(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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10-1.Thermometry, Thermal Expansion and Calorimetry

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Thermal Expansion for Solid

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Two rods of different metals have coefficients of linear expansion $\alpha_1$ and $\alpha_2$ respectively. Their respective lengths are $L_1$ and $L_2$. At all temperatures,$(L_2 - L_1)$ remains the same. The correct relation is:

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$A$ silver rod of $100 \; cm$ at $0^{\circ} C$ is heated to $100^{\circ} C$. Its length increases by $0.19 \; cm$. The coefficient of cubical expansion of the silver rod is:

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$A$ metal tape is calibrated at $25^{\circ} C$. On a cold day when the temperature is $-15^{\circ} C$,the percentage error in the measurement of length is (Coefficient of linear expansion of metal $= 1 \times 10^{-5} {}^{\circ} C^{-1}$) (in $\%$)

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$A$ thin copper wire of length $l$ metre increases in length by $2\%$ when heated through $10^{\circ}C$. The percentage increase in area when a square copper sheet of side $l$ metre is heated through $10^{\circ}C$ is:

Medium

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Assertion :- $A$ brass disc is fast fitted in a hole in a steel plate. The system must be cooled to loosen the disc from the hole.
Reason :- The coefficient of linear expansion of brass is greater than the coefficient of linear expansion of steel.

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Two rods of different metals have coefficients of linear expansion $\alpha_1$ and $\alpha_2$ respectively. Their respective lengths are $L_1$ and $L_2$. At all temperatures,$(L_2 - L_1)$ remains the same. The correct relation is:

$A$ silver rod of $100 \; cm$ at $0^{\circ} C$ is heated to $100^{\circ} C$. Its length increases by $0.19 \; cm$. The coefficient of cubical expansion of the silver rod is:

$A$ metal tape is calibrated at $25^{\circ} C$. On a cold day when the temperature is $-15^{\circ} C$,the percentage error in the measurement of length is (Coefficient of linear expansion of metal $= 1 \times 10^{-5} {}^{\circ} C^{-1}$) (in $\%$)

$A$ thin copper wire of length $l$ metre increases in length by $2\%$ when heated through $10^{\circ}C$. The percentage increase in area when a square copper sheet of side $l$ metre is heated through $10^{\circ}C$ is:

Assertion :- $A$ brass disc is fast fitted in a hole in a steel plate. The system must be cooled to loosen the disc from the hole.Reason :- The coefficient of linear expansion of brass is greater than the coefficient of linear expansion of steel.

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Assertion :- $A$ brass disc is fast fitted in a hole in a steel plate. The system must be cooled to loosen the disc from the hole.
Reason :- The coefficient of linear expansion of brass is greater than the coefficient of linear expansion of steel.