First slide

Elastic behaviour of solids

Question

A steel wire of length 4 m and diameter 5 mm is stretched by 5 kg-wt. The increase in its length, if the Young’s modulus of steel wire is $2 .4 \times 10^{12} dyne {cm}^{- 2} is$

Moderate

A

0.003 cm
B

0.0041 cm
C

0.00041 cm
D

0.005 cm

Solution

Here, $l = 4 m = 400 cm, 2 r = 5 mm$
or $r = 2.5 mm = 0.25 cm$
$F = 5 kg - wt = 5000 g - wt = 5000 \times 980 dyne Δl = ?, Y = 2 .4 \times 10^{12} dyne {cm}^{- 2}$
As $Y = \frac{F}{{πr}^{2}} \times \frac{l}{Δl} or Δl = \frac{Fl}{{πr}^{2} Y} = \frac{(5000 \times 980) \times 400}{(22 / 7) \times (0 .25)^{2} \times 2 .4 \times 10^{12}} = 0.0041 cm$

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