Advertisements
Advertisements
प्रश्न
Write down the expression for the elastic potential energy of a stretched wire.
Advertisements
उत्तर
The work done in stretching the wire by dl,
dW = F.dl
The total work done in stretching the wire from 0 to l is
W = `int_0^"l" "F"."dl"` ..........(1)
From Young’s modulus of elasticity, force becomes,
Y = `"F"/"A" xx "L"/"l" = "YAl"/"L"` ........(2)
Substituting equation (2) in (1) we get,
W = `int_0^"l" "YAl"/"L" "dl"`
Since l is the dummy variable in the integration, we can change l to lʹ(not in limits).
Therefore W = `int_0^"l" "YAlʹ"/"L" "dlʹ" = "YA"/"L"["lʹ"^2/2]_0^"l" = "YA"/"L" "l"^2/2 = 1/2["YAl"/"L"] "l" = 1/2 "Fl"`
W = `1/2` Fl
This work done is known as the elastic potential energy of a stretched wire.
APPEARS IN
संबंधित प्रश्न
If a wire is stretched to double of its original length, then the strain in the wire is __________.
For a given material, the rigidity modulus is `(1/3)^"rd"` of Young’s modulus. Its Poisson’s ratio is
Two wires are made of the same material and have the same volume. The area of cross-sections of the first and the second wires are A and 2A respectively. If the length of the first wire is increased by ∆l on applying a force F, how much force is needed to stretch the second wire by the same amount?
The Young’s modulus for a perfect rigid body is __________.
If the temperature of the wire is increased, then Young’s modulus will __________.
Define strain.
State Hooke’s law of elasticity
State Hooke’s law and verify it with the help of an experiment.
Derive an expression for the elastic energy stored per unit volume of a wire.
We use the straw to suck soft drinks, why?
