9. Calculate the force required to double the length of a steel wire of area of cross section 5 × 10??m².(Y=2 × 10¹¹ Pa) (a) 10?? N (b) 10? (c) 10? N (d) 2 N
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The area of a cross-section of steel wire is $0.1 \mathrm{~cm}^{2}$ and Young's modulus of steel is $2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$. The force required to stretch by $0.1 \%$ of its length is (a) $\frac{\left[\left(Y_{1} Y_{2}\right) A\right]}{\left[2\left(Y_{1} L_{2}+Y_{2} L_{1}\right)\right]}$ (b) $\frac{\left[\left(Y_{1} Y_{2}\right) A\right]}{\left(L_{1} L_{2}\right)^{1 / 2}}$ (c) $4000 \mathrm{~N}$ (d) $5000 \mathrm{~N}$
On increasing the length by $0.5 \mathrm{~mm}$ in a steel wire of length $2 \mathrm{~m}$ and area of cross-section $2 \mathrm{~mm}^{2}$ the force required is..... $\left[\mathrm{Y}\right.$ for steel $\left.=2.2 \times 10^{11} \mathrm{~N} / \mathrm{m}\right]$ (A) $1.1 \times 10^{5} \mathrm{~N}$ (B) $1.1 \times 10^{4} \mathrm{~N}$ (C) $1.1 \times 10^{3} \mathrm{~N}$ (D) $1.1 \times 10^{2} \mathrm{~N}$
On increasing the length by $0.5 \mathrm{~mm}$ in a steel wire of length $2 \mathrm{~m}$ and area of cross-section $2 \mathrm{~mm}^{2}$, the force required is $\left[Y\right.$ for steel $\left.=2.2 \times 10^{11} \mathrm{Nm}^{-2}\right]$ (a) $1.1 \times 10^{5} \mathrm{~N}$ (b) $1.1 \times 10^{4} \mathrm{~N}$ (c) $1.1 \times 10^{3} \mathrm{~N}$ (d) $1.1 \times 10^{2} \mathrm{~N}$
Properties of Solids
Round 1
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Physics: Principles with Applications
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