Heat and Thermodynamics 6 Question 30
35. A $5 m$ long cylindrical steel wire with radius $2 \times 10^{-3} m$ is suspended vertically from a rigid support and carries a bob of mass $100 kg$ at the other end. If the bob gets snapped, calculate the change in temperature of the wire ignoring losses. (For the steel wire : Young’s modulus $=2.1 \times 10^{11} Pa ;$ Density $=7860 kg / m^{3} ;$ Specific heat $=420$ $J / kg-K$ ).
(2001, 5M)
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Answer:
Correct Answer: 35. $4.568 \times 10^{-3}{ }^{\circ} \mathrm{C}$
Solution:
- Given,
Length of the wire, $l=5 m$
Radius of the wire, $r=2 \times 10^{-3} m$
Density of wire, $\rho=7860 kg / m^{3}$
Young’s modulus,
$$ Y=2.1 \times 10^{11} N / m^{2} $$
and specific heat, $s=420 J / kg-K$
Mass of wire, $m=$ (density) (volume)
$$ \begin{aligned} & =(\rho)\left(\pi r^{2} l\right) \\ & =(7860)(\pi)\left(2 \times 10^{-3}\right)^{2}(5) kg=0.494 kg \end{aligned} $$
Elastic potential energy stored in the wire,
$$ \begin{aligned} & U=\frac{1}{2} \text { (stress) (strain) (volume) } \\ & {\left[\because \frac{\text { Energy }}{\text { Volume }}=\frac{1}{2} \times \text { stress } \times \text { strain }\right]} \\ & \text { or } \\ & U=\frac{1}{2}\left(\frac{M g}{\pi r^{2}}\right)\left(\frac{\Delta l}{l}\right)\left(\pi r^{2} l\right) \\ & =\frac{1}{2}(M g) \cdot \Delta l \quad\left(\Delta l=\frac{F l}{A Y}\right) \\ & =\frac{1}{2}(M g) \frac{(M g l)}{\left(\pi r^{2}\right) Y}=\frac{1}{2} \frac{M^{2} g^{2} l}{\pi r^{2} Y} \end{aligned} $$
Substituting the values, we have
$$ \begin{aligned} U & =\frac{1}{2} \frac{(100)^{2}(10)^{2}(5)}{(3.14)\left(2 \times 10^{-3}\right)^{2}\left(2.1 \times 10^{11}\right)} J \\ & =0.9478 J \end{aligned} $$
When the bob gets snapped, this energy is utilised in raising the temperature of the wire.
$$ \begin{aligned} & \text { So, } \\ & \therefore \quad \Delta \theta=\frac{U}{m s}=\frac{0.9478}{0.494(420)}{ }^{\circ} C \text { or K } \\ & \Delta \theta=4.568 \times 10^{-3}{ }^{\circ} C \end{aligned} $$
