Physics · Thermal Properties Of Matter · NEET
Each rod grows by delta L = alpha x L x delta T. The difference (L2 - L1) changes only if the two rods grow by different amounts. So we set the two growths equal: alpha2 L2 delta T = alpha1 L1 delta T. The delta T cancels, leaving alpha1 L1 = alpha2 L2. The rods can have different alpha as long as the shorter rod (larger alpha) and the longer rod (smaller alpha) balance so their absolute expansions match.
It means (L2 - L1) is a fixed number no matter how hot or cold you make the rods. Since both rods start with fixed original lengths, this can only happen if the extra length each gains on heating is exactly equal. Equal absolute expansion means alpha1 L1 = alpha2 L2. The original lengths themselves do not need to be equal.
Yes. 'Difference independent of temperature' and 'difference constant at all temperatures' mean the exact same thing in NEET problems. Both need alpha1 L1 = alpha2 L2. The 2019 NEET copper-aluminium question used the phrase 'increase in length independent of temperature' and the answer used this same condition.
The rod with the smaller alpha must be longer. Since alpha1 L1 = alpha2 L2, a small alpha needs a large L to keep the product equal. Example: copper (alpha = 1.7e-5) is 88 cm and aluminium (alpha = 2.2e-5, bigger) comes out shorter at 68 cm. Bigger alpha means shorter rod.
The condition keeps the difference in the total lengths (L2 - L1) constant. This is achieved by making the difference in expansions zero, that is delta L1 = delta L2. Do not confuse the two: we force the expansions to be equal so the length gap never changes.
The coefficients of linear expansion of brass and steel rods are alpha1 and alpha2; their lengths are l1 and l2 respectively. If (l2 - l1) is to remain the same at all temperatures, the required condition is:
A copper rod of length 88 cm and an aluminium rod of unknown length have their increase in length independent of the increase in temperature. The length of the aluminium rod is: (alpha_Cu = 1.7 x 10^-5 K^-1, alpha_Al = 2.2 x 10^-5 K^-1)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
alpha1 L1 = alpha2 L2, where alpha is the coefficient of linear expansion and L is the original length of each rod. This makes both rods expand by the same amount, so their length difference never changes.
No. The delta T cancels from both sides, so the condition alpha1 L1 = alpha2 L2 holds for any temperature change, hot or cold, small or large.
Only if they have equal lengths. Same material means alpha1 = alpha2, so alpha1 L1 = alpha2 L2 forces L1 = L2 and the difference is zero. To have a non-zero constant difference you need two different materials.
For the difference to stay constant both rods must expand equally (alpha1 L1 = alpha2 L2). For the total length (L1 + L2) to stay constant one rod must expand while the other contracts by the same amount, which needs one negative coefficient - a rarer case.
delta L = alpha x L x delta T. This linear expansion formula is the base of the whole condition; setting delta L equal for both rods gives alpha1 L1 = alpha2 L2.
