An alumium rod when measured with a steel scale both being at 25 degree celcius appears to be 1 meter long. If the scale is correct at 0 degree celcius, what is the true length of the rod at 25 degree celcius? What wull be the length of rod at 0 degree celcius?
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An almunium rod when measured with a steel scale,both being at 25°C appears to be 1meter long. If the scale is correct at 0°C?
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Solution
Asumption
α
z
=
26
×
1
0
−
6
/
°
C
α
z
=26×10
−6
/°C
α
s
=
11
×
1
0
−
6
/
°
C
α
s
=11×10
−6
/°C
So length of scale at 25°C
100
=
l
0
(
1
+
α
z
Δ
t
)
100=l
0
(1+α
z
Δt)
l
0
=
99.93
c
m
l
0
=99.93cm
So
Δ
l
Z
=
99.93
×
1
0
−
6
×
26
×
25
Δl
Z
=99.93×10
−6
×26×25
Δ
l
Z
=
0.064
c
m
Δl
Z
=0.064cm
So lenght of rod at 25°c is
l
r
o
d
l
rod
At 25°C =100.064cm
Actual lenght of rod is
l
r
o
d
(
m
e
a
s
u
r
e
d
)
=
l
a
c
t
u
a
l
(
1
+
(
(
α
Z
n
−
α
r
o
d
)
Δ
t
)
l
rod (measured)
=l
actual
(1+((α
Zn
−α
rod
)Δt)
Asumption
α
z
=
26
×
1
0
−
6
/
°
C
α
z
=26×10
−6
/°C
α
s
=
11
×
1
0
−
6
/
°
C
α
s
=11×10
−6
/°C
So length of scale at 25°C
100
=
l
0
(
1
+
α
z
Δ
t
)
100=l
0
(1+α
z
Δt)
l
0
=
99.93
c
m
l
0
=99.93cm
So
Δ
l
Z
=
99.93
×
1
0
−
6
×
26
×
25
Δl
Z
=99.93×10
−6
×26×25
Δ
l
Z
=
0.064
c
m
Δl
Z
=0.064cm
So lenght of rod at 25°c is
l
r
o
d
l
rod
At 25°C =100.064cm
Actual lenght of rod is
l
r
o
d
(
m
e
a
s
u
r
e
d
)
=
l
a
c
t
u
a
l
(
1
+
(
(
α
Z
n
−
α
r
o
d
)
Δ
t
)
l
rod (measured)
=l
actual
(1+((α
Zn
−α
rod
)Δt)