On increasing the length by $0.5\, mm$ in a steel wire of length $2\, m$ and area of cross-section $2\,m{m^2}$, the force required is $[Y$ for steel$ = 2.2 \times {10^{11}}\,N/{m^2}]$
$1.1 \times {10^5}\,N$
$1.1 \times {10^4}\,N$
$1.1 \times {10^3}\,N$
$1.1 \times {10^2}\,N$
Young's modules of material of a wire of length ' $L$ ' and cross-sectional area $A$ is $Y$. If the length of the wire is doubled and cross-sectional area is halved then Young's $modules$ will be :
A $5\, m$ long aluminium wire ($Y = 7 \times {10^{10}}N/{m^2})$ of diameter $3\, mm$ supports a $40\, kg$ mass. In order to have the same elongation in a copper wire $(Y = 12 \times {10^{10}}N/{m^2})$ of the same length under the same weight, the diameter should now be, in $mm.$
A copper wire of length $2.2 \;m$ and a steel wire of length $1.6\; m ,$ both of diameter $3.0 \;mm ,$ are connected end to end. When stretched by a load, the net elongation is found to be $0.70 \;mm$. Obtain the load applied in $N$.
Young's modulus depends upon
Which of the following curve represents the correctly distribution of elongation $(y)$ along heavy rod under its own weight $L \rightarrow$ length of rod, $x \rightarrow$ distance of point from lower end?