How to do 4th question?

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Answer 1 · 2018-03-15T09:44:50

See below.

Period is given by

#T approx 2pi sqrt(L/g)#

now taking natural log to both sides

#lnT = ln(2pi)+1/2 (ln L - ln g)#

assuming #g# constant and deriving

#(deltaT)/T = 1/2 (deltaL)/L#

now assuming an isotropic material the volumetric thermal expansion coefficient is three times the linear coefficient: or

#alpha_V = 3 alpha_L# and then

#delta L = alpha_L (40^@-20^@) = 1/3 alpha_V(40^@-20^@) =1/2(36/3) xx 20 xx10^-6# [s]

and finally

#(delta T)/T approx 120 xx 10^-6# [s/s]

per day it gives

#delta T = 24 xx 60 xx 60 xx 120 xx 10^-6 = 10.38# [s] delay/day.