Question : A wire with radius of 5 mm is hung freely from the ceiling . A load of 5 N is applied to its free end. Find the elongation in the wire if its volume is 7.85×10-5 m3 & Young's Modulus is 1011 N/m2.
(a) 6.21×10-7 m
(b) 7.00×10-7 m
(c) 6.36×10-7 m
(d) 8.00×10-9 m
Doubt by Pulkit
Solution :
r=5 mm
= 5×10-3 m
F=5 N
Volume of the wire
V=7.85×10-5 m3
Young's Modulus of the Wire
Y=1011 N/m2
Solution :
r=5 mm
= 5×10-3 m
F=5 N
Volume of the wire
V=7.85×10-5 m3
Young's Modulus of the Wire
Y=1011 N/m2
Area × Length = Volume
Length = Volume / Area
Length = Volume / Area
L=V/A
L=V/πr2
L=7.85×10-5/3.14×(5×10-3)2
L=2.5×10-5/25×10-6
L=[25/250]×10
L=250/250
L=1 m
L=2.5×10-5/25×10-6
L=[25/250]×10
L=250/250
L=1 m
We know,
Y=FL/AΔL
Y=FL/AΔL
ΔL=FL/AY
Area × Length = Volume
Length = Volume / Area
Area × Length = Volume
Length = Volume / Area
L=V/A
ΔL=F[V/A]/AY
ΔL=FV/A2Y
ΔL=FV/(πr2)2Y
ΔL=FV/π2r4Y
ΔL=(5)(7.85×10-5)/(3.14)2(5×10-3)4(1011)
ΔL=[(5)(7.85)/(3.14×3.14×625)]×[10-4]
ΔL=[7.85/3.14×3.14×125]×[10-4]
ΔL=[2.5/3.14×125]×[10-4]
ΔL=F[V/A]/AY
ΔL=FV/A2Y
ΔL=FV/(πr2)2Y
ΔL=FV/π2r4Y
ΔL=(5)(7.85×10-5)/(3.14)2(5×10-3)4(1011)
ΔL=[(5)(7.85)/(3.14×3.14×625)]×[10-4]
ΔL=[7.85/3.14×3.14×125]×[10-4]
ΔL=[2.5/3.14×125]×[10-4]
ΔL=[1/3.14×50]×[10-4]
ΔL=[2/314]×[10-4
ΔL=[1/157]×[10-4]
ΔL=[1000/157]×[10-7]
ΔL=6.369×10-7
ΔL=[2/314]×[10-4
ΔL=[1/157]×[10-4]
ΔL=[1000/157]×[10-7]
ΔL=6.369×10-7
Hence, (c) 6.36×10-7 m, would be the correct option.