# The breaking force of a wire is 200N.What will be the breaking force if radius of wire is doubled??

2
by sami123

• Brainly User
2015-01-10T12:02:43+05:30
The breaking of a wire depends on the stress on the plane of failure. If the stress exceeds the ultimate strength of the wire, it breaks.

initial breaking force (F1)= 200N
initial area (A1) = πr² = A
final area(A2)= π(2r)² = 4πr² = 4A
final breaking force = F2

Since ultimate strength is a material property, it does not change.
So ultimate strength = F1/A1 = F2/A2
or 200/A = F2/4A
or F2= 200*4 = 800N

(in case of any doubt, comment below)
thanks dude
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2015-01-12T03:26:09+05:30

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Breaking force depends on the stress (tensile) on the wire.  Breaking stress is same in both cases of radius.  It is the property of the material.

If radius is doubled then  area of cross section becomes four times.  It means the force has to be 4 times to result in the same stress (tensile) on the wire.

Hence  breaking force = 4 * 200 N