A silver rod of 100 ; cm at 0^{circ} C is hea | vedclass

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(A) First,we calculate the coefficient of linear expansion $(\alpha)$:$\alpha = \frac{\Delta L}{L_{0} \Delta 	heta}$$\alpha = \frac{0.19 \; cm}{100 \

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

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(A) First,we calculate the coefficient of linear expansion $(\alpha)$:$\alpha = \frac{\Delta L}{L_{0} \Delta 	heta}$$\alpha = \frac{0.19 \; cm}{100 \; cm 	imes (100^{\circ} C - 0^{\circ} C)}$$\alpha = \frac{0.19}{100 	imes 100} = \frac{0.19}{10000} = 1.9 	imes 10^{-5} /{ }^{\circ} C$The coefficient of cubical expansion $(\gamma)$ is related to the coefficient of linear expansion by the formula $\gamma = 3\alpha$.$\gamma = 3 	imes (1.9 	imes 10^{-5} /{ }^{\circ} C)$$\gamma = 5.7 	imes 10^{-5} /{ }^{\circ} C$

$A$ silver rod of $100 \; cm$ at $0^{\circ} C$ is heated to $100^{\circ} C$. Its length increases by $0.19 \; cm$. The coefficient of cubical expansion of the silver rod is:

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