Question

A force of $10\,\text{N$ is required to break a wire of radius $1\,\text{mm}$. The force required to break the wire of same material but radius $3\,\text{mm}$ will be}

• A. $\dfrac{10}{9}\,\text{N}$
• B. $\dfrac{10}{3}\,\text{N}$
• C. $90\,\text{N}$
• D. $30\,\text{N}$

Hint:

For same material, breaking force depends on cross-sectional area.

Answer 1

The Correct Option is C

Step 1: Breaking force relation.
Breaking force is proportional to cross-sectional area:
\[ F \propto r^2 \]
Step 2: Ratio of forces.
\[ \frac{F_2}{F_1} = \left(\frac{r_2}{r_1}\right)^2 \]
Step 3: Substituting values.
\[ \frac{F_2}{10} = \left(\frac{3}{1}\right)^2 = 9 \]
Step 4: Calculating force.
\[ F_2 = 90\,\text{N} \]
Step 5: Conclusion.
The force required is $90\,\text{N}$.