Mechanical Properties Of Solids Question 5
Question 5 - 30 January - Shift 2
A force is applied to a steel wire ‘A’, rigidly clamped at one end. As a result elongation in the wire is $0.2 mm$. If same force is applied to another steel wire ’ $B$ ’ of double the length and a diameter 2.4 times that of the wire ’ $A$ ‘, the elongation in the wire ’ $B$ ’ will be (wires having uniform circular cross sections)
(1) $6.06 \times 10^{-2} mm$
(2) $2.77 \times 10^{-2} mm$
(3) $3.0 \times 10^{-2} mm$
(4) $6.9 \times 10^{-2} mm$
Show Answer
Answer: (4)
Solution:
Formula: Young’s modulus
$ Y=\frac{F / A}{\frac{\Delta \ell}{\ell}} $
$ \Rightarrow F=\frac{YA}{\ell} \Delta \ell $
$(\frac{A \Delta \ell}{\ell})_1=(\frac{A \triangle \ell}{\ell})_2$
$\Rightarrow \frac{\Delta \ell_2}{\Delta \ell_1}=\frac{A_1}{A_2} \times \frac{\ell_2}{\ell_1}$
$\Rightarrow \frac{\Delta \ell_2}{0.2}=\frac{1}{2.4 \times 2.4} \times \frac{2}{1}$
$\Rightarrow \Delta \ell_2=6.9 \times 10^{-2} mm$
