A mild steel wire of cross-sectional area 0.60 x 10^-2 cm2 and length 2 m is stretched ( not beyond its elastic limit ) horizontally between two columns. If a 100g mass is hung at the midpoint of the wire, find the depression at the midpoint.
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Answers
From the above figure,
Let x be the depression at the mid point i.e. CD=x.
In fig.,
AC=CB=l=0.5m ;
m=100g=0.100Kg
AD=BD=(l2+x2)21
Increase in length, △l=AD+DB−AB=2AD−AB
=2(l2+x2)21−2l=2l(1+l2x2)−2l=2l[1+2l2x2]−2l=lx2
∴ strain =2l△l=2l2x2
If T is the tension in the wire, then
2Tcosθ=mg
or, T=2cosθmg
Here, cosθ=(l2+x2)21x=l(1+l2x2)21x=(1+21l2x2)x
As, x<<l, so 1>>2l21x2 and 1+21l2x2≈1
Hence, T=2(1x)mg
As, x<<l, so 1>>2l21x2 and 1+21l2x2≈1
Hence, T=2(1x)mg=2xmgl
stress =AT=2Axmgl
Y=strainstress=2Axmgl×x22l2=Ax3mgl3
x=l[YAmg]31=0.5[20×10
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What is PHOTOSYNTHESIS puch le bhai
From the above figure,
Let x be the depression at the mid point i.e. CD=x.
In fig.,
AC=CB=l=0.5m ;
m=100g=0.100Kg
AD=BD=(l
2
+x
2
)
2
1
Increase in length, △l=AD+DB−AB=2AD−AB
=2(l
2
+x
2
)
2
1
−2l=2l(1+
l
2
x
2
)−2l=2l[1+
2l
2
x
2
]−2l=
l
x
2
∴ strain =
2l
△l
=
2l
2
x
2
If T is the tension in the wire, then
2Tcosθ=mg
or, T=
2cosθ
mg
Here, cosθ=
(l
2
+x
2
)
2
1
x
=
l(1+
l
2
x
2
)
2
1
x
=
(1+
2
1
l
2
x
2
)
x
As, x<<l, so 1>>
2l
2
1x
2
and 1+
2
1
l
2
x
2
≈1
Hence, T=
2(
1
x
)
mg
=
2x
mgl
stress =
A
T
=
2Ax
mgl
Y=
strain
stress
=
2Ax
mgl
×
x
2
2l
2
=
Ax
3
mgl
3
x=l[
YA
mg
]
3
1
=0.5[
20×10
11
×0.5×10
−6
0.1×10
]
3
1
=1.074×10
−2
m
=1.074 cm