Physics, asked by Raj22222, 10 months ago

A metal wire 4m long has radius 0.5 mm the wire is suspended from a rigid support and a load of 10kg wt is attached at the other end what is the extension produced I'd posaiom ratio for the metal is 0.26 what is the lateral compression produced in it (y metal =1.25 × 10 10 N/m²)​

Answers

Answered by ExclusiveEntertainer
5

Data

L=4 m

r = 0.5

g = 9.8 m/s sq

posaions ratio of metal is 0.26

y=12.5×1010 pascles

F= 10 kg

at = 10×9.8

N = 98 N

Solution

1 young modulus Y = F/A/L /1 = FL/AL

where L is the length of the wire = f1/ay

= FL/pai r sq y

the large intention produce

L=9.8 × 4 /3.142 × (0.5 ×10^-3)^2×12.5×10 I p

= 3.992×10^-3

2. possion ratio

posaions ratio = d/d/ L/L

where d is the diameter of the wire

lateral compression I the change in the diameter of the wire

d = posions ratio

d = 2×0.5×10^-3 m

d = 0.26×1×10^-3×3.992 × 10^-3

=2.595×10^-7

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