
An aluminium rod which is correct at \[{5^ \circ }\], measures a certain distance as 88.42cm at ${35^ \circ }$. Determine the error in measuring the distance due to expansion of the rod. \[\alpha = 24 \times {10^{ - 6}}{C^{ - 1}}\]
A. 0.014cm
B. 0.054cm
C. 0.064cm
D. 0.044cm
Answer
568.5k+ views
Hint: There is a change in length of the rod when the temperature changes. The change in length is directly proportional to original length and the change in the temperature and alpha $\alpha $ is the coefficient of linear expansion.
Complete step by step answer:
When any object is heated it will experience a change in its length, it will be increased as through heating it will expand. That change will be directly proportional to the original length ${L_0}$ and change in the temperature $dt$. And is given by the formula
$dl = \alpha {L_o}dt$
Here $dl$is the change in the length through expansion
$\alpha $is the coefficient of linear expansion
${L_0}$is the original length
$dt$is the change in the temperature
Given \[\alpha = 24 \times {10^{ - 6}}{C^{ - 1}}\]
${L_0} = 88.42cm$
$dt = {35^ \circ } - {5^ \circ }$
Putting the values , we get
$dl = 24 \times {10^{ - 6}}{C^{ - 1}} \times 88.42cm \times {35^ \circ } - {5^ \circ }C$
$\therefore dl = 0.064cm$
Error in measuring distance due to expansion is 0.064cm.
So, the correct answer is “Option C”.
Note:
You need to take care about the calculations and its dimensions, and it is a formula based simple question just put in the values and you get the right answer.
Here,the change in length is directly proportional to original length.
Complete step by step answer:
When any object is heated it will experience a change in its length, it will be increased as through heating it will expand. That change will be directly proportional to the original length ${L_0}$ and change in the temperature $dt$. And is given by the formula
$dl = \alpha {L_o}dt$
Here $dl$is the change in the length through expansion
$\alpha $is the coefficient of linear expansion
${L_0}$is the original length
$dt$is the change in the temperature
Given \[\alpha = 24 \times {10^{ - 6}}{C^{ - 1}}\]
${L_0} = 88.42cm$
$dt = {35^ \circ } - {5^ \circ }$
Putting the values , we get
$dl = 24 \times {10^{ - 6}}{C^{ - 1}} \times 88.42cm \times {35^ \circ } - {5^ \circ }C$
$\therefore dl = 0.064cm$
Error in measuring distance due to expansion is 0.064cm.
So, the correct answer is “Option C”.
Note:
You need to take care about the calculations and its dimensions, and it is a formula based simple question just put in the values and you get the right answer.
Here,the change in length is directly proportional to original length.
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