# PHYSICS  helpA metal wire of diameter D stretches by 0.100mm when supporting a weight W. if the same length wire is used to support a weight W. if the same length wire is used to support a weight...

PHYSICS  help

A metal wire of diameter D stretches by 0.100mm when supporting a weight W. if the same length wire is used to support a weight W. if the same length wire is used to support a weight three times as heavy, what would its diameter have to be (in terms of D) so it still stretches only 0.100mm?

krishna-agrawala | College Teacher | (Level 3) Valedictorian

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For a given length of wire the extent by which the wire will stretch when pulled is directly proportional to the stress experienced by the wire. The stress experienced by the wire is equal to the total force by which the wire is stretched divided by the area of the wire.

Thus if two different wires of same length are stretched by the same amount when pulled by two different weights, then the stress in both the wires must be same.

Applying this principle we can solve the given problem as follows.

Let:

D = Diameter of first wire stretched by the weight W.

D' = Diameter of second wire stretched by the weight 3W (three times W).

Then:

Stress in first wire = W/[{pi(D)^2}/4]

Stress in second wire = 3W/[{pi(D')^2}/4]

As bot wire stretch by same amount:

Stress in first wire = Stress in second wire

Therefore:

W/[{pi(D1)^2/}4] = 3W/[{pi(D2)^2}/4]

==> [{pi(D')^2}/4]/[{pi(D)^2}/4] = 3W/W

==> (D')^2/(D)^2 = 3

==> (D'/D)^2 = 3

Therefore:

D'/D = 3^(1/2)

==> D' = D*3^(1/2)

Diameter of second wire = D*3^(1/2)

neela | High School Teacher | (Level 3) Valedictorian

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We know that the Young's modulus E, the coefficient of linear stretch of the material of a wire is given by:

E = (F*L0)/{A( L1-Lo)}, where F is the force tensile force , A is the cross sectional are of the wire and L0 is the length of the wire beore applying the tensile stretch and L1  is the length of the wire after applying the tensile force F.

We know that Young's Modulus  E is fairly constant  for a considerable variation of force.

Therefore , for

E =  FLo/{kd1^2*(L1-L0)}... (1) , as fross sectional area is proportional to diameter, and d1 is the diameter of the wire for which we get L1-L0 = 0.100mm.

E = 3FL0/{kd2^2 *(L2-L0)}....(2), where d2 is the diameter when applied a force of 3 times F. But L2 -L0 = 0.100mm.

Therefore , we cal equate the righ sides of the equations (1) and (2), and L1-L0 = L2=L0 = 0.100mm = 10^(-4) m

Therefore ,

FL0/{kd1^2*(L1-Lo)} = 3FL0/{kd2^2*(L2-L0)}

FL0/{kd1^2 *10^(-4)} = 3FL0/{kd2^2*10^(-4)}

After cross multiplication and cancelltion of common factors , we get:

(d2)^2 = 3(d1)^2.

Therefore d2 = (sqrt3) d1.

So the diameter of the wire should be sqrt 3 times the wire , in order that it stretches the same for 3 times the pulling weigtht.