Physics, asked by BrainlyHelper, 1 year ago

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

Answered by prmkulk1978
11

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


Answered by MRSmartBoy
0

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 .

.

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