Two blocks of masses $1 \,kg$ and $2 \,kg$ are connected by a metal wire going over a smooth pulley. The breaking stress of metal is $\frac{40}{3 \pi} \times 10^6 \,N m^{-2}$. What should be the minimum radius of the wire used if it should not break (in $mm$)? $(g = 10 \,m s^{-2})$

$A$ string $AB$ of unstretched length $L$ is stretched by applying a force $F$ at the mid-point $C$ such that the segments $AC$ and $BC$ make an angle $\theta$ with $AB$ as shown in the figure. The string may be considered as an elastic element with a force to elongation ratio $K$. The force $F$ is given by

$A$ body of mass $10 \ kg$ is attached to a wire of $0.3 \ m$ length. The breaking stress is $4.8 \times 10^7 \ N \ m^{-2}$. The area of cross-section of the wire is $10^{-6} \ m^2$. The maximum angular velocity with which it can be rotated in a horizontal circle is

$A$ substance breaks down by a stress of $10^6 \ N/m^2$. If the density of the material of the wire is $3 \times 10^3 \ kg/m^3$,then the length of the wire of the substance which will break under its own weight when suspended vertically,is ......... $m$.

If the breaking force for a given wire is $F$, and the thickness of the wire is doubled, then the breaking force will be:

$A$ uniform wire of length $l$ and weight $w$ is suspended from the roof with a weight $W$ attached at the other end. The stress in the wire at a distance $l/3$ from the top is given by $(\frac{W}{A} + \gamma \frac{w}{A})$, where $A$ is the cross-sectional area of the wire. The value of $\gamma$ is . . . . . . .