A, B and C are disc, solid sphere and spherical shell respectively with same radii and masses. These masses are placed as shown in figure.

The moment of inertia of the given system about PQ is fracmathrmx15mathrmI,$\frac{\mathrm{x}}{15}\mathrm{I},$ where I is the moment of inertia of the disc about its diameter. The value of x is

### Related Formula Parallel Axis Theorem: I_textaxis = I_textcom + M R^2$$I_{\text{axis}} = I_{\text{com}} + M R^2$$ Standard Moments of Inertia about center of mass: - Disc about diameter: I_textdisc,dia = fracMR^24$I_{\text{disc,dia}} = \frac{MR^2}{4}$ - Solid sphere: I_textsphere = frac25MR^2$I_{\text{sphere}} = \frac{2}{5}MR^2$ - Spherical shell: I_textshell = frac23MR^2$I_{\text{shell}} = \frac{2}{3}MR^2$ ### Core Logic The axis of rotation PQ passes through the center of the top disc (A) along its diameter. - Top disc (A): I_A = fracMR^24$$I_A = \frac{MR^2}{4}$$ - Bottom-left solid sphere (B): Center lies at distance R$R$ from the axis PQ. I_B = I_textcom + M R^2 = frac25MR^2 + MR^2 = frac75MR^2$$I_B = I_{\text{com}} + M R^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$$ - Bottom-right spherical shell (C): Center lies at distance R$R$ from the axis PQ. I_C = I_textcom + M R^2 = frac23MR^2 + MR^2 = frac53MR^2$$I_C = I_{\text{com}} + M R^2 = \frac{2}{3}MR^2 + MR^2 = \frac{5}{3}MR^2$$ ### Step 1: Calculate Total System Moment of Inertia Sum the contributions: I_textPQ = I_A + I_B + I_C$$I_{\text{PQ}} = I_A + I_B + I_C$$ I_textPQ = fracMR^24 + frac75MR^2 + frac53MR^2$$I_{\text{PQ}} = \frac{MR^2}{4} + \frac{7}{5}MR^2 + \frac{5}{3}MR^2$$ To add the fractions, find a common denominator (60$60$): I_textPQ = left( frac15 + 84 + 10060 right) MR^2 = frac19960 MR^2$$I_{\text{PQ}} = \left( \frac{15 + 84 + 100}{60} \right) MR^2 = \frac{199}{60} MR^2$$ ### Step 2: Express in terms of standard Disc Moment We are given I = fracMR^24 implies MR^2 = 4I$I = \frac{MR^2}{4} \implies MR^2 = 4I$. Substitute this in the expression: I_textPQ = frac19960 (4I) = frac19915 I$$I_{\text{PQ}} = \frac{199}{60} (4I) = \frac{199}{15} I$$ Comparing with I_textPQ = fracx15 I$I_{\text{PQ}} = \frac{x}{15} I$ yields x = 199$x = 199$. ### Pattern Recognition Sees: Composite body consisting of three standard symmetric shapes about a tangent/offset axis. Shortcut: Sum the central inertia terms and the offset terms separately. Offset masses are only B and C, so the offset sum is 2MR^2$2MR^2$. The central sum is (1/4 + 2/5 + 2/3)MR^2$(1/4 + 2/5 + 2/3)MR^2$. Adding these directly yields the combined fractional factor of 199/60$199/60$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion