(B) The diameter of the iron ring $d_i = 5 \ m$ at $T_1 = 27^{\circ} C$.The diameter of the wooden wheel $d_w = 5.012 \ m$.We need the iron ring to ex

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(B) The diameter of the iron ring $d_i = 5 \ m$ at $T_1 = 27^{\circ} C$.The diameter of the wooden wheel $d_w = 5.012 \ m$.We need the iron ring to expand to a diameter of $5.012 \ m$.Using the formula for linear expansion: $d_w = d_i(1 + \alpha \Delta T)$.$\Delta T = \frac{d_w - d_i}{d_i \alpha} = \frac{5.012 - 5}{5 \times 1.2 \times 10^{-5}} = \frac{0.012}{6 \times 10^{-5}} = \frac{1.2 \times 10^{-2}}{6 \times 10^{-5}} = 0.2 \times 10^3 = 200^{\circ} C$.The final temperature $T_2 = T_1 + \Delta T = 27^{\circ} C + 200^{\circ} C = 227^{\circ} C$.

Class 11 Physics 10-1.Thermometry, Thermal Expansion and Calorimetry Thermal Expansion for Solid

Medium

$A$ blacksmith fixes a circular iron frame on the wooden wheel of a bullock cart. The diameters of the wooden wheel and the circular iron frame are $5.012 \ m$ and $5 \ m$ respectively at $27^{\circ} C$. The temperature (in $^{\circ} C$) to which the iron ring must be heated so as to fit the wooden wheel is (Coefficient of linear expansion of iron $= 1.2 \times 10^{-5} \ ^{\circ} C^{-1}$).

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$A$ uniform metal bar of length $10 \ m$ with a crack at its midpoint is clamped between two rigid supports. The bar buckles upward due to a temperature rise of $40^{\circ} C$. If the coefficient of linear expansion of the metal is $2.5 \times 10^{-6} {}^{\circ} C^{-1}$,the maximum displacement of the midpoint of the bar is: (in $cm$)

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Two different wires having lengths $L_{1}$ and $L_{2}$ and respective temperature coefficients of linear expansion $\alpha_{1}$ and $\alpha_{2}$ are joined end-to-end. Then the effective temperature coefficient of linear expansion is

The difference in length between an iron rod and a copper rod is $10 \ cm$ at all temperatures. If ${\alpha _{Fe}} = 11 \times {10^{ - 6}} \, ^\circ C^{ - 1}$ and ${\alpha _{Cu}} = 17 \times {10^{ - 6}} \, ^\circ C^{ - 1}$,what are their lengths respectively?

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