Two wires of diameter 0.25 cm, one made of steel and other made of brass are loaded as shown in figure. The unloaded length of steel wire is 1.5 m& and that of brass wire is 1.0 m. Young's modulus of steel is 2.0 × 10¹¹& and Young's modulus & of Brass& 0.91 × 10¹¹ Pa. The elongations of steel and brass wires are respectively

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Two wires of diameter 0.25 cm, one made of steel and other made of brass are loaded as shown in figure. The unloaded length of steel wire is 1.5 m& and that of brass wire is 1.0 m. Young's modulus of steel is 2.0 × 10¹¹& and Young's modulus & of Brass& 0.91 × 10¹¹ Pa. The elongations of steel and brass wires are respectively

Two wires of diameter 0.25 cm, one made of steel and other made of brass are loaded as shown in figure. The unloaded

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Correct answer: B. 1.5 × 10 -4 & m; 1.3 × 10 -4 m

Given, diameter of wires (2r) = 0.25 cm

r = 0.125 cm = 1.25 × 10^-3 m

For steel wire,

Load (F₁) = (4 + 6) kgf = 10 × 9.8 N = 98 N

Length of steel wire (l₁) = 1.5 m

Young's modulus (Y₁) = 2.0 × 10¹¹ Pa

\text{Young's modulus (Y)} = \dfrac{{F}_{1} \times {l }_{1}}{{A}_{1} \times \text{\Delta} {l }_{1}}

\text{Change in length} \left(\text{\Delta} {l }_{1}\right) = \dfrac{{F}_{1} \times {l }_{1}}{{A}_{1} \times {Y}_{1}}

= \dfrac{{F}_{1} \times {l }_{1}}{\text{\pi} {r}_{1}^{2} \times {Y}_{1}}

= \dfrac{\text{98} \times \text{1.5}}{\text{3.14} \times {\left(\text{1.25} \times {10}^{- 3}\right)}^{2} \times \text{2.0} \times {10}^{11}}

= 1.5 × 10^-4~m

For brass wire,

Load (F₂) = 6 kgf = 6 × 9.8 N = 58.8 N

Length of brass wire (l₂) = 1.0 m

Young's modulus (Y₂) = 0.91 × 10¹¹ Pa

\text{Change in length} \left({l }_{2}\right) = \dfrac{{F}_{2} \times {l }_{2}}{\text{\pi} {r}_{2}^{2} \times {Y}_{2}}

= \dfrac{\text{58.8} \times \text{1.0}}{\text{3.14} \times {\left(\text{1.25} \times {10}^{- 3}\right)}^{2} \times \text{0.91} \times {10}^{11}}

= 1.3 × 10^-4 m

Two wires of diameter 0.25 cm, one made of steel and other made of...

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