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A brass wire $1.8 \mathrm{~m}$ long at $27^{\circ} \mathrm{C}$ is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of $-39^{\circ} \mathrm{C}$, what is the tension developed in the wire, if its diameter is $2.0 \mathrm{~mm}$ ? Coefficient of linear expansion of brass $=2.0 \times 10^{-5} \mathrm{~K}^{-1}$, Young's modulus of brass $=0.91 \times 10^{11} \mathrm{~Pa}$.
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Here, $l=1.8 \mathrm{~m}, T_1=27^{\circ} \mathrm{C}, T_2=-39^{\circ} \mathrm{C}$,
$\begin{aligned}&r=1 \mathrm{~mm}=10^{-3} \mathrm{~m} \\&\alpha=2 \times 10^{-5} /{ }^{\circ} \mathrm{C}\end{aligned} \quad \mathrm{Y}=0.91 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
$\because \quad \mathrm{Y}=\frac{\mathrm{Fl}}{\mathrm{A} \Delta l} \Rightarrow \Delta l=\frac{\mathrm{Fl}}{\mathrm{AY}}$
Also, $\Delta l=\alpha l \Delta T \quad \therefore \quad \frac{\mathrm{Fl}}{\mathrm{AY}}=\alpha l \Delta T$
$\therefore \quad F=A Y \alpha \Delta T=\pi \mathrm{r}^2 \mathrm{Y} \alpha\left(T_2-T_1\right)$
$F=\frac{22}{7} \times\left(10^{-3}\right)^2 \times 0.91 \times 10^{11} \times 2$
$=-3.77 \times 10^2 \mathrm{~N} \quad \times 10^{-5}(-39-27)$
The negative sign indicates that the force is inwards leading to contraction of the wire.
$\begin{aligned}&r=1 \mathrm{~mm}=10^{-3} \mathrm{~m} \\&\alpha=2 \times 10^{-5} /{ }^{\circ} \mathrm{C}\end{aligned} \quad \mathrm{Y}=0.91 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
$\because \quad \mathrm{Y}=\frac{\mathrm{Fl}}{\mathrm{A} \Delta l} \Rightarrow \Delta l=\frac{\mathrm{Fl}}{\mathrm{AY}}$
Also, $\Delta l=\alpha l \Delta T \quad \therefore \quad \frac{\mathrm{Fl}}{\mathrm{AY}}=\alpha l \Delta T$
$\therefore \quad F=A Y \alpha \Delta T=\pi \mathrm{r}^2 \mathrm{Y} \alpha\left(T_2-T_1\right)$
$F=\frac{22}{7} \times\left(10^{-3}\right)^2 \times 0.91 \times 10^{11} \times 2$
$=-3.77 \times 10^2 \mathrm{~N} \quad \times 10^{-5}(-39-27)$
The negative sign indicates that the force is inwards leading to contraction of the wire.
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