A sheet of steel is 40 cm long and 5 cm bro | vedclass

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(C) Initial area $A_1 = 40 \ cm 	imes 5 \ cm = 200 \ cm^2$.Change in temperature $\Delta T = 100^{\circ} C - 0^{\circ} C = 100^{\circ} C$.Change in a

Vedclass · Accepted Answer

$3.5 	imes 10^{-5} /^{\circ} C$

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(C) Initial area $A_1 = 40 \ cm 	imes 5 \ cm = 200 \ cm^2$.Change in temperature $\Delta T = 100^{\circ} C - 0^{\circ} C = 100^{\circ} C$.Change in area $\Delta A = 1.4 \ cm^2$.The coefficient of superficial expansion $\beta$ is given by $\beta = \frac{\Delta A}{A_1 \Delta T}$.Substituting the values: $\beta = \frac{1.4}{200 	imes 100} = \frac{1.4}{20000} = 0.7 	imes 10^{-4} = 7 	imes 10^{-5} /^{\circ} C$.Since the coefficient of linear expansion $\alpha = \frac{\beta}{2}$,we have $\alpha = \frac{7 	imes 10^{-5}}{2} = 3.5 	imes 10^{-5} /^{\circ} C$.

$A$ 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:

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