A copper wire $(Y = 10^{11} {Nm}^{- 2})$ of length 8 m and steel wire $(Y = 2 \times 10^{11} {Nm}^{- 2})$ of length 4 m and each of cross-sectional area 0.5 ${cm}^{2}$ are fastened end to end and stretched with a tension of 500 N. Match the given columns and select the correct option from the codes given below.
Column-I Column-II
i. elongation in copper wire(in mm) p. 0.2
ii. elongation in steel wire(in mm) q. 1.0
iii. total elongation (in mm) r. 0.8
s. 0.1
t. 0.9
Codes

Moderate

Solution

$∆ l_{copper} = \frac{Fl}{AY} = \frac{500 \times 8}{0.5 \times 10^{- 4} \times 10^{11}} = 0.8 mm$
$∆ l_{steel} = \frac{Fl}{AY} = \frac{500 \times 4}{0.5 \times 10^{- 4} \times 2 \times 10^{11}} = 0.2 mm$
${(∆ l)}_{net} = ∆ l_{copper} + ∆ l_{steel} = 0.8 + 0.2 = 1.0 mm$

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	Column-I		Column-II
i.	elongation in copper wire(in mm)	p.	0.2
ii.	elongation in steel wire(in mm)	q.	1.0
iii.	total elongation (in mm)	r.	0.8
		s.	0.1
		t.	0.9