
Two metal rods have coefficients of linear expansion \[1.1 \times {10^{ - 5}}{/^ \circ }C\] and \[1.65 \times {10^{ - 5}}{/^ \circ }C\] respectively. The difference in lengths is \[10cm\] at all temperatures. Their initial lengths must be respectively:
$(A)$ \[40cm\]and \[50cm\]
$(B)$ \[40cm\]and \[30cm\]
$(C)$ \[50cm\] and \[60cm\]
$(D)$ \[30cm\] and \[20cm\]
Answer
504.6k+ views
Hint: The concept of thermal expansion is applied to solving this type of problem. When heat is applied to a solid, especially a metal it results in expansion. The expansion depends on the coefficient of thermal expansion given by \[\alpha \] and the temperature difference. The coefficient of thermal expansion varies from solid to solid.
Formula used:
The increase in length of a metallic rod on the application of temperature is given by:
\[\Delta l = l \times \alpha \times \Delta T\]
Where, \[\Delta l\] is the change in length of the material, \[l\] is the original length of the material, \[\alpha \] is the linear coefficient of expansion, and \[\Delta T\] is the temperature change.
Complete step-by-step solution:
Given:
\[{\alpha _1} = 1.1 \times {10^{ - 5}}{/^ \circ }C\]
\[{\alpha _2} = 1.65 \times {10^{ - 5}}{/^ \circ }C\]
\[{l_1} - {l_2} = 10cm\]
We have already given the information that both rods are expanding by the same amount at the same temperature. So, we can equate a change in the length of each rod.
\[{l_1}{\alpha _1}\Delta T = {l_2}{\alpha _2}\Delta T\]
Since the temperature is the same, we can rewrite this equation as,
${l_1}{\alpha _1} = {l_2}{\alpha _2}$
$\Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{\alpha _2}}}{{{\alpha _1}}} $
$\Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{1.65 \times {{10}^{ - 5}}}}{{1.1 \times {{10}^{ - 5}}}} $
$\Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = 1.5 $
$\Rightarrow{l_1} = 1.5{l_2}$
We know that, the difference in lengths\[{l_1} - {l_2} = 10cm\].
\[{l_1} - {l_2} = 10cm\]
\[ \Rightarrow 1.5{l_2} - {l_2} = 10cm\]
\[ \Rightarrow 0.5{l_2} = 10cm\]
\[ \Rightarrow {l_2} = \dfrac{{10}}{{0.5}} = 20cm\]
And,
\[{l_1} - 20 = 10\]
\[ \Rightarrow {l_1} = 10 + 20 = 30cm\]
Their initial lengths must be \[30cm\] and \[20cm\]. Hence, option D is correct.
Note: We must keep in mind that the coefficient of thermal expansion changes with material changes. The change in dimension of the material can easily be calculated if we know the original dimension of the material and the change in temperature along with the coefficient of thermal expansion. In this case, the coefficient of linear expansion (\[\alpha \]) is considered. Likewise, for area and volume expansions as a result of heat, the coefficient of area expansion (\[\beta \]) and volume expansion \[\left( \gamma \right)\]are used.
Formula used:
The increase in length of a metallic rod on the application of temperature is given by:
\[\Delta l = l \times \alpha \times \Delta T\]
Where, \[\Delta l\] is the change in length of the material, \[l\] is the original length of the material, \[\alpha \] is the linear coefficient of expansion, and \[\Delta T\] is the temperature change.
Complete step-by-step solution:
Given:
\[{\alpha _1} = 1.1 \times {10^{ - 5}}{/^ \circ }C\]
\[{\alpha _2} = 1.65 \times {10^{ - 5}}{/^ \circ }C\]
\[{l_1} - {l_2} = 10cm\]
We have already given the information that both rods are expanding by the same amount at the same temperature. So, we can equate a change in the length of each rod.
\[{l_1}{\alpha _1}\Delta T = {l_2}{\alpha _2}\Delta T\]
Since the temperature is the same, we can rewrite this equation as,
${l_1}{\alpha _1} = {l_2}{\alpha _2}$
$\Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{\alpha _2}}}{{{\alpha _1}}} $
$\Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{1.65 \times {{10}^{ - 5}}}}{{1.1 \times {{10}^{ - 5}}}} $
$\Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = 1.5 $
$\Rightarrow{l_1} = 1.5{l_2}$
We know that, the difference in lengths\[{l_1} - {l_2} = 10cm\].
\[{l_1} - {l_2} = 10cm\]
\[ \Rightarrow 1.5{l_2} - {l_2} = 10cm\]
\[ \Rightarrow 0.5{l_2} = 10cm\]
\[ \Rightarrow {l_2} = \dfrac{{10}}{{0.5}} = 20cm\]
And,
\[{l_1} - 20 = 10\]
\[ \Rightarrow {l_1} = 10 + 20 = 30cm\]
Their initial lengths must be \[30cm\] and \[20cm\]. Hence, option D is correct.
Note: We must keep in mind that the coefficient of thermal expansion changes with material changes. The change in dimension of the material can easily be calculated if we know the original dimension of the material and the change in temperature along with the coefficient of thermal expansion. In this case, the coefficient of linear expansion (\[\alpha \]) is considered. Likewise, for area and volume expansions as a result of heat, the coefficient of area expansion (\[\beta \]) and volume expansion \[\left( \gamma \right)\]are used.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

What is a periderm How does periderm formation take class 11 biology CBSE

