$A$ uniform rod of length $200 \, cm$ and mass $500 \, g$ is balanced on a wedge placed at $40 \, cm$ mark. $A$ mass of $2 \, kg$ is suspended from the rod at $20 \, cm$ and another unknown mass $'m'$ is suspended from the rod at $160 \, cm$ mark as shown in the figure. Find the value of $'m'$ such that the rod is in equilibrium. $(g = 10 \, m/s^2)$

$A$ metre scale is supported on a wedge at its centre of gravity. $A$ body of weight '$W$' is suspended from the $20 \text{ cm}$ mark and another weight of $25 \text{ g-wt}$ is suspended from the $74 \text{ cm}$ mark to balance it, and the metre scale remains perfectly horizontal. Neglecting the weight of the metre scale, the weight of the body is: (in $\text{ g-wt}$)

$A$ uniform rod of mass $15\,kg$ and length $5\,m$ is held stationary with the help of a light string as shown. The tension in the string is ......... $N.$

$A$ physical balance has its arms of unequal length. $A$ body weighs $18 \,kg$ if kept in one pan and weighs $8 \,kg$ if kept in the other pan. The true weight of the body is ............. $kg$

$A$ uniform meter scale of mass $m$ is suspended by two vertical strings at its ends as shown in the figure. $A$ mass $m$ is placed at the $80 \ cm$ mark. Find the ratio of the tension forces in the strings $(T_1/T_2)$.

$A$ rod of mass $M = 1 \ kg$ and length $L = 1 \ m$ is suspended horizontally by two ideal strings as shown in the figure. The strings are attached at a distance of $1/3 \ m$ from each end. First, a mass $m_1$ is suspended from the left end while keeping the rod horizontal. Then, a second mass $m_2$ is suspended from the right end, again keeping the rod horizontal. What is the maximum total mass $(m_1 + m_2)$ that can be suspended in this way while maintaining the horizontal orientation of the rod?