A steel scale measures the length of a copper wire as `80.0 cm` when both area at `20^(@)C` (the calibration temperature for scale). What would be the scale read for the length of the wire when both are at `40^(@)C`? (Given `alpha_("steel") = 11 xx 10^(-6) per ^(@)C` and `alpha_("copper") = 17 xx 10^(-6) per^(@)C`)

Options

A. 80.0096 cm
B. 80.0272 cm
C. 1 cm
D. 25.2 cm

Answer

A

Explanation

`L_(app) = L._(cu) - L._("steel")`
`= L_(0) (1 + alhpa_(cu) Delta theta )- L_(0) (1 + alhpa_(s) Delta theta) = L_0 (alpha_(cu) = alpha_(s) )Delta theta`
`= 80 (17 xx 10^(-6) - 11 xx 10^(-6)) xx (40 - 20)`
`= 0.0096 cm`
`L = 80 + 0.0096 = 80.0096 cm`.