Q133 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Physics

An object of mass $15 \mathrm{~kg}$ is attached to the end of a metal wire of unstretched length $1.0 \mathrm{~m}$. The object is then whirled in a vertical circle with an angular velocity of 4 $\mathrm{rad} / \mathrm{s}$ at the bottom of the circle. If the cross sectional area of the wire is $0.05 \mathrm{~cm}^2$ and Young's modulus of metal is $2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$, then the elongation of the wire when the mass is at the lowest point of its path (Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

  • A
    $0.27 \mathrm{~mm}$
  • Correct answer
    B
    $0.39 \mathrm{~mm}$
  • C
    $0.55 \mathrm{~mm}$
  • D
    $0.25 \mathrm{~mm}$

Solution

When the mass is placed at the position of the vertical circle the force on the mass is $\mathrm{F}=\mathrm{mg}+\mathrm{m} \omega^2 1$ $$ \begin{aligned} & \text { Young's modulus }=\frac{\text { Stress }}{\text { Strain }} \\ & \mathrm{Y}=\frac{\mathrm{F} / \mathrm{A}}{\frac{\Delta \mathrm{l}}{\mathrm{l}}} \\ & \mathrm{Y}=\frac{\mathrm{F}}{\mathrm{A}} \times \frac{1}{\Delta \mathrm{l}} \end{aligned} $$ The elongation of the wire, $$ \begin{aligned} & \Delta \mathrm{l}=\frac{\mathrm{Fl}}{\mathrm{AY}}=\frac{\mathrm{mgl} \times \mathrm{m}^2 \mathrm{l}^2}{\mathrm{AY}}=\frac{15 \times 10 \times 1+10 \times(4)^2 \times(1)^2}{0.05 \times 2 \times 10^{11} \times 10^{-4}} \\ & \approx 0.31 \mathrm{~mm} \end{aligned} $$

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

Relevant Questions