A bar of iron is 10 , cm at 20^{circ}C. At 19 | vedclass

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Question

(C) The change in length $\Delta L$ is given by the formula: $\Delta L = L_0 \alpha \Delta T$.Here,$L_0 = 10 \, cm$,$\alpha = 11 	imes 10^{-6} /^{\ci

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$11 	imes 10^{-5} \, cm$ shorter

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(C) The change in length $\Delta L$ is given by the formula: $\Delta L = L_0 \alpha \Delta T$.Here,$L_0 = 10 \, cm$,$\alpha = 11 	imes 10^{-6} /^{\circ}C$,and $\Delta T = 19^{\circ}C - 20^{\circ}C = -1^{\circ}C$.Substituting the values:$\Delta L = 10 	imes (11 	imes 10^{-6}) 	imes (-1) \, cm$.$\Delta L = -110 	imes 10^{-6} \, cm = -1.1 	imes 10^{-4} \, cm = -11 	imes 10^{-5} \, cm$.The negative sign indicates that the bar becomes shorter.Therefore,the bar is $11 	imes 10^{-5} \, cm$ shorter.

$A$ bar of iron is $10 \, cm$ at $20^{\circ}C$. At $19^{\circ}C$ it will be ($\alpha$ of iron $= 11 \times 10^{-6}/^{\circ}C$)

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