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A student records the initial length $l$, change in temperature $\Delta T$ and change in length $\Delta l$ of a rod as follows :
| S.No. | $l(m)$ | $\Delta T{(^o}C)$ | $\Delta l(m)$ |
| $(1)$ | $2$ | $10$ | $4\times 10^{-4}$ |
| $(2)$ | $1$ | $10$ | $4\times 10^{-4}$ |
| $(3)$ | $2$ | $20$ | $2\times 10^{-4}$ |
| $(4)$ | $3$ | $10$ | $6\times 10^{-4}$ |
If the first observation is correct, what can you say about observations $2,\,3$ and $4$.
Solution
From first observation,
$\alpha=\frac{\Delta l}{l \Delta \mathrm{T}}$
$=\frac{4 \times 10^{-4}}{2 \times 10}$
$=2 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$
From second observation,
$\Delta l= \alpha l \Delta \mathrm{T}$
$\therefore \Delta l=2 \times 10^{-5} \times 1 \times 10$
$=2 \times 10^{-4} \mathrm{~m} \neq 4 \times 10^{-4} \mathrm{~m}$
$\therefore$ Thus, the observation is incorrect.
From third observation,
$\Delta l=\alpha l \Delta \mathrm{T}$
$=2 \times 10^{-5} \times 2 \times 20$
$=8 \times 10^{-4} \mathrm{~m} \neq 2 \times 10^{-4} \mathrm{~m}$
$\therefore$ Thus, the observation is incorrect.
From fourth observation,
$\Delta l=\alpha l \Delta \mathrm{T}$
$=2 \times 10^{-5} \times 3 \times 10$
$=6 \times 10^{-4} \mathrm{~m}=6 \times 10^{-4} \mathrm{~m}$
Thus, the observation is true.
