Column$-II$ is related to Column$-I$. Join them appropriately :
Column $-I$ | Column $-II$ |
$(a)$ When temperature raised Young’s modulus of body | $(i)$ Zero |
$(b)$ Young’s modulus for air | $(ii)$ Infinite |
$(iii)$ Decreases | |
$(iv)$Increases |
$(a-ii),(b-i)$
$(a-iii),(b-i)$
$(a-ii),(b-iv)$
$(a-iii),(b-ii)$
The force required to stretch a steel wire of $1\,c{m^2}$ cross-section to $1.1$ times its length would be $(Y = 2 \times {10^{11}}\,N{m^{ - 2}})$
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 $N$.
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}]$
Four identical hollow cylindrical columns of mild steel support a big structure of mass $50 \times 10^{3} {kg}$, The inner and outer radii of each column are $50\; {cm}$ and $100 \;{cm}$ respectively. Assuming uniform local distribution, calculate the compression strain of each column. [Use $\left.{Y}=2.0 \times 10^{11} \;{Pa}, {g}=9.8\; {m} / {s}^{2}\right]$
A wire elongates by $l$ $mm$ when a load $W$ is hanged from it. If the wire goes over a pulley and two weights $W$ each are hung at the two ends, the elongation of the wire will be (in $mm$)