$A$ force $F$ is needed to break a copper wire having radius $R$. The force needed to break a copper wire of radius $2R$ will be

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 $mm$)? $(g = 10 \,m s^{-2})$

Two identical wires of rubber and iron are stretched by the same weight. The number of atoms in the iron wire will be:

Two blocks of masses $1 \, kg$ and $2 \, kg$ are connected by a metal wire going over a smooth pulley as shown in the figure. The breaking stress of the metal is $2 \times 10^9 \, N/m^2$. What should be the minimum radius of the wire used if it is not to break? Take $g = 10 \, m/s^2$.

If the breaking stress of a wire is $3.18 \times 10^{10} \, N/m^2$,what should be its minimum radius so that the wire does not break? (Refer to the figure for the masses attached to the pulley system,assume $g = 10 \, m/s^2$)

In the given figure,two elastic rods $A$ and $B$ are rigidly joined to end supports. $A$ small mass $m$ is moving with velocity $v$ between the rods. All collisions are assumed to be elastic and the surface is frictionless. The time period of the small mass $m$ will be: [Here,an elastic rod may be treated as a spring of spring constant $k = \frac{YA}{L}$]