A steel wire of length 7 m and cross section 1 mm2 is hung from a rigid support with a steel weight of volume 1000 c.c. hanging from the other end. Find the decrease in the length of wire, when steel weight is completely immersed in water. ( sted = 2 x 10¹¹N / m²) Density of water = 1 g / c.c) (Ana : 0.343 mm)
Answers
Given :
L = 7 m,
A = 1 mm² = 10⁻⁶ m² ,
V = 1000 cm³ = 1000 × 10⁻⁶ m³ = 10⁻³ m³ ,
ρwater = 1 g/cm3 = 10³ kg/m³ ,
g = 9.8 m/s²
Y = 2 × 10¹¹ N/m²
Y = FL/ Al
∴ l = FL /AY
weight of suspended steel is in air, l 1 = F1 L /AY 1
When the suspended steel weight is immersed in water, l 2 = F2 L/ AY 2
upthrust due to water = F1 – F2 = Vρg
∴ The decrease in the length of the wire is l1 – l 2 = (F1 – F2) L/AY
= V ρ gL/ AY
= 10⁻³x 10 ³x9.8x 7 /10 ⁻⁶x2x10¹¹
=34.3 × 10⁻⁵ m
∴ ∆l = 0.343 mm
Answer:
Given :
L = 7 m,
A = 1 mm² = 10⁻⁶ m² ,
V = 1000 cm³ = 1000 × 10⁻⁶ m³ = 10⁻³ m³ ,
ρwater = 1 g/cm3 = 10³ kg/m³ ,
g = 9.8 m/s²
Y = 2 × 10¹¹ N/m²
Y = FL/ Al
∴ l = FL /AY
weight of suspended steel is in air, l 1 = F1 L /AY 1
When the suspended steel weight is immersed in water, l 2 = F2 L/ AY 2
upthrust due to water = F1 – F2 = Vρg
∴ The decrease in the length of the wire is l1 – l 2 = (F1 – F2) L/AY
= V ρ gL/ AY
= 10⁻³x 10 ³x9.8x 7 /10 ⁻⁶x2x10¹¹
=34.3 × 10⁻⁵ m
∴ ∆l = 0.343 .