JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
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 ______.
- A \(\Delta L_1[1+2\alpha^2(\Delta T)^2]\)
- B \(\Delta L_1[1+\alpha^2(\Delta T)^2]\)
- C \(\Delta L_1[1+2\alpha\Delta T]\)
- D \(\Delta L_1[1+\alpha\Delta T]\)
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:…
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