TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

A s both steel rod and metal scale expands at 25°C, we use
True value of length of rod at 25°C
= Scale reading ×(1+α(θ′−θ))
where, α = coefficient of linear expansion of scale
(=20×10−6/‌∘C)
θ′= temperature at which observation is taken (=25∘Cr)
and θ = temperature at which metre scalc reads correctly (= 0°C)
Substituting values in Eq. (i), we get
Length of rod at 25°C
=1×[1+20×10−6(25−0)]
=1+5×10−4=1.0005m
Now, if l2= length of steel rod at 25∘C and
l1= length of steel rod at 0∘C, then we have
l2=l1(1+α∆θ)
Substituting values, we have