Question

A wire can be broken by a load of $20\mathrm{kg}-\mathrm{wt}$. The force required to break wire of same material with twice the diameter will be

• A. $5\mathrm{kg}-\mathrm{wt}$
• B. $80\mathrm{kg}-\mathrm{wt}$
• C. $20\mathrm{kg}-\mathrm{wt}$
• D. $160\mathrm{kg}-\mathrm{wt}$

Answer 1

The correct option is B

$80\mathrm{kg}-\mathrm{wt}$

Step 1: Given information

Load is given as ${\mathrm{F}}_1=20\mathrm{kg}-\mathrm{wt}$.

Breaking stress is the characteristic of the material.

Hence, two wires of the same material have different thicknesses.

Step 2: Determine the force required to break the wire of the same material.

Since both wires have the same length and the same material ${\mathrm{F}}_1\propto {\mathrm{A}}_1$ and ${\mathrm{F}}_2\propto {\mathrm{A}}_2$
$\Rightarrow \frac{{\mathrm{F}}_1}{{\mathrm{F}}_2}=\frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}$
Now, substitute ${\mathrm{A}}_1=\mathrm{\pi }{{\mathrm{r}}_1}^2$and ${\mathrm{A}}_2=\mathrm{\pi }{{\mathrm{r}}_2}^2$in the above equation,
$$\begin{gathered} \frac{{\mathrm{F}}_1}{{\mathrm{F}}_2}=\frac{\mathrm{\pi }{{\mathrm{r}}_1}^2}{\mathrm{\pi }{{\mathrm{r}}_2}^2} \\ =\frac{{{\mathrm{r}}_1}^2}{{{\mathrm{r}}_2}^2} \end{gathered}$$

Substituting for ${\mathrm{r}}_2=2{\mathrm{r}}_1$(As given in the question)

$$\begin{gathered} \frac{{\mathrm{F}}_1}{{\mathrm{r}}_1^2}=\frac{{\mathrm{F}}_2}{{\mathrm{r}}_2^2} \\ \frac{{\mathrm{F}}_1}{{\mathrm{r}}_1^2}=\frac{{\mathrm{F}}_2}{{\left(2{\mathrm{r}}_1\right)}^2} \\ \frac{20}{{\mathrm{r}}_1^2}=\frac{{\mathrm{F}}_2}{\left(4{{\mathrm{r}}_1}^2\right)} \end{gathered}$$

$$\begin{gathered} {\mathrm{F}}_2=\frac{20\times 4{{\mathrm{r}}_1}^2}{{{\mathrm{r}}_1}^2} \\ {\mathrm{F}}_2=80\mathrm{kg}-\mathrm{wt} \end{gathered}$$

Therefore, The force required to break wire of same material with twice the diameter will be $\mathrm{is}80\mathrm{kg}-\mathrm{wt}$.

Hence, option (B) is the correct answer.