There is some change in length when a 33000 ~ | vedclass

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(C) The Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta l/l}$,where $F$ is the force,$A$ is the area,and $\Delta l/l$ is the strain.Given $F =

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$10$

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(C) The Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta l/l}$,where $F$ is the force,$A$ is the area,and $\Delta l/l$ is the strain.Given $F = 33000 ~N$,$A = 10^{-3} ~m^2$,and $Y = 3 	imes 10^{11} ~N/m^2$.Substituting the values: $3 	imes 10^{11} = \frac{33000 / 10^{-3}}{\Delta l/l}$.Thus,the strain $\frac{\Delta l}{l} = \frac{33000}{10^{-3} 	imes 3 	imes 10^{11}} = \frac{33 	imes 10^3}{3 	imes 10^8} = 11 	imes 10^{-5}$.Thermal expansion is given by $\frac{\Delta l}{l} = \alpha \Delta T$,where $\alpha = 1.1 	imes 10^{-5} /{ }^{\circ} C$.Equating the two expressions for strain: $11 	imes 10^{-5} = 1.1 	imes 10^{-5} 	imes \Delta T$.Solving for $\Delta T$: $\Delta T = \frac{11 	imes 10^{-5}}{1.1 	imes 10^{-5}} = 10^{\circ} C$.

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