a clock of pendulum gains 5 secs in a day at 15°C and lost 10 sec in a day at 30°C. calculate the coefficient of linear expansion of the material.
Answers
Explanation:
fractional increment in time with Temperature change
t
Δt
=
2
1
αΔT eq(1)
where
t
Δt
= fractional increment
α = coefficient of linear expansion
ΔT= change in temperature
assume @ T = T
0
, there is no gain and loss in time
case 1:
t
Δt
= gain 5 sec per day
=
24×60×60
5
ΔT = T
0
- 15
substitute values back in eq(1)
24×60××60
5
=
2
1
×α×(T
0
−15) eq(2)
case 2:
t
Δt
= loss 10 sec per day
= -
24×60×60
10
ΔT = T
0
- 30
substitute values back in eq(1)
-
24×60××60
10
=
2
1
×α×(T
0
−30) eq(3)
eq(2)
eq(3)
-2 =
T
0
−15
T
0
−30
T
0
−30 = -2 ×(T
0
−15)
T
0
−30 = −2T
0
+30
3T
0
= 60
T
0
= 20
o
C
substitute this value of T
0
in eq(2)
24×60×60
5
=
2
1
×α×(20−15)
24×60×60
5
=
2
1
×α×5
solve for α
α = 2.31 ×10
−5
o
C
−1