When a meter rod made of silver at 0^{circ} C | vedclass

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(C) Given: Length of silver rod,$L = 1 \ m = 100 \ cm$.Initial temperature,$T_1 = 0^{\circ}C$.Final temperature,$T_2 = 100^{\circ}C$.Change in tempera

Vedclass · Accepted Answer

$5.7 	imes 10^{-5} \ ^{\circ}C^{-1}$

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(C) Given: Length of silver rod,$L = 1 \ m = 100 \ cm$.Initial temperature,$T_1 = 0^{\circ}C$.Final temperature,$T_2 = 100^{\circ}C$.Change in temperature,$\Delta T = T_2 - T_1 = 100^{\circ}C$.Increase in length,$\Delta L = 0.19 \ cm$.The formula for linear expansion is $\Delta L = L \alpha \Delta T$,where $\alpha$ is the coefficient of linear expansion.Substituting the values: $0.19 = 100 	imes \alpha 	imes 100$.$\alpha = \frac{0.19}{10000} = 0.19 	imes 10^{-4} \ ^{\circ}C^{-1} = 1.9 	imes 10^{-5} \ ^{\circ}C^{-1}$.The coefficient of volume expansion $\gamma$ is related to $\alpha$ by $\gamma = 3\alpha$.$\gamma = 3 	imes 1.9 	imes 10^{-5} \ ^{\circ}C^{-1} = 5.7 	imes 10^{-5} \ ^{\circ}C^{-1}$.

When a meter rod made of silver at $0^{\circ} C$ is heated to $100^{\circ} C$,its length increases by $0.19 \ cm$. Find the coefficient of volume expansion of the silver.

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