A wire of diameter 1 mm breaks under a tension of 1000 N. Another wire of same material as that of the first one, but of diameter 2 mm breaks under a tension of

The correct option is 4000 N.

Step-by-Step Explanation:

The breaking stress of a wire is a characteristic property of its material. Since both wires are made of the same material, their breaking stress remains the same.

Breaking stress ($\sigma$) is defined as the maximum breaking force (or tension, $F$) per unit cross-sectional area ($A$):

$\sigma =\frac{F}{A}$

For a wire with a circular cross-section of diameter $d$, the area is given by:

$A=\frac{\pi d^2}{4}$

Substituting the area formula into the stress equation, we get:

$\sigma =\frac{F}{\left(\frac{\pi d^2}{4}\right)}=\frac{4F}{\pi d^2}$

Rearranging the equation to solve for the breaking tension $F$:

$F=\frac{\sigma \pi d^2}{4}$

Since the breaking stress ($\sigma$) and $\pi$ are constants, the breaking tension is directly proportional to the square of the diameter of the wire:

$F\propto d^2$

Therefore, we can write the relation for the two wires as:

$\frac{F_2}{F_1}={\left(\frac{d_2}{d_1}\right)}^2$

Given data:
First wire diameter, $d_1=1\text{mm}$
First wire breaking tension, $F_1=1000\text{N}$
Second wire diameter, $d_2=2\text{mm}$

Substitute these values into the ratio equation:

$\frac{F_2}{1000}={\left(\frac{2}{1}\right)}^2$

$\frac{F_2}{1000}=4$

$F_2=4\times 1000=4000\text{N}$

Thus, the second wire will break under a tension of 4000 N.