Q133 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Physics

One end of a steel rod of radius $10.0 \mathrm{~mm}$ and length $50.0 \mathrm{~cm}$ is clamped on a horizontal table. The other end of the rod is pulled with a force of magnitude $10.0 \times \pi \mathrm{kN}$. This force is uniform across the flat surface of the rod and is perpendicular to it. The change in the length of the rod due to this applied force is (Use Young's modulus $=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$ )

  • Correct answer
    A
    0.25 mm
  • B
    0.75 mm
  • C
    0.50 mm
  • D
    1.0 mm

Solution

We have $\Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{YA}}$ $$ =\frac{10 \pi \times 10^3 \times 0.5}{2 \times 10^{11} \times \pi \times(0.01)^2}=2.5 \times 10^{-4} \mathrm{~m}=0.25 \mathrm{~mm} $$

Hence, the correct answer is option: 'A'.

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