Question

If the breaking strength of a rod of diameter $2$ cm is $2\times {10}^5$ N, then that for a rod of same material and diameter $1$ cm will be

• A. $2\times {10}^5$ N
• B. $1\times {10}^5$ N
• C. $0.5\times {10}^5$ N
• D. $0.25\times {10}^5$ N

Answer 1

The correct option is C $0.5\times {10}^5$ N

$\begin{array}{cc}{d}_1=2cm& d_2=1cm\\ F_1=2\times {10}^{5}N& F_2=?\end{array}$

$\begin{array}{cc}\frac{F_1}{A_1}& =Y\frac{\Delta l}{l}\\ \Rightarrow & \frac{Y\Delta l}{l}=\frac{2\times {10}^5}{\pi {\left({10}^{-2}\right)}^2}\end{array}$

$\text{Now,}$

$\begin{array}{cc}\frac{F_2}{A_2}& =Y\frac{\Delta l}{l}\\ \Rightarrow F_2& =\frac{A_2{Y}^{}L}{l}\\ \Rightarrow F_2& =\frac{\pi {(0.5\times {10}^{-2})}^2\times 2\times {10}^5}{\pi {\left({10}^{-2}\right)}^2}\\ \Rightarrow & F_2=0.5\times {10}^{5}N\end{array}$