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A $100\,m$ long wire having cross-sectional area $6.25 \times 10^{-4}\,m ^2$ and Young's modulus is $10^{10}\,Nm ^{-2}$ is subjected to a load of $250\,N$, then the elongation in the wire will be :
To determine Young's modulus of a wire, the formula is $Y = \frac{F}{A}.\frac{L}{{\Delta L}}$ where $F/A$ is the stress and $L/\Delta L$ is the strain. The conversion factor to change $Y$ from $CGS$ to $MKS$ system is
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 $mm.$
One end of a uniform wire of length $L$ and of weight $W$ is attached rigidly to a point in the roof and a weight ${W_1}$ is suspended from its lower end. If $S$ is the area of cross-section of the wire, the stress in the wire at a height $3L/4$ from its lower end is
An iron rod of length $2m$ and cross section area of $50\,m{m^2}$, stretched by $0.5\, mm$, when a mass of $250\, kg$ is hung from its lower end. Young's modulus of the iron rod is
The ratio of lengths of two rods $A$ and $B$ of same material is $1 : 2$ and the ratio of their radii is $2 : 1$, then the ratio of modulus of rigidity of $A$ and $B$ will be