$A$ 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 $(T_1/T_2)$.

$A$ metre scale is balanced on a knife edge at its centre. When two coins,each of mass $10\, g$,are placed one on top of the other at the $10.0\, cm$ mark,the scale is found to be balanced at the $40.0\, cm$ mark. The mass of the metre scale is found to be $x \times 10^{-2}\, kg$. The value of $x$ is

$A$ $3\; m$ long ladder weighing $20\; kg$ leans on a frictionless wall. Its feet rest on the floor $1\; m$ from the wall as shown in the figure. Find the reaction forces of the wall and the floor.

For a system to be in equilibrium,the torques acting on it must balance. This is true only if the torques are taken about

$A$ car weighs $1800 \, kg$. The distance between its front and back axles is $1.8 \, m$. Its centre of gravity is $1.05 \, m$ behind the front axle. The force exerted by the level ground on each back wheel is ....... $N$.

$A$ 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?