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

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

(C) The change in length for a rod is given by $\Delta l = l \alpha \Delta T$.Given that the increase in length for both rods is the same,we have $\Delta l_a = \Delta l_s$.Substituting the formula,we get $l_1 \alpha_a \Delta T = l_2 \alpha_s \Delta T$.Since $\Delta T$ is the same for both,we can cancel it out: $l_1 \alpha_a = l_2 \alpha_s$.This implies $\frac{l_1}{l_2} = \frac{\alpha_s}{\alpha_a}$.To find $\frac{l_1}{l_1 + l_2}$,we use the property of ratios: $\frac{l_1}{l_1 + l_2} = \frac{\alpha_s}{\alpha_s + \alpha_a}$.Thus,the correct option is $C$.

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 together to form a single rod of length $(l_1 + l_2)$. If the coefficients of linear expansion for aluminum and steel are $\alpha_a$ and $\alpha_s$ respectively,and the increase in length for both rods is the same when the temperature of both is increased by $t \ ^\circ C$,then $\frac{l_1}{l_1 + l_2} = \dots$

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

$A$ steel tape is calibrated at $20^{\circ} C$. On a cold day when the temperature is $-15^{\circ} C$,the percentage error in the tape will be ........... $\%$ $\left[\alpha_{\text{steel}} = 1.2 \times 10^{-5} {}^{\circ} C^{-1}\right]$

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Two uniform brass rods $A$ and $B$ of length $l$ and $2l$ and their radii $2r$ and $r$ respectively are heated to the same temperature. The ratio of the increase in the volume of rod $A$ to that of rod $B$ is

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$A$ steel tape of $300 \,cm$ length is graduated at $27^{\circ} C$. The length of a steel rod measured using the tape is found to be $110 \,cm$ at $50^{\circ} C$. The actual length of the steel rod at $50^{\circ} C$ is $(\alpha_{\text{steel}} = 1.2 \times 10^{-5} \,K^{-1})$. (in $\,cm$)

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The volume of a metal sphere increases by $0.15\%$ when its temperature is raised by $24\,^{\circ}C$. The coefficient of linear expansion of the metal is:

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If two rods of lengths $L$ and $2L$,having coefficients of linear expansion $\alpha$ and $2\alpha$ respectively are connected end-to-end,then find the average coefficient of linear expansion of the composite rod.

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$A$ steel tape is calibrated at $20^{\circ} C$. On a cold day when the temperature is $-15^{\circ} C$,the percentage error in the tape will be ........... $\%$ $\left[\alpha_{\text{steel}} = 1.2 \times 10^{-5} {}^{\circ} C^{-1}\right]$

Two uniform brass rods $A$ and $B$ of length $l$ and $2l$ and their radii $2r$ and $r$ respectively are heated to the same temperature. The ratio of the increase in the volume of rod $A$ to that of rod $B$ is

$A$ steel tape of $300 \,cm$ length is graduated at $27^{\circ} C$. The length of a steel rod measured using the tape is found to be $110 \,cm$ at $50^{\circ} C$. The actual length of the steel rod at $50^{\circ} C$ is $(\alpha_{\text{steel}} = 1.2 \times 10^{-5} \,K^{-1})$. (in $\,cm$)

The volume of a metal sphere increases by $0.15\%$ when its temperature is raised by $24\,^{\circ}C$. The coefficient of linear expansion of the metal is:

If two rods of lengths $L$ and $2L$,having coefficients of linear expansion $\alpha$ and $2\alpha$ respectively are connected end-to-end,then find the average coefficient of linear expansion of the composite rod.

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