Find the value of $T_1$ (in $N$),if the system is in equilibrium and the mass of the rod is $10 \ kg$ and its length is $10 \ m$. $[g = 10 \ m/s^2]$ (in $N$)

$A$ uniform meter scale is balanced at the $30 \ cm$ mark on a knife edge,when weights of $60 \ N$ and $10 \ N$ are suspended at the $10 \ cm$ and $80 \ cm$ marks,respectively. The mass of the meter scale is: $(g = 10 \ m/s^2)$ (in $kg$)

$A$ false balance has unequal arms. An object weighs $X$ when placed in one pan and $Y$ when placed in the other pan. The true weight $W$ of the object is equal to:

In an experiment with a beam balance,an unknown mass $m$ is balanced by two known masses of $16\, kg$ and $4\, kg$ as shown in the figure. The value of the unknown mass $m$ is ....... $kg$.

$A$ 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 $\text{ g-wt}$)

$A$ wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However,the particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium?
How would you set a half wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?