A rod is found to be 0.05 cm longer at 40^{c | vedclass

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(D) Let $L_0$ be the length of the rod at $0^{\circ} C$ and $\alpha$ be the coefficient of linear expansion.The length at temperature $T$ is given by

Vedclass · Accepted Answer

$111.1$

Vedclass · Answer

(D) Let $L_0$ be the length of the rod at $0^{\circ} C$ and $\alpha$ be the coefficient of linear expansion.The length at temperature $T$ is given by $L_T = L_0(1 + \alpha T)$.Length at $10^{\circ} C$: $L_{10} = L_0(1 + 10\alpha)$.Length at $40^{\circ} C$: $L_{40} = L_0(1 + 40\alpha)$.Given that $L_{40} - L_{10} = 0.05 \ cm$.Substituting the expressions: $L_0(1 + 40\alpha) - L_0(1 + 10\alpha) = 0.05$.$L_0(40\alpha - 10\alpha) = 0.05$.$30 L_0 \alpha = 0.05$.$L_0 = \frac{0.05}{30 \alpha}$.Substituting $\alpha = 1.5 	imes 10^{-5} \ {}^{\circ} C^{-1}$:$L_0 = \frac{0.05}{30 	imes 1.5 	imes 10^{-5}} = \frac{0.05}{45 	imes 10^{-5}} = \frac{0.05 	imes 10^5}{45} = \frac{5000}{45} \approx 111.1 \ cm$.

$A$ rod is found to be $0.05 \ cm$ longer at $40^{\circ} C$ than it is at $10^{\circ} C$. The length of the rod at $0^{\circ} C$ is (coefficient of linear expansion of the material of the rod $= 1.5 \times 10^{-5} \ {}^{\circ} C^{-1}$) (in $cm$)

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