
A steel tape 1m long is correctly calibrated for a temperature of $27.0^\circ C$. The length of a steel rod measured by this tape is found to be 63.0cm on a hot day when the temperature is $45.0^\circ C$. What is the actual length of the steel rod on that day? What is the length of the same steel road on a day when the temperature is $27.0^\circ C$ ? Coefficient of linear expansion of steel $ = 1.20 \times {10^{ - 5}}{K^{ - 1}}$.
Answer
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Hint: Increase in length during linear thermal expansion is given by ${l_2} = {l_1}\left( {1 + \alpha \Delta T} \right)$, where ${l_1}$ and ${l_2}$ are initial and final lengths, $\Delta T$ is change in temperature and $\alpha $ is coefficient of linear expansion.
Actual length of the rod is determined after calculating the length of steel tape at $45.0^\circ C$.
Complete step by step solution:
As given in the question,
Length of the steel tape at temperature $T = 27^\circ C$ is ${l_1} = 1m = 100cm$
At temperature ${T_1} = 45^\circ C$, the length of the steel rod, $l = 63cm$
Coefficient of linear expansion of steel, $\alpha = 1.2 \times {10^{ - 5}}{K^{ - 1}}$
Let ${l_1}$ be the actual length of the steel tape and ${l_2}$ be the length of the steel tape at $45^\circ C$
Now, as we know that increase in length during linear thermal expansion is given by ${l_2} = {l_1}\left( {1 + \alpha \Delta T} \right)$ where, ${l_1}$ and ${l_2}$ are initial and final lengths, $\Delta T$ is change in temperature and $\alpha $ is coefficient of linear expansion.
Here, $\Delta T = {T_1} - T$ i.e. $\Delta T = 45^\circ C - 27^\circ C = 18^\circ C$.
Now, substituting the values of ${l_1}$, $\Delta T$ and $\alpha $ in the equation ${l_2} = {l_1}\left( {1 + \alpha \Delta T} \right)$, we get,
${l_2} = 100\left( {1 + 1.2 \times {{10}^{ - 5}} \times 18} \right) = 100.0216cm$
Length of $1cm$ mark at $27^\circ C$ on this scale, at $45^\circ C = 100.0216cm$
Length of $63cm$ measured by this tape at $45^\circ C = \dfrac{{100.0216}}{{100}} \times 63 = 63.0136cm$
Hence, the actual length of the steel rod on that day was $63.0136cm$.
$\therefore$ The length of the same steel rod on a day when the temperature is $27.0^\circ C = 63 \times 1 = 63cm$, as the steel tape is correctly calibrated for a temperature of $27.0^\circ C$.
Note:
Be careful while calculating the temperature difference. Always subtract initial temperature from the final temperature to avoid the mistake of sign change.
Actual length of the rod is determined after calculating the length of steel tape at $45.0^\circ C$.
Complete step by step solution:
As given in the question,
Length of the steel tape at temperature $T = 27^\circ C$ is ${l_1} = 1m = 100cm$
At temperature ${T_1} = 45^\circ C$, the length of the steel rod, $l = 63cm$
Coefficient of linear expansion of steel, $\alpha = 1.2 \times {10^{ - 5}}{K^{ - 1}}$
Let ${l_1}$ be the actual length of the steel tape and ${l_2}$ be the length of the steel tape at $45^\circ C$
Now, as we know that increase in length during linear thermal expansion is given by ${l_2} = {l_1}\left( {1 + \alpha \Delta T} \right)$ where, ${l_1}$ and ${l_2}$ are initial and final lengths, $\Delta T$ is change in temperature and $\alpha $ is coefficient of linear expansion.
Here, $\Delta T = {T_1} - T$ i.e. $\Delta T = 45^\circ C - 27^\circ C = 18^\circ C$.
Now, substituting the values of ${l_1}$, $\Delta T$ and $\alpha $ in the equation ${l_2} = {l_1}\left( {1 + \alpha \Delta T} \right)$, we get,
${l_2} = 100\left( {1 + 1.2 \times {{10}^{ - 5}} \times 18} \right) = 100.0216cm$
Length of $1cm$ mark at $27^\circ C$ on this scale, at $45^\circ C = 100.0216cm$
Length of $63cm$ measured by this tape at $45^\circ C = \dfrac{{100.0216}}{{100}} \times 63 = 63.0136cm$
Hence, the actual length of the steel rod on that day was $63.0136cm$.
$\therefore$ The length of the same steel rod on a day when the temperature is $27.0^\circ C = 63 \times 1 = 63cm$, as the steel tape is correctly calibrated for a temperature of $27.0^\circ C$.
Note:
Be careful while calculating the temperature difference. Always subtract initial temperature from the final temperature to avoid the mistake of sign change.
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