Two exactly similar wires of steel and copper are stretched by equal forces. If the total elongation is $2 \,cm$,then how much is the elongation in steel and copper wire respectively? Given,$Y_{\text{steel}} = 20 \times 10^{11} \,dyne/cm^2$,$Y_{\text{copper}} = 12 \times 10^{11} \,dyne/cm^2$.

The length of a wire becomes $l_1$ and $l_2$ when $100\,N$ and $120\,N$ tensions are applied respectively. If $10l_2 = 11l_1$,the natural length of the wire will be $\frac{1}{x} l_1$. Here,the value of $x$ is ........

In steel,the Young's modulus and the strain at the breaking point are $2 \times 10^{11} \, N/m^2$ and $0.15$ respectively. The stress at the breaking point for steel is therefore:

Two wires each of radius $0.2\,cm$ and negligible mass, one made of steel and the other made of brass, are loaded as shown in the figure. The elongation of the steel wire is $.........\times 10^{-6}\,m$. [Young's modulus for steel $= 2 \times 10^{11}\,N/m^2$ and $g = 10\,m/s^2$]

$A$ metallic rod breaks when the strain produced is $0.2 \%$. The Young's modulus of the material of the rod is $7 \times 10^9 \,N/m^2$. The area of cross-section required to support a load of $10^4 \,N$ is:

$A$ brass rod of length $2\,m$ and cross-sectional area $2.0\,cm^2$ is attached end to end to a steel rod of length $L$ and cross-sectional area $1.0\,cm^2$. The compound rod is subjected to equal and opposite pulls of magnitude $5 \times 10^4\,N$ at its ends. If the elongations of the two rods are equal,then the length of the steel rod $(L)$ is ........... $m$ $(Y_{Brass}=1.0\times 10^{11}\,N/m^2$ and $Y_{Steel} = 2.0 \times 10^{11}\,N/m^2)$.