest} = I_{orig} - I_{small} = \frac{64}{5}mR^2 - \frac{7}{5}mR^2 = \frac{57}{5}mR^2$।
Step VI: छोटे गोले के MOI और शेष भाग के MOI का अनुपात $\frac{I_{small}}{I_{rest}} = \frac{\frac{7}{5}mR^2}{\frac{57}{5}mR^2} = \frac{7}{57}$ है।