Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires. [Young’s modulus of steel is 2.0 x 1011 Pa. $$(1 Pa = 1 N m^2$$)]

Asked by Abhisek | 1 year ago |  92

##### Solution :-

Diameter of the two wires, d =0.25m
Radius of the wires, r= d/2 =0.125cm
Unloaded length of the steel wire, l1 =1.5m
Unloaded length of the brass wire, l2 =1.0m

Force exerted on the steel wire:
F1 =(4+6)g=10×9.8=98N

Cross-section area of the steel wire, a=πr12

Change in length of the steel wire = Δl1

Young’s modulus for steel= 2.0 x 1011 Pa

$$Y_{1}=\dfrac{F_{1}}{a_{1}}\dfrac{l_{1}}{\Delta l_{1}}​$$

$$Y_{1}=\dfrac{F_{1}}{\pi r_{1}^{2}}\dfrac{l_{1}}{\Delta l_{1}}$$

$$\Delta l_{1}=\dfrac{F_{1}}{\pi r_{1}^{2}}\dfrac{l_{1}}{Y_{1}}$$

$$\Delta l_{1}=\dfrac{9.8}{\pi(0.125\times 10^{-2})^{2}}\dfrac{1.5}{2.0\times 10^{11}}$$

= 1.49 x 10-4 m

Force of the brass wire ,F2 = 6 x 9.8 = 58. 8 N

Cross-section area of the brass wire, a=πr22

Change in length of the brass wire = Δl2

Young’s modulus of the brass wire = 0.91 x 1011 Pa

$$Y_{2}=\dfrac{F_{2}}{a_{2}}\dfrac{l_{2}}{\Delta l_{2}}$$

$$\Delta l_{2}=\dfrac{F_{2}}{\pi r_{2}^{2}}\dfrac{l_{2}}{Y_{2}}$$

$$\Delta l_{2}=\dfrac{58.8}{\pi(0.125\times 10^{-2})^{2}}\dfrac{1}{0.91\times 10^{11}}$$

= 1.3 x 10-4 m

Elongation of the steel wire is 1.49 x 10-4 m and that of brass is 1.3 x 10-4 m.

Answered by Pragya Singh | 1 year ago

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