Question

Assertion (A): A bar tapers from a diameter of ${}^{\prime }{d}_1^{\prime }$ to a diameter of '$d_2^{\prime }$ over its length $L$ and is subjected to a tensile force $P$. If extension is calculated based on treating it as a bar of average diameter, the calculated extension will be more than the actual extension.
Reason (R): The actual extension in such bars is given by, $\Delta =\frac{4PL}{\pi d_1{d}_{2}E}$

• A. both A and R are true and R is the correct explanation of A
• B. both A and R are true but R is not a correct explanation of A
• C. A is true but R is false
• D. A is false but R is true

Answer 1

The correct option is D A is false but R is true

Actual extension
${\Delta }_1=\frac{4PL}{\pi d_1{d}_{2}E}$
Extension based on average diameter
${\Delta }_2=\frac{4PL}{\pi {\left(\frac{d_1+d_2}{2}\right)}^{2}E}$
Assuming ${\Delta }_1>{\Delta }_2$
So, ${\Delta }_1=\frac{4PL}{\pi d_1{d}_{2}E}>{\Delta }_2=\frac{4PL}{\pi {\left(\frac{d_1+d_2}{2}\right)}^{2}E}$
$\Rightarrow {\left(\frac{d_1+d_2}{2}\right)}^2>d_1{d}_2$
$\Rightarrow \frac{d_1^2+d_2^2+2d_1{d}_2}{4}-d_1{d}_2=0$
$\Rightarrow (d_1-d_2{)}^2>0$
Which is true.