For a thin non-uniform disc of mass M and radius R the area density of mass, $σ$ as a function of distance r from its centre is given as $σ = σ_{0} (1 - \frac{r}{R})$ . This disc is placed on a rough horizontal surface (coefficient of friction $= μ$ ) and a constant horizontal force $F = Mg$ , applied along the center of mass. The disc rolls without slipping on the surface, then choose the correct statement(s).[Moment of inertia of disc about its own axis is $\frac{3 M R^{2}}{10}]$

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The angular acceleration of the disc is $\frac{5 g}{13 R}$

The value of $μ$ may be 0.3

The linear acceleration of the disc is $\frac{5 g}{13}$

The linear acceleration of the disc is $\frac{10 g}{13}$

answer is A, C.

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Detailed Solution

$\begin{array}{l} M = \int_{0}^{R} σ_{0} (1 - \frac{r}{R}) 2 π r d r = {2 π σ_{0} (\frac{r^{2}}{2} - \frac{r^{3}}{3 R})|}_{0}^{R} \\ = \frac{2 π σ_{0} R^{2}}{6} = \frac{π σ_{0} R^{2}}{3} = M \\ I = \int_{0}^{R} σ_{0} (1 - \frac{r}{R}) 2 π r^{3} d r = 2 π σ_{0} {(\frac{r^{4}}{4} - \frac{r^{5}}{R})}_{0}^{R} \\ = 2 π σ_{0} \frac{R^{4}}{20} = \frac{π σ_{0} R^{4}}{10} I = \frac{3 M}{R^{2}} \cdot \frac{R^{4}}{10} = \frac{3 M R^{2}}{10} F - f = M a \end{array}$

$\begin{array}{l} f R = \frac{3}{10} M R^{2}, \frac{a}{R} f = \frac{3}{10} M a \Rightarrow F = (1 + \frac{3}{10}) M a = \frac{13}{10} M a \\ \Rightarrow a = \frac{10 F}{13 M} = \frac{10 g}{13} f = \frac{3}{10} M \times \frac{10 F}{13} = \frac{3 M g}{13} \\ μ M g > \frac{3 M g}{13} \Rightarrow μ > \frac{3}{13} \end{array}$