TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Given, mass, radius of rigid body be $m$ and $R$ respectively, $g$ is acceleration due to gravity and $\theta$ is inclination. Let $I$ be the moment of inertia, then $I = n m R^{2}$ Taking force along inclined $m g \sin\theta - f = m a$. . . (i) where, $f$ is friction force and $a$ is acceleration. $\tau = I \alpha = f R$. . . (ii) where, $\tau =$ torque, $I = \text{ moment of inertia }$ $\text{ and } \alpha = \text{ angular acceleration } = \frac{a}{R} \text{ . }$ Substituting in Eq. (ii), we get $I \frac{a}{R} = f R$ $n m R^{2} \cdot \frac{a}{R} = f R$ $n m a = f$ . . . (iii) Substituting in Eq. (i), we get $\Rightarrow m g \sin\theta - n m a = m a$ $\Rightarrow g \sin\theta = a (n+1)$ $\Rightarrow a = \frac{g \sin\theta}{n+1}$ Substituting in Eq. (iii), we get $f = n m a$ $= n m \frac{g \sin\theta}{n+1}$ . . . (iv) (A) For ring, $I = m R^{2}$ $\therefore n = 1$ $\text{ and } f_{\text{ring}} = \frac{m g \sin\theta}{1+1} = \frac{m g \sin\theta}{2}$ (B) For solid sphere, $I = \frac{2}{5} m R^{2}$ $\therefore n = \frac{2}{5}$ $f_{\text{solid sphere}} = \frac{ \frac{2}{5} m g \sin\theta }{ \frac{2}{5} + 1 }$ $= \frac{2 m g \sin\theta}{7}$ $= \frac{m g \sin\theta}{3.5}$ $\therefore n = \frac{2}{5}$ $f_{\text{solid sphere}} = \frac{ \frac{2}{5} m g \sin\theta }{ \frac{2}{5} + 1 }$ $= \frac{2 m g \sin\theta}{7}$ $= \frac{m g \sin\theta}{3.5}$ $\text{ (C) For solid cylinder, } I = \frac{m R^{2}}{2}$ $\therefore n = \frac{1}{2}$ $\therefore f_{\text{solid cylinder}} = \frac{ \frac{1}{2} m g \sin\theta }{ \frac{1}{2} + 1 }$ $= \frac{m g \sin\theta}{3}$ $\therefore$ $n = \frac{1}{2}$ $f_{\text{solid cylinder}} = \frac{ \frac{1}{2} m g \sin\theta }{ \frac{1}{2} + 1 }$ $= \frac{m g \sin\theta}{3}$ (D) For hollow cylinder $I = M R^{2}$ $\therefore$ $n = 1$ $f_{\text{hollow cylinder}} = \frac{m g \sin\theta}{1+1}$ $= \frac{m g \sin\theta}{2}$ $\therefore$ Hence, $A \rightarrow 4, B \rightarrow 3, C \rightarrow 2, D \rightarrow 4$ is the correct.