(b)
Two wires of diameter 0.25 cm, one made of steel and the other made of
loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m
brass wire is 1.0 m. Compute the elongations of the steel and the brass wir
1/ / / LLL
1.5 m
Steel
4.0 kg
1.0 m
Brass
6.0 kg
Answers
Answer:
Diameter of the wires, d =0.25m
Hence, the radius of the wires, r=
2
d
=0.125cm
Length of the steel wire, L
1
=1.5m
Length of the brass wire, L
2
=1.0m
Total force exerted on the steel wire:
F
1
=(4+6)g=10×9.8=98N
Young’s modulus for steel:
Y
1
=
(
L
1
△L
1
)
(
A
1
F
1
)
Where,
△L
1
= Change in the length of the steel wire
A
1
= Area of cross-section of the steel wire =πr
1
2
Young’s modulus of steel, Y
1
=2.0×10
11
Pa
∴△L
1
=
(A
1
×Y
1
)
F
1
×L
1
=(98×1.5)/[π(0.125×10
−2
)
2
×2×10
11
]=1.49×10
−4
m
Total force on the brass wire:
F
2
=6×9.8=58.8N
Young’s modulus for brass:
Y
2
=
(
L
2
△L
2
)
(
A
2
F
2
)
Where,
△L
2
= Change in the length of the brass wire
A
2
= Area of cross-section of the brass wire =πr
1
2
∴△L
2
=
(A
2
×Y
2
)
F
2
×L
2
=(58.8×1)/[(π×(0.125×10
−2
)
2
×(0.91×10
11
)]=1.3×10
−4
m
Elongation of the steel wire =1.49×10
–4
m
Elongation of the brass wire =1.3×10
−4
m