Two exactly similar wires of steel and copper are stretched by equal forces. If the difference in their elongations is $0.5$ cm, the elongation $(l)$ of each wire is ${Y_s}({\rm{steel}}) = 2.0 \times {10^{11}}\,N/{m^2}$${Y_c}({\rm{copper}}) = 1.2 \times {10^{11}}\,N/{m^2}$

One end of a horizontal thick copper wire of length $2 L$ and radius $2 R$ is welded to an end of another horizontal thin copper wire of length $L$ and radius $R$. When the arrangement is stretched by a applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is :

A wire of length $L$ and radius $r$ is clamped at one end. If its other end is pulled by a force $F$, its length increases by $l$. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become.

A wire of length $L$ and radius $r$  is clamped rigidly at one end. When the other end of the wire is pulled by a force $f$ its length increases by $l$. Another wire of the same material of length $2L$ and radius $2r$  is pulled by a force $2f$. Then find the increase in length of this wire.

A rod of length $1.05\; m$ having negligible mass is supported at its ends by two wires of steel (wire $A$) and aluminium (wire $B$) of equal lengths as shown in Figure. The cross-sectional areas of wires $A$ and $B$ are $1.0\; mm ^{2}$ and $2.0\; mm ^{2}$. respectively. At what point along the rod should a mass $m$ be suspended in order to produce $(a)$ equal stresses and $(b)$ equal strains in both steel and alumintum wires.