In the figure,two blocks M and m are tied tog | vedclass

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(D) For the block $m$ hanging vertically,the forces acting are gravity $mg$ downwards and tension $T$ upwards. If the system is in equilibrium,$T = mg

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Both $(A)$ and $(C)$

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(D) For the block $m$ hanging vertically,the forces acting are gravity $mg$ downwards and tension $T$ upwards. If the system is in equilibrium,$T = mg$.For the block $M$ on the horizontal surface,the maximum static friction force is $f_{max} = \mu N = \mu Mg$.Case $1$: If $m Case $2$: If $m > \mu M$,the pulling force $mg$ is greater than the maximum static friction $f_{max} = \mu Mg$. The system will accelerate. The equations of motion are: $mg - T = ma$ and $T - \mu Mg = Ma$. Adding these,$mg - \mu Mg = (M + m)a$,so $a = \frac{(m - \mu M)g}{M + m}$. Substituting $a$ into the first equation,$T = m(g - a) = m(g - \frac{(m - \mu M)g}{M + m}) = \frac{mMg + m\mu Mg}{M + m} = \frac{mM(1 + \mu)g}{M + m}$. Since $a > 0$,$T = mg - ma 0$,$T > \mu Mg$. Thus,$\mu Mg < T < mg$.

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