Properties of Matter 1 Question 21
21. A thin rod of negligible mass and area of cross-section $4 \times 10^{-6} m^{2}$, suspended vertically from one end, has a length of $0.5 m$ at $100^{\circ} C$. The rod is cooled to $0^{\circ} C$, but prevented from contracting by attaching a mass at the lower end. Find
(a) this mass and
(b) the energy stored in the rod, given for the rod. Young’s modulus $=10^{11} N / m^{2}$, coefficient of linear expansion $=10^{-5} K^{-1}$ and $g=10 m / s^{2}$.
(1997 C, 5M)
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Solution:
- (a) The change in length due to decrease in temperature,
$$ \begin{aligned} \Delta l _1 & =L \alpha \Delta \theta=(0.5)\left(10^{-5}\right)(0-100) \\ \Delta l _1 & =-0.5 \times 10^{-3} m \end{aligned} $$
Negative sign implies that length is decreasing. Now, let $M$ be the mass attached to the lower end. Then, change in length due to suspension of load is
$$ \begin{array}{rlrl} \Delta l _2=\frac{(M g) L}{A Y} & =\frac{(M)(10)(0.5)}{\left(4 \times 10^{-6}\right)\left(10^{11}\right)} \\ \Delta l _2 & =\left(1.25 \times 10^{-5}\right) M \\ \Delta l _1+\Delta l _2 & =0 \\ \text { or }\left(1.25 \times 10^{-5}\right) M & =\left(0.5 \times 10^{-3}\right) \\ \therefore & & M & =\left(\frac{0.5 \times 10^{-3}}{1.25 \times 10^{-5}}\right) kg \\ & \text { or } & M & =40 kg \end{array} $$
(b) Energy stored,
At $0^{\circ} C$ the natural length of the wire is less than its actual length; but since a mass is attached at its lower end, an elastic potential energy is stored in it. This is given by
$$ U=\frac{1}{2}\left(\frac{A Y}{L}\right)(\Delta l)^{2} $$
Here, $\Delta l=\left|\Delta l _1\right|=\Delta l _2=0.5 \times 10^{-3} m$
Substituting the values,
$$ U=\frac{1}{2}\left(\frac{4 \times 10^{-6} \times 10^{11}}{0.5}\right)\left(0.5 \times 10^{-3}\right)^{2}=0.1 J $$
NOTE
Comparing the equation
$$ Y=\frac{F / A}{\Delta / / L} \text { or } F=\left(\frac{A Y}{L}\right) \Delta l $$
with the spring equation $F=K \cdot \Delta x$, we find that equivalent spring constant of a wire is $k=\left(\frac{A Y}{L}\right)$
Therefore, potential energy stored in it should be
$$ U=\frac{1}{2} k(\Delta I)^{2}=\frac{1}{2}\left(\frac{A Y}{L}\right)(\Delta I)^{2} $$
