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Mechanical Properties of Solids

Question : 1
Total: 21
Solution:
Here, for steel wire,
Area of cross-section, A1=3.0×105m2
Stretching, Δl1=Δl (say)
Stretching force on steel, F1=F
For copper wire, length of wire, l2=3.5m
Area of cross-section, A2=4.0×105m2
Stretching, Δl2=Δl (given);
Stretching force on copper, F2=F
Let Y1 and Y2 be the Young's modulus of steel and copper wire respectively.
Y1=
F1A1
Δl1l1
=
F1×l1
A1×Δl1
.
.
.(i)
and Y2=
F2×A2
Δl2l2

=
F2×l2
A2×Δl2

=
F×3.5
4×105×Δl
.
.
.(ii)

Dividing (i) by (ii), we get
Y1
Y2
=
F×4.7
3×105×Δl
×
4×105×Δl
F×3.5
=
18.8
10.5
=1.79
=1.8

Y1:Y2=1.8:1
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