A brass wire of cross-sectional area 2mm2 is suspended from rigid support and a body of volume 100cm3is attached to its other end. If the decrease in the length of the wire is 0.11 mm, when the body is completely immersed in water, find the natural length of the wire.
(Ybrass=0.91×1011N/m2,ρwater=103kgm−3 )
A brass wire of cross-sectional area 2mm2 is suspended from rigid support and a body of volume 100cm3is attached to its other end. If the decrease in the length of the wire is 0.11 mm, when the body is completely immersed in water, find the natural length of the wire.
(Ybrass=0.91×1011N/m2,ρwater=103kgm−3 )
The correct option is B 2.043m
Given, cross sectional area , A = 2mm² = 2 ×
10^-6 m²
Volume of the body , V = 100cm³ = 100 × 10^-6
m³ = 10^-4 m³
decrease in the length of wire, ∆L = 0.11mm =
11 × 10^-5 m
Young's modulus of brass, Y = 0.91 × 10¹¹N/m²
And density of water , d = 10³ kg/m³
Body completely immersed in water . So, a
upward force acts on the body. e.g., buoyancy force .
Buoyancy force, F = volume of body × density
of water × acceleration due to gravity
= 10^-4 × 10³ × 9.8
= 0.98 N
Now, Young's modulus , Y = FL/A∆L
so, L = YA∆L/F
= 0.91 × 10¹¹ × 2 × 10^-6 × 11 × 10^-5/0.98
= 2.043m
Given, cross sectional area , A = 2mm² = 2 × 10^-6 m²
Volume of the body , V = 100cm³ = 100 × 10^-6 m³ = 10^-4 m³
decrease in the length of wire, ∆L = 0.11mm = 11 × 10^-5 m
Young's modulus of brass, Y = 0.91 × 10¹¹N/m²
And density of water , d = 10³ kg/m³
Body completely immersed in water . So, a upward force acts on the body. e.g., buoyancy force .
Buoyancy force, F = volume of body × density of water × acceleration due to gravity
= 10^-4 × 10³ × 9.8
= 0.98 N
Now, Young's modulus , Y = FL/A∆L
so, L = YA∆L/F
= 0.91 × 10¹¹ × 2 × 10^-6 × 11 × 10^-5/0.98
= 2.043m
