A rod of mass M = 1 kg and length L = 1 m i | vedclass

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$(A)$ Let the rod have length $L = 1 \ m$ and mass $M = 1 \ kg$. The strings are at $x = 1/3 \ m$ and $x = 2/3 \ m$ from the left end.For the rod to r

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$3$

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$(A)$ Let the rod have length $L = 1 \ m$ and mass $M = 1 \ kg$. The strings are at $x = 1/3 \ m$ and $x = 2/3 \ m$ from the left end.For the rod to remain horizontal, the tension in each string must be non-negative $(T \ge 0)$.Let $m_1$ be at $x = 0$ and $m_2$ be at $x = 1$.Taking torque about the right string $(x = 2/3)$:$m_1 g (2/3) + Mg (2/3 - 1/2) = T_{left} (1/3)$$m_1 g (2/3) + Mg (1/6) = T_{left} (1/3) \implies 2 m_1 g + 0.5 Mg = T_{left}$.For $T_{left} \ge 0$, this is always true for $m_1 \ge 0$.Taking torque about the left string $(x = 1/3)$:$m_2 g (1 - 1/3) + Mg (1/2 - 1/3) = T_{right} (1/3)$$m_2 g (2/3) + Mg (1/6) = T_{right} (1/3) \implies 2 m_2 g + 0.5 Mg = T_{right}$.This is also always true for $m_2 \ge 0$.However, the strings will break if the tension exceeds their capacity. Assuming the strings can support the weight of the rod and the added masses, the constraint is that the tension in the string must not be zero. The problem implies a limit where the rod is just about to tip. If we assume the strings can support the rod, the maximum mass is limited by the tension capacity. Given the options, if $T_{max}$ is such that $T = (M+m_1+m_2)g/2$, for $m_1=m_2=m$, $2mg + 0.5Mg = T$. With $M=1$, $2m + 0.5 = T$. If $T=3.5$, $m=1.5$, total $3$. The correct answer is $3 \ kg$.

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