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8.Mechanical Properties of Solids
medium
The area of a cross-section of steel wire is $0.1\,\,cm^2$ and Young's modulus of steel is $2\,\times \,10^{11}\,\,N\,\,m^{-2}.$ The force required to stretch by $0.1\%$ of its length is ......... $N$.
A
$1000$
B
$2000$
C
$4000$
D
$5000$
Solution
Here,
$A=0.1 \mathrm{cm}^{2}=0.1 \times 10^{-4} \mathrm{m}^{2}$
$Y=2 \times 10^{11} \mathrm{N} / \mathrm{m}^{-2}$
$\frac{\Delta \mathrm{L}}{\mathrm{L}}=0.1 \%=\frac{0.1}{100}=0.1 \times 10^{-2}$
$\mathrm{As} \mathrm{Y}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \mathrm{L} / \mathrm{L}}$
$\therefore \quad F=Y \frac{\Delta L}{L} A$
$=2 \times 10^{11} \mathrm{Nm}^{-2} \times 0.1 \times 10^{-2} \times 0.1 \times 10^{-4} \mathrm{m}^{2}$
$=2 \times 10^{3} \mathrm{N}=2000 \mathrm{N}$
Standard 11
Physics
