Question

Two wires of equal length are suspended vertically at a distance of 40 cm as shown in the figure below. Their upper ends are fixed to the ceiling while their lower ends support a rigid horizontal bar which carries a central load of 1t midway between the wires. Details of the two wires are given below:

Wire No.	Area (cm${}^2$)	Material	Modulus of Elasticity (kg/cm${}^2$)	Elongation
1	4	Copper	1 $\times$10${}^6$	${\Delta }_C$
2	2	Steel	2 $\times$ 10${}^6$	${\Delta }_S$

The ratio of the elongation of the two wire, ${\Delta }_C/{\Delta }_S$ is

• A. $0.25$
• B. $0.5$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

Since load of 1t is applied mid-way so force in both wires will be same and equals to 0.5 t.
${\Delta }_c=\frac{500\times 4}{1\times {10}^6\times 4}=5\times {10}^{-4}=m=0.5mm$
${\Delta }_s=\frac{500\times 4}{2\times {10}^6\times 2}=5\times {10}^{-4}m=0.5mm$
$\therefore \frac{{\Delta }_c}{{\Delta }_s}=1.0$