Question-74358

Question Number 74358 by jagannath19 last updated on 23/Nov/19

Commented by mr W last updated on 23/Nov/19

(4) is correct length of copper rod at temperature Δθ: L_0 +L_0 α_S Δθ length of copper rod at room temperatur: L_0 +L_0 α_S Δθ−L_0 α_C Δθ =L_0 [1+(α_S −α_C )Δθ]

$$\left(\mathrm{4}\right)\:{is}\:{correct} \\ $$$${length}\:{of}\:{copper}\:{rod}\:{at}\:{temperature}\:\Delta\theta: \\ $$$${L}_{\mathrm{0}} +{L}_{\mathrm{0}} \alpha_{{S}} \Delta\theta \\ $$$${length}\:{of}\:{copper}\:{rod}\:{at}\:{room}\:{temperatur}: \\ $$$${L}_{\mathrm{0}} +{L}_{\mathrm{0}} \alpha_{{S}} \Delta\theta−{L}_{\mathrm{0}} \alpha_{{C}} \Delta\theta \\ $$$$={L}_{\mathrm{0}} \left[\mathrm{1}+\left(\alpha_{{S}} −\alpha_{{C}} \right)\Delta\theta\right] \\ $$

Commented by jagannath19 last updated on 23/Nov/19

$${sir}\:{i}\:{didnt}\:{understand}\:{please}\:{provide}\:{detailed}\:{soln}\:{sir} \\ $$

Commented by mr W last updated on 23/Nov/19

when the rod is measured with the steel tape, the tape shows a length L_0 . but L_0 is only the scale on the tape. and the scale shows only the true length of the tape at room temperature. the true length of the tape at Δθ is in fact L_0 +L_0 α_S Δθ. this is also the measured true length of the rod at Δθ, the length of the rod at room temperature is thus L_0 +L_0 α_S Δθ−L_0 α_C Δθ.

$${when}\:{the}\:{rod}\:{is}\:{measured}\:{with}\:{the} \\ $$$${steel}\:{tape},\:{the}\:{tape}\:{shows}\:{a}\:{length}\:{L}_{\mathrm{0}} . \\ $$$${but}\:{L}_{\mathrm{0}} \:{is}\:{only}\:{the}\:{scale}\:{on}\:{the}\:{tape}. \\ $$$${and}\:{the}\:{scale}\:{shows}\:{only}\:{the}\:{true} \\ $$$${length}\:{of}\:{the}\:{tape}\:{at}\:{room}\:{temperature}.\: \\ $$$${the}\:{true}\:{length}\:{of}\:{the}\:{tape}\:{at}\:\Delta\theta\:{is} \\ $$$${in}\:{fact}\:{L}_{\mathrm{0}} +{L}_{\mathrm{0}} \alpha_{{S}} \Delta\theta.\:{this}\:{is}\:{also}\:{the}\: \\ $$$${measured}\:{true}\:{length}\:{of}\:{the}\:{rod}\:{at}\:\Delta\theta, \\ $$$${the}\:{length}\:{of}\:{the}\:{rod}\:{at}\:{room}\:{temperature} \\ $$$${is}\:{thus}\:{L}_{\mathrm{0}} +{L}_{\mathrm{0}} \alpha_{{S}} \Delta\theta−{L}_{\mathrm{0}} \alpha_{{C}} \Delta\theta. \\ $$

Commented by mr W last updated on 23/Nov/19