A wire of area of cross-section ${10^{ - 6}}{m^2}$ is increased in length by $0.1\%$. The tension produced is $1000 N$. The Young's modulus of wire is
${10^{12}}\,N/{m^2}$
${10^{11}}\,N/{m^2}$
${10^{10}}\,N/{m^2}$
${10^9}\,N/{m^2}$
The ratio of diameters of two wires of same material is $n : 1$. The length of wires are $4\, m$ each. On applying the same load, the increase in length of thin wire will be
The area of cross section of a steel wire $(Y = 2.0 \times {10^{11}}N/{m^2})$ is $0.1\;c{m^2}$. The force required to double its length 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
What is Young’s modulus ? Explain. and Give its unit and dimensional formula.
What is the percentage increase in length of a wire of diameter $2.5 \,mm$, stretched by a force of $100 \,kg$ wt is .................. $\%$ ( Young's modulus of elasticity of wire $=12.5 \times 10^{11} \,dyne / cm ^2$ )