(C) The change in length for a rod is given by $\Delta \ell = \ell \alpha \Delta T$. Given that the increase in length for both rods is the same: $\De

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$\frac{\alpha_s}{\alpha_a + \alpha_s}$

(C) The change in length for a rod is given by $\Delta \ell = \ell \alpha \Delta T$. Given that the increase in length for both rods is the same: $\Delta \ell_1 = \Delta \ell_2$. Therefore,$\ell_1 \alpha_a t = \ell_2 \alpha_s t$. This simplifies to $\ell_1 \alpha_a = \ell_2 \alpha_s$,which means $\frac{\ell_1}{\ell_2} = \frac{\alpha_s}{\alpha_a}$. To find the ratio $\frac{\ell_1}{\ell_1 + \ell_2}$,we use the property of ratios: $\frac{\ell_1}{\ell_1 + \ell_2} = \frac{\alpha_s}{\alpha_a + \alpha_s}$.

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

Difficult

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 $\frac{ℓ_1}{ℓ_1 + ℓ_2}$.

A
$\frac{\alpha_s}{\alpha_a}$
B
$\frac{\alpha_a}{\alpha_s}$
C
$\frac{\alpha_s}{\alpha_a + \alpha_s}$
D
$\frac{\alpha_a}{\alpha_a + \alpha_s}$

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Thermal Expansion for Solid

Two large holes are cut in a metal sheet. If this sheet is heated,what happens to the distances $AB$ and $BC$ as shown in the figure?

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$A$ clock with an iron pendulum keeps correct time at $15^{\circ} C$. If the room temperature is $20^{\circ} C$,the error in seconds per day will be nearly (coefficient of linear expansion of iron is $\alpha = 1.2 \times 10^{-5} /{ }^{\circ} C$) (in $s$)

MediumMHT CET 2022

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When the temperature of a body increases from $T$ to $T+\Delta T$, its moment of inertia increases from $I$ to $I+\Delta I$. If $\alpha$ is the coefficient of linear expansion of the material of the body, then $\frac{\Delta I}{I}$ is (neglect higher orders of $\alpha$ ).

DifficultAP EAMCET 2012

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$A$ meter scale made of steel reads accurately at $25^{\circ} C$. Suppose in an experiment an accuracy of $0.06 \,mm$ in $1 \,m$ is required, the range of temperature in which the experiment can be performed with this meter scale is (Coefficient of linear expansion of steel is $11 \times 10^{-6} /{ }^{\circ} C$)

MediumTS EAMCET 2017

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The coefficient of apparent expansion of mercury in a glass vessel is $153 \times 10^{-6} /{ }^{\circ} C$ and in a steel vessel is $144 \times 10^{-6} /{ }^{\circ} C$. If $\alpha$ for steel is $12 \times 10^{-6} /{ }^{\circ} C$, then the coefficient of linear expansion $\alpha$ for glass is:

MediumAIIMS 2019

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Two large holes are cut in a metal sheet. If this sheet is heated,what happens to the distances $AB$ and $BC$ as shown in the figure?

$A$ clock with an iron pendulum keeps correct time at $15^{\circ} C$. If the room temperature is $20^{\circ} C$,the error in seconds per day will be nearly (coefficient of linear expansion of iron is $\alpha = 1.2 \times 10^{-5} /{ }^{\circ} C$) (in $s$)

When the temperature of a body increases from $T$ to $T+\Delta T$, its moment of inertia increases from $I$ to $I+\Delta I$. If $\alpha$ is the coefficient of linear expansion of the material of the body, then $\frac{\Delta I}{I}$ is (neglect higher orders of $\alpha$ ).

The coefficient of apparent expansion of mercury in a glass vessel is $153 \times 10^{-6} /{ }^{\circ} C$ and in a steel vessel is $144 \times 10^{-6} /{ }^{\circ} C$. If $\alpha$ for steel is $12 \times 10^{-6} /{ }^{\circ} C$, then the coefficient of linear expansion $\alpha$ for glass is:

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