Physics · Thermal Properties Of Matter · NEET
The hole gets bigger. This confuses many students. A heated solid expands as one whole piece, so every length grows in the same ratio, including the empty gap inside. Think of the ring as a photograph being enlarged: the hole in the picture also gets bigger. So both the outer diameter and the inner diameter of the ring increase. That is why the heated ring can slip over the larger wheel.
When hot, the ring's inner diameter is a little larger than the wheel. The blacksmith fits it on quickly. As the ring cools, it tries to shrink back to its original smaller size. But the solid wooden wheel is in the way and stops it from shrinking fully. So the ring squeezes onto the wheel with a large inward force. This locked-in squeeze is why the joint stays firm for years. It is the same idea as thermal stress in a bar that is not allowed to contract freely.
For the diameter (a length), use linear expansion. The change in any length L is delta L = L times alpha times delta T. The diameter D changes as delta D = D times alpha times delta T. If a question instead asks for the change in the area of the hole, use area expansion with beta = 2 alpha. For the change in the volume of the metal, use gamma = 3 alpha. For fitting the ring, the diameter (linear) is what matters.
A slightly larger ring would fit loosely and slip off the wheel while the cart moves. The whole point is a tight grip. By heating, the blacksmith makes the ring expand just enough to slide on, and cooling brings it back to a size smaller than the fit, so it clamps the wheel with real force. A pre-cut larger ring cannot create that clamping force.
Use delta D = D times alpha times delta T. Example: an iron ring of diameter D = 1 m, alpha of iron about 12 times 10 to the power minus 6 per degree C, heated by delta T = 200 degree C. Then delta D = 1 times 12e-6 times 200 = 2.4e-3 m = 2.4 mm. So a 1 m ring grows by about 2.4 mm across, which is enough to slip over a wheel that is 1 to 2 mm larger.
A metallic bar (Young's modulus Y = 0.5 x 10^11 N/m^2, coefficient of linear expansion alpha = 10^-5 per degree C, length 1 m, area of cross-section 10^-3 m^2) is heated from 0 degree C to 100 degree C while clamped so it cannot expand or bend. The compressive force developed in it is:
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:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is called a shrink fit. The ring is heated to enlarge it, placed over the part, then cooled so it shrinks and clamps tightly. Blacksmiths use it to fix iron rims onto wooden cart wheels.
Yes. Every length in a heated solid grows in the same ratio, so the inner diameter uses the same alpha as the outer diameter. The relation is delta D = D alpha delta T for both.
Use the area (superficial) expansion coefficient beta, and beta = 2 alpha. So the change in the hole's area is delta A = A times 2 alpha times delta T.
It can develop large thermal stress. If that stress stays below the metal's strength, it just grips tightly. Blacksmiths choose the ring size so the stress is safe. If the fit is too tight, the ring could crack, which is why the amount of oversize is kept small.
The physics is the same thermal expansion, but the goal is opposite. Rail gaps ALLOW free expansion to avoid buckling. The ring uses PREVENTED contraction on purpose to create a tight grip.
