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An iron rod appears to be $1m$ long when measured by a brass scale that is correct at both the rods at the time of observation being at ${20^0}C$ . Find the length of the iron rod at ${100^0}C$. ( ${\alpha _{iron}} = 1.2 \times {10^{ - 5}}{C^{ - 1}}$ and ${\alpha _{brass}} = 2 \times {10^{ - 5}}{C^{ - 1}}$ )

Last updated date: 25th Jul 2024
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Hint:On heating or raising the temperature of an iron rod its length gets increased due to heating effect on its molecular level and the variation of change in length with temperature is given as $\Delta L = L\alpha \Delta T$ where, $\alpha $ is called coefficient of thermal linear expansion.

Complete step by step answer:
Let $L$ be the length of rod at ${20^0}C$ and increment in length be denoted as $\Delta L = L\alpha \Delta T$ where we know,
$\Delta T = {20^0}C$
And total appeared length of rod is given by:
$L + \Delta L = 1$
Put the value of $\Delta L = L\alpha \Delta T$ in above equation with given value of ${\alpha _{iron}} = 1.2 \times {10^{ - 5}}{C^{ - 1}}$
We get,
$L(1 + 1.2 \times {10^{ - 5}} \times 20) = 1$
$\Rightarrow L = \dfrac{1}{{(1 + 1.2 \times {{10}^{ - 5}} \times 20)}}$
$\Rightarrow L = 0.9997\,m$
Hence, the magnitude of length of rod at ${0^0}C$ is $L = 0.9997\,m$

Now, we will find the length of rod at ${100^0}C$
Since we know,
$L' = L(1 + \alpha \Delta T)$ Where,
$\Rightarrow L = 0.9997\,m$
$\Rightarrow \Delta T = 100 - 20 = {80^0}C$
$\Rightarrow {\alpha _{brass}} = 2 \times {10^{ - 5}}{C^{ - 1}}$
Putting the values of above parameters in the equation $L' = L(1 + \alpha \Delta T)$
We get,
$L' = 0.9997(1 + 2 \times {10^{ - 5}} \times 80)$
$\therefore L' = 1.00136\,m$

Hence, the magnitude of length of rod at ${100^0}C$ is $L' = 1.00136\,m$.

Note: It should be remembered that the rod expands linearly so the linear expansion of coefficient is used and the change in temperature is taken from initial temperature and final temperature. In case of expanding the rod the net increment of the length of the rod is directly proportional to the increase in temperature.