JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
- JEE Mains 2026
- Medium
- MCQ
The temperature of a metal strip having coefficient of linear expansion \(\alpha\) is increased from \(T_1\) to \(T_2\) resulting in increase of its length by \(\Delta L_1\). The temperature is further increased from \(T_2\) to \(T_3\) such that the increase in its length is \(\Delta L_2\). Given \(T_3+T_1=2T_2\) and \(T_2-T_1=\Delta T\), the value of \(\Delta L_2\) is ______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\Delta L_1[1+\alpha\Delta T]\)
Step-by-step Solution
Detailed explanation
Let the length of the metal strip at temperature \(T_1\) be \(L_1\).
When the temperature is increased from \(T_1\) to \(T_2\), the new length \(L_2\) is given by:
\(L_2 = L_1(1 + \alpha \Delta T)\)
where \(\Delta T = T_2 - T_1\).
The increase in length is:
\(\Delta L_1 = L_2 - L_1 = L_1 \alpha \Delta T\)
When the temperature is further increased from \(T_2\) to \(T_3\), the new length \(L_3\) is:
\(L_3 = L_2(1 + \alpha \Delta T')\)
where \(\Delta T' = T_3 - T_2\).
Given \(T_3 + T_1 = 2T_2\), we have \(T_3 - T_2 = T_2 - T_1 = \Delta T\).
Thus, \(\Delta T' = \Delta T\).
The increase in length for this second temperature change is:
\(\Delta L_2 = L_3 - L_2 = L_2 \alpha \Delta T\)
Substituting \(L_2 = L_1(1 + \alpha \Delta T)\) into the equation for \(\Delta L_2\):
\(\Delta L_2 = L_1(1 + \alpha \Delta T) \alpha \Delta T\)
\(\Delta L_2 = (L_1 \alpha \Delta T)(1 + \alpha \Delta T)\)
Since \(\Delta L_1 = L_1 \alpha \Delta T\), substituting this gives:
\(\Delta L_2 = \Delta L_1(1 + \alpha \Delta T)\)
Answer: \(\Delta L_1[1+\alpha\Delta T]\)
When the temperature is increased from \(T_1\) to \(T_2\), the new length \(L_2\) is given by:
\(L_2 = L_1(1 + \alpha \Delta T)\)
where \(\Delta T = T_2 - T_1\).
The increase in length is:
\(\Delta L_1 = L_2 - L_1 = L_1 \alpha \Delta T\)
When the temperature is further increased from \(T_2\) to \(T_3\), the new length \(L_3\) is:
\(L_3 = L_2(1 + \alpha \Delta T')\)
where \(\Delta T' = T_3 - T_2\).
Given \(T_3 + T_1 = 2T_2\), we have \(T_3 - T_2 = T_2 - T_1 = \Delta T\).
Thus, \(\Delta T' = \Delta T\).
The increase in length for this second temperature change is:
\(\Delta L_2 = L_3 - L_2 = L_2 \alpha \Delta T\)
Substituting \(L_2 = L_1(1 + \alpha \Delta T)\) into the equation for \(\Delta L_2\):
\(\Delta L_2 = L_1(1 + \alpha \Delta T) \alpha \Delta T\)
\(\Delta L_2 = (L_1 \alpha \Delta T)(1 + \alpha \Delta T)\)
Since \(\Delta L_1 = L_1 \alpha \Delta T\), substituting this gives:
\(\Delta L_2 = \Delta L_1(1 + \alpha \Delta T)\)
Answer: \(\Delta L_1[1+\alpha\Delta T]\)
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