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physics ncert solutions class 11th Physics Class 11th Mechanical Properties of Solids

9.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. 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.

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Vishal Baghel | Contributor-Level 10

5 months ago

Diameter of wires, d = 0.25 cm, radius, r = 0.125 cm

Cross-sectional area, $A_{1}$ = $A_{2}$ = $π \times r^{2} = 0.049 {c m}^{2}$ = 4.908 $\times 10^{- 6}$ $m^{2}$

Length of the steel wire, $L_{1} = 1.5 m$ , length of the brass wire, $L_{2} = 1.0 m$

Change in length of the steel wire $= ? L_{1},$ Change in length of the copper wire $= ? L_{2}$

Total force exerted on the steel wire, $F_{1}$ = ( 4+6) kg = 10 kg = 98 N

Young’s modulus of steel , $Y_{1}$ = $\frac{\frac{F_{1}}{A_{1}}}{\frac{? L_{1}}{L_{1}}}$ = 2.0 $\times 10^{11}$ Pa

$? L_{1}$ = $\frac{F_{1}}{A_{1}} \times \frac{L_{1}}{Y_{1}} = \frac{98 \times 1.5}{4.908 \times 10^{- 6} \times 2.0 \times 10^{11}}$ = 1.497 $\times 10^{- 4}$ m

Similarly for brass wire, $F_{2}$ = 6 kg = 58.8 N, $Y_{2}$ = $0.91 \times 10^{11} P a$

$? L_{2}$ = $\frac{F_{2}}{A_{2}} \times \frac{L_{2}}{Y_{2}} = \frac{58.8 \times 1.0}{4.908 \times 10^{- 6} \times 0.91 \times 10^{11}}$ = 1.316 $\times 10^{- 4}$ m

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9.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. 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.

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