Question

A wire of length L and radius r is fixed at one end. When a stretching force F is applied at free end, the elongation in the wire is l. When another wire of same material but of length 2L and radius 2r, also fixed at one end is stretched by a force 2F applied at free end, then elongation in the seconds wire will be

A. 1/2
B. 1
C. 21
D. 1/4

Answer 1

The elongation in the second wire will be half of the elongation in the first wire.

Step-by-step explanation:

The elongation in a wire when a stretching force is applied depends on factors such as the force, the material of the wire, the original length of the wire, and the cross-sectional area of the wire.

In this scenario, the first wire has a length L and radius r, and it elongates by l when a force F is applied. The second wire is twice as long with a length of 2L and twice as thick with a radius of 2r. When a force of 2F is applied to the second wire, the elongation can be calculated using the same principles.

Using the equation for elongation, we have:

l = FL / (A₁E)

and

l₂ = (2F)(2L) / (A₂E)

where A₁ and A₂ are the cross-sectional areas of the first and second wires, and E is the modulus of elasticity of the material.

Since the second wire is twice as long and twice as thick as the first wire, its cross-sectional area is four times larger.

Therefore, the elongation in the second wire (l₂) will be half of the elongation in the first wire (l).