Two wires of the same material and having the same volume, are fixed from one end. A mass (m1 = 3 kg) is hanged to the first wire and a mass (m2) is hanged to the other wire. If the radius of the first wire is half that of the second wire and the elongation in the two wires are the same, find (m2).
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Answer:
L1= L/2. d1=2d
l2=L. d2=d
Y and F is same for both the wires:
increase in length=
FL/AY
1st wire:
increase in length=
F(L/2)/(π2d)²Y
4FL/8πd²Y
FL/2πd²Y
2nd wire : increase in length.
FL/πd²Y
Ratio:
FL/2πd²Y: FL/πd²y
= 1:2
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