Question 449: Thermodynamics & KTG

Question

A pendulum clock loses $12$ $s$ a day if the temperature is ${40^ \circ }C$ and gains $4$ $s$ a day if the temperature is ${20^ \circ }C.$ The temperature at which the clock will show correct time, and the co-efficient of linear expansion $\left( \alpha  \right)$  of the metal of the pendulum shaft are respectively :

Accepted Answer

Time lost/gained per day $ = {1 \over 2} \propto \Delta 	heta  	imes 86400$ second
$12 = {1 \over 2}\alpha \left( {40 - 	heta } ight) 	imes 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i ight)$
$4 = {1 \over 2}\alpha \left( {	heta  - 20} ight) 	imes 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} ight)$
On dividing we get, $\,\,\,3 = {{40 - 	heta } \over {	heta  - 20}}$
$3	heta  - 60 = 40 - 	heta $
$4	heta  = 100 \Rightarrow 	heta  = {25^ \circ }C$

Answer

${30^ \circ }C;\,\,\alpha  = 1.85 	imes {10^{ - 3}}/{}^ \circ C$

Answer

${55^ \circ }C;\,\,\alpha  = 1.85 	imes {10^{ - 2}}/{}^ \circ C$

Answer

${25^ \circ }C;\,\,\alpha  = 1.85 	imes {10^{ - 5}}/{}^ \circ C$

Answer

${60^ \circ }C;\,\,\alpha  = 1.85 	imes {10^{ - 4}}/{}^ \circ C$

Thermodynamics & KTG

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