NCERT Solutions
Class 11 Physics
Mechanical Properties of Solids
Q. 8.5
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. compute the elongations of the steel and the brass wires.
Elongation of the steel wire = (1.49 × 10–4) m
Elongation of the brass wire = (1.3 × 10–4) m
Diameter of the wires, d = 0.25 m
Hence, the radius of the wires, r = (d/2) = 0.125 cm
Length of the steel wire, L1 = 1.5 m
Length of the brass wire, L2 = 1.0 m
Total force exerted on the steel wire:
F1 = (4 + 6) g = (10 × 9.8) = 98 N
Young’s modulus for steel:
Y1 = (F1/A1)/ (ΔL1/L1)
Where,
ΔL1 = Change in length of the steel wire.
A1 =Area of cross section of the steel wire = π r12
Young’s modulus of steel Y1 = (2.0 x 1011) Pa
Therefore, (ΔL1) = (F1 x L1)/ (A1 x Y1)
= (F1 x L1)/ (π r12 x Y1)
= (98 x 1.5)/ (π (0.125 x 10-2)2 x 2 x 1011)
= 1.49 x 10-4 m
Total force on brass wire = F2 = (6 x 9.8)
(ΔL1) =58.8 N
Young’s modulus for brass:
Y2 = (F2/A2)/ (ΔL2/L2)
Where,
ΔL2 = Change in length of the brass wire.
A2 =Area of cross section of the brass wire = π r22
Therefore, (ΔL2) = (F2 x L2)/ (A2 x Y2)
= (F2 x L2)/ (π r22 x Y2)
= (58.8 x 1.0)/ (π x (0.125 x 10-2)2 x (0.91 x 1011)
(ΔL2) = 1.3 x 10-4
Elongation of the steel wire = 1.49 × 10–4 m
Elongation of the brass wire = 1.3 × 10–4 m