(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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$A$ uniform metal rod is used as a bar pendulum. If the room temperature rises by $10^{\circ}C$,and the coefficient of linear expansion of the metal of the rod is $2 \times 10^{-6} /^{\circ}C$,the period of the pendulum will have a percentage increase of:

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$A$ clock which keeps correct time at $20^\circ C$ is subjected to $40^\circ C$. If the coefficient of linear expansion of the pendulum is $12 \times 10^{-6} /^\circ C$,how much time will it gain or lose per day?

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Two rods,one of aluminum and another of steel,have initial lengths $ℓ_1$ and $ℓ_2$ respectively. They are joined to form a single rod of length $ℓ_1 + ℓ_2$. The coefficients of linear expansion for aluminum and steel are $\alpha_a$ and $\alpha_s$ respectively. If the increase in length of both rods is the same when the temperature is increased by $t \ ^\circ C$,find the ratio $\frac{ℓ_1}{ℓ_1 + ℓ_2}$.

Difficult

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The difference in length between two rods $A$ and $B$ is $60 \ cm$ at all temperatures. If $\alpha_A = 18 \times 10^{-6} /^{\circ}C$ and $\alpha_B = 27 \times 10^{-6} /^{\circ}C$,then the lengths of rod $A$ and rod $B$ at $0^{\circ}C$ are respectively:

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If a bimetallic strip is heated,it will:

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$A$ uniform metal rod is used as a bar pendulum. If the room temperature rises by $10^{\circ}C$,and the coefficient of linear expansion of the metal of the rod is $2 \times 10^{-6} /^{\circ}C$,the period of the pendulum will have a percentage increase of:

$A$ clock which keeps correct time at $20^\circ C$ is subjected to $40^\circ C$. If the coefficient of linear expansion of the pendulum is $12 \times 10^{-6} /^\circ C$,how much time will it gain or lose per day?

The difference in length between two rods $A$ and $B$ is $60 \ cm$ at all temperatures. If $\alpha_A = 18 \times 10^{-6} /^{\circ}C$ and $\alpha_B = 27 \times 10^{-6} /^{\circ}C$,then the lengths of rod $A$ and rod $B$ at $0^{\circ}C$ are respectively:

If a bimetallic strip is heated,it will:

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