Question

The diameter of a brass rod is $4mm$ and Young's modulus of brass is $9\times {10}^{10}\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.$. The force required to stretch it by $0.1\%$ of its length is

• A. $360\mathrm{\pi N}$
• B. $36N$
• C. $144\times {10}^{3}N$
• D. $36\mathrm{\pi }\times {10}^5\mathrm{N}$

Answer 1

The correct option is A

$360\mathrm{\pi N}$

Step 1. Given data:

Young's modulus of brass $\left(Y\right)$ = $9\times {10}^{10}\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.$.

Strain $\frac{∆l}{l}=0.001$

Diameter= $4mm$

Radius $\left(r\right)$=$2mm$

Step 2. Formula used:

$F=\raisebox{1ex}{$YA\times ∆l$}\!\left/ \!\raisebox{-1ex}{$l$}\right.$…………$\left(1\right)$

Where $F$=force, $Y$= Youngs modulus, $A$=area,$\raisebox{1ex}{$∆l$}\!\left/ \!\raisebox{-1ex}{$l$}\right.$ = Strain

Step 3. Calculations:

Now Area= $={\mathrm{\pi r}}^2=\mathrm{\pi }{\left(2\times {10}^{-3}\right)}^2=4\mathrm{\pi }\times {10}^{-6}{\mathrm{m}}^2$

Putting all values in $\left(1\right)$, we get

$$\begin{gathered} F=9\times {10}^{10}\times 4\mathrm{\pi }\times {10}^{-6}\times 0.001 \\ =360\mathrm{\pi }\mathrm{N} \end{gathered}$$

Thus, the correct option is option A.