Question

A particular force(f) applied on a wire increase its length by 2×10^-3 to increase the wires length by 4×10^-3 by the applied force will be​

Answer 1

Answer:

A particular force(f) applied on a wire increase its length by 2×10^-3 to increase the wires length by 4×10^-3 by the applied force will be

Answer 2

Given:

Length by which the wire's length is increased by applying force F is =$2 \times 10^{-3}$

To Find:

What is the value of applied force when the length of the wire increases by $4\times 10^{-3}$ =?

Solution:

We know that according to Hook's law we have:

Stress $\propto$ strain

i.e. $\frac{F}{A} \propto \frac{\Delta l}{l}$  

Or, $\frac{F}{A} = \gamma \frac{\Delta l}{l}$

Here, F is applied force, A is sectional area of the wire, $\gamma$ is Young's Modulus of the material of the wire.  

In the 1st case we have:

$\frac{F}{A} = \gamma \times \frac{2 \times 10^{-3}}{l}$ ................(1)

In the 2nd case, we have:

$\frac{F'}{A} =\gamma \times \frac{4 \times 10^{-3}}{l}$...............(2)

Here, F' is the force required to increase the length of the wire by $4 \times 10^{-3}$ .

On dividing equation (2) by (1), we get:

$\frac{(\frac{F'}{A})}{(\frac{F}{A})}=\frac{\gamma (\frac{4\times 10^{-3}}{l})}{\gamma (\frac{2\times 10^{-3}}{l})}$

Or, $\frac{F'}{F}=2$

So, F'= 2F

Hence, twice of the initial force that 2F is required.