A wire of length L and cross-section A is made of material of Young...
Given, length of wire = L, cross-sectional area = A, Young's modulus = Y, and stretch in the wire = x.
Work done in stretching the wire is given by the formula:
Work done = Force × Distance
When a wire is stretched, a restoring force is created due to the deformation in the wire. This restoring force is proportional to the amount of stretch, and is given by Hooke's law:
F = kx
where F is the restoring force, k is the spring constant (which depends on the material properties of the wire), and x is the amount of stretch.
The work done in stretching the wire can be calculated by integrating the restoring force over the distance stretched:
Work done = ∫ F dx
Substituting F = kx, we get:
Work done = ∫ kx dx
Integrating both sides, we get:
Work done = ½ kx² + C
where C is the constant of integration. We can determine the value of C by setting the initial stretch to zero, i.e., when x = 0, the work done is also zero. This gives:
C = 0
Substituting the value of k from Hooke's law, we get:
Work done = ½ (Y A / L) x²
Simplifying, we get:
Work done = x² YA / 2L
Thus, the correct option is (C).
A wire of length L and cross-section A is made of material of Young...
Hiii Harshita so In case of rods with given youngs modulus The value of K=YA/2l therefore work done is 1/2kx^2 put value of x which I have given.And here u go with option c Regards