An aluminum rod (Young's modulus $ = 7 \times {10^9}\,N/{m^2})$ has a breaking strain of $0.2\%$. The minimum cross-sectional area of the rod in order to support a load of ${10^4}$Newton's is

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 rigid bar of mass $15\,kg$ is supported symmetrically by three wire each of $2 \,m$ long. These at each end are of copper and middle one is of steel. Young's modulus of elasticity for copper and steel are $110 \times 10^9 \,N / m ^2$ and $190 \times 10^9 \,N / m ^2$ respectively. If each wire is to have same tension, ratio of their diameters will be …………

Young's moduli of the material of wires $A$ and $B$ are in the ratio of $1: 4$, while its area of cross sections are in the ratio of $1: 3$. If the same amount of load is applied to both the wires, the amount of elongation produced in the wires $A$ and $B$ will be in the ratio of

[Assume length of wires $A$ and $B$ are same]

An area of cross-section of rubber string is $2\,c{m^2}$. Its length is doubled when stretched with a linear force of $2 \times {10^5}$dynes. The Young's modulus of the rubber in $dyne/c{m^2}$ will be