(D) Change in length of the first rod is given by $\Delta \ell_1 = L(2 \alpha) \Delta T = 2 \,L \alpha \Delta T$. Change in length of the second rod i

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(D) Change in length of the first rod is given by $\Delta \ell_1 = L(2 \alpha) \Delta T = 2 \,L \alpha \Delta T$. Change in length of the second rod is given by $\Delta \ell_2 = (2 \,L)(\alpha)(\Delta T) = 2 \,L \alpha \Delta T$. Total change in length of the composite rod is $\Delta l = \Delta l_1 + \Delta l_2 = 2 \,L \alpha \Delta T + 2 \,L \alpha \Delta T = 4 \,L \alpha \Delta T$. For the composite rod of length $3 \,L$, the change in length is $\Delta l = (3 \,L) \alpha_C \Delta T$. Equating the two expressions for $\Delta l$: $4 \,L \alpha \Delta T = 3 \,L \alpha_C \Delta T$. Solving for $\alpha_C$, we get $\alpha_C = \frac{4 \alpha}{3}$.

Class 11 Physics 1 Thermal Properties of Matter: Thermometry, Expansion and Calorimetry Thermal Expansion for Solid

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Two rods of same area of cross-section have lengths $L$ and $2 \,L$ and coefficients of linear expansion $2 \alpha$ and $\alpha$ respectively. If they are welded to form a composite rod of length $3 \,L$, then the coefficient of linear expansion of the composite rod is

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A
$\frac{3 \alpha}{2}$
B
$3 \alpha$
C
$\frac{3 \alpha}{4}$
D
$\frac{4 \alpha}{3}$

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Two rods of same area of cross-section have lengths $L$ and $2 \,L$ and coefficients of linear expansion $2 \alpha$ and $\alpha$ respectively. If they are welded to form a composite rod of length $3 \,L$, then the coefficient of linear expansion of the composite rod is

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$A$ sheet of steel is $40 \ cm$ long and $5 \ cm$ broad at $0^{\circ} C$. The surface area of the sheet increases by $1.4 \ cm^2$ at $100^{\circ} C$. The coefficient of linear expansion of steel is:

Two rods of different materials have lengths $\ell_1$ and $\ell_2$ whose coefficients of linear expansion are $\alpha_1$ and $\alpha_2$ respectively. If the difference between the two lengths is independent of temperature,then:

The moment of inertia $I$ of a uniform rod about a perpendicular bisector increases to $I+\Delta I$, if the temperature is increased slightly by $\Delta T$. If the coefficient of linear expansion is $\alpha$, then $\frac{\Delta I}{I}$ is (Assume $\frac{\Delta T}{T} \ll 1$)

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